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] .