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
kz = 0, then from (2.3) and (2.18) we obtain
av(l) - av(l)
L....t av(l) aDz on aDz.
= -1/2, we have
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
=constant in D1, which means that D1 is a perfect conductor.
Observe that the harmonic function
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
Let U be the solution without the
inclusion D, i.e., the solution to
av an- ,
Then U can be written as
N(x, y)g(x) da(x).
2.10. The solution u of (2.17) can be represented as
U(x)- L:Nv1 /J(l)(x), x
where¢!) is defined in (2.19).
Recall that the Neumann-to-Dirichlet (NtD) map Av :
corresponding to the conductivity distribution
where u is the unique weak solution to (2.17). The space Wr(an) is defined by
Wr(aO) if and only iff
L2 (a0) and
da(x) da(y) +oo.
lan lan ix- Yid
See for instance