Memoirs on Differential Equations and Mathematical Physics
Volume 46, 2009, 129–146
L. Giorgashvili and D. Natroshvili
REPRESENTATION FORMULAS OF GENERAL
SOLUTIONS TO THE STATIC EQUATIONS
OF THE HEMITROPIC ELASTICITY THEORY
Abstract. We consider the differential equations of statics of the theory
of elasticity of hemitropic materials. We derive general representation formulas for solutions, i.e., for the displacement and microrotation vectors by
means of three harmonic and three metaharmonic functions. These formulas are very convenient and useful in many particular problems for domains
with concrete geometry. Here we demonstrate an application of these formulas to the Neumann type boundary value problem for a ball. We construct
explicit solutions in the form of absolutely and uniformly convergent series.
2000 Mathematics Subject Classification. 74H20, 74H45.
Key words and phrases. Elasticity theory, hemitropic materials,
boundary value problems, general representation of solutions, transmission
problems.
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Representation Formulas of General Solutions to the Static Equations
131
1. Introduction
Technological and industrial developments as well as great success in biological and medical sciences require to use more generalized and refined
models for elastic bodies. In a generalized solid continuum, the usual displacement field has to be supplemented by a microrotation field. Such
materials are called micropolar or Cosserat solids. They model composites with a complex inner structure whose material particles have 6 degrees
of freedom (3 displacement components and 3 microrotation components).
Recall that the classical elasticity theory allows only 3 degrees of freedom
(3 displacement components).
Experiments have shown that micropolar materials possess quite different properties in comparison with the classical elastic materials (see, e.g.,
[1]–[6] and the references therein). For example, in noncentrosymmetric
micropolar materials (which are called also hemitropic or chiral materials)
there propagate the left-handed and right-handed elastic waves. Moreover,
the twisting behaviour under an axial stress is a purely hemitropic (chiral)
phenomenon and has no counterpart in classical elasticity.
Hemitropic solids are not isotropic with respect to inversion, i.e., they
are isotropic with respect to all proper orthogonal transformations but not
with respect to mirror reflections.
Materials may exhibit chirality on the atomic scale, as in quartz and
biological molecules – DNA, as well as on a large scale, as in composites
with helical or screw–shaped inclusions, certain types of nanotubes, bone,
fabricated structures such as foams, chiral sculptured thin films and twisted
fibers. For more details see the references [1], [2], [4], [7]–[15].
Mathematical models describing chiral properties of elastic hemitropic
materials have been proposed by Aero and Kuvshinski [1], [2] (for historical
notes see also [4], [12], [13] and the references therein).
In the mathematical theory of hemitropic elasticity there are introduced
the asymmetric force stress tensor and the moment stress tensor which
are kinematically related with the asymmetric strain tensor and torsion
(curvature) tensor via the constitutive equations. All these quantities are
expressed in terms of the components of the displacement and microrotation vectors. In turn, the displacement and microrotation vectors satisfy
a coupled complex system of second order partial differential equations of
dynamics. When the mechanical characteristics (displacements, microrotations, body force and body couple vectors) do not depend on the time
variable t, we have the differential equations of statics. These equations
generate a 6 × 6 strongly elliptic, formally self-adjoint differential operator
involving 9 material constants and have very complex form.
The Dirichlet, Neumann and mixed type boundary value problems
(BVPs) corresponding to this model are well investigated for general domains of arbitrary shape and the uniqueness and existence theorems are
132
L. Giorgashvili and D. Natroshvili
proved. Regularity results for solutions are also established by potential as
well as by variational methods (see [13], [16]–[19] and the references therein).
The main goal of this paper is to derive general representation formulas for the displacement and microrotation vectors by means of harmonic and metaharmonic functions. That is, we can represent solutions to
the very complicated coupled system of simultaneous differential equations
of hemitropic elasticity with the help of solutions of a simpler canonical
equations (similar formulas in the classical elastostatics are well known as
Papkovich–Neuber representation formulas).
Namely, we prove that the six components of the field vectors (three
displacement and three microrotation components) can be expressed linearly
by three harmonic and three metaharmonic scalar functions. Moreover, we
show that this correspondence is one-to-one. The representation formulas
obtained have proved to be very useful in the study of many problems for
domains with concrete geometry.
In particular, here we apply these representation formulas to construct
explicit solutions to the Neumann type boundary value problem for a ball.
We represent the solution in the form of Fourier–Laplace series and show
their absolute and uniform convergence along with their derivatives of the
first order if the boundary data satisfy appropriate smoothness conditions.
The motivation for the choice of the transmission problems treated in the
paper is that by the same approach one can construct explicit solutions to
transmission problems for layered composites with finitely many spherical
interfaces.
Moreover, the representations obtained can be applied to some generalizations of the classical Eshelby type inclusion problems for hemitropic
materials (see [20], [21]). For a wider overview of the subject concerning
different areas of application we refer to the references [5], [7], [9], [10], [15],
and [22]–[24].
2. General Representation of Solutions
The basic equations of statics of the hemitropic elasticity read as follows
[1], [2]
(µ + α)∆u(x) + (λ + µ − α) grad div u(x) + (κ + ν)∆ω(x)+
+(δ + κ − ν) grad div ω(x) + 2α curl ω(x) = 0,
(κ + ν)∆u(x) + (δ + κ − ν) grad div u(x) + 2α curl u(x)+
(2.1)
+(γ + ε)∆ω(x) + (β + γ − ε) grad div ω(x)+
+4ν curl ω(x) − 4αω(x) = 0,
where ∆ = ∂12 +∂22 +∂32 is the Laplace operator, ∂j = ∂/∂xj , u = (u1 , u2 , u3 )>
and ω = (ω1 , ω2 , ω3 )> are the displacement vector and the micro-rotation
vector, respectively; α, β, γ, δ, λ, µ, ν, κ and ε are the material constants;
here and in what follows the symbol ( ·)> denotes transposition.
133
Representation Formulas of General Solutions to the Static Equations
The material constants satisfy the following inequalities
µ > 0, α > 0, γ > 0, ε > 0, λ+2µ > 0, µγ −κ 2 > 0, αε−ν 2 > 0,
(λ + µ)(β + γ) − (δ + κ)2 > 0,
(3λ + 2µ)(3β + 2γ) − (3δ + 2κ)2 > 0,
d1 := (µ + α)(γ + ε) − (κ + ν)2 > 0,
(2.2)
2
d2 := (λ + 2µ)(β + 2γ) − (δ + 2κ) > 0,
µ[(λ + µ)(β + γ) − (δ + κ)2 ] + (λ + µ)(µγ − κ 2 ) > 0,
µ[(3λ + 2µ)(3β + 2γ) − (3δ + 2κ)2 ] + (3λ + 2µ)(µγ − κ 2 ) > 0.
Now we formulate our basic assertion.
Theorem 2.1. A vector u = (u, ω)> is a solution to the system (2.1) if
and only if it is representable in the form
u(x) = grad Φ1 (x) − a grad r2 (r∂r + 1)Φ2 (x) + rot rot[xr2 Φ2 (x)]+
+ rot[xΦ3 (x)] + (δ + 2κ) grad Φ4 (x)+
+
2 h
X
j=1
i
rot rot[xΨj (x)] + kj rot[xΨj (x)] ,
ω(x) = σ grad (2r∂r + 3)(r∂r + 1)Φ2 (x) − rot[x(2r∂r + 3)Φ2 (x)]+
+ 2−1 rot rot[xΦ3 (x)] − (λ + 2µ) grad Φ4 (x)−
−
2
X
j=1
h
i
ηj rot rot[xΨj (x)] + kj rot[xΨj (x)] ,
(2.3)
where x = (x1 , x2 , x3 )> , r = |x|, r∂r = x · grad,
∆Φj (x) = 0, j = 1, 2, 3, (∆ − λ21 )Φ4 (x) = 0,
(∆ + kj2 )Ψj (x) = 0, j = 1, 2,
1 a = µ/(λ + 2µ), σ = −
a(δ + 2κ) − κ + ν , λ21 = 4α(λ + 2µ)/d2 ,
2α
h
i
p
−1
k1,2 = 2d1 µν − ακ ± i (µ + α)[µ(αε − ν 2 ) + α(µγ − κ 2 )] ,
i
1
d1 h
,
α(γ + ε) − 2ν(κ + ν) , ηj = kj σ1 kj2 − σ2 kj −
σ1 =
2α∆1
2
h
i
1
σ2 =
α(κ + ν)d1 + 2(µν − ακ)(α(γ + ε) − 2ν(κ + ν)) ,
α∆1
∆1 = 4α(αγ + κ 2 + αε − ν 2 ) > 0,
d1 and d2 are defined in (2.2).
Proof. Let u and ω solve the system (2.1). Let us show that then they admit
the representation (2.3). Apply the divergence operation to both equations
134
L. Giorgashvili and D. Natroshvili
of the system (2.1)
(λ + 2µ)∆ div u + (δ + 2κ)∆ div ω = 0,
(δ + 2κ)∆ div u + (β + 2γ)∆ − 4α div ω = 0.
From this system we get
(∆ − λ21 ) div ω = 0, ∆(∆ − λ21 ) div u = 0,
where
(2.4)
λ21 = 4α(λ + 2µ)/d2 , d2 = (λ + 2µ)(β + 2γ) − (δ + 2κ)2 .
Now apply the curl operation to both equations of the system (2.1)
(µ + α)∆ rot u + (κ + ν)∆ + 2α rot rot ω = 0,
(κ + ν)∆ + 2α rot rot u + (γ + ε)∆ + 4ν rot −4α rot ω = 0.
(2.5)
With the help of the identities rot rot = grad div −∆ and div rot = 0 this
system implies
d1 rot rot +4(ακ − µν) rot +4αµ (∆ rot u, ∆ rot ω)> = 0.
(2.6)
Rewrite the first equation of the system (2.1) in the following form
where
µ∆u + (λ + µ) grad div u = v(x) − (δ + 2κ) grad div ω,
v(x) = α rot(rot u − 2ω) + (κ + ν) rot rot ω.
From the second equation of the system (2.1) we get
(2.7)
(2.8)
2α(rot u − 2ω) = − (κ + ν)∆u − (δ + κ − ν) grad div u − (γ + ε)∆ω−
− (β + γ − ε) grad div ω − 4ν rot ω.
In view of this equality, from (2.8) we obtain
γ+ε
κ+ν
∆ rot u −
∆ rot ω + (κ − ν) rot rot ω.
v(x) = −
2
2
The equation (2.5) yields
µ+α
κ+ν
rot rot ω = −
∆ rot u −
∆ rot ω.
2α
2α
We can rewrite the equation (2.9) as
v(x) = a1 ∆ rot u + a2 ∆ rot ω,
(2.9)
(2.10)
where
i
i
1 h
1 h
α(κ+ν)+(µ + α)(κ−ν) , a2 = −
α(γ +ε)+κ 2 −ν 2 .
2α
2α
From the equalities (2.10) and (2.6) we derive
d1 rot rot +4(ακ − µν) rot +4αµ v(x) = 0,
a1 = −
i.e.,
(rot −k1 )(rot −k2 )v(x) = 0,
(2.11)
Representation Formulas of General Solutions to the Static Equations
135
where k1 and k2 are the roots of the quadratic equation
d1 z 2 + 4(ακ − µν)z + 4αµ = 0.
The discriminant of this equation is
D = −16(µ + α) µ(αε − ν 2 ) + α(µγ − κ 2 ) < 0.
Therefore
k1,2 =
i
p
2h
µν − ακ ± i (µ + α)(µ(αε − ν 2 ) + α(µγ − κ 2 )) .
d1
A solution of the equation (2.11) can be represented as
v(x) = −
where
2
X
µkj2 vj (x),
(2.12)
j=1
(rot −kj )vj (x) = 0, div vj (x) = 0.
(2.13)
Notice that k2 = k 1 and v2 = v 1 (the over-bar means complex conjugation).
Moreover, remark that for a vector v = (v1 , v2 , v3 )> to be a solution of the
system
rot v(x) ∓ kv(x) = 0, div v(x) = 0,
necessary and sufficient conditions read as follows
v(x) = rot rot(xΨ(x)) ± k rot(xΨ(x)),
where Ψ is a scalar function satisfying the Helmholtz equation (∆ + k 2 )Ψ =
0. Due to this remark, we can represent a solution to the system (2.13) as
vj (x) = rot rot(x Ψj (x)) + kj rot(x Ψj (x)), j = 1, 2,
where (∆ + kj2 )Ψj (x) = 0. The functions Ψ1 and Ψ2 are mutually complex
conjugate functions. Substituting the expressions of the vectors vj into
(2.12), we get
v(x) = −
2
X
j=1
h
i
µkj2 rot rot(xΨj (x)) + kj rot(xΨj (x)) .
(2.14)
Introduce the function Φ4 by the equation
div ω(x) = −λ21 (λ + 2µ)Φ4 (x).
(2.15)
From the equations (2.4) and (2.15) it follows that
(∆ − λ21 )Φ4 (x) = 0.
Taking into consideration the equations (2.14) and (2.15), we get from (2.7)
µ∆u+(λ+µ) grad div u =
2
h
i
X
= λ21 (λ+2µ)(δ+2κ) grad Φ4 (x)− µkj2 rot rot(xΨj (x))+kj rot(xΨj (x)) .
j=1
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L. Giorgashvili and D. Natroshvili
The general solution of this equation is written as
u(x) = u0 (x) + (δ + 2κ) grad Φ4 (x)+
2 h
i
X
rot rot(xΨj (x)) + kj rot(xΨj (x)) ,
+
(2.16)
j=1
where u0 is the general solution of the equation
µ∆u0 (x) + (λ + µ) grad div u0 (x) = 0.
(2.17)
Now let us express the vector ω from (2.1) in terms of u and div ω
ω = σ1 ∆ rot u + σ2 rot rot u +
1
rot u + σ3 grad div u + σ4 grad div ω, (2.18)
2
where
d1 α(γ + ε) − 2ν(κ + ν) ,
2α∆1
i
1 h
σ2 =
α(κ + ν)d1 + 2(µν − ακ) α(γ + ε) − 2ν(κ + ν) ,
α∆1
δ + 2κ
λ + 2µ α(γ + ε)(κ + 3ν) − 4ν 2 (κ + ν) ,
σ3 =
−
4α
α∆1
β + 2γ
δ + 2κ σ4 =
−
α(γ + ε)(κ + 3ν) − 4ν 2 (κ + ν) .
4α
α∆1
σ1 =
Further, substitute (2.16) into (2.18) and take into consideration the
equations
∆ rot u0 = 0, λ21 (δ + 2κ)σ3 − (λ + 2µ)σ4 = −(λ + 2µ)
to obtain
1
rot u0 (x) + σ3 grad div u0 (x)−
2
2
h
i
X
− (λ + 2µ) grad Φ4 (x) −
ηj rot rot(xΨj (x)) + kj rot(xΨj (x)) , (2.19)
ω(x) = σ2 rot rot u0 (x) +
j=1
where ηj = kj (σ1 kj2 − σ2 kj − 1/2). It is known that a solution of the system
(2.17) is representable in the form [28]
u0 (x) = grad Φ1 (x) − a grad r 2 r∂r + 1 Φ2 (x)+
+ rot rot(xr2 Φ2 (x)) + rot(xΦ3 (x)),
where
x = (x1 , x2 , x3 )> , r∂r = x · grad, r = |x|,
∆Φj (x) = 0, j = 1, 2, 3, a = µ/(λ + 2µ).
(2.20)
Representation Formulas of General Solutions to the Static Equations
137
Substitution of (2.20) into (2.16) and (2.19) completes the proof of the
first part of the theorem. The sufficiency easily follows from the identities
i
σ
1 h
a(δ + 2κ) − κ + ν = − ,
σ2 + aσ3 =
4α
2
i
h 4(µν − ακ)
d1
(κ + ν) σ1
− σ2 − 2ασ1 = −
,
d1
4α
4αµ
ακ − µν
1
− 2ασ2 =
(κ + ν)
σ1 +
,
d1
2
α
i
4αµ
µν − ακ h
1
− 4νσ2 − 4ασ1 =
4ασ3 − (δ + 2κ) a,
σ1 +
(γ + ε)
d1
2
αµ
h 4(µν − ακ)
i
i
d1 h
(γ + ε) σ1
4ασ3 − (δ + 2κ) a,
− σ2 − 4νσ1 =
d1
4αµ
h
i
4ασ2 + κ − ν = −a 4ασ3 − (δ + 2κ) .
The proof is complete.
3. Basic Boundary Value Problems
Let Ω1 be a ball whose boundary is the sphere ∂Ω1 having radius R and
centered at the origin:
Ω1 = x : x ∈ R3 , |x| < R}, ∂Ω = x : x ∈ R3 , |x| = R .
Denote Ω2 = R3 \ Ω1 . Assume that the regions Ω1 and Ω2 are filled by
isotropic hemitropic materials.
Problem (I)± . Find a regular vector U = (u, ω)> satisfying the differential equations (2.1) in Ω1 (Ω2 ) and the boundary condition
[U (z)]± = F (z), z ∈ ∂Ω,
where
F (z) = f (1) (z), f (2) (z)
>
,
(j)
(j)
(j)
f (j) (z) = f1 (z), f2 (z), f3 (z) , j = 1, 2.
In the case of the exterior domain Ω2 , the vector U has to satisfy the
following decay conditions at infinity
U (x) = O(|x|−1 ),
∂
U (x) = O(|x|−2 ), j = 1, 2, 3.
∂xj
(3.1)
Problem (II)± . Find a regular vector U = (u, ω)> satisfying the differential equations (2.1) in Ω1 (Ω2 ) and the boundary condition
±
T (∂z, n(z))U (z) = F (z), z ∈ ∂Ω,
(3.2)
where n(z) is the exterior normal vector to ∂Ω, T (∂z , n(z))U is the generalized stress vector [17]
h
i>
(3.3)
T (∂x, n)U (x) = H (1) (∂x, n)U (x), H (2) (∂x, n)U (x) ,
138
L. Giorgashvili and D. Natroshvili
where
H (j) (∂x, n)U (x) = T (2j−1) (∂x, n)u(x) + T (2j) (∂x, n)ω(x),
T (2j−1) (∂x, n)u(x) =
∂u(x)
+ ηj0 n(x) div u(x) + ζj0 [n(x) × rot u(x)],
(3.4)
∂n(x)
∂ω(x)
+
T (2j) (∂x, n)ω(x) = ξj00
∂n(x)
+ηj00 n(x) div ω(x) + ζj00 [n(x) × rot ω(x)] + τj [n(x) × ω(x)], j = 1, 2,
= ξj0
with
(
2µ,
2κ,
l = 1,
l = 2;
(
λ, l = 1,
δ, l = 2;
(
(
µ − α, l = 1,
2κ, l = 1,
ζl0 =
ξl00 =
κ − ν, l = 2;
2γ, l = 2;
(
(
(
δ,
l
=
1,
κ
−
ν,
l
=
1,
2α, l = 1,
ηl00 =
ζl00 =
τl =
β, l = 2;
γ − ε, l = 2;
2ν, l = 2.
ξl0 =
ηl0 =
Here and in what follows the symbol a × b denotes the cross product of
two vectors a, b ∈ R3 .
In the case of the exterior domain Ω2 , the vector U has to satisfy the
decay conditions (3.1).
Transmission Problem (A). Assume that the domains Ωj , j = 1, 2, are
filled by isotropic hemitropic solids with material constants αj , βj , . . . , κj .
Find a pair of regular vectors U (j) = (u(j) , ω (j) )> , j = 1, 2, satisfying the
differential equations
(µj + αj )∆u(j) + (λj + µj − αj ) grad div u(j) + (κj + νj )∆ω (j) +
+ (δj + κj − νj ) grad div ω (j) 2αj rot ω (j) = 0,
(κj + νj )∆u(j) + (δj + κj − νj ) grad div u(j) + (γj + εj )∆ω (j) +
+(βj+γj−εj ) grad div ω (j)+2αj rot u(j)+4νj rot ω (j)−4αj ω (j) = 0, j = 1, 2,
in Ωj and the following transmission conditions on ∂Ω
(1) + (2) −
U (z) − U (z) = F (1) (z),
1
+ −
T (∂z, n)U (1) (z) − T 2 (∂z, n)U (2) (z) = F (2) (z).
Again the vector U (2) has to satisfy the decay conditions (3.1). Here
>
>
F (1) (z) = f (1) (z), f (2) (z) , F (2) (z) = f (3) (z), f (4) (z) ,
>
(j)
(j)
(j)
f (j) (z) = f1 (z), f2 (z), f3 (z) , j = 1, 2, 3, 4
Representation Formulas of General Solutions to the Static Equations
139
are given vector functions. The stress vector is defined by the formulas (3.3)
with the appropriate material constants equipped with superscript (j) and
with U (j) for U .
There hold the following uniqueness theorems [17].
Theorem 3.1. Problems (I)± , (II)− and (A) have at most one solution.
Solutions to Problem (II)+ are defined modulo the rigid displacement vectors
of the type
u(x) = [a × x] + b, ω(x) = a,
where x = (x1 , x2 , x3 )> , and a and b are arbitrary three dimensional constant vectors.
4. Explicit Solutions of the Boundary Value Problems
Here we demonstrate our approach for Problem (II)+ , since other problems can be treated quite similarly.
We look for a solution to Problem (II)+ in the form (2.3), where
Φj (x) =
∞ X
k
r k
X
(m)
(j)
Yk (ϑ, ϕ)Amk , j = 1, 2, 3,
R
k=0 m=−k
Φ4 (x) =
∞ X
k
X
(m)
(ϑ, ϕ)Amk ,
(m)
(ϑ, ϕ)Bmk , j = 1, 2.
gk (λ1 r)Yk
(4)
(4.1)
k=0 m=−k
Ψj (x) =
∞ X
k
X
hk (kj r)Yk
(j)
k=0 m=−k
Here
while
(j)
Amk ,
(j)
j = 1, 2, 3, 4, and Bmk , j = 1, 2, are unknown coefficients,
r
r
R Ik+1/2 (λ1 r)
R Ik+1/2 (kj r)
gk (λ1 r) =
, hk (kj r) =
, j = 1, 2,
r Ik+1/2 (λ1 R)
r Ik+1/2 (kj R)
s
2k + 1 (k − m)! (m)
(m)
P
(cos ϑ)eimϕ .
Yk (ϑ, ϕ) =
4π (k + m)! k
(m)
Pk (cos ϑ) are the Legendre associated polynomials, Ik+1/2 is the Bessel
function of real argument and Ik+1/2 is the Bessel function of complex (pure
imaginary) argument [29].
We assume that the functions Φj , j = 1, 3, and Ψj , j = 1, 2, satisfy the
conditions
Z
Z
Ψj (z) ds = 0, j = 1, 2,
(4.2)
Φj (z) ds = 0, j = 1, 3,
∂Ω0
where
∂Ω0
∂Ω0 = x : x ∈ R3 , |x| = R0 < R .
140
L. Giorgashvili and D. Natroshvili
Substituting Φj , j = 1, 3, and Ψj , j = 1, 2, into (4.2) and taking into account
the equalities
( √
Z
2 πR2 for k = 0, m = 0,
(m)
Yk (ϑ, ϕ) ds =
0
otherwise,
∂Ω
(j)
(j)
we get that A00 = 0, j = 1, 3, and B00 = 0,, j = 1, 2.
Substitute Φj , j = 1, 3, and Ψj , j = 1, 2, into (2.3) and use the relations [28]
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),
k(k + 1)
(m)
a(r)Xmk (ϑ, ϕ)+
rot rot xa(r)Yk (ϑ, ϕ) =
r
d
p
1
a(r)Ymk (ϑ, ϕ),
+
+ k(k + 1)
dr
r
to get
u(x) = u00 (r)X00 (ϑ, ϕ) +
∞ X
k
n
X
umk (r)Xmk (ϑ, ϕ)+
k=1 m=−k
+
h
io
p
k(k + 1) vmk (r)Ymk (ϑ, ϕ) + ωmk (r)Zmk (ϑ, ϕ) ,
ω(x) = u
e00 (r)X00 (ϑ, ϕ) +
+
where [25], [26]
∞ X
k
n
X
k=1 m=−k
(4.3)
u
emk (r)Xmk (ϑ, ϕ)+
h
io
p
k(k + 1) e
vmk (r)Ymk (ϑ, ϕ) + ω
emk (r)Zmk (ϑ, ϕ) ,
(m)
Xmk (ϑ, ϕ) = er Yk (ϑ, ϕ), k ≥ 0,
eϕ ∂ (m)
∂
1
Y
(ϑ, ϕ), k ≥ 1,
+
eϑ
Ymk (ϑ, ϕ) = p
∂ϑ sin ϑ ∂ϕ k
k(k + 1)
Zmk (ϑ, ϕ) =
1
∂ (m)
eϑ ∂
=p
Y
(ϑ, ϕ), k ≥ 1, |m| ≤ k,
− eϕ
∂ϑ k
k(k + 1) sin ϑ ∂ϕ
er , eϑ and eϕ are the unit vectors
er = (cos ϕ sin ϑ, sin ϕ sin ϑ, cos ϑ)> ,
eϑ = (cos ϕ cos ϑ, sin ϕ cos ϑ, − sin ϑ)> ,
eϕ = (− sin ϕ, cos ϕ, 0)> .
(4.4)
Representation Formulas of General Solutions to the Static Equations
141
The system of vectors {Xmk (ϑ, ϕ), Ymk (ϑ, ϕ), Zmk (ϑ, ϕ)}, |m| ≤ k, k =
1, ∞, is orthogonal and complete in L2 (Σ1 ), where Σ1 is the unit sphere,
r k+1
k r k−1 (1)
(2)
Amk + (k + 1)(bk − 2a)R
Amk +
umk (r) =
R R
R
2
k(k + 1) X
d
(j)
(4)
hk (kj r)Bmk , k ≥ 0,
gk (λ1 r)Amk +
+ (δ+2κ)
dr
r
j=1
r k+1
k−1
1 r
(1)
(2)
vmk (r) =
Amk + (b(k + 1) + 2)R
Amk +
R R
R
2
d 1 X
(δ+2κ)
(4)
(j)
gk (λ1 r)Amk +
+
hk (kj r)Bmk , k ≥ 1,
+
r
dr r j=1
ωmk (r) =
r k
R
(3)
Amk +
2
X
j=1
(j)
kj hk (kj r)Bmk , k ≥ 1,
σk(k+1)(2k+3) r k−1 (2) k(k+1) r k−1 (3)
u
emk (r) =
Amk +
Amk−
R
R
2R
R
2
k(k+1) X
d
(j)
(4)
ηj hk (kj r)Bmk , k ≥ 0,
− (λ+2µ) gk (λ1 R)Amk −
dr
r
j=1
σ(k + 1)(2k + 3) r k−1 (2)
k + 1 r k−1 (3)
vemk (r) =
Amk +
Amk −
R
R
2R
R
2
d 1 X
λ+2µ
(4)
(j)
−
gk (λ1 r)Amk −
+
ηj hk (kj r)Bmk , k ≥ 1,
r
dr r j=1
ω
emk (r) = −(2k + 3)
r k
R
(4.5)
2
X
(2)
(j)
Amk − ηj kj hk (kj r)Bmk , k ≥ 1, b = 1−a.
j=1
Now substitute the vectors u and ω into (3.4) and employ the equalities
er × Xmk (ϑ, ϕ) = 0, er × Ymk (ϑ, ϕ) = −Zmk (ϑ, ϕ),
er × Zmk (ϑ, ϕ) = Ymk (ϑ, ϕ),
d
2
(m)
a(r)Yk (ϑ, ϕ),
+
div a(r)Xmk (ϑ, ϕ) =
dr
r
p
a(r) (m)
div a(r)Ymk (ϑ, ϕ) = − k(k + 1)
Yk (ϑ, ϕ),
r
div a(r)Zmk (ϑ, ϕ) = 0,
p
a(r)
Zmk (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
d 1
p
a(r)
rot a(r)Zmk (ϑ, ϕ) = k(k+1)
Xmk (ϑ, ϕ)+
+ a(r)Ymk (ϑ, ϕ)
r
dr r
142
L. Giorgashvili and D. Natroshvili
to obtain
∞ X
k n
X
H (1) (∂x, n)U (x) = a00 (r)X00 (ϑ, ϕ)+
amk (r)Xmk (ϑ, ϕ)+
k=1 m=−k
H
(2)
h
io
p
+ k(k+1) bmk (r)Ymk (ϑ, ϕ)+cmk (r)Zmk (ϑ, ϕ) ,
∞ X
k n
X
(∂x, n)U (x) = e
a00 (r)X00 (ϑ, ϕ)+
e
amk (r)Xmk (ϑ, ϕ)+
(4.6)
k=1 m=−k
h
io
p
+ k(k+1) ebmk (r)Ymk (ϑ, ϕ)+e
cmk (r)Zmk (ϑ, ϕ) ,
where
h
h
d
2λ i
2δ i
d
umk (r) + (δ + 2κ)
u
emk (r)−
+
+
amk (r) = (λ + 2µ)
dr
r
dr
r
i
k(k + 1) h
λvmk (r) + δe
vmk (r) , k ≥ 0,
−
r
i
1h
bmk (r) =
(µ − α)(umk (r) − vmk (r)) + (κ − ν)(e
umk (r) − vemk (r)) +
r
d
d
+ (µ + α)
vmk (r) + (κ + ν)
vemk (r) + 2αe
ωmk (r),
dr
dr
i
h
1
d
− (µ − α) ωmk (r)+
cmk (r) = (µ + α)
dr
r
h
1i
d
emk (r) − 2αe
vmk (r),
− (κ − ν) ω
+ (κ + ν)
dr
r
h
i
h
d
d
2δ
2β i
e
amk (r) = (δ + 2κ)
umk (r) + (β + 2γ) +
u
emk (r)−
+
dr
r
dr
r
i
k(k + 1) h
−
δvmk (r) + βe
vmk (r) ,
r
h
i
1
ebmk (r) =
(κ − ν)(umk (r) − vmk (r)) + (γ − ε)(e
umk (r) − vemk (r)) +
r
d
d
vmk (r) + (γ + ε)
vemk (r) + 2ν ω
emk (r),
+ (κ + ν)
dr
dr
h
i
d
1
cmk (r) = (κ + ν)
e
− (κ − ν) ωmk (r)+
dr
r
h
1i
d
emk (r) − 2νe
vmk (r).
− (γ − ε) ω
+ (γ + ε)
dr
r
Here the vectors umk , vmk , . . . , ω
emk are given by (4.5).
Represent the boundary data as the Fourier–Laplace series with respect
to the system (4.4)
(j)
f (j) (z) = α00 X00 (ϑ, ϕ) +
+
p
∞ X
k
n
X
(j)
αmk Xmk (ϑ, ϕ)+
k=1 m=−k
h
(j)
(j)
k(k + 1) βmk Ymk (ϑ, ϕ) + γmk Zmk (ϑ, ϕ)
io
, j = 1, 2,
143
Representation Formulas of General Solutions to the Static Equations
(j) p
(j) p
(j)
where αmk , k(k + 1) βmk , k(k + 1) γmk are the Fourier coefficients.
In view of the boundary condition (3.2), with the help of the equalities
(4.6) we arrive at the system of linear algebraic equations
(1)
(2)
a00 (R) = α00 , e
a00 (R) = α00 ;
(1)
(1)
(1)
am1 (R) = αm1 , bm1 (R) = βm1 , cm1 (R) = γm1 ,
(2)
(2)
(2)
e
am1 (R) = αm1 , ebm1 (R) = βm1 , e
cm1 (R) = γm1 ;
amk (R) =
e
amk (R) =
(1)
αmk ,
(2)
αmk ,
bmk (R) =
ebmk (R) =
(1)
βmk ,
(2)
βmk ,
cmk (R) =
(4.7)
(1)
γmk ,
(2)
cmk (R) = γmk , k ≥ 2.
e
Let us analyze the solvability of this system. To this end, we prove the
following assertion.
Lemma 4.1. Let the conditions (4.2) be fulfilled. Then the vectors u
and ω represented the by formulas (2.3) vanish identically if and only if Φ j ,
j = 1, 2, 3, 4, and Ψj , j = 1, 2, vanish.
Proof. The sufficiency is trivial. Let us show the necessity. From the equations (2.3) we get
1
div ω(x),
λ21 (λ + 2µ)
∂
∂
i
δ + 2κ
1 h
2r
div u(x) +
+3 r
+ 1 Φ2 (x) =
div ω(x) .
∂r
∂r
2a
λ + 2µ
Φ4 (x) = −
If u and ω vanish, then by the formulas (4.1) we see that
Φ2 (x) = 0, Φ4 (x) = 0.
(4.8)
Again from (2.3) by (4.8) we have
ω(x) −
2 i
X
1
1 h
rot u(x) = −
ηj + kj rot rot(xΨj (x)) + kj rot(xΨj (x)) .
2
2
j=1
Whence, if u and ω vanish, we get
rot rot(xΨj (x)) + kj rot(xΨj (x)) = 0, j = 1, 2.
Taking scalar product of this equation by x leads to
h ∂ ∂
i
r
r
+ 1 + r2 kj2 Ψj (x) = 0, j = 1, 2.
∂r
∂r
Whence by (4.1) and (4.2)
Ψj (x) = 0, j = 1, 2.
(4.9)
The equations (4.8) and (4.9) along with (2.3) imply
u(x) = grad Φ1 (x) + rot(xΦ3 (x)),
1
ω(x) = rot rot(xΦ3 (x)).
2
(4.10)
144
L. Giorgashvili and D. Natroshvili
From (4.10) we get
∂ ∂
r
+ 1 Φ3 (x) = 2(x · ω(x)).
∂r
∂r
If we substitute in these equations ω(x) = 0 and apply (4.1) and (4.2),
we obtain Φ3 (x) = 0. In accordance with this equation we get
r
grad Φ1 (x) = u(x).
Whence Φ1 (x) = const, and by (4.2) we conclude that Φ1 (x) = 0.
Necessary and sufficient conditions for Problem (II)+ to be solvable read
as follows
Z
Z
(1)
z × f (1) (z) + f (2) (z) ds = 0.
(4.11)
f (z) ds = 0,
∂Ω
∂Ω
Substituting f
equations
(j)
, j = 1, 2, into (4.11) and taking into consideration the
q 2π


 3 (δ−1m − δ1m )e1 − i(δ−1m + δ1m )e2 +
√
Xmk (ϑ, ϕ) ds =
+ 2 δ0m e3 R2 , k = 1, m = ±1,


0 otherwise;
∂Ω
 p π

δ1m )e1 − i(δ−1m + δ1m )e2 +
Z
2 3 (δ−1m − √
Ymk (ϑ, ϕ) ds =
+ 2 δ0m e3 R2 , k = 1, m = ±1,


0 otherwise;
∂Ω
Z
Zmk (ϑ, ϕ) ds = 0 for all k and m,
Z
∂Ω
where δkj is the Kronecker symbol, e1 = (1, 0, 0)> , e2 = (0, 1, 0)> , e3 =
(0, 0, 1)> , we get (4.12)
(1)
(1)
(2)
(2)
(1)
αm1 + 2βm1 = 0, αm1 + 2βm1 + 2Rγm1 = 0, m = 0, ±1.
(4.12)
Theorem 3.1 and Lemma 4.1 imply that the system (4.7) is solvable.
Moreover, from (4.12) it follows that we can define all unknown coefficients
(1)
(3)
but Am1 and Am1 , m = 0, ±1. This is natural and reflects the fact that the
solution is defined modulo a rigid displacement vector.
In our analysis we need the following technical results [27].
Theorem 4.2. For k ≥ 0 the following inequalities are true
r
Xmk (ϑ, ϕ) ≤ 2k + 1 , k ≥ 0,
4π
r
2k(k
+ 1)
Ymk (ϑ, ϕ) <
, k ≥ 1,
2k + 1
r
Zmk (ϑ, ϕ) < 2k(k + 1) , k ≥ 1.
2k + 1
Representation Formulas of General Solutions to the Static Equations
145
Theorem 4.3. If f (j) ∈ C l (∂Ω), where ∂Ω is a sphere, then the coeffi(j)
(j)
(j)
cients αkj , βkj and γkj admit the bounds
(j)
(j)
(j)
αmk = O(k −l ), βmk = O(k −l−1 ), γmk = O(k −l−1 ).
Applying the above theorems and the asymptotic behavior of the Bessel
functions, we can show that for x ∈ Ω1 the series are absolutely and uniformly convergent in the interior domain.
Let now x ∈ ∂Ω1 , i.e., r = R. In this case, due to Theorem 4.2 the
majorizing number series for (4.3) and (4.6) is
∞
X
k=k0
1
k2
2 h
X
j=1
i
(j)
(j)
(j) |αmk | + k |βmk | + |γmk | .
By Theorem 4.3 this series is convergent if f (j) ∈ C 2 (∂Ω1 ).
Thus the series (4.3) and (4.6) are absolutely and uniformly convergent
if f (j) ∈ C 2 (∂Ω1 ).
Acknowledgements
This research was supported by the Georgian National Scientific Foundation (GNSF) grant No. GNSF/ST06/3-001.
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(Received 1.03.2008)
Authors’ address:
Department of Mathematics
Georgian Technical University
77, Kostava St., Tbilisi 0175, Georgia
E-mail: natrosh@hotmail.com