Kroner

or, defining:
we can write:
∫
u i (x) = u 0 i (x) +
Now calculate the strains.
G ik,l (x − x ′ ) =
V
∂
∂x l
(
Gik (x − x ′ ) ) (55)
dx ′( G ik,l (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (56)
ɛ ij = 1 2 (u i,j + u j,i ) (57)
Since only the Green tensor function depends on x:
∫
u i,j = u 0 i,j + dx ′( G ik,lj (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (58)
V
∫
u j,i = u 0 j,i + dx ′( G jk,li (x − x ′ ) ) [C] klmn (x ′ )ɛ mn (x ′ ) (59)
V
∫
ɛ ij (x) = ɛ 0 ij(x) + dx ′ G ijkl (x − x ′ )[C] lkmn (x ′ )ɛ mn (x ′ ) (60)
where:
V
G ijkl (x) = 1 2 (G ik,lj(x) + G jk,li (x)) (61)
Since the moduli differ only over the inclusion W , Eq. 60 reduces to:
∫
ɛ ij (x) = ɛ 0 ij(x) + dx ′ G ijkl (x − x ′ )[C] lkmn (x ′ )ɛ mn (x ′ ) (62)
W
Markov’s [6] treatment of this integral equation begins here. Markov proceeds
as follows. ”If ɛ 0 is constant, then from Eq. 62 the strain within the inhomogeneity
will also be constant provided that the moduli differ but by constant
values and the following tensor field is constant within the inhomogeneity”:
∫
P ijkl (x) = − G ijkl (x − x ′ )dx ′ (63)
So that:
But:
so that:
W
ɛ ij = ɛ 0 ij − P ijkl [C] lkmn ɛ mn (64)
I ijmn ɛ mn = ɛ ij (65)
(I ijmn + P ijkl [C] lkmn )ɛ mn = ɛ 0 ij (66)