12 HABIB AMMARI AND HYEONBAE KANG
We note that the formulae (2.15) and (2.19) are valid even for extreme conduc-
tivities k = 0, oo
[88].
If
kz = 0, then from (2.3) and (2.18) we obtain
au
I -
aH
~
*
(!)
""a(Sv.P(s))
I
av(l) - av(l)
+
(2/
+
Kvz)¢
+
L....t av(l) aDz on aDz.
+
8~
Since
.1
= -1/2, we have
a~~)
L
=
0 on aD1,
which means that aD1 is insulated. On the other hand, if k1 = oo, then by the
same argument, we can see that
a~~),_=
0
on aD1,
and hence
u
=constant in D1, which means that D1 is a perfect conductor.
Observe that the harmonic function
H
depends on
f
and hence on the inclusion
D. We now present another decomposition of the solution where the harmonic part
is independent of D, which was derived in
[15].
Let U be the solution without the
inclusion D, i.e., the solution to
(2.21)
tiU
=
0 inn,
aU' _
9
av an- ,
f
U(x) da(x)
=
0.
lan
Then U can be written as
U(y)
=
Nng(y)
=
f
N(x, y)g(x) da(x).
lan
THEOREM
2.10. The solution u of (2.17) can be represented as
m
(2.22) u(x)
=
U(x)- L:Nv1 /J(l)(x), x
E
an,
1=1
where¢!) is defined in (2.19).
Recall that the Neumann-to-Dirichlet (NtD) map Av :
L~(an)
--t
Wr(an)
2
corresponding to the conductivity distribution
x(n \
u::1
Ds)
+ E::l
ksx(Ds) is
defined, forgE
L~(an),
by
Av(g) :=ulan,
where u is the unique weak solution to (2.17). The space Wr(an) is defined by
2
f
E
Wr(aO) if and only iff
E
L2 (a0) and
2
f f
if(x)-
f(y)i2
da(x) da(y) +oo.
lan lan ix- Yid
See for instance
[75] .
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