UNIQUENESS ESTIMATES FOR THE GENERAL COMPLEX CONDUCTIVITY EQUATION AND THEIR APPLICATIONS TO INVERSE PROBLEMS

CONDUCTIVITY EQUATION AND THEIR APPLICATIONS TO INVERSE PROBLEMS

CĂTĂLIN I. CÂRSTEA, TU NGUYEN, AND JENN-NAN WANG Dedicated to Professor Carlos Kenig on the occasion of his 65th birthday

Abstract. The aim of the paper is twofold. Firstly, we derive quantitative unique- ness estimates for solutions of the general complex conductivity equation. It is still not known whether the strong unique continuation property holds for such equa- tions. Nonetheless, we show that the unique continuation property in the form of three-ball inequalities holds if the leading coefficients are Lipschitz. Secondly, we study the problem of estimating the size of an inclusion embedded inside of a con- ductive body with anisotropic complex admittivity by one boundary measurement.

Practical motivations for studying such inverse problem are also given.

1. Introduction

The study in this paper is inspired by a problem arising in Electric impedance tomography (EIT). EIT is a non-invasive method for inferring the interior structure of a specimen by injecting currents and measuring the resulting potential differences on the surface of the specimen. Implementing EIT as a medical imaging technique was first done by Henderson and Webster [12]. The mathematical study of EIT was initiated by Calderon [6]. There have been a lot of clinical and non-clinical applications of EIT, see [7] and references therein for more detailed descriptions.

Ideally, one would like to determine the full information of the interior structure by making boundary measurements. However, according to the uniqueness result of Sylvester and Uhlmann [17], we need infinitely many measurements, which is prac- tically impossible. In some cases of medical diagnosis, one is more interested in determining the characteristic properties of an abnormality surrounded by normal or healthy tissues, for example, breast cancer detection [7] and detectability of the de- graded tissues [11]. It turns out EIT with one measurement can provide a preliminary assessment of the size of the abnormality [1].

In [1], the authors considered a real conductivity equation. In studying inverse problems related to biological tissues, it is, however, more appropriate to use a com- plex conductivity equation. Unlike metallic materials, the electrical conductance within biological tissues is due to the ions in intracellular and extracellular fluids.

The intracellular ions are separated by cell membranes which act like capacitors.

Key words and phrases. Carleman estimate, three-ball inequalities, size estimate.

The second author is funded by Vietnam National Foundation for Science and Technology De- velopment (NAFOSTED) under grant number 101.02-2015.21.

The third author was supported in part by MOST 105-2115-M-002-014-MY3.

1

When a constant voltage (DC) is applied, a conductance current I_cflows through the extracellular fluid. At the same time, a constant amount of charges will be stored in the cell membrane. If now an alternating voltage (AC) is applied, the charges stored in the cell membranes will change according to the frequency, resulting in a displacement current I_d through the intracellular fluid. From Ohm’s law, I_cdepends on the material’s conductivity σ while I_d depends on its permittivity ε. Due to the 90 degrees phase difference, we see that the total current I = I_c+ I_d is proportional to the product of the complex-valued admittivity σ + iωε and the applied voltage, where ω = 2πf and f is the frequency of the applied signal [14].

In addition to the biological tissues, conductivity equations with complex coeffi- cients also arise in the modeling of electromagnetic waves propagation. Let Ω be a conducting body with an anisotropic background with current-voltage relation (or Ohm’s law) is given by

(1.1) γ₀(∇u) = (σ₀(x) + iε₀(x))∇u(x),

where σ₀, ε₀ are symmetric matrix valued functions. We will show below that in the presence of a chiral inclusion D ⊂ Ω, the current-voltage relation becomes

(1.2) γ₁(∇u) = (σ₁(x) + iε₁(x))∇u(x) + ζ₁(x)∇u(x),

where σ1, ε1, ζ1 are real symmetric matrix-valued functions such that the supports of σ1− σ0, ε1− ε0, ζ1 are contained in D. Given a Neumann boundary data h ∈ L²(∂Ω) satisfying ´

∂Ωh = 0, let u₀ be the unique solution of the unperturbed equation

(1.3)











∇ · (γ₀(∇u₀)) = 0 in Ω, γ₀∇u₀· ν = h on ∂Ω,

ˆ

Ω

u₀ = 0,

and u1 be the solution of the perturbed equation with the same boundary data, i.e.

(1.4)











∇ · (γ₁(∇u₁)) = 0 in Ω, γ₁(∇u₁) · ν = h on ∂Ω,

ˆ

Ω

u₁ = 0.

The inverse problem studied here is to estimate the size of the inclusion D by one pair of boundary measurement {u|_∂Ω, h}. More precisely, one tries to estimate |D|

using the power gap δW = W₀− W₁, where W₁ =

ˆ

∂Ω

u₁h, W₀ = ˆ

∂Ω

u₀h.

We now provide a physical reasoning that gives rise to the relation (1.2). We recall the Drude-Born-Fedorov constitutive relations (see for instance [16, (2.21)]):

(1.5) D = ε(E + β∇ × E), B = µ(H + β∇ × H),

where E, H are the electric and magnetic fields, and D, B are the electric and mag- netic flux densities, respectively; ε is the electric permittivity, µ is the magnetic permeability, and β is the chirality. The source-free time-harmonic equations are (1.6) ∇ × E − iωB = 0, ∇ × H + iωD = 0.

Substituting (1.5) into (1.6) leads to

(1.7) ∇ × E − iωµ(H + β∇ × H) = 0, ∇ × H + iωε(E + β∇ × E) = 0, which can be combined to obtain

(1.8) ∇ × E = ηβE + iω µ

(1 − k²β²)H, ∇ × H = ηβH − iω ε

(1 − k²β²)E, where η = k²(1 − k²β²)⁻¹ and k² = ω²εµ. Here we consider the case where β is real and ε, µ are complex-valued satisfying that εµ is real and |µ| is small.

A possible choice of ε and µ is ε = ρe^−iθ and µ = ˜ρe^iθ, where θ ∈ (0, 2π), ρ > 0,

˜

ρ > 0 with | ˜ρ|  1. Suppose that for δ > 0 sufficiently small, ρ = O(1), ˜ρ = O(δ), and θ = O(δ). So it is not unreasonable to ignore the imaginary part of ξ. Since we assume |µ|  1, we may consider k²β² < 1.

Using the Bohren decomposition [16, p.87], we write E and H in terms of the Betrami fields QL, QR, as

(1.9) E = Q_L− iξQ_R, H = 1

iξQ_L+ Q_R,

where ξ = pµ/ε is the intrinsic impedance of the chiral medium and QL, QR are Beltrami fields satisfying

∇ × Q_L= γ_LQ_L, γ_L = k(1 − kβ)⁻¹,

∇ × Q_R= −γ_RQ_R, γ_G = k(1 + kβ)⁻¹.

Here γ_L and γ_R are the wave numbers of the Beltrami fields Q_L and Q_R, respec- tively. Since γ_L and γ_R are assumed to be real, it suffices to choose real Q_L, Q_R. Note that in the case we are considering, εµ = O(δ) and ξ = O(δ^1/2) + iO(δ^3/2), hence, the imaginary part of ξ is negligible and ξ could be considered as real-valued.

Consequently, we can approximately set H = −i

ξE.

Substituting this into the second equation of (1.8) gives

∇ × H = −iηβ

ξ E − iωε (1 − k²β²)E, which gives

(1.10) ∇ · ηβ

ξ E + ωε

(1 − k²β²)E



= 0.

Taking into account of the fact that |µ| is small, from the first equation of (1.7), we set ∇ × E = 0. In other words, we can find a potential function u such that E = ∇u.

Replacing E by ∇u in (1.10) gives the current-voltage relation (1.2).

In [5], Beretta, Francini, and Vessella studied the same problem when current- voltage relation is the usual one, i.e. γ₁(∇u) = (σ₁+ iε₁)∇u. They noted that their derivation of energy inequalities only works under the assumption that either σ₀+ iε₀ and σ₁+ iε₁ are constants, or ε₀ ≡ 0. When ε₀ is a nonzero function, an example with δW = 0, but |D| 6= 0, is constructed in [5]. Such an example indicates that when γ₀ is a complex-valued variable function and the current-voltage relation in D satisfies the usual Ohm’s law, proving the energy inequalities (3.6) seems not to be realistic.

One of the contributions of this paper is to restore the energy inequalities (3.6) when γ₀ is a complex-valued variable function by introducing a new constitutive law in the inclusion. In particular, our result applies to the case of a non-chiral medium with a chiral inclusion having real-valued chirality β. Thus, our result and those in [5] are complementary. Our proof of the energy inequalities exploits the Cherkaev-Gibiansky convex variational form [8] and cannot be used in the setting of [5].

We would like to mention that there are other approaches to size estimate in the case the admittivities of the inclusion and the surrounding body are known. These include the splitting method of Thaler and Milton [18] and the translation method of Kang, Kim, Lee, Li, Milton [13].

Similar to the arguments used in [1], our method relies on some quantitative unique- ness estimates of the solution u₀. This is the subject of Secion 2. The main result is stated and proved in Section 3.

2. Carleman estimate and consequences

In this section we will derive the needed three ball inequality. First, we recall the following Carleman estimate of [9] (see also [10]). Let r = |x| and denote ϕ(x) = ϕ(r) = r^−s.

Proposition 1. Let P =P

jkp_jk(x)∂_jk be a second order elliptic operator with sym- metric Lipschitz coefficients p_jk = a_jk + ib_jk such that

(2.1) λ |ξ|² ≤X

jk

a_jk(x)ξ_jξ¯_k ≤ λ⁻¹|ξ|², |p_jk(x)| ≤ λ⁻¹, ∀ x ∈ Rⁿ, ξ ∈ Cⁿ

for some λ < 1. Assume also that aij(0) = δij. Suppose that k∇pkL^∞(Rⁿ) ≤ L < ∞.

Then there exist positive constants s0,β0 and C depending only on n and λ, and R depending on n, λ, L, so that if β ≥ β₀, s ≥ s₀, for any u ∈ H₀¹(B_R\{0}) the following inequality holds

(2.2) β³s³ ˆ

r^−3se^2βϕ|u|²dx + βs ˆ

r^−s+2e^2βϕ|∇u|²dx ≤ C ˆ

r⁴e^2βϕ|P u|²dx.

Remark 2. In [9], the main concern is the strong unique continuation property for the operator P with Gevrey coefficients p_jk(x). There, the upper bound of the expo- nent s in (2.2) is controlled by the Gevrey degree and its lower is determined by the eigenvalues of a_jk(0) and b_jk(0). Here we are interested in the quantitative estimate of the unique continuation property. Thus, s can be chosen arbitrarily large. We only need to make sure that the lower bound s₀ is a general constant, which can be easily

verified from the proof of (2.2) in [9]. Finally, note that (2.2) follows immediately from applying the Carleman estimate of [9] to the function r²u.

Carleman estimates for general elliptic equations with complex coefficients were also derived in [3] and [4] using Hörmander’s approach. However, it does not seem possible to derive three-ball inequalities like (2.4) from their estimates.

We now prove the three-ball inequality, which is the quantitative form of the (weak) unique continuation property. When P has real coefficients, qualitative and quantita- tive forms of strong unique continuation for (2.3) are well-known. These results cease to hold in the general case of complex coefficients, as Alinhac [2] has constructed such operators of any order in R² without the strong unique continuation property.

Theorem 3. Assume that (2.1) holds and k∇pk_L^∞_(Rⁿ₎ ≤ L. Then there exist positive constants R = R(n, λ, L) and s = s(λ, K) such that if 0 < r0 < r1 < λr2/2 <√

λR/2 with Br2(x0) ⊂ Ω and u ∈ H_loc¹ (Ω) solves

(2.3) |P u| ≤ K(|∇u| + |u|),

then (2.4)

ˆ

B_r1(x0)

|u|²dx ≤ C ˆ

B_r0(x0)

|u|²dx

!τ ˆ

B_r2(x0)

|u|²dx

!1−τ

, where C depends on λ, K, r₁, r₂, s and

τ = (2r1/λ)^−s− r^−s₂ r^−s₀ − r^−s₂ .

Proof. In the proof, C is a constant depending on λ, K, r₁ and r₂, whose value may change from line to line. We also fix s = max{2K², s₀} with s₀ being given in Proposition 1. Without loss of generality, we assume x0 = 0. To apply the Carleman estimate, we first need to make a change of variables. Let S be a matrix satisfies SA(0)S^t = I and let v(y) = u(S⁻¹y), ˜p(y) = SP (S⁻¹y)S^t. Then v satisfies ˜P v = P

jkp˜_jk∂_yjykv ≤ K(|v| + |∇v|). Note that we have <˜p_jk(0) = δ_jk. Furthermore, the eigenvalues of S are bounded by λ^−1/2 and λ^1/2, hence if we let ρ0 = √

λr0, ρ1 = r₁/√

λ, and ρ₂ =√

λr₂ then B_r0 ⊃ S⁻¹B_ρ0, B_r1 ⊂ S⁻¹B_ρ1 and B_r2 ⊃ S⁻¹B_ρ2, where B_ρj are balls in the y variables. Thus, it suffices to show that

(2.5)

ˆ

|y|<ρ1

|v|²dy ≤ C

ˆ

|y|<ρ0

|v|²dy

τˆ

|y|<ρ2

|v|²dy

1−τ

, where

τ = (2ρ₁)^−s− ρ^−s₂ ρ^−s₀ − ρ^−s₂ . Let 0 ≤ ζ ≤ 1 be a cut-off function satisfying

• ζ(y) = 1 if 2ρ0/3 < |y| < ρ2/2,

• ζ(y) = 0 if |y| ≤ ρ₀/2 or |y| ≥ 2ρ₂/3, and,

• |∂^αζ(y)| ≤ 10|y|^−|α|, ∀y, for |α| = 1, 2.

Note that we have 0 < ρ₀ < ρ₁ < ρ₂/2 ≤ R/2, where R is the constant appears in Proposition 1. Denote ˜v = ζv, E = {ρ₀/2 < |y| < 2ρ₀/3} ∪ {ρ₂/2 < |y| < 2ρ₂/3} and χ_E the characteristic function of E. Then

| ˜P ˜v| =|ζ ˜P v + 2X

jk

˜

p_jk(∂_jζ)(∂_kv) + vX

jk

˜ p_jk∂_jkζ|

≤Kζ(|v| + |∇v|) + C(|y|⁻¹|∇v| + |y|⁻²|v|)χ_E

≤K(|˜v| + |∇˜v|) + C(|y|⁻¹|∇v| + |y|⁻²|v|)χ_E.

In the last inequality we have used ζ|∇v| ≤ |∇˜v| + |∇ζ||v| ≤ |∇˜v| + C|y|⁻¹|v|χ_E. Applying (2.2) to ˜v and using the above observations, we have that

β³s³ ˆ

|y|^−3se^2βϕ|˜v|²dy + βs ˆ

|y|^−s+2e^2βϕ|∇˜v|²dy

≤ 2K² ˆ

e^2βϕ |˜v|²+ |∇˜v|²
dy + C

ˆ

E

e^2βϕ |v|²+ |y|²|∇v|²
dy.

(2.6)

By our choice of s, the first term on the right hand side of (2.6) can be absorbed by its left hand side. Consequently, we obtain

e^2βρ−s1 ˆ

2ρ0/3<|y|<ρ1

|v|²dy (2.7)

≤ Ce^{2β(ρ0/2)−s} ˆ

ρ0/2<|y|<2ρ0/3

(|v|²+ |y|²|∇v|²)dy + Ce^{2β(ρ2/2)−s}

ˆ

ρ2/2<|y|<2ρ2/3

(|v|² + |y|²|∇v|²)dy.

Adding e^2βρ−s1 ´

|y|<2ρ0/3|v|²dy to both sides of (2.7) and using the Caccioppoli-type estimate, we have

ˆ

B_ρ1

|v|²dy ≤ Ce^{2β[(ρ0/2)−s−ρ−s1 ]} ˆ

B_ρ0

|v|²dy +Ce^{2β[(ρ2/2)−s−ρ−s1 ]}

ˆ

B_ρ2

|v|²dy, (2.8)

for all β ≥ β0. If´

B_ρ0|v|²dy = 0 then by letting β → ∞, we see that´

B_ρ1|v|²dy = 0 as well, hence (2.5) trivially holds. Otherwise, let

β = maxn β₀,

log

^´

Bρ2|v|²

´

Bρ0|v|²

 2[(ρ₀/2)^−s− (ρ₂/2)^−s]

o ,

and consider two cases: if β > β₀, then from (2.8) we get ˆ

|y|<ρ₁

|v|²dy ≤ C

ˆ

|y|<ρ₀

|v|²dy

τˆ

|y|<ρ₂

|v|²dy

1−τ

, while, if β = β₀, i.e.,

log

´

Bρ2|v|²

´

Bρ0|v|²



2[(ρ₀/2)^−s− (ρ₂/2)^−s] ≤ β0,

then ˆ

|y|<ρ2

|v|²dy ≤ e^{2β0[(ρ0/2)−s−(ρ2/2)−s]} ˆ

|y|<ρ0

|v|²dy.

Raising this to the power τ we obtain ˆ

|y|<ρ2

|v|²dy ≤ e^{2β0[ρ−s1 −(ρ2/2)−s]}

ˆ

|y|<ρ0

|v|²dy

τˆ

|y|<ρ2

|v|²dy

1−τ

,

which clearly implies (2.5). The proof is complete.  3. Size estimate

In this section, we study the size estimate problem described in the Introduction.

Recall that the background admittivity is given in (1.1) while with the inclusion, the current-voltage relation is given in (1.2). For convenience, let ζ₀ = 0. We will make the following assumptions on the symmetric real matrices σ_j, ε_j and ζ_j:

• (Lipschitz property) There exists L > 0 so that (3.1) k∇(σ₀+ iε₀)k_L^∞_(Ω) ≤ L.

• (Boundedness and Ellipticity) for some α ≥ 1,

α⁻¹I ≤ σ_j ± ζ_j ≤ αI, kε_jk_L^∞_(Ω) ≤ α, j = 0, 1.

(3.2)

• (Jump condition) There exists % > 0 such that either (i) ζ₁ ≤ ±(σ₁− σ₀) − %I a.e. x ∈ D, or

(ii) ζ₁ ≥ ±(σ₁− σ₀) + %I a.e. x ∈ D.

(3.3)

Morever, for δ = δ(α, %) > 0 sufficiently small, we have

(3.4) kε₁− ε₀k_L^∞_(D) ≤ δ.

Let us remark that these assumptions are rather standard in this area, see e.g. [1]

and [5]. The main difference between ours and those in previous work is the jump condition which is adapted to our physical model (1.2). It plays an important role in the derivation of the energy gap (see Proposition 5).

Given a set A and s > 0, we define

A` = {x ∈ A : dist(x, ∂A) > `}.

Our main result on size estimate is the following.

Theorem 4. Let Ω be a bounded domain in Rⁿ. Assume that for some `<sub>0</sub>, `₁ > 0, the inclusion D ⊂ Ω satisfies dist(D, ∂Ω) ≥ `₀ and

(3.5) |D_`1| ≥ 1

2|D|.

Suppose that (3.1)-(3.4) are satisfied. Then there exist two positive constants K₁, and K₂ such that

K1

|<δW |

<W₀ ≤ |D| ≤ K2

|<δW |

<W₀ ,

where K₁depends on `<sub>0</sub>, α and K<sub>2</sub> depends on α, L, %, |Ω|, `₁, and khk_L²_(∂Ω)/khk_H−1/2(∂Ω). We will first derive an energy inequality which is crucial in the proof.

Proposition 5. Assume that (3.1), (3.2) and (3.4) hold. Then there exist positive constants C₁ = C₁(α, %) and C₂ = C₂(α) such that

• If (3.3)(i) holds then

(3.6) C1

ˆ

D

|∇u0|² ≤ <δW ≤ C2

ˆ

D

|∇u0|²

• If (3.3)(ii) holds then

(3.7) C₁

ˆ

D

|∇u₀|² ≤ −<δW ≤ C₂ ˆ

D

|∇u₀|²

Proof. In this proof, C denotes a positive constant whose value may change from line to line. We have, for j = 0, 1,

(3.8) <γ_j(∇u_j)

=γ_j(∇u_j)



=σ_j+ ζ_j −ε_j ε_j σ_j − ζ_j

 <∇u_j

=∇u_j

 . Following Cherkaev and Gibiansky [8], we rewrite (3.8) as

(3.9)

 <∇uj

=γj(∇uj)



= (σj + ζj)⁻¹ (σj + ζj)⁻¹εj

εj(σj+ ζj)⁻¹ σj − ζj + εj(σj + ζj)⁻¹εj

 <γj(∇uj)

=∇uj

 . We denote the square matrix on the right hand side B_j and v_j = (<γ_j(∇u_j), =∇u_j)^t. From our assumptions, it can be easily checked that B_j is positive-definite. Using (3.9), we have, for j, k ∈ {0, 1},

(3.10)

ˆ

Ω

B_jv_j· v_k = ˆ

Ω

<∇u_j · <γ_k(∇u_k) + =γ_j(∇u_j) · =∇u_k

= <

ˆ

Ω

div(γ_k(∇u_k)<u_j) + = ˆ

Ω

div(γ_j(∇u_j)=u_k)

= ˆ

∂Ω

(<u_j<h + =u_k=h).

From this and B₁ = B₁^T, we can deduce that

<δW = <

ˆ

∂Ω

(u₀− u₁)h = ˆ

Ω

(B₀− B₁)v₀· v₁ = ˆ

D

(B₀− B₁)v₀ · v₁. (3.11)

Using (3.10) and B₀ = B₀^T, we obtain

(3.12)

ˆ

Ω

B₀(v₀− v₁) · (v₀− v₁) = ˆ

Ω

(B₀v₀· v₀− 2B₀v₀· v₁+ B₁v₁· v₁) +

ˆ

Ω

(B₀− B₁)v₁· v₁

= <

ˆ

∂Ω

(u1− u0)h + ˆ

D

(B0− B1)v1· v1. Thus,

(3.13) <δW = ˆ

Ω

B₀(v₀− v₁) · (v₀− v₁) + ˆ

D

(B₁− B₀)v₁· v₁. Swapping the indices 0 and 1, we obtain a similar identity

(3.14) <δW = − ˆ

Ω

B₁(v₀− v₁) · (v₀− v₁) + ˆ

D

(B₁− B₀)v₀· v₀. Here we have used the fact that supp (B₁− B₀) ⊂ ¯D.

Case 1: ζ₁ ≤ ±(σ₁− σ₀) − %I

The positivity of B1 and (3.14) give the second inequality of (3.6). Using (3.13) and the triangle inequality, we see that the first inequality of (3.6) follows if B1− B0

is positive-definite. We have B1− B₀ = P + Q, where P = M M ε₁

ε₁M N + ε₁M ε₁



, Q =

 0 σ⁻¹₀ (ε₁ − ε₀)

(ε₁− ε₀)σ₀⁻¹ (ε₁− ε₀)σ₀⁻¹ε₁+ ε₀σ⁻¹₀ (ε₁ − ε₀)

 , and M = (σ1+ ζ1)⁻¹− σ₀⁻¹, N = (σ1− ζ1) − σ0. From (3.3)(i), it follows that M and N are positive-definite.

For X = (p, q)^T ∈ R²ⁿ, we have

P X · X = M (p + ε1q) · (p + ε1q) + N q · q and

|QX · X| ≤ Ckε₁− ε₀k_L^∞_(D)|X|.

The required positivity then follows from condition (3.4) for δ small depending on α, %.

Case 2: ζ₁ ≥ ±(σ₁− σ₀) + %I

Similar argument as in the previous case shows that B₀− B₁ is positive-definite.

Thus, the first inequality of (3.7) follows from (3.14). To prove the second inequality of (3.7), we first deduce from (3.13) and (3.14) that

ˆ

D

(B₀− B₁)v₁ · v₁ = −<δW + ˆ

Ω

B₀(v₀− v₁) · (v₀− v₁)

≤ −<δW + C ˆ

Ω

B₁(v₀ − v₁) · (v₀− v₁)

= −(C + 1)<δW + C ˆ

D

(B₁− B₀)v₀· v₀. (3.15)

Using (3.11) and Cauchy-Schwarz we have

−<δW ≤ 2(C + 1) ˆ

D

(B₀− B₁)v₀· v₀+ 1 2(C + 1)

ˆ

D

(B₀− B₁)v₁· v₁. (3.16)

Eestimating the last term of (3.16) by (3.15), we obtain

−<δW ≤ 4(C + 1) ˆ

D

(B₀− B₁)v₀· v₀. (3.17)

The proof is complete. 

Another tool we need is the following propagation of smallness which is a conse- quence of the Carleman estimate.

Proposition 6. Let u₀ be the solution of (1.3). Assume that (3.2) and (3.1) hold.

Then for any ρ > 0 and every x ∈ Ω_4ρ, we have (3.18)

ˆ

Bρ(x)

|∇u₀|² ≥ C_ρ ˆ

Ω

|∇u₀|², where C_ρ depends on α, L, |Ω|, ρ, khk_L²_(∂Ω)/khk_H−1/2(∂Ω).

Proof. We follow the arguments presented in [1, Lemma 2.2]. We first observe that it suffices to consider the case ρ is small, so we can assume that Ω_ρis connected. Using Caccioppoli and Poincaré inequalities, we can deduce from Theorem 3 that

(3.19) k∇u₀k_L²_(B3r(x))≤ Ck∇u₀k^τ_L2(Br(x))k∇u₀k^1−τ_L2(B_4λ−1r(x)),

where C depends on α, β₀. Given x, y ∈ Ω_4λ⁻¹_ρ, let γ be a curve in Ω_4λ⁻¹_ρ joining x and y. We define a sequence x_k’s as follows: Let x₁ = x. For k > 1, let x_k = γ(t_k) where t_k = max{t : |γ(t) − x_k−1| = 2ρ} if |x_k−1− y| > 2ρ; otherwise let x_k= y, N = k and stop the process. Note that since the balls B_ρ(x_k) are disjoint, N ≤ N₀ = _ω^|Ω|

nρⁿ. Using (3.19) with x = x_k and r = ρ, noting that B_ρ(x_k+1) ⊂ B_3ρ(x_k) because

|x_k+1− x_k| ≤ 2ρ, we can deduce that k∇u₀k_L²_(Bρ(xk+1))

k∇u0k_L²_(Ω) ≤ C

k∇u₀k_L²_(Bρ(xk)) k∇u0k_L²_(Ω)

^τ . Here C depends on α, β0, ρ. By induction, we obtain

k∇u₀k_L²_(Bρ(y))

k∇u₀k_L²_(Ω) ≤ C^{1/(1−τ )}

k∇u₀k_L²_(Bρ(x)) k∇u₀k_L²_(Ω)

τ^N

.

Since we can cover Ω_(4λ⁻¹_+1)ρ by no more than ^|Ω|n₂nρ^n/2n balls of radius ρ, we obtain k∇u₀k_L²_(Ω

(4λ−1+1)ρ)

k∇u₀k_L²_(Ω) ≤ C

k∇u₀k_L²_(Bρ(x)) k∇u₀k_L²_(Ω)

τ^N0

, where C depends on α, β₀, |Ω|, and ρ.

Repeating the arguments in [1, page 60-61], we see that there exists ¯ρ, depending on α, β₀, |Ω|, khk_L²_(∂Ω)/khk_H−1/2(∂Ω), such that

(3.20) 1

2 ≤ k∇u₀k_L²_(Ω

(4λ−1+1)ρ)

k∇u₀k_L²_(Ω)

for all 0 < ρ ≤ ¯ρ. Thus the estimate (3.18) holds for all 0 < ρ ≤ ¯ρ and for large ρ,

(3.18) is then obvious. 

Finally, we give the proof of our main result.

Proof of Theorem 4. We follow the proof of Theorem 2.1 in [1, page 61]. By the standard elliptic estimate, we have the interior estimate

sup

D

|∇u₀| ≤ C||∇u₀||_L²_(Ω) ≤ C (<W₀)^1/2. Applying the second inequality of Proposition 5 we obtain

<δW ≤ C₂|D|

 sup

D

|∇u₀|

2

≤ K₁⁻¹|D|<W₀, which is the first of our needed inequalities.

Let ` = min (`₀/2, `<sub>1</sub>/2) and cover D<sub>`1</sub> with squares {Q_k}_k=1,...,N of side length ` and disjoint interiors. It is clear that N ≥ `⁻²|D_{`1</sub>| ≥ <sub>2`}¹2|D|. Applying Proposition 6 with ρ = `/2 we see that ´

Qk|∇u₀|² ≥ C<W₀, hence ˆ

D

|∇u₀|² ≥

N

X

k=1

ˆ

Qk

|∇u₀|² ≥ N C<W₀ ≥ C|D|<W₀.

Combining this with the first inequality of Proposition 5, the proof is finished.  We note that (3.5) is only used to obtain an upper bound for |D|. In [5], an estimate was obtained without assuming (3.5). This was possible because under their conditions, u₀ satisfies a doubling inequality, which is not known in the case considered here.

Acknowledgement

The authors would like to express their gratitude to Professor Carlos Kenig for his generosity and support throughout their careers.

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School of Mathematics, Sichuan University, Chengdu, Sichuan, 610064, P.R.China Email address: catalin.carstea@gmail.com

Institute of Mathematics, Vietnam Academy of Science and Technology, 18 Hoang Quoc Viet, Cau Giay, Hanoi, Vietnam

Email address: natu@math.ac.vn

Institute of Applied Mathematical Sciences, NCTS, National Taiwan University, Taipei 106, Taiwan

Email address: jnwang@math.ntu.edu.tw