Three-region inequalities for the second order elliptic equation with discontinuous coefficients and size estimate

Three-region inequalities for the second order elliptic equation with discontinuous coefficients and

size estimate

E. Francini

C.-L. Lin

S. Vessella

J.-N. Wang

Abstract

In this paper, we would like to derive a quantitative uniqueness estimate, the three-region inequality, for the second order elliptic equation with jump discontinuous coefficients. The derivation of the inequality relies on the Carle- man estimate proved in our previous work [5]. We then apply the three-region inequality to study the size estimate problem with one boundary measurement.

1 Introduction

In this work we aim to study the size estimate problem with one measurement when the background conductivity has jump interfaces. A typical application of this study is to estimate the size of a cancerous tumor inside an organ by the electric impedance tomography (EIT). In this case, considering discontinuous medium is typical, for instance, the conductivities of heart, liver, intestines are 0.70 (S/m), 0.10 (S/m), 0.03 (S/m), respectively. Previous works on this problem assumed that the conductivity of the studied body is Lipschitz continuous, see, for example, [3, 4]. The first result on the size estimate problem with a discontinuous background conductivity was given in [17], where only the two dimensional case was considered. In this paper, we will study the problem in dimension n ≥ 2.

The main ingredients of our method are quantitative uniqueness estimates for

div(A∇u) = 0 Ω ⊂ Rⁿ. (1.1)

Those estimates are well-known when A is Lipschitz continuous. The derivation of the estimates is based on the Carleman estimate or the frequency function method. For n = 2 and A ∈ L^∞, quantitative uniqueness estimates are obtained via the connection

∗Universit´a di Firenze, Italy. Email: francini@math.unifi.it

†National Cheng Kung University, Taiwan. Email: cllin2@mail.ncku.edu.tw

‡Universit´a di Firenze, Italy. Email: sergio.vessella@dmd.unifi.it

§National Taiwan University, Taiwan. Email: jnwang@ntu.edu.tw

between (1.1) and quasiregular mappings. This is the method used in [17]. For n ≥ 3, the connection with quasiregular mappings is not true. Hence we return to the old method – the Carleman estimate, to derive quantitative uniqueness estimates when A is discontinuous. Precisely, when A has a C^1,1 interface and is Lipschitz away from the interface, a Carleman estimate was obtained in [5] (see [10, 11, 12] for related results). Here we will derive three-region inequalities using this Carleman estimate.

The three-region inequality provides us a way to propagate ”smallness” across the interface (see also [11] for similar estimates). Relying on the three-region inequality, we then derive bounds of the size of an inclusion with one boundary measurement.

For other results on the size estimate, we mention [1] for the isotropic elasticity, [14, 15,16] for the isotropic/anisotropic thin plate, [7, 6] for the shallow shell.

2 The Carleman estimate

In this section, we would like to describe the Carleman estimate derived in [5]. We first denote H± = χ_Rⁿ_± where Rⁿ± = {(x, y) ∈ Rⁿ⁻¹ × R : y ≷ 0} and χRⁿ± is the characteristic function of Rⁿ±. Let u± ∈ C^∞(Rⁿ) and define

u = H₊u₊+ H−u−=X

±

H±u±,

hereafter,P

±a± = a₊+ a−, and L(x, y, ∂)u :=X

±

H_±div_x,y(A_±(x, y)∇_x,yu_±), (2.1)

where

A±(x, y) = {a^±_ij(x, y)}ⁿ_i,j=1, x ∈ Rⁿ⁻¹, y ∈ R (2.2) is a Lipschitz symmetric matrix-valued function satisfying, for given constants λ0 ∈ (0, 1], M₀ > 0,

λ₀|z|² ≤ A±(x, y)z · z ≤ λ⁻¹₀ |z|², ∀(x, y) ∈ Rⁿ, ∀ z ∈ Rⁿ (2.3) and

|A±(x⁰, y⁰) − A±(x, y)| ≤ M₀(|x⁰− x| + |y⁰− y|). (2.4) We write

h₀(x) := u₊(x, 0) − u₋(x, 0), ∀ x ∈ Rⁿ⁻¹, (2.5) h₁(x) := A₊(x, 0)∇_x,yu₊(x, 0) · ν − A−(x, 0)∇_x,yu−(x, 0) · ν, ∀ x ∈ Rⁿ⁻¹, (2.6) where ν = −e_n.

For a function h ∈ L²(Rⁿ), we define ˆh(ξ, y) =

Z

Rⁿ⁻¹

h(x, y)e^−ix·ξdx, ξ ∈ Rⁿ⁻¹.

As usual H^1/2(Rⁿ⁻¹) denotes the space of the functions f ∈ L²(Rⁿ⁻¹) satisfying Z

Rⁿ⁻¹

|ξ|| ˆf (ξ)|²dξ < ∞, with the norm

||f ||²_H1/2(Rⁿ⁻¹) = Z

Rⁿ⁻¹

(1 + |ξ|²)^1/2| ˆf (ξ)|²dξ. (2.7) Moreover we define

[f ]_1/2,Rⁿ⁻¹ =

Z

Rⁿ⁻¹

Z

Rⁿ⁻¹

|f (x) − f (y)|²

|x − y|ⁿ dydx

1/2

,

and recall that there is a positive constant C, depending only on n, such that C⁻¹

Z

Rⁿ⁻¹

|ξ|| ˆf (ξ)|²dξ ≤ [f ]²_1/2,Rn−1 ≤ C Z

Rⁿ⁻¹

|ξ|| ˆf (ξ)|²dξ,

so that the norm (2.7) is equivalent to the norm ||f ||_L²_(Rⁿ⁻¹₎+ [f ]_1/2,Rⁿ⁻¹. From now on, we use the letters C, C0, C1, · · · to denote constants (depending on λ0, M0, n).

The value of the constants may change from line to line, but it is always greater than 1. We will denote by B_r(x) the (n − 1)-ball centered at x ∈ Rⁿ⁻¹ with radius r > 0.

Whenever x = 0 we denote Br = Br(0).

Theorem 2.1 Let u and A±(x, y) satisfy (2.1)-(2.6). There exist L, β, δ₀, r₀, τ₀ pos- itive constants, with r₀ ≤ 1, depending on λ₀, M₀, n, such that if α₊ > Lα−, δ ≤ δ₀ and τ ≥ τ0, then

X

± 2

X

|k|=0

τ^3−2|k|

Z

Rⁿ±

|D^ku±|²e^{2τ φδ,±(x,y)}dxdy +X

± 1

X

|k|=0

τ^3−2|k|

Z

Rⁿ⁻¹

|D^ku±(x, 0)|²e^2φδ(x,0)dx

+X

±

τ²[e^{τ φδ(·,0)}u±(·, 0)]²_1/2,Rn−1+X

±

[D(e^{τ φδ,±}u±)(·, 0)]²_1/2,Rn−1

≤C X

±

Z

Rⁿ±

|L(x, y, ∂)(u±)|²e^{2τ φδ,±(x,y)}dxdy + [e^{τ φδ(·,0)}h₁]²_1/2,Rn−1

+[∇_x(e^{τ φδ}h₀)(·, 0)]²_1/2,Rn−1+ τ³ Z

Rⁿ⁻¹

|h₀|²e^{2τ φδ(x,0)}dx + τ Z

Rⁿ⁻¹

|h₁|²e^{2τ φδ(x,0)}dx

 . (2.8) where u = H+u+ + H−u−, u± ∈ C^∞(Rⁿ) and supp u ⊂ Bδ/2 × [−δr0, δr0], and φ_δ,±(x, y) is given by

φδ,±(x, y) =









 α+y

δ +βy²

2δ² − |x|²

2δ , y ≥ 0, α−y

δ + βy²

2δ² −|x|²

2δ , y < 0,

(2.9)

and φ_δ(x, 0) = φ_δ,+(x, 0) = φ_δ,−(x, 0).

Remark 2.2 It is clear that (2.8) remains valid if can add lower order terms P

±H_±(W · ∇_x,yu_±+ V u_±), where W, V are bounded functions, to the operator L.

That is, one can substitute L(x, y, ∂)u =X

±

H±div_x,y(A±(x, y)∇_x,yu±) +X

±

H±(W · ∇_x,yu±+ V u±) (2.10)

in (2.8).

3 Three-region inequalities

Based on the Carleman estimate given in Theorem 2.1, we will derive three-region inequalities across the interface y = 0. Here we consider u = H₊u₊+H−u−satisfying

L(x, y, ∂)u = 0 in Rⁿ, where L is given in (2.10) and

kW k_L^∞_(Rⁿ₎+ kV k_L^∞_(Rⁿ₎≤ λ⁻¹₀ . Fix any δ ≤ δ₀, where δ₀ is given in Theorem 2.1.

Theorem 3.1 Let u and A±(x, y) satisfy (2.1)-(2.6) with h₀ = h₁ = 0. Then there exist C and R, depending only on λ₀, M₀, n, such that if 0 < R₁, R₂ ≤ R, then

Z

U2

|u|²dx ≤ (e^τ0R2 + CR⁻⁴₁ )

Z

U1

|u|²dxdy

_2R1+3R2^R2 Z

U3

|u|²dxdy

^2R1+2R2_2R1+3R2

, (3.1) where τ₀ is the constant derived in Theorem 2.1,

U₁ = {z ≥ −4R₂, R₁

8a < y < R₁ a }, U₂ = {−R₂ ≤ z ≤ R1

2a, y < R1

8a}, U₃ = {z ≥ −4R₂, y < R₁

a }, a = α₊/δ, and

z(x, y) = α−y

δ +βy²

2δ² − |x|²

2δ . (3.2)

Proof. To apply the estimate (2.8), u needs to satisfy the support condition. Also, we can choose α+ and α− in Theorem2.1 such that α+ > α−. We can choose r ≤ r0

satisfying

r ≤ min 13α−

8β , 2δ

19α−+ 8β



. (3.3)

y

x

z=−4R2

z=−R2

z=^R1_2a

y=^R1_a

y=^R1_8a

U₁

U₂

Figure 1: U₁ and U₂ are shown in green and red, respectively. U₃ is the region enclosed by blue boundaries.

Note that the choices of δ, r also depend on λ₀, M₀, n. We then set R = α−r

16 . It follows from (3.3) that

R ≤ 13α²₋

128β. (3.4)

Given 0 < R1 < R2 ≤ R. Let ϑ1(t) ∈ C₀^∞(R) satisfy 0 ≤ ϑ¹(t) ≤ 1 and ϑ₁(t) =

(1, t > −2R2, 0, t ≤ −3R₂. Also, define ϑ₂(y) ∈ C₀^∞(R) satisfying 0 ≤ ϑ2(y) ≤ 1 and

ϑ₂(y) =







0, y ≥ R₁ 2a, 1, y < R₁

4a.

Finally, we define ϑ(x, y) = ϑ₁(z(x, y))ϑ₂(y), where z is defined by (3.2).

We now check the support condition for ϑ. From its definition, we can see that suppϑ is contained in







z(x, y) = α−y

δ + βy²

2δ² −|x|²

2δ > −3R₂, y < R₁

2a.

(3.5)

In view of the relation

α₊ > α₋ and a = α₊ δ , we have that

R₁ 2a < δ

2α−

· R1 < δ α−

· α₋r 16 < δr, i.e., y < δr ≤ δr₀. Next, we observe that

−3R₂ > −3R = −3α−r 16 > α−

δ (−δr) + β

2δ²(−δr)²,

which gives −δr < y due to (3.3). Consequently, we verify that |y| < δr. One the other hand, from the first condition of (3.5) and (3.3), we see that

|x|²

2δ < 3R₂+ α−y

δ + βy²

2δ² ≤ 3α−r 16 + α−

δ · δr + β

2δ² · δ²r² ≤ δ 8, which gives |x| < δ/2.

Since h₀ = 0, we have that

ϑ(x, 0)u₊(x, 0) − ϑ(x, 0)u₋(x, 0) = 0, ∀ x ∈ Rⁿ⁻¹. (3.6) Applying (2.8) to ϑu and using (3.6) yields

X

± 2

X

|k|=0

τ^3−2|k|

Z

Rⁿ_±

|D^k(ϑu±)|²e^{2τ φδ,±(x,y)}dxdy

≤CX

±

Z

Rⁿ_±

|L(x, y, ∂)(ϑu±)|²e^{2τ φδ,±(x,y)}dxdy

+ Cτ Z

Rⁿ⁻¹

|A₊(x, 0)∇_x,y(ϑu₊(x, 0)) · ν − A−(x, 0)∇_x,y(ϑu−)(x, 0) · ν|²e^{2τ φδ(x,0)}dx + C[e^{τ φδ(·,0)} A₊(x, 0)∇_x,y(ϑu₊)(x, 0) · ν − A₋(x, 0)∇_x,y(ϑu₋)(x, 0) · ν
]²_1/2,Rn−1.

(3.7) We now observe that ∇_x,yϑ₁(z) = ϑ⁰₁(z)∇_x,yz = ϑ⁰₁(z)(−^x_δ,^α_δ⁻+^βy_δ2) and it is nonzero only when

−3R₂ < z < −2R₂. Therefore, when y = 0, we have

2R₂ < |x|²

2δ < 3R₂. Thus, we can see that

|∇x,yϑ(x, 0)|² ≤ CR⁻²₂  6R₂ δ +α²₋

δ²



≤ CR₂⁻². (3.8)

By h₀(x) = h₁(x) = 0, (3.8), and the easy estimate of [5, Proposition 4.2], it is not hard to estimate

τ Z

Rⁿ⁻¹

|A₊(x, 0)∇_x,y(ϑu₊(x, 0)) · ν − A−(x, 0)∇_x,y(ϑu−)(x, 0) · ν|²e^{2τ φδ(x,0)}dx + [e^{τ φδ(·,0)} A₊(x, 0)∇_x,y(ϑu₊)(x, 0) · ν − A−(x, 0)∇_x,y(ϑu−)(x, 0) · ν
]²_1/2,Rn−1

≤ CR⁻²₂ e^{−4τ R2}

 τ

Z

{√

4δR2≤|x|≤√ 6δR2}

|u₊(x, 0)|²dx + [u₊(x, 0)]²_1/2,{^√_4δR

2≤|x|≤√ 6δR2}



+ Cτ²R⁻³₂ e^{−4τ R2} Z

{√

4δR2≤|x|≤√ 6δR2}

|u₊(x, 0)|²dx

≤ Cτ²R⁻³₂ e^{−4τ R2}E,

(3.9) where

E = Z

{√

4δR2≤|x|≤√ 6δR2}

|u₊(x, 0)|²dx + [u₊(x, 0)]²_1/2,{^√_4δR

2≤|x|≤√ 6δR2}.

Expanding L(x, y, ∂)(ϑu±) and considering the set where Dϑ 6= 0, we can estimate X

± 1

X

|k|=0

τ^3−2|k|

Z

{−2R₂≤z≤^R1

2a, y<^R1_4a}

|D^ku±|²e^{2τ φδ,±(x,y)}dxdy

≤ CX

± 1

X

|k|=0

R^2(|k|−2)₂ Z

{−3R2≤z≤−2R2, y<^R1_2a}

|D^ku±|²e^{2τ φδ,±(x,y)}dxdy

+ C

1

X

|k|=0

R^2(|k|−2)₁ Z

{−3R₂≤z,^R1

4a<y<^R1_2a}

|D^ku+|²e^{2τ φδ,+(x,y)}dxdy

+ Cτ²R⁻³₂ e^{−4τ R2}E

≤ CX

± 1

X

|k|=0

R^2(|k|−2)₂ e^{−4τ R2}e^{2τ(α+−α−)δ R14a} Z

{−3R2≤z≤−2R2, y<^R1_4a}

|D^ku±|²dxdy

+

1

X

|k|=0

R^2(|k|−2)₁ e^{2τα+δ R12a}e^{2τ2δ2β (R12a)2} Z

{z≥−3R₂,^R1_4a<y<^R1_2a}

|D^ku+|²dxdy + Cτ²R⁻³₂ e^{−4τ R2}E.

(3.10)

Let us denote U₁ = {z ≥ −4R₂, ^R_8a¹ < y < ^R_a¹}, U₂ = {−R₂ ≤ z ≤ ^R_2a¹, y < ^R_8a¹}.

From (3.10) and interior estimates (Caccioppoli’s type inequality), we can derive that τ³e^{−2τ R2}

Z

U2

|u|²dxdy

≤ τ³e^{−2τ R2} Z

{−R2≤z≤^R1_2a, y<^R1_8a}

|u|²dxdy

≤X

±

τ³ Z

{−2R2≤z≤^R1_2a, y<^R1_4a}

|u±|²e^{2τ φδ,±(x,y)}dxdy

≤ CX

± 1

X

|k|=0

R^2(|k|−2)₂ e^{−4τ R2}e^{2τ(α+−α−)δ R14a} Z

{−3R2≤z≤−2R2, y<^R1_4a}

|D^ku±|²dxdy

+

1

X

|k|=0

R^2(|k|−2)₁ e^{2τα+δ R12a}e^{2τ2δ2β (R12a)2} Z

{z≥−3R2,^R1_4a<y<^R1_2a}

|D^ku₊|²dxdy + Cτ²R⁻³₂ e^{−4τ R2}E

≤ CR⁻⁴₁ e^{−3τ R2} Z

{−4R2≤z≤−R2, y<^R1_a }

|u|²dxdy + Cτ²R₂⁻³e^{−4τ R2}E

+ CR⁻⁴₁ e⁽¹⁺

βR1 4α2−

)τ R1Z

{z≥−4R2,^R1_8a<y<^R1_a }

|u|²dxdy

≤CR⁻⁴₁

 e^{2τ R1}

Z

U1

|u|²dxdy + τ²e^{−3τ R2}F

 ,

(3.11)

where

F = Z

{z≥−4R₂, y<^R1_a }

|u|²dxdy and we used the inequality _4α^βR2¹

− < 1 due to (3.4).

Dividing τ³e^{−2τ R2} on both sides of (3.11) implies that Z

U2

|u|²dxdy ≤ CR⁻⁴₁



e^{2τ (R1+R2)} Z

U1

|u|²dxdy + e^{−τ R2}F



. (3.12)

Now, we consider two cases. If R

U1|u|²dxdy 6= 0 and e^2τ0(R1+R2)

Z

U1

|u|²dxdy < e^−τ0R2F,

then we can pick a τ > τ₀ such that e^{2τ (R1+R2)}

Z

U1

|u|²dxdy = e^{−τ R2}F.

Using such τ , we obtain from (3.12) that Z

U2

|u|²dxdy ≤ CR⁻⁴₁ e^{2τ (R1+R2)} Z

U1

|u|²dxdy

= CR⁻⁴₁

Z

U1

|u|²dxdy

_2R1+3R2^R2

(F )^{2R1+2R22R1+3R2}.

(3.13)

If R

U1|u|²dxdy = 0, then letting τ → ∞ in (3.12) we have R

U2|u|²dxdy = 0 as well.

The three-regions inequality (3.1) obviously holds.

On the other hand, if

e^−τ0R2F ≤ e^2τ0(R1+R2) Z

U1

|u|²dxdy, then we have

Z

U2

|u|²dx ≤ (F )^2R1+3R2R2 (F )^{2R1+2R22R1+3R2}

≤ exp (τ₀R₂)

Z

U1

|u|²dxdy

_2R1+3R2^R2

(F )^{2R1+2R22R1+3R2} .

(3.14)

Putting together (3.13), (3.14), we arrive at Z

U2

|u|²dx ≤ (exp (τ₀R₂) + CR⁻⁴₁ )

Z

U1

|u|²dxdy

 ^R2

2R1+3R2

(F )^{2R1+2R22R1+3R2} . (3.15)

2

4 Size estimate

We will apply the three-region inequality (3.1) to estimate the size of embedded inclusion in this section. Here we denote Ω a bounded open set in Rⁿ with C^1,α boundary ∂Ω with constants s₀, L₀, where 0 < α ≤ 1. Assume that Σ is a C² hypersurface with constants r₀, K₀ satisfying

dist(Σ, ∂Ω) ≥ d₀ (4.1)

for some d₀ > 0. We divide Ω into three sets, namely, Ω = Ω+∪ Σ ∪ Ω−

where Ω± are open subsets. Note that Ω− = ∂Ω ∪ Σ and ∂Ω₊ = Σ. We also define Ω_h = {x ∈ Ω : dist(x, ∂Ω) > h}.

Definition 4.1 [C^1,α regularity] We say that Σ is C² with constants r₀, K₀ if for any P ∈ Σ there exists a rigid transformation of coordinates under which P = 0 and

Ω±∩ B(0, r₀) = {(x, y) ∈ B(0, r₀) ⊂ Rⁿ: y ≷ ψ(x)}, where ψ is a C² function on B_r0(0) satisfying ψ(0) = 0 and

kψk_C²_(Br0(0)) ≤ K₀.

The definition of C^1,α boundary is similar. Note that B(a, r) stands for the n-ball centered at a with radius r > 0. We remind the reader that B_r(a) denotes the (n − 1)-ball centered at a with radius r > 0.

Assume that A± = {a^±_ij(x, y)}ⁿ_i,j=1satisfy (2.3) and (2.4). Let us define H± = χ_Ω±, A = H+A++ H−A−, u = H+u++ H−u−. We now consider the conductivity equation

div(A∇u) = 0 in Ω. (4.2)

It is not hard to check that u satisfies homogeneous transmission conditions (2.5), (2.6) (with h₀ = h₁ = 0), where in this case ν is the outer normal of Σ. For φ ∈ H^1/2(∂Ω), let u solve (4.2) and satisfy the boundary value u = φ on ∂Ω.

Next we assume that D is a measurable subset of Ω. Suppose that ˆA is a sym- metric n × n matrix with L^∞(Ω) entries. In addition, we assume that there exist η > 0, ζ > 1 such that

(1 + η)A ≤ ˆA ≤ ζA a.e. in Ω (4.3)

or η > 0, 0 < ζ < 1 such that

ζA ≤ ˆA ≤ (1 − η)A a.e. in Ω. (4.4)

Now let v = H₊v₊+ H−v− be the solution of

(div((Aχ_{Ω\ ¯D} + ˆAχ_D)∇v) = 0 in Ω,

v = φ on ∂Ω. (4.5)

The inverse problem considered here is to estimate |D| by the knowledge of {φ, A∇v · ν|_∂Ω}. In this work we would like to consider the most interesting case where

D ⊆ ¯¯ Ω₊. (4.6)

In practice, one could think of Ω₊ being an organ and D being a tumor. The aim is to estimate the size of D by measuring one pair of voltage and current on the surface of the body.

We denote W₀ and W the powers required to maintain the voltage φ on ∂Ω when the inclusion D is absent or present. It is easy to see that

W₀ = Z

∂Ω

φA∇u · ν = Z

Ω

A∇u · ∇u

and

W = Z

∂Ω

φ(Aχ_{Ω\ ¯D}+ ˆAχ_D)∇v · ν = Z

Ω

(Aχ_{Ω\ ¯D} + ˆAχ_D)∇v · ∇v.

The size of D will be estimate by the power gap W − W0. To begin, we recall the following energy inequalities proved in [4].

Lemma 4.1 [4, Lemma 2.1] Assume that A satisfies the ellipticity condition (2.3).

If either (4.3) or (4.4) holds, then C₁

Z

D

|∇u|² ≤ |W₀− W | ≤ C₂ Z

D

|∇u|², (4.7)

where C₁, C₂ are constants depending only on λ, η, and ζ.

The derivation of bounds on |D| will be based on (4.7) and the following Lipschitz propagation of smallness for u.

Proposition 4.1 (Lipschitz propagation of smallness) Let u ∈ H¹(Ω) be the solution of (4.2) with Dirichlet data φ. For any B(x, ρ) ⊂ Ω₊, we have that

Z

B(x,ρ)

|∇u|² ≥ C Z

Ω

|∇u|², (4.8)

where C depends on Ω±, d0, λ0, M0, r0, K0, s0, L0, α, α⁰, ρ, and kφ − φ₀k_C1,α0(∂Ω)

kφ − φ₀k_H1/2(∂Ω)

,

for φ₀ = |∂Ω|⁻¹R

∂Ωφ. Here α⁰ satisfies 0 < α⁰ < _(α+1)n^α .

Before proving Proposition4.1, we need to adjust the three-region inequality (3.1) for the C² interface Σ. Let 0 ∈ Σ and the coordinate transform (x⁰, y⁰) = T (x, y) = (x, y − ψ(x)) for x ∈ B_s0(0). Denote ˜U = T (B(0, s₀)) and ˜A± = {˜a^±_i,j}ⁿ_i,j=1 the coefficients of A± in the new coordinates (x⁰, y⁰). It is easy to see that ˜A± satisfies (2.3) and (2.4) with possible different constants ˜λ₀, ˜M₀, depending on λ₀, M₀, r₀, K₀. Then there exist C and ˜R, depending on ˜λ₀, ˜M₀, n, such that for

0 < R1 < R2 ≤ ˜R (4.9)

and U₁, U₂, U₃ defined as in Theorem 3.1, we have that U₃ ⊂ ˜U (so U₁, U₂ are con- tained in ˜U as well) and (3.1) holds. Now let ˜U_j = T⁻¹(U_j), j = 1, 2, 3, then (3.1) becomes

Z

U˜2

|u|²dxdy ≤ C

Z

U˜1

|u|²dxdy

_2R1+3R2^R2 Z

U˜3

|u|²dxdy

^2R1+2R2_2R1+3R2

, (4.10)

where C depends on λ₀, M₀, r₀, K₀, n, R₁, R₂. Furthermore, by Caccioppoli’s inequal- ity and generalized Poincar´e’s inequality (see (3.8) in [2]), we obtain from (4.10) that

Z

U˜2

|∇u|²dxdy ≤ C

Z

U˜1

|∇u|²dxdy

_2R1+3R2^R2 Z

U˜3

|∇u|²dxdy

^2R1+2R2_2R1+3R2

(4.11) with a possibly different constant C.

Since A₊ (respectively A₋) is Lipschitz in Ω₊ (respectively Ω₋), the following three-sphere inequality is well-known. Let u± be a solution to div(A±∇u±) = 0 in Ω±. Then for B(x₀, ¯r) ⊂ Ω₊ (or B(x₀, ¯r) ⊂ Ω−) and 0 < r₁ < r₂ < r₃ < ¯r, we have that

Z

B(x0,r2)

|∇u±|²dxdy ≤ C

Z

B(x0,r1)

|∇u±|²dxdy

θZ

B(x0,r3)

|∇u±|²dxdy

1−θ

, (4.12) where 0 < θ < 1 and C depend on λ₀, M₀, n, r₁/r₃, r₂/r₃.

Now we are ready to prove Proposition 4.1.

Proof of Proposition 4.1. It suffices to study the case where ρ is small. Since Σ ∈ C², it satisfies both the uniform interior and exterior sphere properties, i.e., there exists a₀ > 0 such that for all z ∈ Σ, there exist balls B ⊂ Ω₊ and B⁰ ⊂ Ω− of radius a₀ such that B ∩ Σ = B⁰∩ Σ = {z}. Next let ν_z be the unit normal at z ∈ Σ pointing into Ω₊ (inwards) and L = {z + tν_z ⊂ Rⁿ : t ∈ [ρ₀, −3ρ₀]}. We then fix R1, R2 satisfying (4.9) and choose ρ0 > 0 so that

S_z = ∪_y∈LB(y, ρ₀) ⊂ ˜U₂.

Denote κ = R₂/(2R₁+ 3R₂). Note that we move the construction of the three-region inequality from 0 to z.

Let x ∈ Ω+ and consider B(x, ρ) ⊂ Ω+, where ρ ≤ min{a0, ρ0}. For any y ∈ Ω2ρ, we discuss three cases.

(i) Let y ∈ Ω_+,ρ, then by (4.12) and the chain of balls argument, we have that R

B(y,ρ)|∇u|² R

Ω|∇u|² ≤ C R

B(x,ρ)|∇u|² R

Ω|∇u|²

!θ^N1

, (4.13)

where N₁ depends on Ω₊ and ρ.

(ii) Let y ∈ {y ∈ Ω₊ : dist(y, Σ) ≤ ρ} ∪ {y ∈ Ω− : dist(y, Σ) ≤ 3ρ}, then B(y, ρ) ⊂ S_z for some z ∈ Σ. Note that ˜U1 ⊂ Ω+,ρ (taking ρ even smaller if necessary). We then apply (4.13) iteratively to estimate

R

U˜1|∇u|² R

Ω|∇u|² ≤ C R

B(x,ρ)|∇u|² R

Ω|∇u|²

!θ^N1

, (4.14)

where C depends on ˜U₁ and ρ. Combining estimates (4.14) and (4.11) yields R

B(y,ρ)|∇u|² R

Ω|∇u|² ≤ C R

B(x,ρ)|∇u|² R

Ω|∇u|²

!κθ^N1

. (4.15)

(iii) Finally, we consider the case where y ∈ Ω−∩Ω_2ρand dist(y, Σ) > 3ρ. We observe that if y∗ = z + (−3ρ)νz, then (4.15) implies

R

B(y∗,ρ)|∇u|² R

Ω|∇u|² ≤ C R

B(x,ρ)|∇u|² R

Ω|∇u|²

!κθ^N1

. (4.16)

Again using (4.12) and the chain of balls argument (starting with (4.16)), we obtain that

R

B(y,ρ)|∇u|² R

Ω|∇u|² ≤ C R

B(x,ρ)|∇u|² R

Ω|∇u|²

!κθ^N1θ^N2

. (4.17)

Putting together (4.13), (4.15), and (4.17) gives R

B(y,ρ)|∇u|² R

Ω|∇u|² ≤ C R

B(x,ρ)|∇u|² R

Ω|∇u|²

!s

(4.18) for all y ∈ Ω_2ρ, where 0 < s < 1 and C depends on λ₀, M₀, n, r₀, K₀, ρ, Ω±.

In view of (4.18) and covering Ω_3ρ with balls of radius ρ, we have that R

Ω3ρ|∇u|² R

Ω|∇u|² ≤ C R

B(x,ρ)|∇u|² R

Ω|∇u|²

!s

. (4.19)

Note that u − φ₀ is the solution to (4.2) with Dirichlet boundary value φ − φ₀. By Corollary 1.3 in [13], we have that

k∇uk²_L∞(Ω) = k∇(u − φ₀)k²_L∞(Ω)≤ Ckφ − φ₀k²_{C1,α0(∂Ω)} with 0 < α⁰ < _(α+1)n^α , which implies

Z

Ω\Ω3ρ

|∇u|² ≤ C|Ω \ Ω_5ρ|kφ − φ₀k²_C1,α0(∂Ω) ≤ Cρkφ − φ₀k²_C1,α0(∂Ω). (4.20) Here we have used |Ω\Ω_5ρ| . ρ proved in [3]. Using the Poincar´e inequality, we have

kφ − φ₀k²_H1/2(∂Ω) ≤ Cku − φ₀k²_H1(Ω) ≤ Ck∇uk²_L2(Ω).

Combining this and (4.20), we see that if ρ is small enough depending on Ω±, d₀, λ₀, M₀, r₀, K₀, s₀, L₀, α, α⁰, ρ, and kφ − φ₀k_C1,α0(∂Ω)/kφ − φ₀k_H1/2(∂Ω), then

k∇uk²_L2(Ω3ρ)

k∇uk²_L2(Ω)

≥ 1 2.

The proposition follows from this and (4.19).

2

We now have enough tools to derive bounds on |D|.

Theorem 4.2 Suppose that the assumptions of this section hold.

(i) If, moreover, there exists h > 0 such that

|D_h| ≥ 1

2|D| (fatness condition). (4.21) Then there exist constants K₁, K₂ > 0 depending only on Ω_±, d₀, h, λ₀, M₀, r₀, K₀, s₀, L₀, α, α⁰, and kφ − φ₀k_C1,α0(∂Ω)/kφ − φ₀k_H1/2(∂Ω), such that

K₁    

W₀ − W W0

≤ |D| ≤ K₂    

W₀− W W0

    .

(ii) For a general inclusion D contained strictly in Ω₊, we assume that there exists d₁ > 0 such that

dist(D, ∂Ω₊) ≥ d₁.

Then there exist constants K₁, K₂⁰, p > 1, depending only on Ω_±, d₀, d₁, h, λ₀, M₀, r₀, K₀, s₀, L₀, α, α⁰, and kφ − φ₀k_C1,α0(∂Ω)/kφ − φ₀k_H1/2(∂Ω), such that

K₁    

W₀ − W W₀

≤ |D| ≤ K₂⁰    

W₀ − W W₀

1 p

. (4.22)

Proof. The proof follows closely the arguments in [4] and [17]. The lower bound can be obtained by basic estimates. Let c = ¹

|^Ωd/4| R

Ωd/4u. By the gradient estimate of [13, Theorem 1.1], the interior estimate of [9, Theorem 8.17] and the Poincar´e inequality for the domain Ω_d/4, we have

k∇uk_L^∞_(Ωd/2) ≤ Cku − ck_L^∞_(Ωd/3) ≤ Cku − ck_L²_(Ωd/4)≤ Ck∇uk_L²_(Ω).

From this, the trivial estimate k∇uk²_L2(D) ≤ C|D|k∇uk²_L∞(Ω_d/2) and the second in- equality of (4.7), the lower bound follows.

Next, we prove the upper bounds.

(i) Let ρ = ^h₄ and cover D_h with internally nonoverlapping closed squares {Q_k}^J_k=1 of side length 2ρ. It is clear that Qk ⊂ D, hence

Z

D

|∇u|²dx ≥ Z

∪^J_k=1Q_k

|∇u|²dx ≥ |D_h| ρ² min

k

Z

Q_k

|∇u|²dx.

≥ C|D|

ρ² Z

Ω

|∇u|²dx.

Here we have used Proposition 4.1 and the fatness condition at the last inequality.

The upper bound of |D| follows from this and the first inequality of (4.7).

(ii) To prove the upper bound without the fatness condition, we need the fact that

|∇u|² is an A_p weight which an easy consequence of the doubling condition for ∇u.

It turns out when D is strictly contained in Ω₊ where the coefficient A₊ is Lipschitz.

The well-known theorem guarantees that |∇u|² is an A_p weight in Ω₊ (see [8] or [4]), i.e., for any ¯r > 0, there exists B > 0 and p > 1 such that

 1

|B(a, r)|

Z

B(a,r)

|∇u|²

  1

|B(a, r)|

Z

B(a,r)

|∇u|^−p−12

p−1

≤ B

for any ball B(a, r) ⊂ Ω_+,¯r, where B and p depends on various constants listed in Proposition4.1. To derive the upper bound of (4.22), we choose ¯r = d1/2 and follow exactly the same lines as in the proof of Theorem 2.2 [4].

2

Remark 4.3 We point out that part (i) of Theorem4.2 still holds if the assumption (4.6) is replaced by

dist(D, ∂Ω) ≥ d₂ > 0.

Acknowledgements

EF and SV were partially supported by GNAMPA - INdAM. EF was partially sup- ported by the Research Project FIR 2013 Geometrical and qualitative aspects of PDE’s. JW was supported in part by MOST102-2115-M-002-009-MY3.

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