16 HABIB AMMARI AND HYEONBAE KANG
THEOREM
2.15. For each (F, G)
E
W[(aD) x L
2
(aD), there exists a unique
solution (!,g)
E
L
2
(aD) x L
2
(aD) of the integml equation
{
Sffd-S~g=F
(2.31) - A
A
on aD.
v · AY'Svfl-- v · AY'SvYI+ = G
Moreover, there exists a constant C depending only on the largest and smallest
eigenvalues of
A,
A, and
A-
A, and the Lipschitz chamcter of D such that
llfiiP(aD)
+
IIYIIP(aD) :::; C(IIFIIw?(aD)
+
IIGIIP(aD))·
We can easily see that if G E
L~(aD),
then the solution
g
of (2.31) lies in
L~(aD).
Moreover, if G = 0 and F =constant, then g = 0. We summarize these
facts in the following lemma.
LEMMA
2.16. Let (!,g) be the solution to (2.31}. If G
E
L~(aD),
then g
E
L~(aD).
Moreover, ifF is constant and G
=
0, then g
=
0.
Let 0 be a bounded Lipschitz domain in JRd,
d
= 2, 3. Suppose that 0 contains
an inclusion
D.
Suppose that the conductivity of the background 0 \
D
is
A
and
that of D is
A.
The conductivity profile of the body 0 is given by
'Yn
:=
Xn\vA
+
xvA,
where
XD
is the characteristic function corresponding to
D.
For a given g
E L~(aO),
let
u
denote the steady-state voltage in the presence
of the conductivity anisotropic inclusion
D, i.e.,
the solution to
(2.32)
Let
V'.
'Yn(x)V'u
=
0
in 0,
v·AY'ul =g,
an
f
u(x) da(x)
=
0.
lan
(2.33)
HA(x)
:=
-Sit(g)(x)
+
Dii(f)(x), x
E
0,
f
:=ulan.
The following representation formula from [84] holds.
THEOREM
2.17. LetHA be defined by (2.33). Then the solution u to (2.32}
can be represented as
{
HA(x)
+
S~¢(x),
u(x)
= -
S~'¢,
XED,
x
E
0\D,
where the pair ( ¢,
'1/J)
is the unique solution in
£
2
(aD)
X
£
2
(aD) to the system of
integml equations
{
St'l/J-
S~¢
=
HA
- A
A A
v · AV'Sv'l/JI-- v · AY'Sv¢1+
=
v · AV'H
on aD.
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