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Size estimates for the inverse boundary value problems of isotropic elasticity and complex

conductivity in 3D

C˘ at˘ alin I. Cˆ arstea

Jenn-Nan Wang

Abstract

In the inverse boundary value problems of isotropic elasticity and complex conductivity, we derive estimates for the volume fraction of an inclusion whose physical parameters satisfy suitable gap conditions.

In the elasticity case we require one measurement for the lower bound and another for the upper one. In the complex conductivity case we need three measurements for the lower bound and three for the upper. We accomplish this with the help of the “translation method”

which consists of perturbing the minimum principle associated with the equation by either a null-Lagrangian or a quasi-convex quadratic form.

1 Introduction

An inverse boundary value problem is the question of identifying the interior physical properties of an object using measurements taken on the boundary of the object. One interesting sub-problem, in the case when the object consists of two homogeneous phases, would be to estimate the volume fraction of one of those phases. This is known as the size estimate problem. In this paper we consider the size estimate problem for the isotropic elasticity equation and for the isotropic complex conductivity equation. In the case of elasticity, we seek to estimate the size of the inclusion using boundary measurements

National Center for Theoretical Sciences Mathematics Division, Taipei 106, Taiwan.

Email: cicarstea@ncts.ntu.edu.tw

Institute of Applied Mathematical Sciences, NCTS, National Taiwan University, Taipei 106, Taiwan. Email: jnwang@math.ntu.edu.tw. Partially supported by MOST 105-2115-M-002-014-MY3.

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of displacements and tractions. In the case of the complex conductivity equation, our method will use boundary measurements of potentials and current densities.

Unlike more general reconstruction procedures, which require a large number of measurements, size estimate methods usually rely on only a few measurements. One known approach to deriving size estimates is a PDE method using quantitative uniqueness estimates, the other, which we use here, is a variational method which is related to the estimates of effec- tive properties of the composites. In the case of the conductivity equa- tion estimates for the size of the inclusion have been derived in the work of Alessandrini-Rosset [2], Alessandrini-Rosset-Seo [1], Ikehata [5], Kang- Seo-Sheen [10], Milton [14], and others. The translation method which was introduced by Murat-Tartar [17], [18], [16] and Lurie-Cherkaev [11], [12], was used by Kang-Kim-Milton [6], Kang-Milton [8] to obtain lower and upper bounds for the relative volume of an inclusion. The same approach was also used to bound the volume fraction of two-phase materials in the 2D elasticity case [15] and in the shallow shell case [9].

One practical example where size estimates in the complex conductivity case may be useful is the estimation of the size of tumors or other anomalous tissue in biological samples. Biological tissue, when probed using alternating currents, may be modeled using complex conductivities. This is due to the fact that cell membranes act as capacitors, thus introducing an imaginary component to the conductivity (see [13]). There are a number of results known in this case such as the one of Beretta-Francini-Vessella [3], Kang- Kim-Lee-Li-Milton [7], and Milton-Thaler [19]. We want to point out that [3]

is based on the PDE method, while both [7] and [16] use the variational approach. To put the current work in perspective, we remark that [7] uses the translation method for the 2D problem and [16] considers 2D and 3D problems using the splitting method.

1.1 Main results – elasticity

Let Ω ⊂ R3 be a bounded domain and let C be a two-valued elastic tensor defined on this domain. We will assume that C is isotropic, i.e., that its components are

Cijkl = λδijδkl+ µ(δikδjl+ δilδjk),

where λ and µ are known as the Lam´e coefficients. The ellipticity condition for C in terms of the Lam´e coefficients is

λ(x) + 2µ(x) > 0, µ(x) > 0, ∀x ∈ Ω.

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We will make a stronger assumption, namely that

3λ(x) + 2µ(x) > 0, µ(x) > 0, ∀x ∈ Ω.

This is known as the strong convexity condition (see [20, Section 2.9]).

We denote by Ω1 and Ω2 the domains corresponding to the two phases and will use corresponding lower indices to identify the values the Lam´e coef- ficients and any other quantity derived from them take on these two domains.

We also define the volume fractions f1 := |Ω1|/|Ω| and f2 = |Ω2|/|Ω|. In or- der to obtain our results it will be necessary to assume that the values of the Lam´e coefficients in the two phases satisfy the gap conditions

µ1 > µ2, 3λ1+ 2µ1 > 3λ2+ 2µ2. (1) A function u : Ω → R3 is a solution of the linear elasticity equation with elastic tensor C provided

j(Cijklkul) = 0, i = 1, 2, 3.

We will denote by φi = ui|∂Ω, i = 1, 2, 3, the Dirichlet data of such a solution (displacement), and by ψi = njCijklkul|∂Ω i = 1, 2, 3, the Neumann data (traction), where n = (n1, n2, n3)tis the outer normal on ∂Ω. We will consider two cases:

i) the solution u has Dirichlet data φi = 13xi, i = 1, 2, 3.

ii) the solution u has Neumann data ψi = ni, i = 1, 2, 3.

Our main result is

Theorem 1.1. There exists constants Ci, Cii≥ 0, Ci ≤ 1 depending (explic- itly) on Ω, the values of C and the boundary Dirichlet and Neumann data of the solutions in case i) for Ci and case ii) for Cii, such that Cii ≤ f1 ≤ Ci. Furthermore, Ci = 1 only if f1 = 1 and Cii = 0 only if f1 = 0. The explicit forms of the constants Ci and Cii are the ones appearing in equations (4) and (5) respectively.

The method of proof is inspired by the work of Kang-Milton [8] for the 3D conductivity equation. In section 2.1 we will ”translate” the minimum principle for the elasticity equation by a null-Lagrangian, i.e. a quantity that may be determined form boundary data. We will then extend the class of fields over which we minimize to obtain inequalities that would lead to an upper estimate of f1. In section 2.2 we will proceed similarly, except that in this case we will ”translate” the minimum principle by a quasi-convex

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quadratic. This quadratic form is not determined by the boundary data, but it’s size can nevertheless be controlled by it. The result will be the lower estimate on f1.

We would like to compare this to Milton’s result in [14]. With the same special boundary conditions, Milton also derives bounds of the volume frac- tion for the two-phase isotropic elasticity in two and three dimensions. Some of these bounds require multiple measurements, though bounds derived from single measurements are also derived. The bounds, which are given in im- plicit form and may be dificult to be put into explicit forms, are obtained from the bounds of elastic response tensors.

1.2 Main results – complex conductivity

Let Ω ⊂ R3 be a bounded domain and let σ : Ω → {σ1, σ2} ⊂ C be a conductivity function with two values. For any complex valued quantity a we will write a0 = Re a and a00= Im a. Then, for example, we have

σ = σ0+√

−1σ00.

We will assume that the following gap conditions hold σ10 > σ20 > 0,

σ1021002

σ10 > σ202σ0 2002

2 . (2)

As above, let Ω1, Ω2, be the domains of the two phases and let f1 := |Ω1|/|Ω|, f2 := |Ω2|/|Ω|.

Let u1, u2, u3 be three R-linearly independent solutions of

∇ · (σ∇u) = 0, in Ω. (3)

Denote φj := u00j|∂Ω, ψj = n · (σ∇u)0|∂Ω, j = 1, 2, 3. We will consider three cases

i) φj = 0 (i.e. real Dirichlet data) and the matrix

Z

∂Ω

xiψj



i,j∈{1,2,3}

is invertible. Since the ψj can be any set of three linearly independent Neumann data, this condition is generic.

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ii) ψj = 0 (i.e. imaginary Neumann data) and the matrix

Z

∂Ω

niφj



i,j∈{1,2,3}

is invertible. Likewise, this condition is generic.

iii) φj = 0, ψj = nj, j = 1, 2, 3. This is a particular instance of the first case.

Our main theorem is stated as follows.

Theorem 1.2. In each of the three cases, there exist constants c, C ≥ 0, depending (explicitly) on Ω, σ1, σ2, uj|∂Ω, n·(σ∇uj)|∂Ωsuch that c ≤ f1 ≤ C.

For the first case the bounds are the ones in (9), for the second case they are the ones in (10), and for the third case the ones in (11).

This result is obtained using two different versions of the translation method, adapted from the work of Kang-Milton [8]. In both instances we begin with a variational principle of Cherkaev-Gibiansky (see [4]). In the first and second case, we perturb the minimized quantity by a null-Lagrangian.

Extending the class of fields over which we are taking the minimum, we obtain inequalities that eventually lead to upper and lower bounds for f1. We carry out the detailed computations in section 3.1 for the first case and section 3.2 for the second case.

In the third case we perturb the minimum principle by a quasi-convex quadratic form. This method is explained in detail in section 3.3.

2 Estimates for elasticity

2.1 Upper bound

Using the notation

h · i := 1

|Ω|

Z

· dx, we define

W = h∇u : C : ∇ui = h∂iujCijklkuli.

It is known that

W = min

u|∂Ωh∇u : C : ∇ui.

We note that, using integration by parts, W = 1

|Ω|

Z

∂Ω

uinjCijklkul,

(6)

that is, W is determined by the Dirichlet data φ and the Neumann data ψ.

We denote by T the tensor with coefficients Tijkl= δijδkl− δilδjk.

Note that, for u that shares the same Dirichlet data as u, (∇T : ∇u)j = −∂iTijklkul = −∂jkuk+ ∂ijui = 0, and therefore

h∇u : T : ∇ui = 1

|Ω|

Z

∂Ω

niuikuk− niukkui = 1

|Ω|

Z

∂Ω

ui(nik− nki)uk. As the vector vj := niδjk − nkδij satisfies njvj = 0, we conclude that nik− nki is a tangential derivative. It follows that h∇u : T : ∇ui is determined by the boundary data φ, i.e. it is a null-Lagrangian.

We define, for c > 0,

Wc:= h∇u : C : ∇ui + ch∇u : T : ∇ui.

Since the extra term is determined by φ, we have that Wc= min

u|∂Ω(h∇u : C : ∇ui + ch∇u : T : ∇ui) .

Any 3 × 3 matrice may be written in vector form by ordering the pairs of indices (ij) as follows: (11), (21), (31), (12), . . . . Then ∇u becomes

v := ∂1u12u13u11u22u23u21u32u33u3t

. Using this notation we can write

Wc= v · Lv, where

L =

λ + 2µ 0 0 0 λ + c 0 0 0 λ + c

0 µ 0 µ − c 0 0 0 0 0

0 0 µ 0 0 0 µ − c 0 0

0 µ − c 0 µ 0 0 0 0 0

λ + c 0 0 0 λ + 2µ 0 0 0 λ + c

0 0 0 0 0 µ 0 µ − c 0

0 0 µ − c 0 0 0 µ 0 0

0 0 0 0 0 µ − c 0 µ 0

λ + c 0 0 0 λ + c 0 0 0 λ + 2µ

 .

(7)

Let J be the orthogonal matrix

J :=

1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0

 .

Then, we have

JtLJ =

λ + 2µ λ + c λ + c 0 0 0 0 0 0

λ + c λ + 2µ λ + c 0 0 0 0 0 0

λ + c λ + c λ + 2µ 0 0 0 0 0 0

0 0 0 µ µ − c 0 0 0 0

0 0 0 µ − c µ 0 0 0 0

0 0 0 0 0 µ µ − c 0 0

0 0 0 0 0 µ − c µ 0 0

0 0 0 0 0 0 0 µ µ − c

0 0 0 0 0 0 0 µ − c µ

 .

The upper left block is of particular interest. We denote it by

B :=

λ + 2µ λ + c λ + c λ + c λ + 2µ λ + c λ + c λ + c λ + 2µ

.

This block may be fully diagonalized by conjugation with the orthogonal matrix

S =

1 2

1 6

1 3

1

2

1 6

1 3

0 −2

6

1 3

, that is

C := StBS =

2µ − c 0 0

0 2µ − c 0

0 0 3λ + 2µ + 2c

.

Here we may also note that each block

 µ µ − c µ − c µ



can be diagonalized by

√1 2

1 1 1 −1

t

µ µ − c µ − c µ

 1

√2

1 1 1 −1



=2µ − c 0

0 c

 .

(8)

We see then that as long as 0 < c < 2µ, then L > 0. By (1), the condition is that 0 < c < 2µ2.

If u : Ω → R3 is such that u|∂Ω= φ, then h∂iuji = 1

|Ω|

Z

∂Ω

niφj. It then follows that

Wc≥ min

hvi=hvihv · Lvi > 0.

Let ˆv be an element for which the minimum is realized. Then for any ϕ s.t.

hϕi = 0,

hϕ · Lˆvi = 1 2

d dt

t=0

h(ˆv + tϕ) · L(ˆv + tϕ)i = 0, so there must be a constant m ∈ R9 such that

(Lˆv)(x) = µ, x ∈ Ω.

Let L1 and L2 be the values L takes on Ω1 and Ω2 respectively. Clearly ˆv is constant on Ω1 and Ω2. It follows that

Wc≥ min

f1v1+f2v2=hvi(f1v1· L1v1+ f2v2· L2v2) ,

where v1, v2 ∈ R9 are constants. By [7, Lemma 3.1] we conclude that Wc≥ hvi · hL−1i−1hvi,

where

hL−1i = f1L−11 + f2L−12 . Since C is diagonal it is easy to determine that

hC−1i−111 = hC−1i−122 = (2µ2− c)



1 − f12(µ1− µ2) 2µ1− c

−1

, and

hC−1i−133 = (3λ2+ 2µ2+ 2c)



1 − f13(λ1− λ2) + 2(µ1− µ2) 3λ1+ 2µ1+ 2c

−1

, the other matrix elements being zero.

If we choose Dirichlet data φj = 1

3xj|∂Ω, then we have h∂iuji = 1

3δij. Then

Wc ≥ hvi · hL−1i−1hvi = hC−1i−133.

(9)

If we take the limit as c % 2µ2, we obtain the inequality f1 ≤ 3λ1+ 2µ1+ 4µ2

3(λ1− λ2) + 2(µ1− µ2)

 1 − 1

T



, (4)

where

T := limc%2µ2Wc2+ 6µ2 .

With the choice of Dirichlet data we are making, by the minimum prin- ciple, we have

W2 ≤ 1

3(hI3×3 : C : I3×3i + 2µ2hI3×3: T : I3×3i) ≤ 3λ1+ 2µ1+ 4µ2, with equality attained only if f1 = 1. Then we have that

1 − 1

T ≤ 3(λ1 − λ2) + 2(µ1 − µ2) 3λ1+ 2µ1+ 4µ2 , and it follows that the upper bound in (4) satisfies

1+ 2µ1+ 4µ2 3(λ1 − λ2) + 2(µ1− µ2)

 1 − 1

T



≤ 1,

with equality attained only if f1 = 1. The bound is therefore non-trivial.

Remark 2.1. Here we have used one solution with specifically chosen Dirich- let boundary data, but a similar bound may be obtained for almost any solu- tion. To see this, note that there is an orthonormal basis in which limc%2µ2hL−1i−1

is 

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 hC−1i−133 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 2µ2 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 2µ2 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 2µ2

 .

Then in this basis we would have

hC−1i−133hvi23 ≤ W2 − 2µ2 hvi25+ hvi27+ hvi29 ,

which would produce the same type of bound as (4), as long as hvi3 6= 0.

However, in this case, there is no guarantee that the upper bound obtained for this data is non-trivial.

(10)

2.2 Lower bound

Let

α := − λ

2µ(3λ + 2µ), β := 1 4µ, and D be the (compliance) tensor with components

Dijkl = αδijδkl+ β(δikδjl+ δilδjk).

We note that α and β also satisfy the strong convexity conditions, and that β1 < β2, 3α1+ 2β1 < 3α2+ 2β2.

Let J := C : ∇u, then

J = Jt, ∇J = 0, D : J = 1

2 ∇u + (∇u)t , and therefore we have

W = h∇u : C : ∇ui = hJ : D : Ji.

If J is symmetric, satisfies ∇J = 0 and J n|∂Ω = ψ, then clearly h(J − J) : D : Ji = Tr h(J − J)∇ui = 0.

It then follows easily that

W = min

∇J=0,Jn|∂ΩhJ : D : Ji.

In this section we assume that the solution u has Neumann data ψj = nj. In this case

hJiji = 1

|Ω|

Z

∂Ω

xinkJkj = 1

|Ω|

Z

∂Ω

xiψj = δij.

The tensors Λh, Λs, Λa defined below are orthogonal projections acting on M3×3. Their components are

Λhijkl = 1 3δijδkl, Λsijkl= 1

2(δikδjl+ δilδjk) − 1 3δijδkl, Λaijkl= 1

2(δikδjl− δilδjk).

(11)

With this notation we notice that

D = (3α + 2β)Λh+ 2βΛs. Let

T0 := 2Λs− Λh. It can be shown (see [8]) that

P : T0 : P ≥ 0, ∀ P of rank at most 2.

Using methods borrowed from homogenization theory, in [8, (4.20)] it is shown that, when we choose boundary data as we have done,

g := min

∇J=0, Jn|∂ΩhJ : T0 : J i = −3.

We define

Wc0 := hJ : (D − cT0) : J i = hJ :(3α + 2β + c)Λh+ 2(β − c)Λs : Ji, where c > 0. We can estimate Wc0 from above by

Wc0 ≤ min

∇J=0, Jn|∂ΩhJ : D : Ji−c min

∇J=0, Jn|∂ΩhJ : T0 : J i ≤ hI3×3: D : I3×3i+3c

≤ 3(3α2+ 2β2+ c).

Now if J n|∂Ω= ψ, then hJ i = hJ i. Therefore Wc0 ≥ min

hJi=hJi(3α + 2β + c)J : Λh : J + 2(β − c)J : Λs : J > 0, as long as c < β1. In this case we may drop the second term to obtain

Wc0 ≥ 1 3 min

hJi=hJi(3α + 2β + c)(Tr J)2 . By Jensen’s inequality we have

(3α + 2β + c)(Tr J)2 ≥ (3α1+ 2β1+ c)1 f1

 1

|Ω|

Z

1

Tr J

2

+ (3α2+ 2β2+ c)1 f2

 1

|Ω|

Z

2

Tr J

2

. Let

z := 1

|Ω|

Z

1

Tr J ∈ [0, 3].

(12)

Then

Wc0 ≥ 1 3



(3α1+ 2β1+ c)1

f1z2+ (3α2 + 2β2+ c)1

f2(3 − z)2

 , and minimizing over z ∈ [0, 3] we have that

Wc0 ≥ 3

f1

1+2β1+c + f2

2+2β2+c

. Taking the limit c % β1, we may rewrite this estimate as



1 + f13(α2− α1) + 2(β2− β1) 3(α1+ β1)

−1

≤ T0, where

T0 := Wβ0

1

3(3α2 + 2β2+ β1) ≥ 1, with equality holding only when f1 = 0. We conclude that

f1 ≥ 3(α1 + β1)

3(α2− α1) + 2(β2− β1)

 1 T0 − 1



. (5)

As we have observed above, this lower bound is non-trivial and equals zero only when f1 = 0.

3 Estimates for complex conductivity

3.1 The first case – real general Dirichlet data

We adopt some notations from [7] and [8]. Define ei := −∇ui, ji := σei = −σ∇ui, ei =: e0i +√

−1e00i, ji =: j0i+√

−1j00i, σ =: σ0 +√

−1σ00, E := (Eij)ij, Eij := −∂iuj, J := σE,

E =: E0 +√

−1E00, J =: J0+√

−1J00, V := J0

E00

 . Let d be a 6 × 6 matrix given by

d =d11I3×3 d12I3×3 d21I3×3 d22I3×3

 := 1

σ0

 I3×3 σ00I3×3 σ00I3×302+ σ002)I3×3



(13)

and denote

A := hVtdVi.

It is easy to check that

e0i j00i



= d j0i e00i

 , and therefore, using integration by parts, Aij = h j0i

e00i



·d j0j e00j



i = h j0i e00i



·e0j j00j

 i = 1

|Ω|

Z

∂Ω

u0jn · (σ∇ui)0+ u00in · (σ∇uj)00 . We see then that A is determined by boundary measurements for the three solutions we are considering.

Furthermore, notice that for any V = J0

E00

 , it holds that

VtdV = 1

σ0(J0+ σ00E00)t(J0+ σ00E00) + σ0(E00)tE00 ≥ 0, (6) meaning positive-definite. We again use the tensor T with components

Tijkl= δijδkl− δilδjk. Then

hE00 : T : E00i = 1

|Ω|

Z

∂Ω

niu00iku00k− niu00kku00i = 1

|Ω|

Z

∂Ω

u00i(nik− nki)u00k. As remarked above, nik − nki is a tangential derivative. We can then conclude that hE00 : T : E00i is determined by the boundary data φj for any E00 as in (8). Thus, T defines a null Lagrangian for our problem. We also note that

P : T : P ≥ −Tr (PtP ), ∀P ∈ M3×3. (7) For any k, l, let Ikl be the 3 × 3 matrix given by Ikl := (δikδjl)ij, then (E00Ikl)ij = −∂iu00kδjl and ∂iTijkl(E00Imn)kl = 0. The tensor M with compo- nents

Mijkl := h(E00Iij) : T : (E00Ikl)i = h(E00Iij)αβTαβγδ(E00Ikl)γδi

= h∂ju00ilu00k− ∂lu00iju00ki

(14)

can also be determined from boundary measurements. For any K ∈ M3×3(R), let bibe the product of the normalized eigenvector of KKtand the associated singular value of K, then KKt = P3

i=1bi(bi)t. In view of the variational principle of Cherkaev and Gibiansky [4], we have that

WK := Tr hKtVtdVKi = Tr (KtAK) =

3

X

i=1

(bi)tAbi = min

V∈C1

Tr hKtVtdVKi,

where Tr stands for the trace of a matrix. Here the admissible space C1 is given by

C1 =



V = J0 E00



: E00ij = −∂iu00j, ∂iJ0ij = 0, u00j|∂Ω= φj, J0ijni|∂Ω = ψj

 . (8) We also have that

h(E00K) : T : (E00K)i = K : M : K.

Next, we ”translate” WK by a null Lagrangian Wc,K := Tr hKtVtdVKi + ch(E00K) : T : (E00K)i

= Tr (KtAK) + cK : M : K.

By (6), (7) we have that for any V Tr hVtdVi + chE00 : T : E00i ≥ 1

σ0(J0+ σ00E00) : (J0+ σ00E00) + (σ0− c) E00 : E00, and it is positive as long as 0 < c < σ0. Taking (2) into consideration, this condition reduces to

0 < c < σ02. Also notice that

Wc,K = min

V∈C1

Tr hKtVtdVKi + ch(E00K) : T : (E00K)i .

As we have done above, we will put 3 × 3 matrices in vector form by ordering the pairs of indices (ij) as follows: (11), (21), (31), (12), . . . . K would then become

k := k11 k21 k31 k12 k22 k32 k13 k23 k33t

.

(15)

Extending this idea to 6 × 3 matrices, V becomes

v :=

 j01 j02 j03 e001 e002 e003

 ,

while VK becomes

vK :=

k11j01+ k21j02+ k31j03 k12j01+ k22j02+ k32j03 k13j01+ k23j02+ k33j03 k11e001+ k21e002+ k31e003 k12e001+ k22e002+ k32e003 k13e001+ k23e002+ k33e003

 .

With this notation we have WK = k ·

A 03×3 03×3 03×3 A 03×3 03×3 03×3 A

k and

Wc,K = k · Dck, where

Dc :=

A 03×3 03×3 03×3 A 03×3 03×3 03×3 A

+ cM

and M is defined via k · Mk := K : M : K. Equivalently, we can write Wc,K = hvK· LvKi = min

V∈C1

hvK· LvKi, where

L =

d11 0 0 0 0 0 0 0 0 d12 0 0 0 0 0 0 0 0

0 d11 0 0 0 0 0 0 0 0 d12 0 0 0 0 0 0 0

0 0 d11 0 0 0 0 0 0 0 0 d12 0 0 0 0 0 0

0 0 0 d11 0 0 0 0 0 0 0 0 d12 0 0 0 0 0

0 0 0 0 d11 0 0 0 0 0 0 0 0 d12 0 0 0 0

0 0 0 0 0 d11 0 0 0 0 0 0 0 0 d12 0 0 0

0 0 0 0 0 0 d11 0 0 0 0 0 0 0 0 d12 0 0

0 0 0 0 0 0 0 d11 0 0 0 0 0 0 0 0 d12 0

0 0 0 0 0 0 0 0 d11 0 0 0 0 0 0 0 0 d12

d21 0 0 0 0 0 0 0 0 d22 0 0 0 c 0 0 0 c

0 d21 0 0 0 0 0 0 0 0 d22 0 −c 0 0 0 0 0

0 0 d21 0 0 0 0 0 0 0 0 d22 0 0 0 −c 0 0

0 0 0 d21 0 0 0 0 0 0 −c 0 d22 0 0 0 0 0

0 0 0 0 d21 0 0 0 0 c 0 0 0 d22 0 0 0 c

0 0 0 0 0 d21 0 0 0 0 0 0 0 0 d22 0 −c 0

0 0 0 0 0 0 d21 0 0 0 0 −c 0 0 0 d22 0 0

0 0 0 0 0 0 0 d21 0 0 0 0 0 0 −c 0 d22 0

0 0 0 0 0 0 0 0 d21 c 0 0 0 c 0 0 0 d22

.

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3.1.1 Construction of a lower bound for Wc,K

Note that Wc,K is completely determined by the boundary measurements provided c and K are given. Here we would like to derive a lower bound for Wc,K. If V ∈ C1 integrating by parts we get

hE00iji = − 1

|Ω|

Z

∂Ω

niφj, hJ0iji = 1

|Ω|

Z

∂Ω

xinkJ0kj = 1

|Ω|

Z

∂Ω

xiψj. Let J0, E00 be any 3 × 3 matrix-valued functions. We define the set

C2 =



V = J0 E00



: hVi = hVi



Then it is easily seen that C1 ⊂ C2 and Wc,K ≥ min

hvi=hvihvK· LvKi > 0,

the second inequality holding provided 0 < c < σ20. Let ˆV =

ˆJ000



∈ C2 be an element for which the minimum is realized. Then

0 = d dt

t=0

h(ˆv + tψ)K · L(ˆv + tψ)Ki = 2hψK· LˆvKi

for any ψ s.t. hψi = 0. There is therefore a constant µ ∈ R18 s.t.

(LˆvK)(x) = µ, x ∈ Ω.

We denote by L1and L2the values L takes on Ω1and Ω2 respectively. Clearly ˆ

vK is constant on Ω1 and Ω2. It follows that Wc,K ≥ min

f1v1+f2v2=hvKi(f1v1· L1v1+ f2v2· L2v2) ,

where v1, v2 ∈ R18 are constants. We can use, for example, [7, Lemma 3.1]

to conclude

Wc,K ≥ hvKi · hL−1i−1hvKi, where

hL−1i = f1L−11 + f2L−12 .

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3.1.2 Matrix manipulations

In this subsection, we will derive an interval bound for f1. Let J be the orthogonal matrix

J =

1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0

 Then, denoting d1 := d11, d2 := d12, d3 := d22,

JtLJ =

d1 0 0 d2 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 d1 0 0 d2 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 d1 0 0 d2 0 0 0 0 0 0 0 0 0 0 0 0

d2 0 0 d3 c c 0 0 0 0 0 0 0 0 0 0 0 0

0 d2 0 c d3 c 0 0 0 0 0 0 0 0 0 0 0 0

0 0 d2 c c d3 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 d1 0 d2 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 d1 0 d2 0 0 0 0 0 0 0 0

0 0 0 0 0 0 d2 0 d3 −c 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 d2 −c d3 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 d1 0 d2 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 d1 0 d2 0 0 0 0

0 0 0 0 0 0 0 0 0 0 d2 0 d3 −c 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 d2 −c d3 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 d1 0 d2 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 d1 0 d2

0 0 0 0 0 0 0 0 0 0 0 0 0 0 d2 0 d3 −c

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 d2 −c d3

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In order to investigate hL−1i−1, we may consider the individual blocks separately. Let

B =

d1 0 0 d2 0 0 0 d1 0 0 d2 0 0 0 d1 0 0 d2 d2 0 0 d3 c c

0 d2 0 c d3 c 0 0 d2 c c d3

be the first of the diagonal blocks. We define another orthogonal matrix

S =

1

2 0 1

6 0 1

3 0

1

2 0 1

6 0 1

3 0

0 0 −26 0 13 0 0 1

2 0 1

6 0 1

3

0 −1

2 0 1

6 0 1

3

0 0 0 −2

6 0 1

3

 .

Then

C := StBS =

d1 d2 0 0 0 0

d2 d3− c 0 0 0 0

0 0 d1 d2 0 0

0 0 d2 d3− c 0 0

0 0 0 0 d1 d2

0 0 0 0 d2 d3+ 2c

 .

The inverse of C is

C−1 =

d3−c

1−cd11−cdd2

1 0 0 0 0

1−cdd2

1

d1

1−cd1 0 0 0 0

0 0 1−cdd3−c

11−cdd2

1 0 0

0 0 −1−cdd2

1

d1

1−cd1 0 0

0 0 0 0 1+2cdd3+2c

11+2cdd2

1

0 0 0 0 −1+2cdd2

1

d1

1+2cd1

=:

X1 02×2 02×2

02×2 X2 02×2 02×2 02×2 X3

We will write

δ1 := 1

σ20, δ2 := σ002

σ20 , δ3 := σ202+ σ2002 σ20 ,

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1 := 1

σ01, ∆2 := σ100

σ01, ∆3 := σ021 + σ1002 σ01 .

We want to look into what happens when c % 1/δ1 = σ20. So we can write X1, X2, X3 as

X1 = X2 = 1 1 − cd1

d22

d1 + 1−cdd 1

1 −d2

−d2 d1

! ,

X3 =

d22+3

3d1 + 3d 2d2

1(1+2cd1)(1 − cd1) −d323(1+2cd2d2

1)(1 − cd1)

d323(1+2cd2d2

1)(1 − cd1) d31 +3(1+2cd2d1

1)(1 − cd1)

! . Then

dethX1i =



f22+ f1f2

δ1− ∆11(∆3δ1− 1) + δ2(∆1δ2− δ12)]



(1−cδ1)−1+O(1−cδ1) and

hX1i−1 =



f2+ f1δ1(∆3δ1− 1) + δ2(∆1δ2− δ12) δ1− ∆1

−1

δ1 δ2 δ2 δδ22

1

!

+O(1−cδ1).

In the case of X3, we can see that dethX3i > 0 and satisfies

dethX3i = f221

3 + f12 δ1 2∆1 + δ1 + f1f2 1

3(2∆1+ δ1)[∆122+ 3) + (∆3δ1+ 2)δ1− 2∆2δ1δ2] + O(1 − cδ1)

= 1

3+ f1 1

3(2∆1 + δ1)[∆122− 1) + ∆3δ12− 2∆2δ1δ2] + f12 1

3(2∆1+ δ1)[2δ1− ∆122+ 1) − ∆3δ21+ 2∆2δ1δ2] + O(1 − cδ1)

=: 1

31 + f1α + f12β + O(1 − cδ1) and

hX3i−1 =1 + f1α + f12β−1

×

δ1 − f1δ12∆1−∆1)

11 δ2− f1



δ22∆3∆2δ1

11

 δ2− f1

δ22∆3∆2δ1

11

 δ2 2+3

δ1 − f1δ2 2+3

δ1 − 32∆3δ1+2

11



+ O(1 − cδ1),

參考文獻

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