P[D,g]
ON STABILITY OF A TRANSMISSION PROBLEM
HYEONBAE KANG AND JIN KEUN SEO
ABSTRACT. We investigate the behavior of the gradient of solutions to the refraction equation div((1
+
(k - I)XD)\7U) = 0 under per- turbation of domainD. IfU and Uh are solutions to the refraction equation corresponding to subdomainsD and Dh of a domainn
in 2 dimensional plane with the same Neumann data onan,
respec-tively, we prove that 1I\7(u - uh)IIL2(fl) ~ CVdist(D,Dh) where dist(D,Dh ) is the Hausdorff distance between D and Dh . We also show that this is the best possible result.
1.
Introduction and statement of results
Let 0 be a simply connected bounded domain in
jRn(n 2: 2) with the C
2smooth boundary and let D be a simply connected C
2subdomain of 0 with closure in O. Let k i= 1 be a positive number. Consider the following Neumann problem
div((l + (k -
l)XD)\7u)= ° in 0
ou
=9 on 00, r 9
=0, 9
EL
2(00)
ov ian
L u=O,
where
v(x)is the unit normal to the boundary,
~~ = v . \7u,and
XDis the characteristic function of D.
In this paper we study stability of the solution to the transmission problem P[D,g] under the perturbation of D. This study is moti- vated in relation to the inverse problem to P[D, g], namely the inverse conductivity problem.
Received May 17, 1997.
1991 Mathematics Subject Classification: 35B35.
Key words and phrases: conductivity problem, stability, layer potential.
The both authors are partially supported by GARC-KOSEF, BSRl'97, and KOSEF 95-0701-01-01-3.
(1.4)
Let Dh be a bounded simply connected subdomain with C
2smooth boundary defined by
aDh : f(s) + hwh(S)V(S) (aD: x
=f(s»
where s is 1 dimensional local parameter,
Wh(s) is a Cl function on aD whose Cl norm is bounded uniformly in h, and v(s) is the outward unit normal to aD. Let U and Uh be solutions to P[D,gJ and P[Dh,gJ, respectively. In [2J and [1J, it is proved that
(1.1) lIu - uhIlLP(o)
~Ch
if p < 4 and aD and aDh are only Cl,l. (This result is for n-dimension (n
=2,3).) In [7J, (1.1) is proved for
p = 00.Both results were used strongly in the study of the inverse problem (see [2, 1, 7]).
In this paper we investigate the behaviour of\7u under perturbation of D. We prove that
1I\7{U - uh)IIL2(o) ~ CVh.
We also prove that Vh is the best possible result one can expect. To be precise, we have the following theorem:
THEOREM
1.1. Let n be a simply connected bounded domain in
]R2and D and Dh be as above. Let Db,.Dh be the symmetric difference of D and Dh. Then, there exists a constant C such that
(1.2) tim
-hI {1V'(u - Uh)/2dx
=0,
h->O lO\Dt::..Dh
(1.3)
limsup ~ { 1\7(u - uh)l2dx ~ C { laa:
12der.
h->O
h
lDt::..Dh lBDv
Here, der is the line element on aD and
a a: (P) = lim (\7u(P ± tv(P», v(P».
v
t->O+Moreover, if
Whconverges to
Wuniformly as h
---+0, then
I 1
1 k-1 au
2lim -h { 1\7(u - uh)1 2 dx
=- k - { a + Iwlder.
h->O
1
Dt::..Dh1
BDv
If the Neumann data 9 is not zero, then cfz,\
isnot zero and hence
we have
COROLLARY
1.2. There exists a constant 0 independent of h such that
1I\7(U -
uh)IIL2(n):S CVh.
If
Whconverges to
Wuniformly as h
---+0, then for small enough h
Corollary 1.2 says that ..,ffi is the best possible.
The proof of Theorem 1.1 is based on our earlier result on the rep- resentation of the solution to P[D,g] ([9]). So, we first review the representation formula in Section 2 and then prove Theorem 1.1 in Section 3.
One comment on a notation: the constants 0 in estimates may differ from one step to another. However, those constants do not depend on the quatities to be estimated.
2. Representation of solutions
For this work, we assume that 0 is a simply connected bounded 0
2domain in R
2and D be a simply connected subdomain with C
2boundary which is compactly contained in O. The single and double layer potentials on D is defined by
SDf(X) = 2 1
{ log IX - Qlf(Q)dl7Q, X
ER
2,1r laD
1 { (vQ,X - Q)
DDf(X) = 21r laD IX _ QI
2f(Q)ooQ, X
ER
2.The following trace formula is well known (see [F] or [8]):
(2.1) where
o~± SDf(P) = (±~I + KiJ)f(P) (P
EaD)
K* f(P) = ~
{(V?, P - Q) f(Q)dl7 .
D
21r laD IP - Qln Q
We denote by KD the dual of KiJ.
Let £5(on) = {f
E£2(on) : fan fdu = D}. Then the representation formula for the solution to the problem P[D, g] is as follows.
Representation Formula [9].
If
Uis the weak solution to the Neumann problem P[D,g], then there are unique harmonic function H
EW
l,2(n) and 'PD
E£5(aD) so that u can be expressed as
(2.2) u(x) = H(x) + SD'PD(X) for x
En.
Moreover, if f = ulan, (2.3)
and
H(x)
=-Sng(x) + 'Dnf(x)
k+ 1 * oH
(2.4) (2(k _1)1 - KD)'PD = av laD on oD.
See [9] for proof. We remark that the representation formula holds for Lipschitz domains in
]Rn, n ~2.
LEMMA
2.1 [9].
Ifu is the weak solution to the Neumann problem P[D,g], then
(2.6) ou k-10u
'PD
=(k - 1 ) -
= - - - - .ov- k av+
3. Proofs
Let D and Dh be as in Section 1. Write oDh as aDh : (+ hWh(()V(() , (
EoD
if slight abuse of notations is allowed, where Wh (() is a Cl function on oD whose Cl norm is uniformly bounded and v(() is the outward unit normal to oD at (. Let u and
Uhbe the weak solutions of P[D, g]
and P[Dh' g], respectively. By the representation formula (2.2), the solutions u and
Uhcan be expressed uniquely as :
(3.1) u=H+SD<PD and uh=Hh+SDh'PDh inn
where H, <PD, H
h ,and 'PDh satisfy the relations (2.3) and (2.4). To make the notations short, we put
<P=<PD, S=SD, K*=K*v, <Ph = <PDh, Sh=SDh'
K~=K*vh·LEMMA
3.1. There is a positive constant C such that (3.2)
Lemma 3.1 is proved in [7].
By identifying the real 2-D vector
(VI,V2) with the complex number
VI
+ iV2, we can see that
(3.3) \7Scp(z)
=-1 r ~((~dl(1
27r l;:w z - (
and
(3.4) JCcp(z)
=~~ r CP(() d(.
27r laD z - (
Here
~is the imaginary part. Let cPh be the diffeomorphism from oD onto oD h defined by cPh(()
=(+ hwh(()V((),
LEMMA
3.2. There is a positive constant C such that Ilcph
0cPh - cpIlL2(aD) ::; Ch
if h is small enough.
Proof. Let A
= 2tk~\).Since (AI - JC*) is invertible on L 2 (oD) [5],
we have from (2.4) that Ilcph
0cPh - cpIlL2(aD)
::; CII(AI - JC*)(CPh
0cPh - cp)IIL2(8D)
::; CII(AI - JC'hcph)
0cPh - (AI - JC*)cpIlL2(8D)
+ CII(JC'hcph)
0cPh - JC*(CPh
0cPh)IIL2(8D)
oHh oH
::; CII oV
0cPh - oV IIL2(8D)
+ CII(JC'hcph)
0cPh - JC*(CPh
0cPh)IIL2(8D)'
Since the first term in the most right hand side of the above in- equalities is O(h) by Lemma 3.1, Lemma 3.2 follows from the following
lemma. 0
SUBLEMMA. For
any
function 9 EL2(8Dh)
11
(Kj;g)
0<Ph - K*(g
0<ph)IIL2(aD) :S Chllgll£2(aDh) if h is small enough.
Proof. By duality and boundedness of
<P~,it suffices to show that II(Khg)
0<Ph - K(g
0<Ph) 11£2 (aD) :S Chllgll£2(aDh)
for any 9
EL2(8Dh). By (3.4), (Khg)
0<Ph(Z) - K(g
0<Ph)(Z)
= ~SS [ r g(() d( _ r (g
0<Ph)(() d(]
27r laDh <Ph(Z) - ( laD Z - (
=
:7r SS lD [<Ph(Z~~(~h(() - Z ~ (] (g
0<Ph) (()d(
= 2~ SS lD [<Ph(Z) ~ <Ph(() - Z ~ (] (g
0<Ph) (()d(
1 (): r h(WhV)'(()
+ 27r ~ laD <Ph(Z) _ <Ph(() (g
0<Ph) (()d(
:=
I(z) + II(z).
Since the Cauchy transform on C
2-curves (in fact, on Lipschitz curves) is bounded on L
2,we have
r III(()1 2 dl(l:s Ch21Iglli2(aDh).
laD .
Note that
I(z) = ~SS r
_1_~ hj ((WhV)(Z) - (WhV)(())j (g
0<Ph)(()d(.
27r 1
flDv Z - (~
J=1 Z - ( Itis proven in [3] that
i II(()!2dl(l:s f(hCII(whv)'IILoo(aD))2jllglli2(aDh)
aD
j=1:S C'h2 I1glli2(aDh)
if h is small enough. This completes the proof. 0
Finally the following lemma leads us to Theorem
1.1.LEMMA
3.3. Tbere exists a constant C sucb tbat (3.5)
lim
-hIf 1\7(Sep - Sheph) (z) !2 dV (z) =
0h-+O In\DADh
(3.6)
limsup
-hIf 1\7(Sep - Sheph) (z)1
2dV(z) :s; C f lep(()1
2dO"(().
h-+O JDADh JaD
Moreover,
ifWh -+Wuniformly as h
-+0, tben (3.7)
lim
-hIf [\7(Sep - Sheph) (z)!2 dV(z)
=f lep((Wlw(()ldO"(().
h-+O JDADh JoD
Proof.
Itis easy to see that for each 8 > 0
lim
-hI~ist(Z,aD»oIV'(Sep-Sheph)(z)12dV(z) =0.
h-+O Jd
zEnSo, we assume, from the beginning, that n = {z =
(+tv(() : It I <
8, ( E oD} for some 8. (8 is chosen so that the normal projection from n onto oD is well-defined.) Let
E> 0 be a fixed number to be determined later and let U be the tubular neighborhood of oD defined by U = {z
=(+tv(() : ItI <
E,(E oD}. If zEn \ U, then by (3.3)
V'(Sh'Ph - S'P)(z)
=
~ [r ~(C)_dICI
- {~h(C2
d1C1]211" laDz-C laDh z-C
= ~ [r
'P(C)-:!h
0 <I>h(C) dlCI+
h { (WhV
)«() 'Ph0 <I>h(C)dICI 211" laD z -C
laD (z - ()(z - <I>h(C))+ (
'Ph0 <I>h(C)[I<I>h(C)I-lldlCI]laD
z -
<I>h(C):=h(z)
+
h(z)+
Is(z).Suppose z = e + tv(e) E n \ D, e E oD. If N is the smallest integer such that 2
Nt> max{lz - (I: (E oD}, then N:S; Clog t and
~ f lep(() - eph
0~h(()1
Ih(z)1 :s; L.J Pj-1t5c1t;-(1<2jt Iz _ (I dl(1
j=l
(EaD
:s; CllogtlM(ep - eph
0 ~h)(e)where M is the Hardy-Littlewood maximal operator on aD. Since M is bounded on L
2(aD) ([10]), it follows from Lemma 3.2 that
1 \II(Z)12dV = J f Ih(~ + tll(~))12dl~ldt
n\u
e<ltl<6laD
~CJ Ilogtl 2 dt f IM(<p-<PhO<Ph)(~)12dl~1
e<ltl<6
laD
(3.8) ~ CII<p - <Ph
0<Phlli2(aD) ~ Ch 2 . For 1
2(z), we have
II2(z)1 ~ Ch laD z - ( f I 1 12dl(1 ~ Chi-I,
and hence
(3.9) 1 n\u II2(z)1 2 dV ~ Ch 2 J
e<ltl<6r 2 dt ~ Ch
2E-I.Since l<ph «()1-1 = O(h), we have
(3.10) II
3(z)1 ~ Ch f I 1 (I dl(1 ~ Ch log !.
laD z - t
Thus, we have
(3.11) 1 nw II
3(z)1
2dV ~ Ch
2•Combining (3.8), (3.9), and (3.11), we have
(3.12) ~ f 1\7(S<p - Sh<Ph) (z)!2 dV ~ ChE-I.
low
Now suppose that z =
~ +tll(~) EU. Put
~h=
~+
hWh(~)lI(~).Put se = {(
EaD: I( -
~I< E} and
S~= {( + hw«()lI«() :
(Ese}. Then,
'l(Sh'Ph -S'P)(z) =
~ [r ~(()-dl(l-
{~h((2dl(l]
211"
1aD\s' z - ( 1aD
h\S~z - (
+ ~ [r
'P(() - 'P(e)dl(1 _ ( 'Ph(() - 'Ph(f,h)dl(l]
~ 1~ z-( 1~ z-(
+
'Ph(f,h)[r ~dl(l-
{ _1_dl(l] .
211"
1
S' z - (1
S~ z - (+
['P(e) - 'Ph(eh)]r _
1 -dJ(1211"
1s. z - (
:=IIr(z) +II2(Z) +II3(Z) +II4(Z).
In the same way to derive (3.8), one can see that (3.13)
Since 'P is CO: for every a < 1 (see [4]), IIh(z)1 ::; CEO: independently of h. Hence, we have
By Lemma 3.2 and the estimate used in (3.10), we have (3.15)
~ lI114(z)12dV = ~ lE ( II4(~ + tv(~))12dl~ldt
U
-EloD
::; ~ LEE ~D I'P(~) - 'Ph(~hWllogtI2dl~ldt
::; Ch.
We now deal with Ih(z). Put
For each z
EU, we may assume that
SEis a graph by taking
Esmall enough if necessary, namely,
SE : ( =
x + ig(x),
-E< x <
E,g(O) = g'(O) = 0, and z = it(ltl < h). Put J() = 11 + ig'(x)1 and J
h () =
11 + ig'(x) + h(WhV)'(x)l.
1 + ig'(x) 1 + ig'(x) + h(WhV)'(X)
Let rh
(r~,resp.) be the straight line connecting the left (right, resp.)
endpoints of
SEand S1;. and let Ch be the positively oriented closed
curve composed of se, Sh' rh, and
r~.Then
By the Cauchy integral formula,
if z
EDflDh if zEn \ DflDh.
It is easy to see that
Note that
IJ(() - 11:::; Clg'(x) I :::; clxl :::; CI(/' (E oD
IJh(() - 11:::; Clg'(x)1 +
Chlw~(x)1:::; C(I(I + h), (E ODh.
It then follows that for z =
~+
tv(~),So far, we proved that
if z
= ~+
tv(~)tt Dl1Dh, and
if z
= ~+
SWh(~)V(~) EDl1Dh.
Itthen follows from (3.16) that
(3.18)
On the other hand, from (3.17), we have (3.19)
I -hI lD~Dh r III3(z)1
2dV - laD r 1<p(~)12Iwh(~)ldl~11
::; * lD l h III3(~ + sWh(~)v(~))12 _1<p(~)121 dslwh(~)ldl~1
::; C laD r l<Ph(~h) - <p(~)12dl~1
+ ~ laDlo r rh IIWh(~ + SWh(~)v(~))12 - Ildsl<p(~)12Iwh(~)ldl~1
2 1
r 1
::; C(h +
E+ hE- + laD Ihwh(~)llog Ihwh(~)1 dl~l)·
Take
E= h
2/3•Then (3.12)-(3.15), (3.18), and (3.19) prove Lemma
3.3. 0
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