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1.Introduction HorstHeck,Jenn-NanWang Optimalstabilityestimateoftheinverseboundaryvalueproblembypartialmeasurements

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Optimal stability estimate of the inverse boundary value problem by

partial measurements

Horst Heck, Jenn-Nan Wang

We dedicate this work to Giovanni Alessandrini for his 60th birthday and for his pioneering contribution in the stability estimates of inverse problems.

Abstract. This manuscript was originally uploaded to arXiv in 2007 (arXiv:0708.3289v1). In the current version, we expand the Introduc- tion and the list of references which are related to the results of this pa- per after 2007. In this work we establish log type stability estimates for the inverse potential and conductivity problems with partial Dirichlet- to-Neumann map, where the Dirichlet data is homogeneous on the inac- cessible part. The proof is based on the uniqueness result of the inverse boundary value problem in Isakov’s work [16].

1. Introduction

In this paper we study the stability question of the inverse boundary value prob- lem for the Schr¨odinger equation with a potential and the conductivity equa- tion by partial Cauchy data. This type of inverse problem with full data, i.e., Dirichlet-to-Neumann map, were first proposed by Calder´on [6]. For three or higher dimensions, the uniqueness issue was settled by Sylvester and Uhlmann [27] and a reconstruction procedure was given by Nachman [25]. For two dimen- sions, Calder´on’s problem was solved by Nachman [26] for W2,pconductivities and by Astala and P¨aiv¨arinta [3] for L conductivities. This inverse problem is known to be ill-posed. A log-type stability estimate was derived by Alessan- drini [1]. On the other hand, it was shown by Mandache [24] that the log-type estimate is optimal.

All results mentioned above are concerned with the full data. Over the last decade, the inverse problems with partial data have received a lot of attention.

We list several earlier results [14], [17], [5], [20], [12], [16], [23], [21], [18] [11]

and refer the reader to the survey article [19] for its detailed development and for related references. After the uniqueness proof comes stability estimates.

We summarize related results in the following.

• log log type: [15], [29], [7], [9], [8], [10], [22].

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• log type: [7], [13], [4], [2].

The method in [15] was based on [5] and a stability estimate for the analytic continuation proved in [30]. We believe that the log type estimate should be the right estimate for the inverse boundary problem, even with partial data. In this paper, motivated by the uniqueness proof in Isakov’s work [16], we prove a log type estimate for the inverse boundary value problem under the same a priori assumption on the boundary as given in [16]. Precisely, the inaccessible part of the boundary is either a part of a sphere or a plane. Also, one is able to use zero data on the inaccessible part of the boundary. The strategy of the proof in [16] follows the framework in [27] where complex geometrical optics solutions are key elements. A key observation in [16] is that when Γ0

is a part of a sphere or a plane, we are able to use a reflection argument to guarantee that complex geometrical optics solutions have homogeneous data on Γ0. Caro in [7] also used Isakov’s idea to derive a log type estimate for the Maxwell equations. The articles [13], [4], [2] have a common feature that the undetermined coefficients are known near the boundary.

Now we would like to describe the results in this work. Let n ≥ 3 and Ω ⊂ Rn be an open domain with smooth boundary ∂Ω. Given q ∈ L(Ω), we consider the boundary value problem:

(∆ − q)u = 0 in Ω

u = f on ∂Ω, (1)

where f ∈ H1/2(∂Ω). Assume that 0 is not a Dirichlet eigenvalue of ∆ − q on Ω. Then (1) has a unique solution u ∈ H1(Ω). The usual definition of the Dirichlet-to-Neumann map is given by

Λqf = ∂νu|∂Ω

where ∂νu = ∇u · ν and ν is the unit outer normal of ∂Ω.

Let Γ0⊂ ∂Ω be an open part of the boundary of Ω. We set Γ = ∂Ω\Γ0. We further set H01/2(Γ) := {f ∈ H1/2(∂Ω) : supp f ⊂ Γ} and H−1/2(Γ) the dual space of H01/2(Γ). Then the partial Dirichlet-to-Neumann map Λq,Γ is defined as

Λq,Γf := ∂νu|Γ∈ H−1/2(Γ)

where u is the unique weak solution of (1) with Dirichlet Data f ∈ H01/2(Γ).

In what follows, we denote the operator norm by kΛq,Γk:= kΛq,ΓkH1/2

0 (Γ)→H−1/2(Γ)

We consider two types of domains in this paper:

(a) Ω is a bounded domain in {xn < 0} and Γ0= ∂Ω ∩ {xn = 0};

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(b) Ω is a subdomain of B(a, R) and Γ0= ∂B(a, R)∩∂Ω with Γ06= ∂B(a, R), where B(a, R) is a ball centered at a with radius R. Denote by ˆq the zero extension of the function q defined on Ω to Rn.

The main result of the paper reads as follows:

Theorem 1.1. Assume that Ω is given as in either (a) or (b). Let N > 0, s > n2 and qj∈ Hs(Ω) such that

kqjkHs(Ω)≤ N (2)

for j = 1, 2, and 0 is not a Dirichlet eigenvalue of ∆ − qj for j = 1, 2. Then there exist constants C > 0 and σ > 0 such that

kq1− q2kL(Ω)≤ C

log kΛq1− Λq2k

−σ (3)

where C depends on Ω, N, n, s and σ depends on n and s.

Theorem 1.1 can be generalized to the conductivity equation. Let γ ∈ Hs(Ω) with s > 3 + n2 be a strictly positive function on Ω. The equation for the electrical potential in the interior without sinks or sources is

div(γ∇u) = 0 in Ω u = f on ∂Ω.

As above, we take f ∈ H01/2(Γ). The partial Dirichlet-to-Neumann map defined in this case is

Λγ,Γ: f 7→ γ∂νu|Γ.

Corollary 1.2. Let the domain Ω satisfy (a) or (b). Assume that γj≥ N−1 >

0, s > n2, and

jkHs+3(Ω)≤ N (4)

for j = 1, 2, and

βνγ1|Γ= ∂νβγ2|Γ on ∂Ω, ∀ 0 ≤ β ≤ 1. (5) Then there exist constants C > 0 and σ > 0 such that

1− γ2kL(Ω)≤ C

log kΛγ1− Λγ2k

−σ (6)

where C depend on Ω, N, n, s and σ depend on n, s.

Remark 1.3. For the sake of simplicity, we impose the boundary identification condition (5) on conductivities. However, using the arguments in [1] (also see [15]), this condition can be removed. The resulting estimate is still in the form of (6) with possible different constant C and σ.

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2. Preliminaries

We first prove an estimate of the Riemann-Lebesgue lemma for a certain class of functions. Let us define

g(y) = kf (· − y) − f (·)kL1(Rn)

for any f ∈ L1(Rn). It is known that lim|y|→0g(y) = 0.

Lemma 2.1. Assume that f ∈ L1(Rn) and there exist δ > 0, C0 > 0, and α ∈ (0, 1) such that

g(y) ≤ C0|y|α (7)

whenever |y| < δ. Then there exists a constant C > 0 and ε0> 0 such that for any 0 < ε < ε0 the inequality

|F f (ξ)| ≤ C(exp(−πε2|ξ|2) + εα) (8) holds with C = C(C0, kf kL1, n, δ, α).

Proof. Let G(x) := exp(−π|x|2) and set Gε(x) := ε−nG(xε). Then we define fε:= f ∗ Gε. Next we write

|F f (ξ)| ≤ |F fε(ξ)| + |F (fε− f )(ξ)|.

For the first term on the right hand side we get

|F fε(ξ)| ≤ |F f (ξ)| · |F Gε(ξ)|

≤ kf k1−nεnF G(εξ)|

≤ kf k1exp(−πε2|ξ|2).

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To estimate the second term, we use the assumption (7) and derive

|F (fε− f )(ξ)| ≤ kfε− f k1

≤ Z

Rn

Z

Rn

|f (x − y) − f (x)|Gε(y) dy dx

= Z

|y|<δ

Z

Rn

|f (x − y) − f (x)|Gε(y) dx dy

+ Z

|y|≥δ

Z

Rn

|f (x − y) − f (x)|Gε(y) dx dy

= I + II.

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In view of (7) we can estimate I =

Z

|y|<δ

g(y)Gε(y) dy

≤ C0

Z

|y|<δ

|y|αGε(y) dy

= C0

Z

Sn−1

Z δ 0

rαε−nexp(−πε−2r2)rn−1dr dψ

= C1

Z δ 0

εαuαε−nexp(−u2n−1uu−1ε du

= C2εα Z δ

0

un+α−1exp(−u2) du = C3εα, where C3= C3(C0, n, δ, α).

As for II, we obtain that for ε sufficiently small II =

Z

|y|≥δ

g(y)Gε(y) dy

≤ 2kf kL1

Z

|y|≥δ

Gε(y) dy

≤ C4kf k1

Z δ

ε−nexp(−πε−2r2)rn−1dr

= C4kf k1 Z

δε−1

un−1exp(−πu2) du

≤ C4kf k1

Z δε−1

exp(−πu) du

≤ C4kf k11

πexp(−πδε−1) ≤ C5εα,

where C5= C5(kf kL1, n, δ, α). Combining the estimates for I, II, and (9), we immediately get (8).

We now provide a sufficient condition on f , defined on Ω, such that (7) in the previous lemma holds.

Lemma 2.2. Let Ω ⊂ Rn be a bounded domain with C1 boundary. Let f ∈ Cα(Ω) for some α ∈ (0, 1) and denote by ˆf the zero extension of f to Rn. Then there exists δ > 0 and C > 0 such that

k ˆf (· − y) − ˆf (·)kL1(Rn)≤ C|y|α for any y ∈ Rn with |y| ≤ δ.

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Proof. Since Ω is bounded and of class C1, there exist a finite number of balls, say m ∈ N, Bi(xi) with center xi∈ ∂Ω, i = 1, . . . , m and associated C1- diffeomorphisms ϕi: Bi(xi) → Q where Q = {x0∈ Rn−1: kx0k ≤ 1} × (−1, 1).

Set d = dist (∂Ω, ∂(Sm

i=1Bi(xi))) > 0 and ˜Ωε=S

x∈∂ΩB(x, ε), where B(x, ε) denotes the ball with center x and radius ε > 0. Obviously, for ε < d, it holds that ˜Ωε⊂Sm

i=1Bi(xi). Let x ∈ ∂Ω and 0 < |y| < δ ≤ d, then for any z1, z2∈ B(x, |y|) ∩ Bi(xi) we get that

i(z1) − ϕi(z2)| ≤ k∇ϕikL|z1− z2| ≤ C|y|

for some constant C > 0. Therefore, ϕi( ˜Ω|y|∩ Bi(xi)) ⊂ {x0 ∈ Rn−1: kx0k ≤ 1} × (−C|y|, C|y|). By the transformation formula this yields vol( ˜Ω|y|) ≤ C|y|.

Since |y| < δ we have ˆf (x − y) − ˆf (x) = 0 for x 6∈ Ω ∪ ˜Ω|y|. Now we write k ˆf (· − y) − ˆf kL1(Rn)=

Z

Ω\ ˜|y|

| ˆf (x − y) − ˆf (x)| dx

+ Z

˜|y|

| ˆf (x − y) − ˆf (x)| dx

≤ C vol(Ω)|y|α+ 2kf kLvol( ˜Ω|y|)

≤ C(|y|α+ |y|) ≤ C|y|α for δ ≤ 1.

Now let q1and q2be two potentials and their corresponding partial Dirichlet- to-Neumann maps are denoted by Λ1,Γ and Λ2,Γ, respectively. The following identity plays a key role in the derivation of the stability estimate.

Lemma 2.3. Let vj solve (1) with q = qj for j = 1, 2. Further assume that v1= v2= 0 on Γ0. Then

Z

(q1− q2)v1v2dx = h(Λ1,Γ− Λ2,Γ)v1, v2i

Proof. Let u2 denote the solution of (1) with q = q2 and u2 = v1 on ∂Ω.

Therefore

Z

∇v1· ∇v2+ q1v1v2dx = h∂νv1, v2i Z

∇u2· ∇v2+ q2u2v2dx = h∂νu2, v2i.

Setting v := v1− u2 and q0= q1− q2 we get after subtracting these identities Z

∇v · ∇v2+ q2vv2+ q0v1v2= h(Λ1− Λ2)v1, v2i.

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Since v2 solves (∆ − q2)v2= 0, v = 0 on ∂Ω and v2= 0 on Γ0, we have Z

∇v · ∇v2+ q2vv2= 0,

h(Λ1− Λ2)v1, v2i = h(Λ1,Γ− Λ2,Γ)v1, v2i, and the assertion follows.

In treating inverse boundary value problems, complex geometrical optics solutions play a very important role. We now describe the complex geometrical optics solutions we are going to use in our proofs. We will follow the idea in [16]. Assume that q1, q2 ∈ L(Rn) are compactly supported and are even in xn, i.e.

q1(x1, · · · , xn−1, xn) = q1(x1, · · · , xn−1, xn) and

q2(x1, · · · , xn−1, xn) = q2(x1, · · · , xn−1, xn).

Hereafter, we denote

h(x1, · · · , xn−1, xn) = h(x1, · · · , xn−1, −xn).

Given ξ = (ξ1, · · · , ξn) ∈ Rn. Let us first introduce new coordinates ob- tained by rotating the standard Euclidean coordinates around the xn axis such that the representation of ξ in the new coordinates, denoted by ˜ξ, satisfies ξ = ( ˜˜ ξ1, 0, · · · , 0, ˜ξn) with ˜ξ1=q

ξ21+ · · · + ξ2n−1and ˜ξn= ξn. In the following we also denote by ˜x the representation of x in the new coordinates. Then we define for τ > 0

˜ ρ1:= (

ξ˜1

2 − τ ˜ξn, i| ˜ξ|(1

4 + τ2)1/2, 0, · · · , 0, ξ˜n

2 + τ ˜ξ1),

˜ ρ2:= (

ξ˜1

2 + τ ˜ξn, −i| ˜ξ|(1

4 + τ2)1/2, 0, · · · , 0, ξ˜n

2 − τ ˜ξ1),

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and let ρ1 and ρ2 be representations of ˜ρ1 and ˜ρ2 in the original coordinates.

Note that xn = ˜xn and Pn

i=1xiyi =Pn

i=1ii. It is clear that, for j = 1, 2, ρj· ρj= 0 as well as ρj· ρj = 0 hold.

The construction given in [27] ensures that there are complex geometrical optics solutions uj= ej·x(1 + wj) of (∆ − qj)uj= 0 in Rn, j = 1, 2, and the functions wj satisfy kwjkL2(K) ≤ CKτ−1 for any compact set K ⊂ Rn. We then set

v1(x) = e1·x(1 + w1) − e1·x(1 + w1)

v2(x) = e−iρ2·x(1 + w2) − e−iρ2·x(1 + w2). (11) From this definition it is clear that these functions are solutions of (∆−qj)vj= 0 in Rn+ with vj = 0 on xn = 0.

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3. Stability estimate for the potential

Now we are in the position to prove Theorem 1.1. We first consider the case (a) where Γ0is a part of a hyperplane. To construct the special solutions described in the previous section, we first perform zero extension of q1and q2 to R+n and then even extension to the whole Rn. As in the last section, we can construct special geometrical optics solutions vj of the form (11) to (∆ − qj)vj = 0 in Ω for j = 1, 2. Note that v1= v2= 0 on Γ0. We now plug in these solutions into the identity (2.3) and write q0= q1− q2. This gives

h(Λ1,Γ− Λ2,Γ)v1, v2i

= Z

q0v1v2dx

= Z

q0(x)

ei(ρ12)·x(1 + w1)(1 + w2) + ei(ρ12)·x(1 + w1)(1 + w2)

− ei(ρ12)·x(1 + w1)(1 + w2) − ei(ρ12)·x(1 + w1)(1 + w2) dx

= Z

q0(x)(eiξ·x+ e·x) dx + Z

q0(x)f (x, w1, w2, w1, w2) dx

− Z

q0(x) ei(ρ12)·x+ ei(ρ12)·x dx,

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where

f = eiξ·x(w1+ w2+ w1w2) + e·x(w1+ w2+ w1w2)

− ei(ρ12)·x(w1+ w2+ w1w2) − ei(ρ12)·x(w1+ w2+ w1w2).

The first term on the right hand side of (12) is equal to Z

Rn

q0(x)eiξ·xdx = F q0(ξ)

because q0 is even in xn. For the second term, we use the estimate kw1k2+ kw1k2+ kw2k2+ kw2k2≤ Cτ−1 to obtain

Z

q0f (x, w1, w2, w1, w2) dx

≤ Ckq0k2τ−1. (13) As for the last term on the right hand side of (12), we first observe that

1+ ρ2) · x = ( ˜ρ1+ ˜ρ2) · ˜x = ˜ξ11+ 2τ ˜ξ1n= ξ0· x0+ 2τ |ξ0|xn

and

1+ ρ2) · x = ( ˜ρ1+ ˜ρ2) · ˜x = ˜ξ11− 2τ ˜ξ1n= ξ0· x0− 2τ |ξ0|xn,

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where ξ0 = (ξ1, · · · , ξn−1) and x0= (x1, · · · , xn−1). Therefore, we can write Z

q0(x)ei(ρ12)·xdx = F q00, 2τ |ξ0|) as well as

Z

q0(x)ei(ρ12)·xdx = F q00, −2τ |ξ0|).

The Sobolev embedding and the assumptions on qj ensure that q0∈ Cα(Ω) for α = s −n2 and therefore q0 satisfies the assumption of Lemma 2.2. Applying Lemma 2.1 to q0 yields that for ε < ε0

|F q00, 2τ |ξ0|)| + |F q00, −2τ |ξ0|)| ≤ C(exp(−πε2(1 + 4τ2)|ξ0|2) + εα). (14) Finally, we estimate the boundary integral

Z

Γ

1,Γ− Λ2,Γ)v1· v2

≤ kΛ1,Γ− Λ2,Γkkv1k

H12(Γ)kv2k

H12(Γ)

≤ kΛ1,Γ− Λ2,Γkkv1kH1(Ω)kv2kH1(Ω)

≤ C exp(|ξ|τ )kΛ1− Λ2k.

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Combining (12), (13), (14), and (15) leads to the inequality

|F q0(ξ)| ≤ C{exp(|ξ|τ )kΛ1− Λ2k+ exp(−πε2(1 + 4τ2)|ξ0|2) + εα+1 τ} (16) for all ξ ∈ Rn and ε < ε0, where C only depends on a priori data on the potentials.

Next we would like to estimate the norm of q0 in H−1. As usual, other estimates of q0 in more regular norms can be obtained by interpolation. To begin, we set ZR = {ξ ∈ Rn : |ξn| < R and |ξ0| < R}. Note that B(0, R) ⊂ ZR ⊂ B(0, cR) for some c > 0. Now we use the a priori assumption on potentials and (16) and calculate

kq0k2H−1 ≤ Z

ZR

|F q0(ξ)|2(1 + |ξ|2)−1dξ + Z

ZRc

|F q0(ξ)|2(1 + |ξ|2)−1

≤ Z

ZR

|F q0(ξ)|2(1 + |ξ|2)−1dξ + CR−2

≤ C{Rnexp(cRτ )kΛ1− Λ2k2+ Rnε+ Rnτ−2+ R−2 +

Z R

−R

Z

B0(0,R)

exp(−2πε2(1 + 4τ2)|ξ0|2) dξ0n},

(17)

(10)

here B0(x0, R) denotes the ball in Rn−1with center x0 and radius R > 0. For the second term on the right hand side of (17), we choose ε = (1 + 4τ2)−1/4 with τ ≥ τ0 1 and integrate

Z R

−R

Z

B0(0,R)

exp(−2πε2(1 + 4τ2)|ξ0|2) dξ0n

= 2R Z

B0(0,R)

exp(−2π(1 + 4τ2)1/20|2) dξ0

= 2R Z

Sn−2

Z R 0

rn−2exp(−2π((1 + 4τ2)1/4r)2) dr dω

≤ CR(1 + 4τ2)−(n−1)/4 Z

0

un−2exp(−2πu2) du

≤ CRτ−(n−1)/2.

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Plugging (18) into (17) with the choice of ε = (1 + 4τ2)−1/4 we get for R > 1 kq0k2H−1 ≤ C{Rnexp(cRτ )kΛ1− Λ2k2+ Rnτ−α+ Rτ−(n−1)/2+ R−2}

≤ C{Rnexp(cRτ )kΛ1− Λ2k2+ Rnτ− ˜α+ R−2}, (19) where ˜α = min{α, (n − 1)/2}.

Observing from (19), we now choose τ such that Rnτ− ˜α = R−2, namely, τ = R(n+2)/ ˜α. Substituting such τ back to (19) yields

kq0k2H−1 ≤ C{Rnexp(cRn+2α˜ +1)kΛ1− Λ2k2+ R−2}. (20) Finally, we choose a suitable R so that

Rnexp(cRn+2α˜ +1)kΛ1− Λ2k2= R−2, i.e., R =

log kΛ1− Λ2k

γ for some 0 < γ = γ(n, ˜α). Thus, we obtain from (20) that

kq1− q2kH−1(Ω)≤ C

log kΛ1− Λ2k

−γ. (21)

The derivation of (21) is legitimate under the assumption that τ is large. To make sure that it is true, we need to take R sufficiently large, i.e. R > R0for some large R0. Consequently, there exists ˜δ > 0 such that if kΛ1− Λ2k < ˜δ then (21) holds. For kΛ1− Λ2k≥ ˜δ, (21) is automatically true with a suitable constant C when we take into account the a priori bound (2).

The estimate (3) is now an easy consequence of the interpolation theorem.

Precisely, let  > 0 such that s = n2 + 2. Using that [Ht0(Ω), Ht1(Ω)]β = Ht(Ω) with t = (1 − β)t0+ βt1 (see e.g. [28, Theorem 1 in 4.3.1]) and the

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Sobolev embedding theorem, we get kq1− q2kL ≤ Ckq1− q2kHn2+≤ Ckq1− q2k(1−β)Ht0 kq1− q2kβHt1. Setting t0= −1 and t1= s we end up with

kq1− q2kL(Ω)≤ Ckq1− q2k

 s+1

H−1(Ω)

which yields the desired estimate (3) with σ = γs+1 .

We now turn to case (b). With a suitable translation and rotation, it suffices to assume a = (0, · · · , 0, R) and 0 /∈ Ω. As in [16], we shall use Kelvin’s transform:

y = 2R

|x|

2

x and x = 2R

|y|

2

y. (22)

Let

˜

u(y) = 2R

|y|

n−2

u(x(y)), then

 |y|

2R

n+2

yu(y) = ∆˜ xu(x).

Denote by ˜Ω the transformed domain of Ω. In view of this transform, Γ0 now becomes ˜Γ0⊂ {yn= 2R} and Γ is transformed to ˜Γ and ˜Γ = ∂ ˜Ω ∩ {yn> 2R}.

On the other hand, if u(x) satisfies ∆u − q(x)u = 0 in Ω, then ˜u satisfies

∆˜u − ˜q ˜u = 0 in Ω,˜ (23)

where

˜

q(y) = 2R

|y|

4

q(x(y)).

Therefore, for (23) we can define the partial Dirichlet-to-Neumann map ˜Λq,˜˜Γ acting boundary functions with homogeneous data on ˜Γ0.

We now want to find the relation between Λq,Γ and ˜Λq,˜˜Γ. It is easy to see that for f, g ∈ H01/2(Γ)

q,Γf, gi = Z

(∇u · ∇v + quv) dx, where u solves

∆u − qu = 0 in Ω, u = f on ∂Γ and v ∈ H1(Ω) with v|∂Ω= g. Defining

f =˜  2R

|y|

n−2 ∂ ˜

f, ˜g = 2R

|y|

n−2 ∂ ˜

g,

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and

˜

v(y) = 2R

|y|

n−2

v(x(y)).

Then we have ˜f , ˜g ∈ H01/2(˜Γ) and

q,Γf, gi = h ˜Λq,˜˜Γf , ˜˜gi, in particular,

h(Λq1− Λq2)f, gi = h( ˜Λq˜

1Γ− ˜Λq˜

2Γ) ˜f , ˜gi. (24) With the assumption 0 /∈ Ω, the change of coordinates x → y by (22) is a diffeomorphism from Ω onto ˜Ω. Note that (2R/|y|)n−2 is a positive smooth function on ∂ ˜Ω. Recall a fundamental fact from Functional Analysis:

q1− Λq2k= sup

(|h(Λq1− Λq2)f, gi|

kf kH1/2

0 (Γ)kgkH1/2 0 (Γ)

: f, g ∈ H01/2(Γ) )

. (25)

The same formula holds for k ˜Λq˜

1Γ− ˜Λq˜

2Γk. On the other hand, it is not difficult to check that kf kH1/2

0 (Γ) and k ˜f kH1/2

0 Γ), kgkH1/2

0 (Γ) and k˜gkH1/2 0 Γ)

are equivalent, namely, there exists C depending on ∂Ω such that 1

Ckf kH1/2

0 (Γ)≤ k ˜f kH1/2

0 Γ)≤ Ckf kH1/2 0 (Γ), 1

CkgkH1/2

0 (Γ)≤ k˜gk

H01/2Γ)≤ CkgkH1/2 0 (Γ).

(26)

Putting together (24), (25), and (26) leads to k ˜Λq˜

1Γ− ˜Λq˜

2Γk≤ CkΛq1− Λq2k (27) with C only depending on ∂Ω.

With all the preparations described above, we use case (a) for the domain Ω with the partial Dirichlet-to-Neumann map ˜˜ Λq,˜˜Γ. Therefore, we immediately obtain the estimate:

k˜q1− ˜q2kL( ˜Ω)≤ C log k ˜Λq˜

1Γ− ˜Λq˜

2Γk

−σ. Finally, rewinding ˜q and using (27) yields the estimate (3).

4. Stability estimate for the conductivity

We aim to prove Corollary 1.2 in this section. We recall the following well- known relation: let q =

γ

γ then

Λq,Γ(f ) = γ−1/2|ΓΛγ,Γ−1/2|Γf ) +1

2(γ−1νγ)|Γf.

(13)

In view of the a priori assumption (5), we have that

q1− Λq2)(f ) = γ−1/2|Γγ1− Λγ2)(γ−1/2|Γf ) where γ−1/2|Γ:= γ1−1/2|Γ= γ2−1/2|Γ, which implies

q1− Λq2k≤ CkΛγ1− Λγ2k (28) for some C = C(N ) > 0. Hereafter, we set qj =

γj

γj , j = 1, 2. The regularity assumption (4) and Sobolev’s embedding theorem imply that q1, q2 ∈ C1(Ω).

Using this and (5), we conclude that ˆq1 − ˆq2 satisfies the assumptions of Lemma 2.2 with α = 1. Therefore, Theorem 1.1 and (28) imply that

kq1− q2kL(Ω)≤ C

log kΛγ1− Λγ2k

−σ1

(29) where C depend on Ω, N, n, s and σ1 depend on n, s. Next, we recall from [1, (26) on page 168] that

1− γ2kL(Ω)≤ Ckq1− q2kσL2(Ω) (30) for some 0 < σ2< 1, where C = C(N, Ω) and σ2 = σ2(n, s). Finally, putting together (29) and (30) yields (6) with σ = σ1σ2 and the proof of Corollary 1.2 is complete.

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Authors’ addresses:

Horst Heck

Bern University of Applied Sciences Engineering and Information Technology Jlcoweg 1

CH-3400 Burgdorf

E-mail: horst.heck@bfh.ch Jenn-Nan Wang

Institute of Applied Mathematical Sciences, NCTS National Taiwan University

Taipei 106 Taiwan

E-mail: jnwang@math.ntu.edu.tw

The second author was supported in part by the MOST (NSC 102-2115-M-002-009-MY3).

Received

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