Depth dependent stability estimate in electrical impedance tomography

Depth dependent stability estimate in electrical impedance tomography

Sei Nagayasu

Gunther Uhlmann

Jenn-Nan Wang

Abstract

We study the inverse problem of determining an electrical inclusion from boundary measurements. We derive a stability estimate for the linearized map with explicit formulae on generic constants that shows that the problem becomes more ill-posed as the inclusion is farther from the boundary. We also show that this estimate is optimal.

1 Introduction

Electrical Impedance Tomography (EIT) is an inverse method that attempts to determine the conductivity distribution inside a body by making voltage and cur- rent measurements at the boundary. The boundary information is encoded in the Dirichlet-to-Neumann map associated to the conductivity equation. More precisely, let Ω be an open bounded domain with smooth boundary in R^dwith d = 2 or 3. As- sume that γ(x) > 0 in Ω possesses a suitable regularity. The conductivity equation is described by the following elliptic equation:

∇ · (γ(x)∇u) = 0 in Ω. (1)

For an appropriate function f defined on ∂Ω, there exists a unique solution u(x) to the boundary value problem for (1) with Dirichlet condition u|∂Ω= f . Thus, one can define a map Λ_γ sending the Dirichlet data to the Neumann data by

Λ_γ(f ) = γ ∂u

∂ν    ∂Ω

.

The map Λ_γ is the Dirichlet-to-Neumann map associated with the conductivity equation (1). It is worth to mention that even though the equation (1) is linear, the

∗Department of Mathematics, Taida Institute of Mathematical Sciences, NCTS (Taipei), Na- tional Taiwan University, Taipei 106, Taiwan. Email:nagayasu@math.ntu.edu.tw

†Department of Mathematics, University of Washington, Box 354305, Seattle, WA 98195-4350, USA. Email:gunther@math.washington.edu

‡Department of Mathematics, Taida Institute of Mathematical Sciences, NCTS (Taipei), Na- tional Taiwan University, Taipei 106, Taiwan. Email:jnwang@math.ntu.edu.tw

map Λ_γ depends nonlinearly on γ. The famous Calder´on problem [3] is to determine γ from the knowledge of Λ_γ. The EIT problem is notoriously known to be ill-posed.

A log-type stability was obtained by Alessandrini [1] and, in fact, this estimate is optimal [8]. A Lipschitz type stability estimate for the values of the conductivity from the Dirichlet-to-Neumann map was proven in [9].

In several practical situations we only need to get partial information on the conductivity. An important example is the determination of electrical inclusions.

In this situation, the conductivity function γ(x) = γ₀(x) + γ₁(x)χ_D, where D b Ω is called an inclusion and χ_D is the characteristic function of D. Here γ₀ is the background medium and γ₁, D are the abnormalities. For this problem, assuming γ₀ is known. We are interested in determining the shape of D by the Dirichlet- to-Neumann map, denoted by Λ_D. Under some natural assumptions on γ₀ and γ₁, uniqueness was shown by Isakov [7]. Numerical methods based on special complex geometrical optics solutions for ∇ · γ₀∇u = 0 are given in [4], [10] (also see [5], [6], [11] for related results). It has been observed numerically that the deeper the inclusion, the worst the numerical reconstruction. See for instance [4], [10], and [11]. In this paper we give a precise quantitative description of this phenomenon in a model case.

We consider the problem in two dimensions, i.e., Ω ⊂ R². Let k > 0, k 6= 1 and define L_Du := ∇ · ((1 + (k − 1)χ_D)∇u). For any f ∈ H^1/2(∂Ω), there exists a unique weak solution to

(LDu = 0 in Ω, u = f on ∂Ω.

The Dirichlet-to-Neumann map is given by Λ_D : H^1/2(∂Ω) → H^−1/2(∂Ω) as Λ_Df = ∂u

∂ν    ∂Ω

,

where ν is the unit outer normal of ∂Ω. The inverse problem is to determine D from Λ_D. As mentioned above, the uniqueness for this problem is known [7]. A log-type stability was obtained in [2]. More precisely, it was proved in [2] that under some minor a priori assumptions on the inclusions, if kΛD1 − ΛD2k_L(H^1/2_,H^−1/2₎ <  with

 > 0, then the Hausdorff distance between ∂D₁ and ∂D₂ satisfies d_H(∂D₁, ∂D₂) < ω(),

where ω(t) is an increasing function in [0, ∞) and satisfies ω(t) ≤ C| log t|^−η for t ∈ (0, 1).

The constants C and 0 < η < 1 depend on the a priori data of the inclusions, but their dependence is not explicitly given in [2]. As a matter of fact, to our best knowledge, we do not know any available stability estimates for inverse problems having explicit descriptions of the data-dependent constants.

Our concern here is to understand how the stability estimate depends on the depth of the inclusion. In this paper, we consider the linearized map of Λ_D around a known inclusion. We believe that, either from numerical or theoretical viewpoint, the stability estimate using Λ_D should behave similarly to the estimate using the linearized map of Λ_D. To set up our problem, we let Ω = {|x| < R} and B = {|x| <

r}, where 0 < r < R. We introduce a smooth function ψ : ∂B → R

in order to describe a perturbation B_s of the domain B, namely, the boundary ∂B_s of the domain B_s is described by the image of

y = Fs(x) := x + sψ(x)νx(x), x ∈ ∂B,

where ν_x(x) is the unit outward normal vector to ∂B at x ∈ ∂B. For f ∈ H^1/2(∂Ω), let u₀ be the solution to the problem

(L_Bu₀ = 0 in Ω,

u₀ = f on ∂Ω. (2)

Likewise, let u_s be the solution to the problem (L_Bsu_s = 0 in Ω,

u_s= f on ∂Ω. (3)

The linearized map of the Dirichlet-to-Neumann map at the direction ψ(x), denoted by dΛB(ψ), is formally defined by

dΛ_B(ψ) = lim

s→0

1

s(Λ_Bs − Λ_B).

We will show that dΛ_B(ψ) is legitimately defined in the later section. We now state our main theorem.

Theorem 1. Let k > 0 satisfy k 6= 1. Let m > 0. Given M₀, r₀ > 0 and X₀ > 1.

Assume that

M ≥ M₀, r ≤ r₀, R

r ≥ X₀. Then for any ψ ∈ H^m(∂B) satisfying

kψk_H^m_(∂B) ≤ M and kdΛ_B(ψ)kL < 1, the following estimate holds:

kψk_L²_(∂B) ≤ CM



log R r

m

log kdΛB(ψ)kL

−m, (4)

where a positive constant C depends only on k, m, M₀, r₀, X₀.

Here k · kL denotes the operator norm on the space of bounded linear operators between H^1/2(∂Ω) and H^−1/2(∂Ω). Moreover, this estimate (4) is optimal in the sense of Propositions 12 and 13 (See Section 4).

Remark 2. Estimate (4) clearly indicates that the determination of an inclusion by boundary measurements is getting more ill-posed when the inclusion is hidden deeper inside of the conductor, i.e., R/r becomes large.

The paper is organized as follows. In Section 2, we discuss the linearized map dΛB(ψ) of the Dirichlet-to-Neumann map. In Section 3, we state some technical lemmas which we need and then we prove our main theorem. In Section 4, we discuss the optimality of the stability estimate.

2 The linearized map

In this section, we discuss the linearized map dΛ_B(ψ) of the Dirichlet-to-Neumann map Λ_D. We first remark that it is known that the map γ 7→ Λ_γ is bounded and analytic in the subset of L^∞(D) consisting of functions which are real and have a positive lower bound (see [3]). We now introduce polar coordinates (ρ, θ), that is, x = ρ(cos θ, sin θ) ∈ R², in order to express the linearized operator explicitly as the solution to some transmission problem. We put eψ(θ) := ψ(r cos θ, r sin θ) and ψ_l := R2π

0 ψ(θ) ee ^−ilθdθ for a function ψ ∈ L²(∂B). We remark that eψ(θ) = (2π)⁻¹P

l∈Zψ_le^ilθ, kψk²_L2(∂B) = r

2π X

l∈Z

|ψl|² and kψk²_Hm(∂B) = r 2π

X

l∈Z

(1 + l²)^m|ψl|². (5)

Let ef (θ) := f (R cos θ, R sin θ) and fl := R2π

0 f (θ) ee ^−ilθdθ for a function f defined on ∂Ω in the same way. Throughout this paper the subscripts + (respectively −) denote the limit from outside (respectively inside) the inclusion.

Lemma 3. The linearized operator dΛB(ψ) satisfies dΛ_B(ψ)(f ) = ∂U

∂ν    ∂Ω

(6) for any f ∈ H^1/2(∂Ω), where U is the solution to the problem























∆U = 0 in Ω \ ∂B, U |₊− U |−= 1 − k

k ψ∂u₀

∂ν    +

on ∂B,

∂U

∂ν    +

− k∂U

∂ν    −

= r⁻²(1 − k) ∂_θ

ψ(θ) ∂e _θu₀|₊

on ∂B, U = 0 on ∂Ω

(7)

and u₀ is the solution to (2).

Proof. The solutions u₀ and u_s to the problems (2) and (3) satisfy



















∆u₀ = 0 in Ω, u0|+ = u0|− on ∂B,

∂u₀

∂ν    +

= k∂u₀

∂ν    ₋

on ∂B, u₀ = f on ∂Ω

and



















∆u_s = 0 in Ω, us|+ = us|− on ∂Bs,

∂u_s

∂ν    +

= k∂u_s

∂ν    ₋

on ∂B_s, u_s= f on ∂Ω,

respectively. Now we put

U (x) := lim

s→0

u_s(x) − u₀(x)

s .

and formula (6) is obvious. Moreover, if we write y = F_s(x), we have 1

s u_s(y)|±− u₀(x)|±
→ U (x)|±+ ψ(x)∂u0

∂ν (x)    _±

and 1

s

 ∂u_s

∂ν_y(y)    _±

− ∂u_s

∂ν_x(x)    _±



→ ∂_ρU (x)|±− r⁻²ψe⁰(θ) ∂_θu₀(x)|±+ eψ(θ) ∂_ρ²u₀(x)|±

on ∂B as s → 0. Thus we prove this lemma by using

∂_ρ²u0(x)|± = −1

r∂ρu0(x)|±− 1

r²∂_θ²u0(x)|±

on ∂B.

Using Fourier series, we can write the linearized operator dΛ_B(ψ) more explicitly.

Lemma 4. For f ∈ H^1/2(∂Ω) we have

dΛ_B(ψ)(f )(R cos θ, R sin θ) = X

l∈Z

λ_le^ilθ,

where we put λ₀ := 0 and λ−l := k − 1

π² (Rr)⁻¹S_l

∞

X

p=1

S_p{(k + 1)ψ−l+pf−p+ (k − 1)ψ−l−pf_p} ,

λ_l := k − 1

π² (Rr)⁻¹S_l

∞

X

p=1

S_p{(k + 1)ψ_l−pf_p+ (k − 1)ψ_l+pf−p} ,

S_l := l

(k − 1)R^−lr^l− (k + 1)R^lr^−l for any positive integer l.

Proof. Note that the solution to the problem (2) is expressed as follows:

u0(ρ cos θ, ρ sin θ)

=













 1 2π

" _∞ X

l=1

S_l

l (k − 1)r^lρ^−l− (k + 1)r^−lρ^l

f−le^−ilθ+ fle^ilθ
+ f0

#

, r < ρ < R, 1

2π

"

−2

∞

X

l=1

S_l

l r^−lρ^l f−le^−ilθ + f_le^ilθ
+ f₀

#

, 0 < ρ < r.

In particular, the right-hand sides of the transmission conditions on ∂B in (7) can be written as:

1 − k k ψ∂u₀

∂ν    +

= −(1 − k)r⁻¹ 2π²

X

p∈Z

∞

X

j=1

S_j(ψ_p+jf−j+ ψ_p−jf_j)e^ipθ,

r⁻²(1 − k) ∂θ



ψ(θ) ∂e θu0|+



= (1 − k)r⁻² 2π²

X

p∈Z

p

∞

X

j=1

Sj(−ψp+jf−j + ψp−jfj)e^ipθ.

Hence, the solution to the problem (7) is given by U (ρ cos θ, ρ sin θ)

=

























































 2

(2π)²(1 − k)r⁻¹

∞

X

l=1

S_l

l (R^lρ^−l− R^−lρ^l)

×

∞

X

p=1

S_p(k + 1)(ψ−l+pf_−pe^−ilθ+ ψ_l−pf_pe^ilθ)

+ (k − 1)(ψ−l−pf_pe^−ilθ+ ψ_l+pf−pe^ilθ) , r < ρ < R,

− 4

(2π)²(1 − k)r⁻¹

∞

X

l=1

Sl

l ρ^l

×

∞

X

p=1

S_pR^−l(ψ−l+pf−pe^−ilθ+ ψ_l−pf_pe^ilθ)

+ R^lr^−2l(ψ_−l−pf_pe^−ilθ + ψ_l+pf_−pe^ilθ)

+ 2

(2π)²(1 − k)r⁻¹

∞

X

p=1

S_p(ψ_pf−p+ ψ−pf_p), 0 < ρ < r.

We then finish the proof of the lemma using formula (6).

With the help of Lemma 4, we can given an estimate of the size of dΛ_B(ψ) in terms of r and R.

Lemma 5. The operator dΛ_B(ψ) : H^1/2(∂Ω) → H^−1/2(∂Ω) is a bounded linear operator. In particular, we have the following estimate:

kdΛ_B(ψ)(f )k_H−1/2(∂Ω)

≤ 2^3/2|k − 1|

(k + 1)π

1

(1 − (r/R)²)²(Rr)⁻¹

×

" _∞ X

l=1

lr R

2l ∞

X

j=1

jr R

2j

|ψ_−l+j|²+ |ψ_−l−j|²+ |ψ_l−j|²+ |ψ_l+j|²

#1/2

× kf k_H1/2(∂Ω). (8)

Proof. We first remark that

|S_l| = l k + 1

r R

l 1

1 − ((k − 1)/(k + 1))(r/R)^2l < l k + 1

1 1 − (r/R)²

r R

l

. So, it follows from Lemma 4 that

|λ±l| ≤ |k − 1|

(k + 1)π²

1

(1 − (r/R)²)²(Rr)⁻¹

× lr R

l ∞

X

p=1

pr R

p

(|ψ±l−p||f_p| + |ψ±l+p||f−p|)

for any positive integer l. Hence we have kdΛB(ψ)(f )k²_H−1/2(∂Ω) = 2πRX

l∈Z

(1 + l²)^−1/2|λl|² ≤ 2πR

∞

X

l=1

l⁻¹(|λl| + |λ−l|)²

≤ 2|k − 1|² (k + 1)²π³

1

(1 − (r/R)²)⁴R⁻¹r⁻²

∞

X

l=1

l

r R

2l

×

" _∞ X

p=1

pr R

p

(|ψ−l−p||f_p| + |ψ−l+p||f−p| + |ψ_l−p||f_p| + |ψ_l+p||f−p|)

#2

.

We immediately obtain this lemma since we have

" _∞ X

p=1

p

r R

p

(|ψ−l−p||fp| + |ψ−l+p||f−p| + |ψl−p||fp| + |ψl+p||f−p|)

#2

≤

" _∞ X

p=1

pr R

2p

|ψ−l−p|²+ |ψ−l+p|²+ |ψ_l−p|²+ |ψ_l+p|²

#

×

" _∞ X

p=1

p |f_p|²+ |f−p|²+ |f_p|²+ |f−p|²

#

=

" _∞ X

p=1

pr R

2p

|ψ−l−p|²+ |ψ−l+p|²+ |ψ_l−p|²+ |ψ_l+p|²

#

× 2

∞

X

p∈Z

|p||f_p|²

≤

" _∞ X

p=1

p

r R

2p

|ψ−l−p|²+ |ψ−l+p|²+ |ψl−p|²+ |ψl+p|²

#

× 4π

Rkf k²_H1/2(∂Ω)

by the Schwarz inequality.

Remark 6. By changing the index, we can write the term on the right-hand side of (8) as follows:

∞

X

l=1

l

r R

2l ∞

X

j=1

j

r R

2j

|ψ−l+j|²+ |ψ−l−j|²+ |ψl−j|²+ |ψl+j|²

= 2(1 − s²)⁻³(1 + s²)s²|ψ0|² +

∞

X

p=1

s^p p³− p

6 + 2(1 − s²)⁻³s²p(1 − s²) + (1 + s²)



|ψ_p|²+ |ψ_−p|²
,

where we put s := (r/R)² for simplicity.

Corollary 7. We have the estimate kdΛ_B(ψ)kL ≤ 8|k − 1|

π^1/2(k + 1)

1

{1 − (r/R)²}⁴r^1/2R⁻³kψk_L²_(∂B).

Proof. We obtain this corollary by Lemma 5 and the estimate |ψ_l|² ≤ (2π/r)kψk²_L2(∂B)

since we haveP∞

j=1jt^j = (1 − t)⁻²t for |t| < 1.

In the following corollary, we consider a particular case, which will be needed in the proof of the optimality of the stability estimate (see Section 4).

Corollary 8. Let a > 0 and µ be a positive integer. Let eψ(θ) = 2a cos µθ. Then we have the following estimate:

kdΛB(ψ)kL ≤ C0aµ^3/2 1

{1 − (r/R)²}^7/2(Rr)⁻¹

r R

µ

, (9)

where the positive constant C₀ depends only on k.

Proof. By Lemma 5 and Remark 6, we get that kdΛB(ψ)kL

≤ 8a|k − 1|

k + 1

1

{1 − (r/R)²}²(Rr)⁻¹r R

µ

×

"

µ³− µ 6 + 2



1 −r R

4⁻³

r R

4 µ



1 −r R

4 +



1 +r R

4#1/2

because of ψ±µ= 2πa and ψl = 0 for l 6= ±µ. This corollary follows from µ³ − µ

6 + 2



1 −r R

4−3

r R

4 µ



1 −r R

4 +



1 +r R

4

≤



1 −r R

4−3

 µ³

6 + 2(µ + 2)



≤ 37 6 µ³



1 −r R

2−3

, where the constant C₀ = (37/6)^1/2· 8|k − 1|/(k + 1).

3 The proof of the stability estimate

In this section, we prove our main theorem. We first state some useful identities.

Lemma 9. For f, g ∈ H^1/2(∂Ω) we have the identity Z

∂Ω

dΛ_B(ψ)(f ) g dσ

= −(1 − k)

 r⁻²

Z

∂B

ψ ∂_θu₀|₊∂_θv₀|₊dσ + 1 k

Z

∂B

ψ ∂u₀

∂ν    +

∂v₀

∂ν    +

dσ



, (10) where u₀ and v₀ are the solutions to the problem (2) with the boundary conditions u₀ = f and v₀ = g, respectively.

Proof. Applying Green’s formula yields 0 =

Z

Ω\B

∆U v₀dx − Z

Ω\B

U ∆v₀dx

= Z

∂Ω

dΛ_B(ψ)(f ) g dσ − Z

∂B

∂U

∂ν    +

v₀|₊dσ + Z

∂B

U |₊ ∂v₀

∂ν    +

dσ and 0 =

Z

B

∆U v₀dx − Z

B

U ∆v₀dx = Z

∂B

∂U

∂ν    ₋

v₀|−dσ − Z

∂B

U |−

∂v₀

∂ν    ₋

dσ, where U is the solution to the problem (7). Using these identities and the transmis- sion conditions for U and v0, we obtain this lemma.

Lemma 10. Let g_j on ∂Ω be given by g_j = e^ijθ for any integer j, where i = √

−1.

Then we have Z

∂Ω

dΛB(ψ)(g±l) g±pdσ = 4(1 − k)²r⁻¹SlSpψ∓(l+p), (11) Z

∂Ω

dΛ_B(ψ)(g±l) g∓pdσ = −4(1 − k)(1 + k)r⁻¹S_lS_pψ∓(l−p) (12) for positive integers l and p.

Proof. We first remark that the solution u₀ to the problem (2) with the boundary condition u₀ = g_±l is

u₀(ρ sin θ, ρ cos θ) =





 S_l

l (k − 1)r^lρ^−l− (k + 1)r^−lρ^l e^±ilθ, r < ρ < R,

− 2S_l

l r^−lρ^le^±ilθ, ρ < r for any positive integer l and in particular we have

∂u₀

∂ρ    +

= −2kr⁻¹S_le^±ilθ and ∂u₀

∂θ    +

= ∓2iS_le^±ilθ

on ∂B. So, by taking f = g±l and g = g±p (or g = g∓p) and applying Lemma 9, we obtain this lemma.

Now we denote X := R/r. It is important to estimate each ψ_j in view of formula (5).

Lemma 11. We have that

|ψ0| ≤ C1r²X³kdΛB(ψ)kL, |ψ±1| ≤ C1r²X⁴kdΛB(ψ)kL

and

|ψ±l| ≤ C₁

l r²X^l+1kdΛ_B(ψ)kL

for any integer l ≥ 2, where the positive constant C₁ depends only on k.

Proof. We first note that kg±lk_H^1/2_(∂Ω) = (1 + l²)^1/4R^1/2 for any positive integer l.

It is easy to see that 

   Z

∂Ω

dΛ_B(ψ)(g_j) g_j⁰dσ    

≤ kdΛ_B(ψ)(g_j)k_H^−1/2_(∂Ω)kg_j⁰k_H1/2(∂Ω)

≤ kdΛ_B(ψ)k_Lkg_jk_H1/2(∂Ω)kg_j⁰k_H1/2(∂Ω)

= (1 + j²)^1/4(1 + (j⁰)²)^1/4R kdΛ_B(ψ)kL

= (1 + j²)^1/4(1 + (j⁰)²)^1/4rX kdΛB(ψ)kL

for any integers j, j⁰ 6= 0. On the other hand, we have 1

|Sl| = k + 1 l X^l

"

1 − k − 1 k + 1

 1 X

2l#

≤ 2(k + 1) l X^l. By taking l = p = 1 in the identity (12), we get

|ψ₀| = r

4|1 − k|(1 + k) 1 S₁²

    Z

∂Ω

dΛ_B(ψ)(g₁) g−1dσ    

≤ 2^1/2(k + 1)

|1 − k| r²X³kdΛ_B(ψ)kL. Likewise, taking l = 2 and p = 1 in the identity (12) gives

|ψ±1| ≤ 10^1/4(k + 1)

2|1 − k| r²X⁴kdΛB(ψ)kL.

On the other hand, taking p = l ≥ 1 in the identity (11), we obtain

|ψ∓2l| = r 4(1 − k)²

1 S_l²

    Z

∂Ω

dΛ_B(ψ)(g±l) g±ldσ    

≤ (k + 1)² (1 − k)²

(1 + l²)^1/2

l² r²X^2l+1kdΛ_B(ψ)kL

≤ 2^3/2(k + 1)² (1 − k)²

1

2lr²X^2l+1kdΛ_B(ψ)k_L.

In the same way, taking l ≥ 1 and p = l + 1 in the identity (11), we get

|ψ±(2l+1)| ≤ 2 · 10^1/4(k + 1)² (1 − k)²

1

2l + 1r²X^(2l+1)+1kdΛB(ψ)kL. The proof of the lemma is complete.

We now prove our main theorem.

Proof of Theorem 1. Note that the a priori assumption kψk_H^m_(∂B) ≤ M is equiva- lent to

X

l∈Z

(1 + l²)^m|ψ_l|² ≤ 2π

r M². (13)

We first consider A := kdΛ_B(ψ)k²_L sufficiently small. Let 0 < t < 2 · 3^−2mπM²r⁻¹ be given. We remark that (2πM²/rt)^1/2m > 3. Let N be the minimum integer satisfying 2πM²N^−2mr⁻¹ ≤ t, namely,

N − 1 <  2πM² rt

1/2m

≤ N. (14)

One can see that N ≥ 4. Using Lemma 11, we have X

|l|≤N −1

|ψ_l|²

≤ C₁²r⁴X⁶kdΛ_B(ψ)k²_L+ 2C₁²r⁴X⁸kdΛ_B(ψ)k²_L+ 2

N −1

X

l=2

C₁²

l² r⁴X^2(l+1)kdΛ_B(ψ)k²_L

≤ C₁²r⁴ X⁶+ 2X⁸+ 2X^2N

N −1

X

l=2

1 l²

!

A ≤ 5C₁²r⁴X^2NA.

On the other hand, we can estimate X

|l|≥N

|ψ_l|² ≤ (1 + N²)^−m X

|l|≥N

(1 + l²)^m|ψ_l|² ≤ (1 + N²)^−m2π

r M² ≤ N^−2m2π

r M² ≤ t by estimate (13). Combining the estimates above and (14), we get that

X

l∈Z

|ψ_l|² ≤ F (t),

where

F (t) := 5C₁²r⁴X²

(^2πM2/r)^{1/2mt−1/2m+1} A + t.

Now we would like to show the estimate

F (t₀) ≤ C₂M²r⁻¹(log X)^2m(− log A)^−2m for some 0 < t₀ < 2 · 3^−2mπM²r⁻¹, (15) where the positive constant C₂ depends only on k and m. We choose t₀ such that

r⁴X²

(^2πM2/r)^1/2mt−1/2m0 +1

A = M²r⁻¹(log X)^2m(− log A)^−2m, i.e., we pick

t₀ = 2^2m+1πM²r⁻¹(log X)^2m(log G(A, M, r, X))^−2m,

where G(A, M, r, X) := A⁻¹(− log A)^−2mM²r⁻⁵(log X)^2mX⁻². Then we have F (t₀) = 5C₁²M²r⁻¹(log X)^2m(− log A)^−2m+ t₀.

Therefore, it is enough to estimate t₀. Now we fix η, η⁰ ∈ (0, 1) small enough such that η + 2mη⁰ < 1 and put

A₀ :=(eη⁰)^2mM²r⁻⁵(log X)^2mX⁻⁸1/(1−η−2mη⁰)

. Let 0 < A < min{A₀, 1}, then we have

A^{1−η−2mη0} ≤ A^{1−η−2mη}₀ ⁰ = (eη⁰)^2mM²r⁻⁵(log X)^2mX⁻⁸.

Consequently, we obtain

G(A, M, r, X) ≥ A^−η(− log A)^−2m(eη⁰)⁻¹A^−η0 2m

X⁶ ≥ A^−ηX⁶ ≥ (X⁶

A^−η since 0 < − log t ≤ (eη⁰)⁻¹t^−η0 for all 0 < t < 1. Thus we deduce that

t0 ≤ 2^2m+1πM²r⁻¹(log X)^2m log A^−η
−2m

= 2^2m+1πη^−2mM²r⁻¹(log X)^2m(− log A)^−2m and

t0 ≤ 2^2m+1πM²r⁻¹(log X)^2m log X^6
−2m

= 2 · 3^−2mπM²r⁻¹.

Summing up, we have proved that if 0 < A < min{A₀, 1} then the estimate (15) holds with C₂ = 5C₁²+ 2^2m+1πη^−2m. In other words, we obtain

kψk_L²_(∂B) = r 2π

X

l∈Z

|ψ_l|²

!1/2

≤ r

2πF (t₀)1/2

≤ C₂ 2π

1/2

M (log X)^m(− log A)^−m for 0 < A < min{A₀, 1}.

Next we consider the case where A₀ ≤ A < 1. Note that A0 =(eη⁰)^2mM²r⁻⁵(log X)^2mX⁻⁸1/(1−η−2mη⁰)

≥ cX−8/(1−η−2mη⁰)

, where c := min(eη⁰)^2mM₀²r₀⁻⁵(log X₀)^2m1/(1−η−2mη⁰)

, 1/2 . We remark that − log c >

0. So we can estimate

kψk_L²_(∂B) ≤ kψk_H^m_(∂B) ≤ M

≤ (− log A₀)^mM (− log A)^−m ≤ C₃^mM (log X)^m(− log A)^−m for A₀ ≤ A < 1 since

− log A₀ ≤ − log(cX−8/(1−η−2mη⁰)) = − log c + 8

1 − η − 2mη⁰ log X ≤ C₃log X, where C3 := (− log c)/(log X0) + 8/(1 − η − 2mη⁰). Thus we obtain estimate (4) with C = 2^−mmin{(C₂/2π)^1/2, C₃^m}.

4 Optimality of the stability estimate

In this section, we discuss the optimality of the stability estimate in the sense that the polylogarithmic order m in estimate (4) can not be improved. We divide our discussion into two parts. For the first part, we fix the constants k, m, R, r, M > 0.

In Theorem 1, we derived the estimate kψk_L²_(∂B) ≤ C∗

log kdΛ_B(ψ)kL

−m (16)

for any ψ ∈ H^m(∂B) satisfying kψk_H^m_(∂B) ≤ M and kdΛ_B(ψ)k_L < 1, where C_∗ is independent of ψ. We now prove that the polylogarithmic order in (16) is optimal.

Proposition 12. Let k, m, R, r, M, ε > 0 be fixed. Assume k 6= 1 and R > r. Then there exists no positive constant C⁰ which is independent of ψ such that the following estimate holds:

kψk_L²_(∂B) ≤ C⁰

log kdΛ_B(ψ)kL

−m−ε (17)

for any ψ ∈ H^m(∂B) satisfying

kψk_H^m_(∂B) ≤ M (18)

and

kdΛ_B(ψ)kL< 1. (19)

Proof. We prove this proposition by contradiction. That is, we assume that there ex- ists C⁰ which is independent of ψ such that (17) holds for all ψ ∈ H^m(∂B) satisfying (18) and (19). Let µ be a positive integer. Put a_µ:= 2⁻¹π^−1/2r^−1/2(1 + µ²)^−m/2M . Define a function ψ on ∂B by eψ(θ) = 2a_µcos µθ. Then we have

kψk_H^m_(∂B) = M and kψk_L²_(∂B) = (1 + µ²)^−m/2M. (20) So, the function ψ satisfies the condition (18) in particular. Moreover, using Corol- lary 8, we can see that

kdΛ_B(ψ)kL≤ C₀a_µµ^3/2 1

{1 − (r/R)²}^7/2(Rr)⁻¹r R

µ

= C₀⁰(1 + µ²)^−m/2µ^3/2r R

µ

< C₀⁰µ^−m+3/2r R

µ

, (21)

where C₀⁰ := 2⁻¹π^−1/2C₀M R⁻¹r^−3/2{1 − (r/R)²}^−7/2. Note that the constant C₀⁰ is independent of µ. Hence the function ψ satisfies the condition (19) when µ is large enough. Consequently, the estimate (17) holds for µ sufficiently large. By (20), (21) and (17), we then obtain

(1 + µ²)^−m/2M = kψk_L²_(∂B) ≤ C⁰

log kdΛ_B(ψ)kL

−m−ε

< C⁰

− logn

C₀⁰µ^−m+3/2r R

µo^−m−ε

= C⁰



− log C₀⁰ +

 m − 3

2



log µ + µ log X

−m−ε

, i.e.,

M ≤ C⁰(1 + µ²)^m/2



− log C₀⁰ +

 m − 3

2



log µ + µ log X

−m−ε

(22) for µ  1. Recall that X := R/r. However, the right-hand side of (22) tends to zero as µ → +∞. This is a contradiction.

In the second part, we discuss the dependency of the constant C∗ in (16) on R and r. Fix r₀ > 0 and X₀ > 1. We have shown in Theorem 1 that C∗ in (16) satisfies

C∗ ≤ C_]



log R r

m

(23) for R, r > 0 with r ≤ r₀ and R/r ≥ X₀, where C_] depends only on k, m, r₀, X₀, M . Similar to Proposition 12, we can prove that the polylogarithmic order in (23) is optimal, at least, when the constant R and the ratio R/r are large.

Proposition 13. Let k > 0 satisfy k 6= 1, m > 0, and M > 0. Given R₀ > 0 and X0 > 1. Let ε > 0. Then there exists no positive constant C⁰⁰, depending only on k, m, R₀, X₀, M and ε, such that for any

R ≥ R0, R

r ≥ X0 (24)

and for any ψ ∈ H^m(∂B) satisfying

kψk_H^m_(∂B) ≤ M and kdΛ_B(ψ)kL < 1, (25) the following estimate holds:

kψk_L²_(∂B) ≤ C⁰⁰



log R r

m−ε

log kdΛ_B(ψ)k_L 

−m. (26)

Proof. We also prove this proposition by contradiction. We assume that there exists a positive constant C⁰⁰ which depends only on k, m, R₀, X₀, M and ε such that for any R, r > 0 satisfying (24) and for any ψ ∈ H^m(∂B) satisfying (25) the estimate (26) holds.

Define a function ψ on ∂B by eψ(θ) = 2a₂cos 2θ, where a₂ := 2⁻¹5^−m/2 × π^−1/2r^−1/2M . Then we have that

kψk_L²_(∂B) = 5^−m/2M, kψk_H^m_(∂B) = M and

kdΛ_B(ψ)k_L ≤ 2^1/25^−m/2C₀

π^1/2 M R^−5/2



1 −r R

2−7/2

r R

1/2

≤ C₀⁰⁰r R

1/2

for r, R > 0 satisfying (24) as in the proof of Proposition 12, where C₀⁰⁰ := 2^1/2π^−1/25^−m/2C₀M R^−5/2₀ (1 − X₀⁻²)^−7/2.

We remark that the constant C₀ is independent of r, R > 0. Thus, the conditions in (25) hold whenever R/r is large enough. By the assumptions, the estimate (26)

holds for R/r  1. It follows that 5^−m/2M = kψk_L²_(∂B) ≤ C⁰⁰



log R r

m−ε

log kdΛ_B(ψ)k_L 

−m

≤ C⁰⁰



log R r

m−ε

− log

 C₀⁰⁰r

R

1/2−m

= C⁰⁰



log R r

m−ε

 1

2log R r



− log C₀⁰⁰

−m

→ 0 as R

r → +∞, which leads to a contradiction. The proposition is now proved.

Acknowledgements

Nagayasu is partly supported by a postdoc fellowship from the Taida Institute of Mathematical Sciences and the National Science Council of Taiwan. Uhlmann is partly supported by NSF and a Walker Family Endowed Professorship. Wang is supported in part by the National Science Council of Taiwan.

References

[1] G. Alessandrini, Stable determination of conductivity by boundary measure- ments, Appl. Anal., 27 (1988), 153–172.

[2] G. Alessandrini and M. Di Cristo, Stable determination of an inclusion by boundary measurements, SIAM J. Math. Anal., 37 (2005), 200–217.

[3] A. P. Calder´on, On an inverse boundary value problem, Seminar on Numer- ical Analysis and its Applications to Continuum Physics, Soc. Brasileira de Matem´atica, Rio de Janeiro, 1980, 65–73.

[4] T. Ide, H. Isozaki, S. Nakata, S. Siltanen, and G. Uhlmann, Probing for elec- trical inclusions with complex spherical waves, Comm. Pure Appl. Math., 60 (2007), 1415–1442.

[5] M. Ikehata and S. Siltanen, Numerical method for finding the convex hull of an inclusion in conductivity from boundary measurements, Inverse Problems, 16 (2000), 1043–1052.

[6] M. Ikehata and S. Siltanen, Electrical impedance tomography and Mittag- Leffler’s function, Inverse Problems, 20 (2004), 1325–1348.

[7] V. Isakov, On uniqueness of recovery of a discontinuous conductivity coefficient, Comm. Pure Appl. Math., 41 (1988), 865–877.

[8] N. Mandache, Exponential instability in an inverse problem for the Schr¨odinger equation, Inverse Problems, 17 (2001), 1435–1444.

[9] J. Sylvester and G. Uhlmann, Inverse boundary value problems at the boundary- continuous dependence, Comm. on Pure and Appl. Math., 41 (1988), 197–219.

[10] G. Uhlmann and J.N. Wang, Reconstructing discontinuities using complex geo- metrical optics solutions, SIAM J. Appl. Math., 68 (2008), 1026–1044.

[11] G. Uhlmann, C.T. Wu, and J.N. Wang, Reconstruction of inclusions in an elastic body, preprint.