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.