209
Internat. J. Math. & Math. Sci.
Vol. 8 No. 2 (1985) 209-230
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE
PROBLEMS IN COMPOSITE ELASTIC AND VISCOUS MEDIA
D. L. JAIN
Department of Mathematics
University of Delhi
Delhi 110007, India
R. P. KANWAL
Department of Mathematics
Pennsylvania State University
University Park, PA 16802
(Received April 5, 1985)
ABSTRACT.
We present the solutions for the boundary value problems of elasticity
when a homogeneous and istropic solid of an arbitrary shape is embedded in an infinite
homogeneous isotropic medium of different properties.
The solutions are obtained
inside both the guest and host media by an integral equation technique.
The
boundaries considered are an oblong, a triaxial ellipsoid and an elliptic cyclinder
of a finite height and their limiting configurations in two and three dimensions.
The exact interior and exterior solutions for an ellipsoidal inclusion and its
limiting configurations are presented when the infinite host medium is subjected to
a uniform strain.
In the case of an oblong or an elliptic cylinder of finite height
the solutions are approximate.
Next, we present the formula for the energy stored
in the infinite host medium due to the presence of an arbitrary symmetrical void in
it.
This formula is evaluated for the special case of a spherical void.
Finally,
we analyse the change of shape of a viscous incompressible ellipsoidal region
embedded in a slowly deforming fluid of a different viscosity.
Two interesting
limiting cases are discussed in detail.
KEY WORDS AND PHRASES.
Isotropic solid, composite media, strain energy, viscous
inhomogeneity, triaxial ellipsoid.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODE,
i.
73C40.
INTRODUCTION.
Composite media problems arise in various fields of mechanics and geophysics.
In this paper we first present the solutions for boundary value problems of
elastostatics when a homogeneous and isotropic solid of an arbitrary shape is
embedded in an infinite homogeneous isotropic medium of different properties.
solutions are obtained inside both the guest and the host media.
The
The boundaries
considered are an oblong, an ellipsoid with three unequal axes, and elliptic
210
D.L. JAIN AND R. P. KANWAL
cylinder of finite height and their limiting configurations in two and three
The exact interior and exterior solutions for an ellipsoidal inclusion
dimensions.
and its limiting configurations are presented when the infinite host media is
For other configurations the solution presented are
subjected to a uniform strain.
Next we present the formula for the energy stored in the infinite
approximate ones.
host medium due to the presence of an arbitrary symmetrical void in it.
formula is evaluated for the special case of a spherical void.
This
Finally, we
analyse the change of shape of a viscous incompressible ellipsoidal region embedded
in a slowly deforming fluid of a different viscosity.
Two interesting limiting
cases are discussed in detail.
The analysis is based on a computational scheme in which we first convert the
boundary value problems to integral equations.
Thereafter, we convert these
integral equations to infinite set of algebraic equations.
scheme then helps us in achieving our results.
A judicial truncation
Interesting feature
of this
computational technique is that the very first truncation of the algebraic system
yields the exact solution for a triaxial ellipsoid and very good approximations
for other configurations.
The main analysis of this article is devoted to three-dimensional problems of
elasticity and viscous fluids.
The limiting results for various two-dimensional
problems can be deduced by taking appropriate limits.
2.
MATHEMATICAL PRELIMINARIES
(x,y,z)
Let
be Cartesian coordinate system.
solid of arbitrary shape of elastic constants
k2
A homogeneous three-dimensional
and
is embedded in an infinite homogeneous isotropic medium of
X1
and
i"
R
the centroid of
1,2
R
R
1
Let
2.
+
S
S
+ R 2.
0
of elastic constants
1
of the coordinate system is situated at
be the boundary of the region
The stiffness tensors
Cijkg(),
R
2
so that the entire
(x,y,z)
R,
are constants and are defined as
XSij6k + a(6ikj + i6jk),
Cijk
where
R
2
The elastic solid is assumed to be symmetrical with respect to the
three coordinate axes and the origin
region is
R
occupying region
2
6’s are Kronecker deltas.
(2.1)
The latin indices have the range 1,2,3.
The integral equation which embodies this boundary value problem is derived in
precisely the same fashion as the one in reference [i].
field
u.j(x)~
()
0
u.j(x)~
Indeed, the displacement
satisfies the integral equation
+
ACigkm /R Gjm,k(X’X~~’)ug,i,(x’)dR,~
2
x~
R,
(2.2)
u0(x)
where subscript comma stands for
is the displacement field
differentiation,
in the infinite host medium occupying the whole region R due to the prescribed
2
1
stressed at infinity,
while Green’s function
ACigkm
the differential equation
Cigkm
Cikm,
Gkm
satisfies
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
c.1ljkgGkm,gj (x,x)
6 (x-x’)
and
6ira 6(x-x’) x,5"
(2.3)
R
Explicitly,
is the Dirac delta function.
Gij (x,x’)
211
Gji(x,x’)
8
Ix-
kl*Z I
ij
x" I.
(2.4)
For the sake of completeness and for future reference we write down briefly
scheme for solving the integral equation (2.2).
the basic steps of the truncation
To obtain the interior solution of the integral equation (2.2) when
differentiate equation (2.2)
uj, Pl---Pn (x)
u
0
j,
n
we
times to get
(-l)n+l ACikm
Pl---Pn
x R2,
I
G
R2
k,p{___pn(X,X
j m
)u
i" (x’)dR 2
(2.5)
R2
p’s have the values 1,2,3.
where
0
Taylor series about the origin
u
x"
R
ug,i,(x’)
in
Thus,
2.
...x"
z
,i
ql---qs(5) }x
where
q’s
5
in both sides we obtain
u
0
’Pl---Pn
(0)
(2.6)
l---Xqs’
Substituting these values in (2.5) and setting
have the values 1,2,3.
uj ’Pl---Pn (0)
u
where
i
s= 0
,
Now we expand the quantities
(-i)
n+l
ACigkm
Tjm’kPl---Pn’ql---qs
s=O
(2.7)
6, iql---qs (Q),
where
/
Tjm’kPl---Pn’ ql---qs
R
Gjm
kp
2
As in reference [i], taking
u),
uj (0)
l---Pn
n
(x,O)x
---x
ql
0,i, s
qs
dR 2,
(2.8)
0, in equation (2.5) we get
and
0
Uj,p )
Uj,p)
ACigkmTjm,kpUg,i (0),
(2.9)
respectively, where
f
Tjm,kp
{I 16 jmtkp
while
M
1
k
I
+ 21
and
tjmkp
R2
+
tjmkp
1
8
y
G
jm,kp (x,0)dR 2
(M i_
I
t
j
mkp}
(2.10)
are the shape factors
4
0 r
R2 Oxj 8XmOXkOX P
dR 2, r
(2.11)
D.L. JAIN AND R. P. KANWAL
212
2
Now we substitute the value
u.
3,P
(0)
uiJ ,P (0)
+
and
from (2.1)
in
(2.9) and get
AkTjk,kpU,(O)
A(Tjm,kpUm,k(O)~ + Tjm,kpUk,m(0)).
u.3,p(O)
When we decompose
i
ACjmkp C-’mkp3 Cjmkp
(2.12)
into the symmetric and antisymmetric parts
Ujp )
respectively, as we did in reference [I] and define
ajp(O)
i
TIjm,kp
--(Tj
2
m,kp- Tpm
kj
we find that relation (2.9) yields the following two relations
ujpCO)
u
ajp(O)
0
ajp (0) AC i gkmTjm,kpU i(O)"
P
AC
(0)
+
i
gkmTjm,kpUgi (0),
(2.13)
(2.14)
Equation (2.13) gives rise to the relation
Ukk(O) [I
+
2A[
Ak
Ull (0), u22(0), u33(0)
The values of
0
u33 (0)
and
u
--0
(2.16)
u
are
Ull (o)
u22 (0)
Ull
u22
o
u33 (o).I
--0
u
LU33(0)
while the elements
(i-2
bij,
1,2,3
i,j
of the matrix
i tiikk)6iJ i tiikk- 2A(I- i
and the suffices
i # j, i,j
i
1,2,3
and
j
are not summed.
B
(2.17)
are given as
(2.18)
tiijj’
Furthermore, the values of
are given in terms of the known constants
relation
[l-A{ll(tjjkk+tiikk
+
4(M; I
0
ui4J (0),
uij (0)
0
aij(0)
A
+l (tjjkk-tiikk)Uij(0)’
is defined by
i # j, i,j
aij(O),
1,2,3,
(2.20)
(2.19).
Finally, substituting the above values of
expans ions
ulj (0),
J by the
(2.19)
Similarly, equation (2.14) yields the values of non-zero components of
in the form
aij(0)
i #
0
ll)tiijj}] -Iuij(0).
i # j
where
0
UlOl(0), u22(0),
are given by the matrix equation
where the column vectors
uij(0)
(2.15)
in terms of the known constants
Bu
bij
0
Ukk (0).
-i -I tllkkUll (O)+t22kkU22 (O)+t33kkU33 (0)
uij (0)
and
aij (0)
in the
213
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
u() + (Uik() + aik())xk,
ui()
E R 2,
yield the required approximate inner solution where
(2.21)
Relation
Xl--X, x2=Y, x3=z.
(2.21) gives the exact solution for an ellipsoidal inclusion and its limiting
configurations when the infinite host medium is subjected to a uniform prescribed
stress.
In the case of elastic inclusions which are symmetrical with respect to the
three coordinate axes and have only one characteristic length as in the case of a
sphere a cube etc., there are only two distinct non-zero shape factors, namely,
Indeed since relation (2.9) yields
tllll ti122.
I
=
/ V 2 (V 2 r )dR
2
R
2
tkkmm 8--
it follows that in this case
t2222
tllll
t3333; ti122
t2233
/
R
[-86()]dR2
(2.22)
-i,
2
t3311; tllkk
t33kk
t22kk
.
i
(2.23)
When we substitute these relations in (2.15) and (2.16), we get the simplified
results,
Ukk(O)
[i +
AK -i
i
0
Ukk (0);
K
k
g
2
+ p,
gK
Ak
+ 2 A,
(2.24)
(2.25)
5
d
where
c
2A-I+c-I
A
I+2AG3 (MI- l)(tllll-tl122)},
-I
-A
+ C-I
d
I
AK
C= I+-M
I
Similarly, in this case, results (2.19) and (2.20) yield
uij(O)
[I+2A{13I 2(MI- I )tiijj }]-l uijO (0),
i # j, i,j
(2.26)
1,2,3,
and
0
aij (0).
aij()
The above values of
uij(0)
and
(2.27)
aij(O)
when substituted in the expansions
(2.21) give rise to the required inner solution in this case.
In order to complete
the analysis of this section we need the values of the shape factors of various
inclusion.
3.
They are presented in the next section.
VALUES OF THE SHAPE FACTORS FOR VARIOUS SOLIDS
(i) Oblong.
so that the region
Let the faces of the oblong be given by
R2
is
Ixl
a,
IYl
b,
Iz
< c.
x
+/-a, y
In this case,
+/-b, z
+/-c
D.L. JAIN AND R. P. KANWAL
214
2
tllll
tan
-i bc
i
ti122
A
where
+
a2+d 2
a2A2+b2
i abc
A
abc
t
(a2+b2)A
(a2+b2+c2) 1/2
C
2
i
abc
and
t
)’a2+c2"A
1 133
tan
-i bc
(3.1)
All the other shape factors can be
is summed.
k
2
1 Ikk
obtained by permutations.
For a cube of edge
2a, the above values reduce to
tllll
t2222
t3333
ti122
t2233
t3311
tllkk t22kk
t33kk
When we take the limit
i
i
+--
I
(3.2)
2V
i
in relations (3.1), we obtain the following
c
.
values of the non-zero shape factors for an infinite rectangular cylinder occupying
R2: Ixl
the region
2
tllll
tan
2
tan
2222
b
Setting
< a,
IYl
-i b
< b,
< z <
ab
ab
--a + (-a2+b2) t1122 t2211
i a
(a2+b 2
(3.3)
ab
+
(a2+b 2)
in the above formulas we obtain the values of the corresponding
a
shape factors for an infinite square cylinder occupying the region
R
xl
2
< a,
i
tllll
Yl
< a,
+1
=.
< z <
These values are
I
t2211
ti122
i
t2222
2---
I
+ 2---
(3.4)
The limiting results (3.3) and (3.4) agree with the ones obtained in reference [i].
(ii) Triaxial Ellipsoid.
Let the equation of the surface of the ellipsoidal
elastic solid be
x
a
a, b
where
In this case
and
R
2
2
2
c
+
2
+
i,
are the lengths of the semi-principal axes of the ellipsoid.
x
is the region
2
non-vanlshlng shape factors are
R
where
and
u
j
c > O,
b
a
c
tiiii
3abc /
4
0
tiijj
ab__c f
4
0
2/a 2 + y2/b 2
udu
udu
2
,i
< 1, and the values of the
(3.5a)
1 2 3
J, i,j
1,2,3,
(3.5b)
(u+ai) (u+a.)R
u
j
[(u+a2)(u+b2)(u+c2)] I/2
are not sued.
i
(u+a)2Ru
2
+ z 2 /c 2
a
I
a, a
2
b, a
3
c
and the suffices
For a prolate spheroid with the semi-principal axes
i
215
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
a,b,b, a >_ b, the foregoing shape factors reduce to
3ab2
tllll
0 (u+a
-
3ab
t3333
t2222
ti122
ab
ti133
t2233-
ab 2
4
f
where
2)
(l-k
LI
In the limit when
L
I
1/5
0 (u+b
2)
3
(u+a
(u+b
i
b
2)
l+k.
(Z)
i
’
i
15’
i
(3.6c)
k
2
k
2
2
b__
I
(3.7)
2
O, in relations (3.6) and (3.7) we find that
k
i # j,
and the suffices
(3-k2)Ll
(3.6d)
a
a, i.e.
(3.6b)
t2 222’
3
i
6 (l-k2) + 6 (3+k2)Ll
(l-k2)
i
3
+
2) 1/’2
2)2(u+a 2)3/2
udu
log
3
udu.
udu
f0
tiijj
1,2,3
i,j,k
f
(3 6a)
and the shape factors for a sphere of radius
I
tiiii
2
2
k
k
2
2) 5/2 (u+b 2)
(u+a 2) 1/2 (u+b
0
(l-k2) (3LI-I)’
i
udu
f
4
I
tiikk
(3.8)
3
and
i
are
a
are not summed.
j
Similarly, the shape factors for the oblate spheroid with seml-principal
axes a,a,b, a >_ b derived from relations (3.5) are
3a2b
t2222
tllll
3a2b
t3333
(u+b)
0
a
t2233
ti133
0 (u+a
udu
2 5/2
f
0
4
3
2) 3(u+b 2) 1/2
i
2
(u+a
1
(u+a2)(u+b2)3/2
i-
i
+
6 (I+<2) + 6 (3-<2)L2
(3L2-i)
udu
udu
0
(-1+<2)
(u+a)
2b=
4
__a2b
ti122
udu
f
4
(1+<2)
(3.9b)
1
(3+<2)L2
(3.9c)
(3.9d)
tllll’
2) 3 (u+b 2) I/2
(3.9a)
where
L2
When
"i+<I<4 ),< tan(
i
i + i <2
<-
),
<2
2
(a 2
i).
(3.10)
b
b
a, we again recover the corresponding results for a spherical inclusion.
In the limit <
in relations (3.9) and (3.10), L
O,
1/3,
2
we find that the values of the non-vanishing shape factors for an extremely
<2(3L2-I)
thin oblate spheroid are
tllll
1/8,
t2222
Similarly, when
-
ti133
1/6,
t2233
in results
a
1/24.
ti122
(3.5), we deduce the shape factors for the
y2/b2 + z 2 /c 2 < i, < x
R
2
infinite elliptic cylinder occupying the region
Then the non-zero values of the shape factors are
t2222
3bc
udu
f
0
(u+b
2) 5/2 (u+c)2 1/2
-c(b+2c)
2
2(b+c)
I e
2
0
sinh
O(2-e Ocosh gO ),
.
D. L. JAIN AND R. P. KANWAL
216
3abc
4
t3333
bc
b/c
/
(u+b
0
udu
2 3/2
I
-b (c+2b)
2
udu
(u+b2)I/2(u+c2) 5/2
t3322
t2233
and
/
0
i
-bc
(u+c
2 3/2
2 (b+c)
e
2
0’ which agree with the known results
coth
0cosh 0(2-e
e
2
2(b+c)
-20 sinh
[i].
When
c
0
s
inh0)
2 0,
(3.11)
b, we recover
0
the shape factors for a circular cylindrical inclusion while in the limit
0
we
htain the corresponding values for an infinite strip.
Let the elliptic cylinder of height
Elliptic Cylinder of Finite Height.
(iii)
2h
occupy the region
2
}a
R2
where
a
and
2
+
Izl
i,
h,
In this case the
are the lengths of its semi principal axes.
b
values of shape factors are
tllll
t2222
t3333
ti122
ti133
/2
3hab
/
2
/2
3hab
J"
[-sin29+
(L)
2
I
0
hab
hab
/2
/
2s in 2q
(h2+a2cos2)
[-cos2l + 3b2sin2qcos2
2
[-sin2
h
+
h
/2
[-sin
/
0
2 .2
oa
h
2
h
3
2 in2qcs 2
+ 3b s
{i-
/2
[_cos2q
(h2+b2sin2)
0
where
2
L2+h
(L2+h 2
(h2+a 2 cos 2
(3.12c)
L2+h
2
dq
}]
(L)
2
2
(3.12d)
/L 2+h 2
dq
L2+h2
(3.12e)
(h2+a 2cos 2) L2+h 2
a2cos2
3
(3.12b)
2
dq
2
b2sin
3(
L2+h2
q I]
3 (L2+h 2)
{i-
(3.12a)
dq
2
(e)
b sin
{I
I-
(L)2
}]
3 (L2+h2)
qcos
2 +(h2+a2cos2q)
dq
}]
2
2
sin
2
(L2+h 2)
3
(L)
0
!
il-
4
i
{i
(L)2
0
3hab
4
a cos q
0
hab
2233
[_cos2q+
(3.12f)
d
(L2+h 2)
(h2+b 2s in2q)
L2+h
2
L2 (a2cos2+b2sin2).
In order to get the corresponding shape factors of a circular cylinder of
a in the above formulas and obtain
radius a and height 2h, we let b
t
iiii
ti133
t
2222
16
1
4
t2233
(+
-
t3333
3 )’
h/
-
a2+h 2.
These results, in turn, yield the shape factors of a circular disc
of radius
small thickness 2h. They are
tllll
t1122
t2222
__3
16
(a +
h
3
2__
tllll’tl133 t2233
t
-i
3333
4
a
2a
+ I
3
(3h
3
5 h
2 3
a
(3.13)
a
and
217
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
EXACT INTERIOR AND EXTERIOR SOLUTIONS FOR AN ELLIPSOIDAL ENCLOSURE OCCUPYING
2
2
2
REGION R
+
< i.
2
a
c
4.
b-2 +
When the infinite host (homogeneous and isotropic) elastic medium occupying the
R
whole region
is subjected to a prescribed uniform stress
of z-axls, the components of the vectors
u
Tl
E1
0
I (x)
o’s
where
x, u
Tl
0
2(x)
E--
have to be defined.
0
0
0
aij
(0)
0
(x)
0
i
UB(X)
y’
and
o
z,
22 (x)
T
(x)
along the direction
are given as
0
x
T
R,
(4.1)
Accordingly, in this case
ToI
Ul[ (0)= u22 (0)
u
0
T
0
u33(0)
El
(4.2)
1 2 3.
i # j,
When this infinite host medium has an isotropic elastic ellipsoidal inclusion of
Lame’s
constants
k2’
occupying the region
2
x
2
+
2
+
a
R2:
2
<i,
c
then the exact inner solution is given by
u
Ull(O), u22(0)
where the constants
(2.17).
(u33(0))z, x
(u22(O))y, u 3(x)
(Ull(0))x, u 2(x)
I()
and
u33(0)
(
(4.3)
R2,
are given by the matrix equation
Thus
Bu =u
b.. of the matrix B
The solution of equation (4.4), after
where the components
(glh2=g2hl ), u22()
Ull(O)
(4.4)
E1
or
are defined in (2.18).
me simplifications, is given as
(hlf2-h2fl ), u33()
(flg2-f2gl),
(4.5)
where
fl
I
’)
i
gl
+ 2
Ak
2kk
2&
i
2
(4.6a)
(4.6b)
i
A
hi
f2
2A_ _ _I tllkk- Mq (tllkk-2kk) 2A(l -i)(i tllll t1122
A___
(tllkk_t22kk)
(I l)(tl
i 122-t2222
I t2
i (tllkk-t22kk) 2Ak(l _i)i (tl133-t2233)’
A
_i)
i tllkk M (tllkk+it33kk) 2A(I i lll+itl13 3)
A
(t I
(4.6c)
(4.6d)
218
D. L. JAIN AND R. P. KANWAL
M (tllkk+it33kk) 2A(I
Ak
g2
h
A
I(I-2 i t33kk)
(I-2Ol)T
o
2
(glh2-g2hl)(l+
+
t’s
where
(4.6e)
(tllkk+t33kk)-2AG(l- l)(tl i133+ t3333)
i
(4.6f)
I
(4.6g)
Ak
2AG
MI
M
(flg2-f2gl)(l
i) (tl122+t2233)
i
t
I
AA
+
llkk
2
(hlf2-h2f)(i
i
Ak
2AI,
+ M
I
M
t22kk
I
(4.6h)
A
t33kk)’
+i- i
are the shape factors of the ellipsoid as given by (3.5), when we set
al=a a2=b, a3=c.
uii(),
When we substitute the values of
i
1,2,3, from (4.5)
in (4.3), we get the exact inner solution.
Limiting Cases
To check these results we take the limits
x
region reduces to the spherical region
b
2+y2+z2
a, so that the ellipsoidal
a, c
a
2
Now for the sphere there are
only two distinct non-zero shape factors, namely,
t3333
r2222
tllll
;i
and consequently
tllkk t22kk t33kk
ti122
.
t
2233
t311j
I
I-
i
Thus, for this limiting case we have
fl
-gl
1
+
2A’(I +5 (I
I)}’
(4.7a)
(4.7b)
(4.7c)
(4.7d)
2Ak.
i(i+ 3TI)
h2
+ Ak
2Az (1
TI(I+<I) +MI i])(1+3oi
(4.7e)
(4.7f)
MI
(i+
where
,K)
11 [gl(2h2-f2-g2) ]’
K
Ak
+ (2A)/3.
(4.7g)
Also,
Tl
u33 (0)
where
i
2Tl (1 )1
i
2
+ T
T
2
i
TI (X + )’
i
i
),
(4.8a)
(4.8b)
219
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
i (7-5I)+2 (8-101)
3I
I
(1-2c1) [2
A<
(i+ .-7-)
C
(4.9)
15 I (1-I
M
I (1-2c2)+ 2
(a
(1+
2
(4.10)
3 I (i-oI) (i-22)
Substituting the values of
from (4.8) to (4.10) in (4.3), we obtain
uii()
the exact inner solution for the spherical inclusion.
ui()
in spherical polar coordinates
r
Ur(X)~ {[u II ()
+
u33 ()] + [u33()
r
[u33 (Q)
u0(x
(r,8,@)
Expressing the components
r < a,
we have for
Ull()]cos
2},
(4.11a)
(4.11b)
28
Ull ()]sin
The corresponding non-vanishing stress components are
rr(X)
+
k2 [2ull(0)
u
+
33(0)
+
2 {[u II(0)
u
33(0)~
+
[u33(0)
Ull(0)]cos
20,
(4.12a)
k212Ull(0)
T06(x)
r0(x)
-2[u33(0)
(x)
k [2u (0)
2
Ii
+
u33(0)]
+
G2{[Ull(0)
u
33
(0)] +
u33(Q)]
u33(0)-u ll(0)]cs
2t,
(4.12b)
(4.12c)
28,
Ull(0)]sin
+
+
(4.12d)
22Uli(0)
As far as the authors are aware [3,4] even these exact interior solutions for a
sphere are new.
Interior solutions for a prolate-spheroidal enclosure of semi-principal axes
a,b,b, a
b
are obtained by appealing to the corresponding shape factors.
values of the shape factors
and
t
t-.
i # j, i
llll
llJJ
are not summed), are given by relations (3.6) while,
tllkk
t22kk
1
2
(l-k2)(3Ll-l) + (3-k2)Ll
t33kk
i
I
k
2
3(I+-)
+
i
k
2
1
(l-k2)
j
1,2,3 (i
The
and
k2Ll
e I.
The values of the shape factors and relations (4.3), (4.5) and (4.6) lead to the
required exact solutions.
Similarly, using the shape factors of oblate spheroid as given by relations
(3.9)
we obtain from equations (4.3), (4.5) and (4.6) the exact solution for this
limiting case.
Formulas for various other configurations such as an elliptic disk
can now also be derived.
In precisely the same manner we use the shape factors of the oblong as given by
(3.1) and that of elliptic cylinder of finite height as given by (3.12) and derive
the first approximation to the interior solutions of these cases from equations
(4.3), (4.5) and (4.6).
These results yield, in the limit, the corresponding
formulas for the configurations such as a cube and a circular cylinder of finite
height.
Let us now discuss the exact outer solution for an ellipsoidal enclosure
2
2 2
i. We have found that the exact
x /a + y
+ z 2 /c
occupying the region
R2:
2/b2
inner solution in this case is given by relation (4.3) where the values of the
D. L. JAIN AND R. P. KANWAL
220
Ull(0),
constants
1,2,3
i
are given explicitly by equations (4.5) and (4.6) in
terms of the known shape factors of the ellipsoid.
Substituting this inner
solution in the governing integral equation (2.2) and setting
(x=x,y,z),
we have
u) +
uj(x)
Xl=X, yl--y, Zl=Z,
+ 2
+
+
AXUll(0)
u22(0)
/
AUll(0)
R
R2
jl ,i
2
+ u 22 (0) /
(x,x’)dR_
R2
Gj2,2(x,x’)dR2
(4.13)
R I,
2
Green’s
where the components of
G
Gjk,k(X,x’)dR2
/
R
Gj3,3 (x,x’)dR, x
/
u33(0)
u33(0)}
+
function
-
are given by (2.4).
Gij(x,x’)
Various
integrals in this relation can be evaluated in the following way.
Gjl,l(X,X )dR$
/
R
2
8
+
i
i
where
{2
8
dR
2
ox J R2/
(x
I
( )
Ml i
I)
(i
8
-(x)
aR
{ 8ij o s lx-x’l
8Xl R2
i
8
2
--(x. (x)
x dR
2
R2/
x-x’
(4.14)
j (x)),
is the Newtonian potential due to the solid ellipsoid of unit density
occupying region
2
R2 x2/a
+
y2/b2+z2/c 2 <1,
at the point
x..
R
1
the Newtonian potential due to a solid ellipsoid of variable density
R I, that is
the region R 2 at the point
2
3
x
k
du
V
@(x)~
V=
and
ala2a3
abe,
2
a.x
Ru
3
S
RI
(u+ak2)
k=l
R2
V
x.
j(x)
is
occupying
x
dR
xdR
-x"
and
Z
{i-
2
x
k
du
x
R I, j
(4 15)
1,2,3, (4.16)
(u+a21) (u+a22)(u+a)}112,
is the positive root of
x
2
a2+
2
2
+
+-b2+ c2+{
i,
x
6 R
I(>0).
(4.13) can be
Similarly, other integrals occuring in the right hand side of equation
obtain
evaluated. Substituting these values of the integrals in (4.13) we
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROLBEMS
A
Q0) +
2
0
)}
(0) +
221
0
+A{-I ujj (0) xj ((x))
__i) (Ull(O) 02
(i- i
+
where
O--l
1,2,3
j
+
x
and
E R
u22(0)
and
I
0
2
02
+
u33()
0
2
is not summed.
j
All the terms on the right
Indeed, the first term is given by (4.1), the functions
side are known.
are known from (4.15) and (4.16) while the quantities
j(x)
are expressed in relations (4.5) and (4.6).
(4.17)
o3)(xj(x)-j(x))},
uii(O),
i
(x)
and
1,2,3
Let us check this formula by
considering the limiting case of a spherical inclusion.
Let
a, c
b
-
(x)
a
j (x)
du
/
2
2
(u+a
xj
/
(i
r -a
=r
u+a 2
Ix
r
> a
(4.18)
5
2
r
(u+a)2)
2 2
r -a
3
a
2
r
(I-
2 3/2
5
ax.
du
(u+a
where we have used the fact that
values of
(4.17) so that
in relations (4.14) to
a
3
a
2
+
r
2
Ix
r
2)5/2 15r3
(4.19)
> a
Substituting these values and the
from (4.8) (for the sphere) in (4.17) we get the required
uii(O)
exterior solution for the spherical enclosure, namely
Ak
(I
+
---i) (Ull
i
02
(0))(x12
+
02
---)
Ox
+
a
3
02
Ox3
Ixl
Ixl
[r + -]r + [(i-’2i)
[2 __C
2
s
ue()
r
6B
+--]sin
r
r
3
--]COSr 2e,
8
(0)
a
r > a, j
3
(r)
5
15r
r > a, and writing
coordinates we obtain
u$(x)
2
+
u33(0) _-] [xj (a___
3r
2
u(x) + u. (x),
uj (x)
Setting
8
+
3 )]}
1,2,3.
u.
(x)
(4.20)
in spherical polar
(4.21a)
(4.21b)
2,
where
A
3
a
5T
24 1
(
I- 2) (5-401
(7-5) l+(8-10Ol)G
1
(1+o
B
5
a
2
2
l (1-2o2)- 2 (l+c l) (1-2o 1)
T
2g
(1+o 2)
6I
I (i-2o2)+ 2
(i-2)
T
8I
(7-5oi) i+(8-10 I) 2
(4.22a)
(4.22b)
-
222
a
5
T
C
(l-2Ol) (l
-
D. L. JAIN AND R. P. KANWAL
2
(4.22c)
8---I (7-5oi)Gi+(8-i0oi) 2
The corresponding stress components are
Srr(X)~ XlA
{[-r +---r
(l+l)C
3
(i-2oi)r
[-
i
s
09 (x)=klA +
5-
12-
4
rsS (x)~
-
2
+ 2I
+ 48
36
(4.23a)
(4.23b)
}sin 26,
1+41
3
-3 - 21B’r
2C
9B
r
r
Icos
28},
(4.23c)
cos 28
(4.23d)
r
5B
C
+- +-] + 3[ (l-2J )r
r
28},
r
2l{-[r--3 + -]r + [(122’i
s (x)=XIA + 2I{-
4oi
[-2(i-219 +---r ]cos
+
3
r
1
where
A
k
When
Spherical Void.
-(--)[1+3
r
and
2
cos
in the above relations we obtain the
0
G2
28].
corresponding interior and exterior solutions for a spherical void of radius
a.
For
example, the components of the stress tensor at the outer surface of the void are
T
s
rr(a’8"4)
(l+cos
T
s
8 (a,8,)
{5(1-2oI
2(7-5Ol)
T
sin 28,
2}
(8+5oi)cos
3T
s (a,6,)
=-
s
r(a’e’)
2e)
28}.
2(7_5Ol) [l+5OlCOS
Relations (4.23) agree with the known results and serve as a check on our formulas.
Finally, we present the outer solutions for the limiting case of the prolate
- -
spheroid where semi-axes are
x
a,b,b, a >_ b, i.e.,
2
+
a
Y2+Z2 2
< i,
b
2
a
2
(l-e2).
b
In this case relation (4.15) reduces to
(x)
2
=--- u+a
ab
du
2
2
2x
3 3
a e
ann
y2+z2
ae
a3e
where
x
(u+b
u+a
2)
.-I
+b
RI,
2
i.e.,
2
u+b
V;+
VK+a
3
Y2+Z2}
2
2
x
{1-
tann
> O.
2
2
/a-b
]’
A/.------’
+a
ab2
/a2_b
[e tanh-i (V+a
2
2
223
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
Similarly, relation (4.16) becomes
el(X)
x[--3 {tanh
4
_"
-/ +a"
+a 2
V
a e
tann
a5e5
+a 2- $+2
2(y2+z 2) {-i ae+a2 + a2b 2
2
5+a
(+b 2)
a5e 5
ab
Cj (x)
[a--e {ae +a
ae+a +
{
4
xj
2
2x
5 5
3
(+b
+b
55
a e
tanh
-I
2
I
a e
i
2
{
2
22
+a
tanh-i
-:a2}
+j(x)
and
j
x
"/’ ’2’
V +a1,2,3
R
I(>0)
j=2,3
in equations (4.17)
from the inner solution for this limiting
uii(O)
and using the limiting values of
-
+a
tanh
(+b)
(x)
Substituting these values of
2
2
2
8
(+b)
a2-b22
3
.-i
tann -/
2
a2-b
3
2
V
-+a2J
3
configuration, we readily derive the exact exterior solution for the prolate
spheroid.
All the other limiting configurations can be handled in the same way.
5.
ARBITRARY SYMMETRICAL CAVITY AND STRAIN ENERGY
By a symmetrical cavity we mean a cavity which is symmetrical with respect to
three coordinate axes.
Observe that this is also true for a symmetrical inclusion
for which the method of finding the interior solution is given in Section 2.
Interior solutions in the case of an arbitrary symmetrical cavity embedded in
an infinite elastic medium are obtained in terms of the shape factors of the
2
inclusion by setting
O,
O, in the analysis of Section 2.
2
This interior
solution yields the values of the displacement field at the outer surface of the
inclusion.
Indeed, due to the continuity of the displacement field across
S
we
have
u
0
(x s)
s
+~u (Xs)
+
u
~(x
s )I +
-"
Thus
ui(s) l+
where the superscript
s
s
cavity, the stress field vanishes inside
tractions across
ni(SS)l+
so that
O, or,
-’
implies the perturbed field.
S
Since the inclusion is a
and due to the continuity of the
S, we have
0
ni(S
+
Sni -,(xs)l +
:sni (Xs)
+
0
_:oni (Xs)"
(5.2)
D. L. JAIN AND R. P. KANWAL
224
Thus from the interior solution derived by us, we can find the components of the
displacement field
u(S) l+
by using formula (5.1).
Formula (5.2) gives the
values of the perturbation in the tractions across the outer surface
i(S
cavity in terms of the known values of
S
of the
due to the prescribed stresses to
which the host medium is subjected.
E
The elastic energy
stored in the host medium due to the presence of the
symmetrical cavity is given by the formula
E
i
s
ui(S)’+ sni (Xs) I+dS
f
S
(5.3)
Note that in the above formula we have dropped the second integral taken over the
sphere of infinite radius because it vanishes when we appeal to the far-field
behavior.
u.(x)
l
0
ij
r
)
O(
as
r
r
of the displacement and the traction fields.
Let us illustrate fromula (5.3) for the spherical cavity embedded in the
R
is r < a. For this purpose we assume
2
that the prescribed stress field is such that we have the uniform tension T in the
infinite host medium so that the region
directions of
x,y,z
axes before the creation of the cavity.
In this case the
components of the displacement field are
0
T
1-21
1
0
(5.4a)
or
u
0
r (x)
1-2o
T
0
u()
(-I+oi)r, Us)
O,
x
The corresponding non-vanishing components of the stress tensor
0
Xll (x)= 022 (x)
0
x33 (x)=
T
(5.4b)
R.
0ij (x)
are
x E R
(5.5a)
x
(5.5b)
or
0rr
3TOl,
0
0
T
tee
i+o
I
R
Accordingly, in this case
0
Ull (0):
0
uij(0)
u20
2
(0)=
0
aij
0,
0
T
u33 (Q)= i
1-21
i+i)’
(5.6)
i # j.
Substituting these values in relations (2.16), we get
Ull(O)
u22(0)
u33(0)
3T
"l-l")’
ii
uij (0)
aij (0)
0
i #
(5.7)
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
225
which yield the required exact interior solution
u
i-i
3T
i(E)
l (ll)xi’
0,
ij(x)
r
where we have used the fact that the region
r < a,
(5.8a)
a,
(5.8b)
R
is void so that
2
k2 2
O.
Hence, from relations (5.1), (5.2), (5.4a) and (5.5a) it follows that
s
T
ISI
(S)I+ i S’
sni(S) l+ _TOniS
E
where
point
n.l
S
(5.9a)
a,
0
-ijS )nj
are the components of the unit normal
of S.
-Tnis )’ ISI
B(Xs) directed
(5.9b)
a,
outwards at the
Finally, we substitute the above values in formula (5.3) and get the required
value of the stored energy
T
E
E
as
T
S(xs).+ (Xs)dS
j.
r=a
f
Tan s)
r=a
3
s)dS T2a
2
i
i
ANALYSIS OF VISCOUS INHOMOGENEITY
6.
The analysis of the displacement fields in elastic composite media can be
applied to solve the problem of the slow deformation of an incompressible homogeneous viscous fluid ellipsoidal inhomogeneity embedded in an infinite homogeneous
viscous fluid of different viscosity which is subjected to a devitorial constant
pure strain rate whose principal axes are parallel to those of the ellipsoidal
This problem is of interest in the theory of the deformation of rocks
inclusion.
and in the theory of mixing and homogenization of viscous fluids
Let an infinite region
of viscosity
R
i
0
U
where
0
I()
U,
u
Then at time
t
viscosity
0
> b
, ,
0(),
0
0
0
e33 (x)=-
e22 (x)
(6 la)
R,
x
is positive constant so that the corresponding velocity components are
Ux,
a
be filled with an incompressible homogeneous fluid
and be subjected to devitorial uniform pure strain rate
with non-zero components:
el()
u
R
[5].
> c
2
O
u
0
2()
y,
u
u
0
3()
z,
div
(0)
()
O,
(6.1b)
R.
0, let an ellipsoidal homogeneous viscous incompressible fluid of
2 2
2 2
x /a +
/b + z /c 0 < i,
0
which occupies the region
y2
R0:
be embedded in the infinite host medium which is subjected to the
0()
devitorial uniform pure strain rate
principal axis of
0()
as described in (6.1) so that the
are parallel to those of the ellipsoidal inclusion.
Due
to this uniform pure strain rate the ellipsoidal inclusion gets deformed to an
ellipsoid at each subsequent instant. Let, at time t, the inclusion occupy the
2
< l, a > b > c, where a,b,c are functions of
region
+ /b 2 + z
time.
R2: x2/a y2
Thus, (4,/3)a0b0c
0
The inner solution
2/c2
(4,/3)abc, i.e., abc
E(E),
R
2
at instant
a0b0c 0.
t
is linear in
x,y,z
and is
D. L. JAIN AND R. P. KANWAL
226
readily obtained from the analysis of the corresponding elastostatic problem of
composite media by taking appropriate limits. The quantity (), which is
displacement vector in the previous analysis, now represents velocity field in
R
region
and
I
R
In both these regions we have to satisfy the equation of
2.
continuity
div
u(x)
x
0,
E R
2
I.
(i/2)(ui,j()+uj,i())
eij()
Secondly, while the tensor
R
or
denotes the pure strain rate in the present case.
is the strain tensor, it
With these changes in the notation
understood, we derive our results in the present case when the guest medium is
2
deformed to the ellipsoid occupying
+ y 2 /b 2 + z 2 /c 2 < i at time t
R2: x2/a
by
taking the appropriate limits in the analysis of Section 2:
and
-\i
div
uCx),
x
-k
div
u(x),
x
0
div
uij(0)
(6.2b)
2
u0(x)
0,
0.
i # j,
,
u33 (0)
0
for all
u
0
0
Ull (0)= u, u22(0)
so that
RI,
u
0
Also
0
aij (0)
-
R2,
In view of relations (6.1) we have
is finite.
(6.2a)
2
p(x):
such that the hydrostatic pressure
p(x)
R
(6.3a)
(6.3b)
i,j.
Let us note from our elastostatic analysis that, since the inner solution
3
E
R
R
is linear in x,y,z, we have div u(x)
2.
2
k=l
xE
u(),
Ukk(0),
Now, we take the limits as explained in (6.2) above in the relations (2.13) and
(2.14) of elastostatics and get
{i-2A(tlx122+tllB3 )}Ull(0
i
2A
2A
Pl
--i tl122u22 (0) + --i tllB3U33(0)
2A
2A( t2233+t2211)u22(0)
--I
+--i t2233u33()
2A
2A
3311Uli (0) + -i t3322u22(O) + {i -I (tBBll+t3322) }u33(0)
t2211Ull(0)
t
+ 2A
+ {i-
Also
uij(0) aij(0)
0,
i # j,
i,j
1,2,3,
(6.4a)
U,
U
’
U
(6.4b)
(6.4c)
(6.4d)
where we have used relation (6.3b) and the quantities tiijj, i # j, are the shape
factors of the ellipsoid occupying the region R
and their values are given by
2
(3.5b), namely
-
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
abc
tiijj
where
a
a,
1
a
udu
f
b,
2
a
(6.4b) and (6.4c) we find that
Ull(O)
u(x)
div
n
+
u22(0
x
0,
+
(6.5)
1,2,3,
i,j
{(u+a 2)(u+b 2)(u+c
R
and
c
3
J,
i #
(u+a)(u+a)R
0
u33(0
227
2)}1/2
Adding (6.4a),
0, i.e.
E R2,
so that the equation of continuity is satisfied
-- -
Solving equations (6.4) simultaneously, we obtain
-
A
U {I-
Ull(0)
U {1-
u22(0)
u33(0)
where
(tl122+tl13 3+4t 2 233)
(6.6a)
2A___ (2t
l133-tl122+2t2233)
(6.6b)
i
1
{1-
{1- 2AG (t
D
4(A)
A
}’
-1 (3tl122-tl133+4t2233)
223
}{1- 2Al
3+2t2 211
The innner solution at time
where
Ull(O)x,
l(x)
Ull(0)
u
t
and
(6.6d)
is
u22(O)y,
2(x)
u22(0)
and
2t3311+t3322 )}
(t2233-t2211) (t3322-t3311),
2
u
(6.6c)
u33(0)
Two Important Limiting Cases.
u
3(x)
E R 2,
u33(O)z,
are given by (6.6).
Case I.
Let
bo ao
Co
i.e., at time
t
so that the guest medium consists of a spherical viscous incompressible fluid of
viscosity
Zx 2
occupying the spherical region
2
<
a
which is embedded in the
infinite host medium of viscous incompressible fluid of viscosity
host medium is subjected to devitorial constant pure strain rate
i"
eli(X)
This host
whose
non-zero components are
0
0
ell (x)
-2e22(x)
0
-2e33(x)
U > 0.
In this particular case, the spherical inclusion gets deformed to prolate spheroid
and at time
t
occupies the region
2
2
x__ + L +
2
2
a
b
R2:
2
zz-
w < i,
b
a > b,
ab
2
3
aj
(6.7)
Accordingly, we can derive the values of the distinct non-zero shape factors from
(6.5) by selling
c
b, and they are given by (3.6), substituting these values in
(6.6) we have, in this case,
0,
D. L. JAIN AND R. P. KANWAL
228
Ull(0)
-2u22(0)
-2u33(0)
6A____
i-
6[2 (l-k 2)
{i-
(A)/I
where
i
i ti122
(6.8)
(3-k2)Ll
(2-i)/I
Finally, to obtain the values of
and
a
b, which are functions of time
t;
we appeal to the partial differential equation
DE
Dt
0,
(6.9)
satisfied by the moving surface
2
-=
F(x,y,z,t)
+
b2+z 2
b
a
t, where
at time
ab
a.
2
O,
Thus
0
i da
Ull ()
T{
where the constant
I
2
0
Ull ()
(6 i0)
is given by equation (6 8) which when substituted in
(6.10) yields the following differential equation for w, defined as w2
U
2 dw
3w dt
{l-aa[
i
i
12w
2
i/w
(2+i/w2)
(i- i/w
where we have used the values of
equation is readily
-i
w cosh w
i
= w2_l
where
(w2_i)3/2
A
I
i
2 2
and
k
2
n
(w+) 4w2+i.)
2
3w
as given by (3.7).
3
(6.11)
This differential
by the method of separation of variables and we have
solved
2
+ log
is the constant of
initial condition that as
L
2
3
a /a 0,
integration.
0, w
t
(6.12)
U++A,
w
I.
To find this constant, we use the
Thus
i
A=--a
(6.13)
and (6.12) becomes
[
where
t
S
i
+
w cosh-lw
i
w2-1 (w2_i)3/2
log(a/ao),
SH
(6.14)
S,
is the natural strain of the inhomogeneity and
Ut, is the natural strain applied at infinity.
known result [5] and gives
w
(a/ao)3/r
0
SH el I
Relation (6.14) agrees with the
in term of time
t
and expresses the
required value of a in terms of t. Substituting this value of
3
2
a O, we obtain the value of b in terms of t.
relation ab
a
in the
INTERIOR AND EXTERIOR SOLUTIONS FOR BOUNDARY VALUE PROBLEMS
Let us now consider a two-dimenslonal limit.
Case II.
b
c
i.e. at time
y2+z 2 < bo2
a
(R),
0
O, the guest medium consists of an infinite circular
t
O,
o
cylinder of an viscous incompressible fluid of viscosity
region
Letting
229
Ixl
occupying the
2
embedded in the infinite host medium of viscous
<
incompressible fluid of viscosity
i which is subjected to devltorlal uniform
0
(x). Its non-zero components
pure strain rate eij~
0
-e33 (x)= u
0
e22 (x)
U
where
In this case, the right circular cylindrical
is a positive constant.
inclusion gets deformed to an infinite elliptic cylinder and at time
the region
b
Ixl<-,
2
nb
nbc
c
b
bc
or
2
The non-zero distinct shape factors in this case
O.
are derived from (6.5) by letting
c (b+2c)
t2222
2 (b+c)
b (c+2b)
t3333
2
and the values are
a
2 (b+c)
where
u33(0)z,
u2(x) u22(O)y, u3()
u22(0) and u33(0 satisfy
0,
bc
t2233
2
In this case, the exact inner solution at time
Ul(X)
occupies
2
2
where
t
R 2,
2 (b+c)
t
R2,
2
(6.15)
is
(6.16)
equations (6.4b) and (6.4c) which, in view of
(6.15), become
{i
Abc
+
’i (b+c)
1 (b+c)
2
2"
i
u22(0
u22(0)
bc
(b+c)
Abc
+ {i +
I (b+c)
u33 (0)
2
2} u33 (0)
U,
(6.17a)
=-U.
(6 17b)
These equations yield
u22(0)
U
-u33(0)
{I +
(6.18)
2Abc
l(b+)
2}
Substituting these values in (6.16), we obtain the required inner solution at time
t.
To find the values of
differential equation
b
DF/Dt
and
c
in terms of
2
F(x,y,z)
t, we appeal to the partial
O, where
2
b-2 +-
1
0,
c
and get
i db
u22(0)
U
{1+ 2cbc
2
(b+c)
(6.19)
D. L. JAIN AND R. P. KANWAL
230
(A)/ I
where
(2-i)/I.
2
6c --b 0, the above relation becomes
Since
U
i db
b dt
1+
2b2b20
(b2+b02) 2
Its solution is
Sn b
B
where
2
b2+bo
is the constant
Ut
-
of integration.
(6.20) that
B-- log b
0
so that
log
+ B,
b+
(b/bo) 2-i
(b/b0 2+1
(6.20)
Since, when
t
0, b
b 0, we find
Ut,
or
S
S
where
S
0
H
e22
t
+
tanh S
log b/b 0 is the natural strain of elliptical inhomogeneity and
Ut is the natural strain applied at infinity. Relation (6.21) agrees
with the known result [5]. It gives b in terms of
2
b
we can determine c in terms of t.
bc
0
i.
(6.21)
SH,
t
and using the relation
REFERENCES
JAIN, D.L. and KANWAL, R.P., Interior and exterior solutions for boundary value
problems in composite media:-two-dimensional problems, J. Math. Phys. 23
(1982) 1433-1443.
2.
CHEN, F.C. and YOUNG, K., Inclusions of arbitrary shape in an elastic medium,
J. Math. Phv.s. 18(1977), 1412-1416.
3.
GOODIER, J.N., Concentration of stress around spherical and cylindrical
inclusions and flaws, J. Appl. Mech. 1(1933), 39-44.
4.
ESHELBY, J.D., Elastic inclusions and inhomogeneities, Prog. Solid Mech.
2 (1961) 88-140.
5.
BILBY, B.A., ESHELBY, J.D. and KUNDU, A.K. The change of shape of a viscous
ellipsoidal region embedded in a slowly deforming matrix having a different
viscosity, Tectonophysics 28(1975), 265-274.