elliptic and parabolic equations in fractured...

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July 9, 2015

Elliptic and parabolic equations in fractured media

Li-Ming Yeh

Department of Applied Mathematics

National Chiao Tung University, Hsinchu, 30050, Taiwan, R.O.C.

[email protected]

The elliptic and the parabolic equations with Dirichlet boundary conditions in frac-tured media are considered. The fractured media consist of a periodic connected highpermeability sub-region and a periodic disconnected matrix block subset with low perme-ability. Let ǫ ∈ (0, 1] denote the size ratio of the matrix blocks to the whole domain andlet ω2

∈ (0, 1] denote the permeability ratio of the disconnected subset to the connectedsub-region. It is proved that the W 1,p norm of the elliptic and the parabolic solutionsin the high permeability sub-region are bounded uniformly in ω, ǫ. However, the W 1,p

norm of the solutions in the low permeability subset may not be bounded uniformly inω, ǫ. For the elliptic and the parabolic equations in periodic perforated domains, it isalso shown that the W 1,p norm of their solutions are bounded uniformly in ǫ.

Keywords: fractured media, permeability, periodic perforated domain, VMO

AMS Subject Classification: 35J05, 35J15, 35J25

1. Introduction

The W 1,p estimates for the solutions of the elliptic and the parabolic equations with

Dirichlet boundary conditions in fractured media are concerned. The problem arises

from two-phase problems, flows in fractured media, and the stress in composite ma-

terials (see [3, 9, 15]). Let Ω be a smooth simply-connected domain in Rn for n ≥ 3,

∂Ω be the boundary of Ω, Y ≡ (0, 1)n consist of a smooth sub-domain Ym completely

surrounded by another connected sub-domain Yf (≡ Y \ Ym), ǫ ∈ (0, 1], Ω(2ǫ) ≡

x ∈ Ω : dist(x, ∂Ω) ≥ 2ǫ, Ωǫm ≡ x : x ∈ ǫ(Ym + j) ⊂ Ω(2ǫ) for some j ∈ Zn be

a disconnected subset of Ω, Ωǫf (≡ Ω\Ωǫ

m) denote a connected sub-region of Ω, and

Kν,ǫ(x) ≡

1 if x ∈ Ωǫ

f

ν if x ∈ Ωǫm

for any ν, ǫ > 0.

The elliptic equation that we consider is−∇ · (Kω2,ǫ∇U +G) = F in Ω,

U = 0 on ∂Ω,(1.1)

where ω, ǫ ∈ (0, 1] and G,F are given functions. If G,F are bounded, a solution of

(1.1) in Hilbert space H1(Ω) exists uniquely for each ω, ǫ by Lax-Milgram Theorem

[12]. The L2 norm of the gradient of the solution of (1.1) in the connected sub-region

Ωǫf is bounded uniformly in ω, ǫ if G,F are small in Ωǫ

m. However, the L2 norm of

1

July 9, 2015

2 Elliptic and parabolic equations

the gradient of the solution of (1.1) in matrix blocks Ωǫm can be very large when ω

closes to 0. The parabolic equation that we consider is, for any ω, ǫ ∈ (0, 1],

∂tU −∇ · (Kω2,ǫ∇U) = F in Ω × (0, T ),

U = 0 on ∂Ω × (0, T ),

U(x, 0) = U0(x) in Ω.

(1.2)

If F,U0 are smooth, a solution of (1.2) in Hilbert space L2([0, T ];H1(Ω)) exists

uniquely for each ω, ǫ. The L2 norm of the gradient of the solution of (1.2) in

the connected sub-region Ωǫf × (0, T ) is bounded uniformly in ω, ǫ if F is small in

Ωǫm× (0, T ). However, the L2 norm of the gradient of the solution of (1.2) in matrix

blocks Ωǫm × (0, T ) can be very large when ω closes to 0. One also notes that for the

elliptic and the parabolic equations in periodic perforated domains, the H1 norm

of their solutions are bounded uniformly in ǫ.

There are some literatures related to this work. Lipschitz estimate andW 2,p esti-

mate for uniform elliptic equations with discontinuous coefficients had been proved

in [15, 18]. Uniform Holder, W 1,p, and Lipschitz estimates for uniform elliptic equa-

tions with Holder periodic coefficients were shown in [4, 5]. Uniform W 1,p estimate

for uniform elliptic equations with continuous periodic coefficients was considered

in [6] and the same problem with VMO periodic coefficients could be found in [22].

Uniform W 1,p estimate for the Laplace equation in periodic perforated domains was

considered in [19] and the same problem in Lipschitz estimate was studied in [21].

Uniform Holder, W 1,p, and Lipschitz estimates in ǫ for uniform parabolic equations

with oscillating periodic coefficients were obtained in [10]. For non-uniform ellip-

tic equations with smooth periodic coefficients, existence of C2,α solution could be

found in [13]. Uniform Holder estimate in ǫ for non-uniform parabolic equations

with discontinuous periodic coefficients was shown in [23].

Here we present uniform W 1,p estimate for the solutions of the non-uniform

elliptic and the non-uniform parabolic equations with Dirichlet boundary conditions

in fractured media. It is proved that the W 1,p norm of the elliptic and the parabolic

solutions in the high permeability sub-region Ωǫf are bounded uniformly in ω, ǫ.

However, the solutions in the low permeability subset may not be bounded uniformly

in ω, ǫ. For the elliptic and the parabolic equations in perforated domains, it is also

shown that the W 1,p norm of their solutions are bounded uniformly in ǫ. A three-

step compactness argument introduced in [4, 5] will be employed to obtain the

uniform estimate for non-uniform elliptic equations. Different from the approach in

[10], we apply semigroup theory and the uniform estimate results for non-uniform

elliptic equations to prove the uniform estimate for non-uniform parabolic equations.

The rest of this work is organized as follows: Notation and main results are stated

in section 2. In section 3, we present a priori estimates for some interface problems

and present some local uniform Lipschitz and local uniform W 1,p estimates in ω, ǫ

for the solutions of elliptic equations in fracture media. Proofs of the main results

are given in section 4. The proof of local uniform Lipschitz estimate for the solutions

July 9, 2015

Elliptic and parabolic equations 3

of elliptic equations in fracture media (claimed in section 3) is given in section 5.

2. Notation and main result

Let Ck,α denote the Holder space with norm ‖ · ‖Ck,α, W s,p the Sobolev space with

norm ‖ · ‖W s,p , and [ϕ]C0,α the Holder semi-norm of ϕ for k ≥ 0, α ∈ [0, 1], s ≥

−1, p ∈ [1,∞] (see [2, 12]). Lp = W 0,p and H1 = W 1,2. C∞0 (D) is the space

of infinitely differentiable functions with support in D and C∞per(R

n) is the space

of infinitely differentiable Y -periodic functions in Rn. W s,p0 (D) is the closure of

C∞0 (D) under the W s,p norm and W s,p

per(Rn) is the closure of C∞

per(Rn) under W s,p

norm and ‖ϕ‖W s,pper(Rn) ≡ ‖ϕ‖W s,p(Y ) for s ≥ 1, p ∈ [1,∞]. Am ≡ x : x ∈ Ym +

j for some j ∈ Zn and Af ≡ Rn \ Am. H1per(R

n) ≡ ϕ ∈ W 1,2per(R

n) :∫

Yfϕ(y)dy =

0 and H1per(Af ) ≡ ϕ|Af

: ϕ ∈ H1per(R

n). Let ‖ϕ1, · · · , ϕm‖B1 ≡ ‖ϕ1‖B1 + · · · +

‖ϕm‖B1 and ‖ϕ‖B1∩B2 ≡ ‖ϕ‖B1 + ‖ϕ‖B2 . Set rD = D/r−1 ≡ x : x/r ∈ D, D be

the closure of D, ∂D be the boundary of D, XD is the characteristic function on D,

and let Br(x) denote a ball centered at x with radius r. For any ϕ ∈ L1(D),

(ϕ)D ≡ −

∫

D

ϕ(y)dy ≡1

|D|

∫

D

ϕ(y)dy.

Kω,ν(x) ≡

1 if x ∈ νAf

ω if x ∈ νAm

for ω ∈ [0, 1], ν ∈ (0,∞). If ~n is an outward normal

vector on ∂Ym, we define, for any function ϕ in Y and x ∈ ∂Ym,

⌊ϕ⌋∂Ym(x) = ϕ,+(x) − ϕ,−(x) where ϕ,±(x) ≡ limt→0+

ϕ(x± t~n). (2.1)

Similarly, if ~nǫ is an outward normal vector on ∂Ωǫm, we define, for any function ϕ

in Ω and x ∈ ∂Ωǫm,

⌊ϕ⌋∂Ωǫm

(x) = ϕ,+(x) − ϕ,−(x) where ϕ,±(x) ≡ limt→0+

ϕ(x± t~nǫ).

Next we give two statements:

A1. Ω is a bounded smooth simply-connected domain in Rn for n ≥ 3,

A2. Ym is a smooth simply-connected sub-domain of Y .

A1–A2 will be assumed throughout this paper except in subsection 5.1. Our main

results are the following:

Theorem 2.1. Suppose A1–A2 and

A3. ω, ǫ ∈ (0, 1], p ∈ (1,∞), G ∈ Lp(Ω), F ∈W−1,p(Ω),

then a W 1,p(Ω) solution of (1.1) exists uniquely and satisfies

‖Kω/ǫ,ǫU,Kω,ǫ∇U‖Lp(Ω)

≤ c(‖K1/ω,ǫG‖Lp(Ω) + ‖F‖W−1,p(Ω) + ω−1‖F‖W−1,p(Ωǫ

m)

)for ω

ǫ ≤ 1,

‖U,Kω,ǫ∇U‖Lp(Ω)

≤ c(‖K1/ω,ǫG‖Lp(Ω) + ‖F‖W−1,p(Ω) + ω−1‖F‖W−1,p(Ωǫ

m)

)for ω

ǫ ≥ 1,

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where c is a constant independent of ω, ǫ.

Theorem 2.1 implies an analogous result for perforated domains.


A4. ǫ ∈ (0, 1], p ∈ (1,∞), G ∈ Lp(Ωǫf ), F ∈W−1,p(Ω), ‖F‖W−1,p(Ωǫ

m) = 0,

then a W 1,p(Ωǫf ) solution of

−∇ · (∇U +G) = F in Ωǫf

(∇U +G) · ~nǫ = 0 on ∂Ωǫm

U = 0 on ∂Ω

(2.2)

exists uniquely and satisfies

‖U‖W 1,p(Ωǫf ) ≤ c(‖G‖Lp(Ωǫ

f ) + ‖F‖W−1,p(Ω)), (2.3)

where ~nǫ is a unit normal vector on ∂Ωǫm and c is a constant independent of ǫ.

For any ω, ǫ ∈ (0, 1] and p ∈ (1,∞), let us define

Bp ≡ϕ : ϕ ∈ W 1,p

0 (Ω) ∩W 2,p(Ωǫf ) ∩W 2,p(Ωǫ

m), ⌊Kω2,ǫ∇ϕ · ~nǫ⌋∂Ωǫm

= 0,

Dp ≡ϕ : ϕ ∈ W 2,p(Ωǫ

f ), ϕ|∂Ω = 0,∇ϕ · ~nǫ|∂Ωǫm

= 0,

where ~nǫ is a normal vector on ∂Ωǫm. By Lemma 3.4 [23], Bp with norm ‖ϕ‖Bp ≡

‖∇·(Kω2,ǫ∇ϕ)‖Lp(Ω) is a Banach space. If Bp denotes the closure of Bp in Lp space,

then Bp = Lp(Ω). Also note Dp with norm ‖ϕ‖Dp ≡ ‖∆ϕ‖Lp(Ωǫf ) is a Banach space.

The function spaces C([0, T ];B), Cσ([0, T ];B) for σ ∈ (0, 1] are defined as those in

pages 1, 3 [17].


A5. ω, ǫ, σ ∈ (0, 1], p ∈ (n,∞), ǫ ≤ ω, F ∈ Cσ([0, T ];Lp(Ω)), U0 ∈ Bp,

then a C([0, T ];W 1,p(Ω)) solution of (1.2) exists uniquely and satisfies

‖U‖C1([0,T ];Lp(Ω)) + ‖Kω,ǫ∇U‖C([0,T ];Lp(Ω)) ≤ c(‖U0‖Bp + ‖F‖Cσ([0,T ];Lp(Ω))

),



A6. ǫ, σ ∈ (0, 1], p ∈ (n,∞), F ∈ Cσ([0, T ];Lp(Ωǫf )), U0 ∈ Dp,

then a C([0, T ];W 1,p(Ωǫf )) solution of

∂tU − ∆U = F in Ωǫf × (0, T )

∇U · ~nǫ = 0 on ∂Ωǫm × (0, T )

U = 0 on ∂Ω × (0, T )

U(x, 0) = U0(x) in Ωǫf

(2.4)

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exists uniquely and satisfies

‖U‖C1([0,T ];Lp(Ωǫf)) + ‖∇U‖C([0,T ];Lp(Ωǫ

f)) ≤ c

(‖U0‖Dp + ‖F‖Cσ([0,T ];Lp(Ωǫ

f))

),

where ~nǫ is a normal vector on ∂Ωǫm and c is a constant independent of ǫ.

3. Preliminaries

From Theorem 2.1 [1], we know

Lemma 3.1. For p ∈ [1,∞) and ǫ ∈ (0, 1], there is a constant c(Yf , p) and a linear

continuous extension operator Pǫ : W 1,p(Ωǫf ) →W 1,p(Ω) such that if ϕ ∈W 1,p(Ωǫ

f ),

then

Pǫϕ = ϕ in Ωǫf ,

‖Pǫϕ‖Lp(Ω) ≤ c‖ϕ‖Lp(Ωǫf ),

‖∇Pǫϕ‖Lp(Ω) ≤ c‖∇ϕ‖Lp(Ωǫf ),

0 < d1 ≤ Pǫϕ ≤ d2 if 0 < d1 ≤ ϕ ≤ d2 for some constants d1, d2,

Pǫϕ = ζ in Ω if ϕ = ζ|Ωǫf

for some linear function ζ in Ω.

Moreover, if ζ(x) ≡ ϕ(rx) in B1(0)∩Ωǫf/r for any r > ǫ, then Pǫ/rζ(x) = Pǫϕ(rx)

in B1/2(0).

Remark 3.1. Tracing the proof of Theorem 7.25 [12], we know that if 0 ∈ ∂Ym and

p, ν ∈ [1,∞), there exist a constant c(Yf ) and a linear continuous extension operator

Pν : W 1,p(B1(0) ∩ νYf ) →W 1,p(B1(0)) such that, for any ϕ ∈ W 1,p(B1(0) ∩ νYf ),

Pνϕ = ϕ in B1(0) ∩ νYf ,

‖Pνϕ‖Lp(B1(0)) ≤ c‖ϕ‖Lp(B1(0)∩νYf ),

‖∇Pνϕ‖Lp(B1(0)) ≤ c‖∇ϕ‖Lp(B1(0)∩νYf ).

Lemma 3.2. Let ω ∈ (0, 1], ν ∈ (0,∞), ϕ ∈ H1(B1(0)), and Pνϕ|νAfdenote the

extension of ϕ|νAfon B1(0). There is a constant c independent of ω, ν such that

∥∥Kω,ν

(ϕ− (Pνϕ|νAf

)B1(0)

)∥∥L2(B1(0))

≤ c‖Kω,ν∇ϕ‖L2(B1(0)).

See section 2 for Kω,ν .

Proof. By Poincare inequality [12], Lemma 3.1, and Remark 3.1, the extension

function Pνϕ|νAf∈ H1(B1(0)) satisfies

∥∥Pνϕ|νAf− (Pνϕ|νAf

)B1(0)

∥∥L2(B1(0))

≤ c∥∥∇Pνϕ|νAf

∥∥L2(B1(0))

≤ c‖∇ϕ‖L2(B1(0)∩νAf ), (3.1)

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where c is independent of ω, ν. (3.1), Lemma 3.1, Remark 3.1, and Poincare in-

equality imply∥∥Kω,ν(ϕ− (Pνϕ|νAf

)B1(0))∥∥

L2(B1(0))

≤∥∥Kω,ν(Pνϕ|νAf

− (Pνϕ|νAf)B1(0))

∥∥L2(B1(0))

+ω∥∥ϕ− Pνϕ|νAf

∥∥L2(B1(0)∩νAm)

≤ c‖∇ϕ‖L2(B1(0)∩νAf ) + cω∥∥∇ϕ−∇Pνϕ|νAf

∥∥L2(B1(0)∩νAm)

≤ c‖Kω,ν∇ϕ‖L2(B1(0)).

If 0 ∈ ∂Ω, by A1 and rotation, there is a smooth function Ψ : Rn−1 → R such

that

Ψ(0) = |∇Ψ(0)| = 0,

B1(0) ∩ Ω/r = B1(0) ∩ (x′, xn) ∈ Rn : rxn > Ψ(rx′) if r ∈ (0, 1].(3.2)

If r = 0, we define B1(0) ∩ Ω/r ≡ B1(0) ∩ (x′, xn) ∈ Rn : xn > 0. Set

Kν,ǫ,r ≡

1 in Ωǫ

f/r

ν in Ωǫm/r

for ν, ǫ, r ∈ (0, 1]. (3.3)

Similar to Lemma 3.2, we also have, by Poincare inequality [12], Lemma 3.1,

and Remark 3.1,

Lemma 3.3. If ω, ǫ, r ∈ (0, 1], 0 ∈ ∂Ω and ϕ ∈ H1(B2(0) ∩ Ω/r) with ϕ|∂Ω/r = 0,

there is a constant c independent of ω, ǫ, r such that

‖Kω,ǫ,rϕ‖L2(B1(0)∩Ω/r) ≤ c‖Kω,ǫ,r∇ϕ‖L2(B1(0)∩Ω/r).

3.1. Interface problem

Let Γ(x−y) denote the fundamental solution of the Laplace equation in Rn, see §6.2

[7]. Define a single-layer and a double-layer potentials as, for any smooth function

ϕ on the boundary ∂Ym of Ym,

S∂Ym(ϕ)(x) ≡

∫

∂Ym

Γ(x− y)ϕ(y)dy

L∂Ym(ϕ)(x) ≡

∫

∂Ym

∇yΓ(x− y)~ny ϕ(y)dyfor x ∈ ∂Ym,

where ~ny is the unit vector outward normal to ∂Ym. By tracing the argument of

Lemma 3.2 [23], we know

Lemma 3.4. For any p ∈ (1,∞) and s ∈ 0, 1, the linear operatorsS∂Ym : W s− 1

p ,p(∂Ym) →W s+1− 1p ,p(∂Ym)

L∂Ym : W s+1− 1p ,p(∂Ym) →W s+2− 1

p ,p(∂Ym)

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are bounded; the operator I − ℓL∂Ym is continuously invertible in W s+1− 1p ,p(∂Ym)

for any ℓ ∈ [−2, 2]; and there is a constant c independent of ℓ so that

‖ϕ‖W

s+1− 1p

,p(∂Ym)

≤ c‖(I − ℓL∂Ym)(ϕ)‖W

s+1− 1p

,p(∂Ym)

for ϕ ∈W s+1− 1p ,p(∂Ym),

where I is the identity operator.

Let us introduce some notations:

∂Y is an open portion of the boundary ∂Y ,

D1,D2 are smooth domains satisfying Ym ⊂ D1 ⊂ D2 ⊂ Y ,

dist(Ym, ∂D1), dist(D1, ∂D2), dist(D2, ∂Y \ ∂Y ) > d0 > 0.

Lemma 3.5. Suppose ω ∈ (0, 1], any solution Φ of−∇ · (Kω2,1∇Φ + V ) = ζ in Y

Φ = 0 on ∂Y(3.4)

satisfies

‖Kωσ,1Φ‖W 1,p(D1\Ym)∩W 1,p(Ym) ≤ c(‖Φ‖L2(Yf )

+‖Kωσ−2,1V ‖Lp(Y ) + ‖Kωσ−2,1ζ‖W−1,p(Yf )∩W−1,p(Ym)

),

‖Φ‖W 2,p(D1\Ym)∩W 2,p(Ym) ≤ c(‖Φ‖L2(Yf )

+‖Kω−2,1V ‖W 1,p(Yf )∩W 1,p(Ym) + ‖Kω−2,1ζ‖Lp(Y )

),

(3.5)

where p ∈ [2,∞), σ ∈ 0, 1, and c is a constant independent of ω.

Proof. Define Iω,σ ≡ ‖Kωσ−2,1V ‖Lp(Y ) + ‖Kωσ−2,1ζ‖W−1,p(Yf )∩W−1,p(Ym) and let c

denote a constant independent of ω. First we prove (3.5)1.

Step 1: Assume V ∈ W 1,p0 (Yf ) ∩W 1,p

0 (Ym) and ζ ∈ Lp(Y ). Consider the fol-

lowing problem−∇ · (Kω2,1∇φ+ V ) = ζ in D2,

φ = 0 on ∂D2.(3.6)

The unique existence of a H1 solution of (3.6) is from Lax-Milgram Theorem [12].

By energy method, the solution satisfies

‖φ‖H1(D2\Ym) ≤ cIω,1. (3.7)

Let η ∈ C∞(D2\Ym), η ∈ [0, 1], η = 1 in D2\D1, η = 0 on ∂Ym, ‖η‖W 1,∞(D2\Ym) ≤

c. Multiply (3.6)1 by η to get−∇ · (∇(ηφ) − φ∇η + ηV ) = ηζ − (∇φ+ V )∇η in D2 \ Ym,

ηφ = 0 on ∂D2 ∪ ∂Ym.(3.8)

By (3.7)–(3.8), [8], Theorem 7.26 [12], and an iterative argument, we have

‖φ‖W 1,p(D2\D1) ≤ cIω,σ. (3.9)

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Let ϕ in Ym satisfy−∇ · (ω2∇ϕ+ V ) = ζ in Ym,

ϕ = 0 on ∂Ym,(3.10)

and ϕ in D2 \ Ym satisfy−∇ · (∇ϕ+ V ) = ζ in D2 \ Ym,

ϕ = 0 on ∂(D2 \ Ym).(3.11)

By [8] again,

‖ϕ‖W 1,p(D2\Ym) + ωσ‖ϕ‖W 1,p(Ym) ≤ cIω,σ. (3.12)

If we define ψ ≡ φ− ϕ in D2, then (3.6) and (3.10)–(3.11) imply

∆ψ = 0 in D2 \ ∂Ym,

⌊ψ⌋∂Ym = 0 on ∂Ym,

⌊Kω2,1∇ψ⌋∂Ym · ~ny = F on ∂Ym,

ψ = 0 on ∂D2,

(3.13)

where ~ny is the unit vector outward normal to ∂Ym. See (2.1) for (3.13)2,3. Since

V ∈W 1,p0 (Yf ) ∩W 1,p

0 (Ym),

F ≡(ω2∇ϕ,− −∇ϕ,+

)· ~ny|∂Ym .

By (3.12),

‖F‖W

−1p

,p(∂Ym)

≤ cIω,σ. (3.14)

By Green’s formula, (3.13), and Theorem 6.5.1 [7],ψ/2 + L∂Ym(ψ) = S∂Ym(∇ψ,− · ~ny|∂Ym)

ψ/2 − L∂Ym(ψ) = −S∂Ym(∇ψ,+ · ~ny|∂Ym) + S∂D2(∂nyψ|∂D2)on ∂Ym,

where ∂nyψ|∂D2 is the normal derivative of ψ on ∂D2. So(I −

2(1 − ω2)

ω2 + 1L∂Ym

)ψ =

2

ω2 + 1

(S∂D2 (∂nyψ|∂D2) − S∂Ym(F)

)on ∂Ym. (3.15)

Then (3.9), (3.12)–(3.15), and Lemma 3.4 imply

‖ψ‖W

1− 1p

,p(∂Ym)

≤ c

(‖F‖

W−

1p

,p(∂Ym)

+ ‖∂nyψ‖W

−1p

,p(∂D2)

)≤ cIω,σ. (3.16)

(3.13) and (3.16) imply

‖ψ‖W 1,p(D2) ≤ cIω,σ.

Together with (3.12), we obtain

‖Kωσ,1φ‖W 1,p(D2\Ym)∩W 1,p(Ym) ≤ cIω,σ. (3.17)

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Note W 1,p0 (Yf ) (resp. W 1,p

0 (Ym)) is dense in Lp(Yf ) (resp. Lp(Ym)) and Lp(Y )

is dense in W−1,p(Y ). By a limiting argument, we see that if V ∈ Lp(Y ) and

ζ ∈W−1,p(Y ), any solution of (3.6) satisfies (3.17).

Step 2: Let η be a smooth function satisfying η ∈ C∞0 (D2), η ∈ [0, 1], η = 1 in

D1, ‖η‖W 1,∞(D2) ≤ c. Multiply (3.4) by η to obtain−∇ · (Kω2,1∇(Φη) − Φ∇η + V η) = ζη − (∇Φ + V )∇η in D2,

Φη = 0 on ∂D2.

By the result of Step 1, we have

‖Kωσ,1Φ‖W 1,p(D1\Ym)∩W 1,p(Ym) ≤ c(‖Φ‖Lp(D2\D1) + Iω,σ). (3.18)

Let η be another smooth function satisfying η ∈ C∞0 (Yf ), η ∈ [0, 1], η = 1 in D2\D1,

‖η‖W 1,∞(Y ) ≤ c. Multiply (3.4) by ηΦ and use energy method and Theorem 7.26

[12] to get

‖Φ‖Lp(D2\D1) ≤ c(‖Φ‖L2(Yf ) + Iω,σ).

Together with (3.18), we obtain (3.5)1. (3.5)2 are proved in a similar way as (3.5)1,

so we skip it.

By a similar argument as Lemma 3.5, we also have the following local estimate:

Lemma 3.6. If ω ∈ (0, 1], ν ∈ (1,∞), x0 ∈ ν∂Ym, and B1(x0) ⊂ νY , then any

solution Φ of

−∇ · (Kω2,ν∇Φ) = 0 in νY (3.19)

satisfies

‖Kωσ,νΦ‖W 2,p(B1/3(x0)∩νYf )∩W 2,p(B1/3(x0)∩νYm) ≤ c‖Kωσ,νΦ‖L2(B1(x0)), (3.20)

where p ∈ [2,∞), σ ∈ 0, 1, and c is a constant independent of ω, ν.

Proof. After translation, we assume x0 is the origin. For each ν > 1, we find a

smooth domain Dν such that

B1/2(x0) ∩ νYm ⊂ Dν ⊂ B1(x0) ∩ νYm and B1/2(x0) ∩ ν∂Ym ⊂ ∂Dν .

Since Dν is smooth, for any z ∈ ∂Dν there is a ball B(z) centered at z and there

is a smooth one-to-one mapping ϕz,ν of B(z) onto ϕz,ν(B(z)) ⊂ Rn satisfying

ϕz,ν(B(z) ∩Dν) ⊂ Rn+, ϕz,ν(B(z) ∩ ∂Dν) ⊂ ∂Rn

+, ϕz,ν(B(z) \ Dν) ⊂ Rn−. (3.21)

Here Rn+ ≡ x = (x1, · · · , xn) : xn > 0, ∂Rn

+ ≡ x : xn = 0,Rn− ≡ x : xn < 0.

Since ∂Dν is compact for each ν > 1, there exist a finite number ℓν of open balls

B(zi)ℓν

i=1 and one-to-one mappings ϕzi,νℓν

i=1 such that

zi ∈ ∂Dν for i ∈ 1, · · · , ℓν,

(3.21) holds for each ball B(zi) and i ∈ 1, · · · , ℓν,

∂Dν ⊂⋃ℓν

i=1 B(zi).

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Since Ym is smooth, it is possible to choose domains Dν for all ν > 1 such that

the number ℓν is bounded above by a constant independent of ν,

‖ϕz,ν‖C3,0(B(z)), ‖ϕ−1z,ν‖C3,0(ϕz,ν(B(z))) ≤ c∗,where c∗ is independent of ν, z.

By assumption x0 = 0 ∈ ν∂Ym, we define Kω2,ν and φ in Rn by

Kω2,ν ≡

ω2 in Dν ,

1 elsewhere,φ ≡

Φ in B1/2(x0),

0 elsewhere.

Let η ∈ C∞0 (B1/2(x0)) be a bell-shaped function satisfying η ∈ [0, 1], η = 1 in

B1/3(x0), ‖∇η‖W 1,∞(B1/2(x0)) ≤ c. Multiply (3.19) by η to get

−∇ ·

(Kω2,ν∇(ηφ) − Kω2,νφ∇η

)= −Kω2,ν∇φ∇η in B1(x0),

ηφ = 0 on ∂B1(x0).

Then we follow the argument of Step 1 of Lemma 3.5 to see that (3.20) holds.

Let X(j)ω,1 ∈ H1

per(Rn) for ω ∈ (0, 1] be a function satisfying

∇ · (Kω2,1(∇X(j)ω,1 + ~ej)) = 0 in Y , (3.22)

and let X(j)0,1 ∈ H1

per(Af )∩H1(Am) be a function satisfying X(j)0,1(x) = 0 in Am and

∇ · (∇X

(j)0,1 + ~ej) = 0 in Yf ,

(∇X(j)0,1 + ~ej) · ~ny = 0 on ∂Ym,

where ~ej , j = 1, · · · , n is the unit vector in the j-th coordinate direction, and ~ny

is a unit normal vector on ∂Ym. By Lax-Milgram Theorem [12], X(j)ω,1 for ω ∈ [0, 1]

and j = 1, · · · , n is uniquely solvable. By Theorem 6.30 [12] and (3.5)2 of Lemma

3.5,

‖X(j)ω,1‖W 2,p(Yf )∩W 2,p(Ym) ≤ c(n, Ym) for ω ∈ [0, 1], p ∈ [2,∞). (3.23)

Define Xω,1 ≡ (X(1)ω,1, · · · ,X

(n)ω,1) and Xω,ǫ(x) ≡ ǫXω,1(

xǫ ) for ω ∈ [0, 1], ǫ ∈ (0, 1].

Denote by Ξω for ω ∈ [0, 1] a n × n matrix function whose (i, j)-component is

∂iX(j)ω,1. By remark in pages 17-19, 94-95 [14],

Kω ≡

∫

Yf∪Ym

Kω2,1(I + Ξω)dy for ω ∈ [0, 1] (3.24)

is a symmetric positive definite matrix dependent only on ω. Here I is the identity

matrix. By (3.23), it is not difficult to see, for ω ∈ [0, 1],d3I ≤ Kω ≤ d4I where d3, d4 are positive constants,

Kω is a continuous function of ω.(3.25)

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3.2. Local Lipschitz and local Lp gradient estimates

We have the following Lipschitz estimate:

Lemma 3.7. Suppose ω, ǫ ∈ (0, 1], any solution of−∇ · (Kω2,ǫ∇Φ) = 0 in B1(0) ∩ Ω

Φ = 0 on B1(0) ∩ ∂Ω

satisfies

‖∇Φ‖L∞(B1/2(0)∩Ω) ≤ c‖Kω,ǫΦ‖L2(B1(0)∩Ω), (3.26)


Proof of Lemma 3.7 is given in section 5. Next is the local Lp gradient estimate:

Lemma 3.8. Let ω, ǫ, r ∈ (0, 1], p ∈ (2,∞), and either B2r(x0) ⊂ Ω or x0 ∈ ∂Ω.

Any solution of−∇ · (Kω2,ǫ∇Φ) = 0 in B2r(x0) ∩ Ω

Φ = 0 on B2r(x0) ∩ ∂Ω(3.27)

satisfies

(−

∫

Br/2(x0)

|Kω,ǫ∇Φ|pXΩ dx

)1/p

≤ c

(−

∫

Br(x0)

|Kω,ǫ∇Φ|2XΩ dx

)1/2

, (3.28)

where c is a constant independent of ω, ǫ, r, x0.

Proof. Let c denote a constant independent of ω, ǫ, r, x0.

Case I: For B2r(x0) ⊂ Ω case. By translation, we move x0 to the origin (that

is, x0 = 0 ∈ Ω). Let d ∈ R and ϕ(y) = Φ(ry). By (3.27), we know

−∇ · (Kω2,ǫ/r∇(ϕ+ d)) = 0 in B2(0).

If ǫ/r ≤ 1 (resp. ǫ/r > 1), Lemma 3.7 (resp. Theorem 9.11 [12] and Lemma 3.6)

implies

‖Kω,ǫ/r∇ϕ‖Lp(B1/2(0)) ≤ c‖Kω,ǫ/r(ϕ+ d)‖L2(B1(0)),

where c is also independent of d. By Lemma 3.2,

‖Kω,ǫ/r∇ϕ‖Lp(B1/2(0)) ≤ c‖Kω,ǫ/r∇ϕ‖L2(B1(0)).

Which implies (3.28).

Case II: For x0 ∈ ∂Ω case. Set x0 = 0 ∈ ∂Ω by translation and set ϕ(y) =

Φ(ry). By (3.3) and (3.27),−∇ · (Kω2,ǫ,r∇ϕ) = 0 in B2(0) ∩ ∂Ω/r,

ϕ = 0 on B2(0) ∩ ∂Ω/r.(3.29)

July 9, 2015


If ǫ/r ≤ 1 (resp. ǫ/r > 1), Lemma 3.7 (resp. Theorem 9.13 [12]) implies that the ϕ

in (3.29) satisfies

‖Kω,ǫ,r∇ϕ‖Lp(B1/2(0)∩Ω/r) ≤ c‖Kω,ǫ,rϕ‖L2(B1(0)∩Ω/r).

By Lemma 3.3, we obtain

‖Kω,ǫ,r∇ϕ‖Lp(B1/2(0)∩Ω/r) ≤ c‖Kω,ǫ,r∇ϕ‖L2(B1(0)∩Ω/r).

Which implies (3.28).

4. Proof of main results

Proof of Theorem 2.1: Suppose ω, ǫ ∈ (0, 1], let us find U ∈ H1(Ω) satisfying−∇ · (Kω2,ǫ∇U + Kω,ǫG) = 0 in Ω,

U = 0 on ∂Ω.(4.1)

By Lax-Milgram Theorem [12], U exists uniquely if G ∈ L2(Ω). If we define T :

L2(Ω) → L2(Ω) by TG = Kω,ǫ∇U , then T is a linear and bounded operator on

L2(Ω) by energy method. Lemma 3.8 implies that the operator T satisfies (1.9) of

Theorem 1.3 [22] for any G ∈ Lp(Ω), p ∈ (2,∞). So T is a bounded and linear

operator in Lp(Ω) for p ∈ (2,∞) by Theorem 1.3 [22]. By Poincare inequality [12]

and Lemma 3.1, the solution of (4.1) satisfies

‖U‖Lp(Ωǫf ) ≤ ‖PǫU |Ωǫ

f‖Lp(Ω) ≤ c‖∇PǫU |Ωǫ

f‖Lp(Ω) ≤ c‖∇U‖Lp(Ωǫ

f ),

‖U‖Lp(Ωǫm) ≤ ‖U − PǫU |Ωǫ

f‖Lp(Ωǫ

m) + ‖PǫU |Ωǫf‖Lp(Ωǫ

m)

≤ cǫ‖∇U −∇PǫU |Ωǫf‖Lp(Ωǫ

m) + c‖U‖Lp(Ωǫf ),

where c is independent of ω, ǫ. Function PǫU |Ωǫf

above denotes the extension of

U |Ωǫf

on Ω. So we have

Lemma 4.1. Under A1–A2, if ω, ǫ ∈ (0, 1], p ∈ [2,∞), and G ∈ Lp(Ω), then a

W 1,p(Ω) solution U of (4.1) exists uniquely and‖Kω/ǫ,ǫU,Kω,ǫ∇U‖Lp(Ω) ≤ c‖G‖Lp(Ω) for ω

ǫ ≤ 1,

‖U,Kω,ǫ∇U‖Lp(Ω) ≤ c‖G‖Lp(Ω) for ωǫ ≥ 1,


By a duality argument, Poincare inequality [12], and Lemmas 3.1, 4.1, we have

Lemma 4.2. Under A1–A2, if ω, ǫ ∈ (0, 1], p ∈ (1, 2], and G ∈ Lp(Ω), then a

W 1,p(Ω) solution U of (4.1) exists uniquely and‖Kω/ǫ,ǫU,Kω,ǫ∇U‖Lp(Ω) ≤ c‖G‖Lp(Ω) for ω

ǫ ≤ 1,

‖U,Kω,ǫ∇U‖Lp(Ω) ≤ c‖G‖Lp(Ω) for ωǫ ≥ 1,


July 9, 2015


If ω, ǫ ∈ (0, 1] and G,F ∈ L∞(Ω), then the H1(Ω) solution of−∇ · (Kω2,ǫ∇U) = F in Ω

U = 0 on ∂Ω(4.2)

and the H1(Ω) solution of−∇ · (Kω2,ǫ∇ϕ− Kω,ǫG) = 0 in Ω

ϕ = 0 on ∂Ω(4.3)

exist uniquely by Lax-Milgram Theorem [12]. Lemma 4.1 and Lemma 4.2 imply

that the solution of (4.3) satisfies‖Kω/ǫ,ǫϕ,Kω,ǫ∇ϕ‖Lr(Ω) ≤ c‖G‖Lr(Ω) for ω

ǫ ≤ 1,

‖ϕ,Kω,ǫ∇ϕ‖Lr(Ω) ≤ c‖G‖Lr(Ω) for ωǫ ≥ 1,

(4.4)

where r ∈ (1,∞) and c is a constant independent of ω, ǫ. Multiply (4.2) by the

solution of (4.3), multiply (4.3) by the solution of (4.2), integrate by part, as well

as employ (4.4), Lemma 3.1, and Holder inequality to get∫

Ω

Kω,ǫ∇U Gdx =

∫

Ω

ϕ Fdy =

∫

Ω

Pǫϕ|ΩǫfFdy +

∫

Ωǫm

(ϕ− Pǫϕ|Ωǫf)Fdy

≤ c‖G‖Lr(Ω)(‖F‖W−1,p(Ω) + ω−1‖F‖W−1,p(Ωǫm)),

where 1r + 1

p = 1 and c is independent of ω, ǫ. Since L∞(Ω) is dense in Lr(Ω) for

any r ∈ (1,∞), we obtain

‖Kω,ǫ∇U‖Lp(Ω) ≤ c(‖F‖W−1,p(Ω) + ω−1‖F‖W−1,p(Ωǫm)),

where 1r + 1

p = 1 and c is a constant independent of ω, ǫ. By Poincare inequality

[12] and Lemma 3.1, it is easy to see that‖Kω/ǫ,ǫU‖Lp(Ω) ≤ c(‖F‖W−1,p(Ω) + ω−1‖F‖W−1,p(Ωǫ

m)) for ωǫ ≤ 1,

‖U‖Lp(Ω) ≤ c(‖F‖W−1,p(Ω) + ω−1‖F‖W−1,p(Ωǫm)) for ω

ǫ ≥ 1,

where p ∈ (1,∞) and c is a constant independent of ω, ǫ. Together with Lemma

4.1 and Lemma 4.2, we see that Theorem 2.1 holds for G ∈ Lp(Ω), F ∈ L∞(Ω). If

G ∈ Lp(Ω), F ∈ W−1,p(Ω), Theorem 2.1 can be proved by a limiting argument.

Proof of Theorem 2.2: Suppose G,F ∈ Lp(Ωǫf ) for p ∈ (1,∞), let us do zero

extension for G,F . That is, set G ≡

G on Ωǫ

f

0 on Ωǫm

and F ≡

F on Ωǫ

f

0 on Ωǫm

. Then

G, F ∈ Lp(Ω). Let Uω,ǫ denote the solution of (1.1) with G,F replaced by G, F

above. By Theorem 2.1,‖Kω/ǫ,ǫUω,ǫ,Kω,ǫ∇Uω,ǫ‖Lp(Ω) ≤ c(‖G‖Lp(Ω) + ‖F‖W−1,p(Ω)) for ω

ǫ ≤ 1,

‖Uω,ǫ,Kω,ǫ∇Uω,ǫ‖Lp(Ω) ≤ c(‖G‖Lp(Ω) + ‖F‖W−1,p(Ω)) for ωǫ ≥ 1,

(4.5)

where c is independent of ω, ǫ. If we fix ǫ, then we see, by (4.5) and Lemma 3.1,

July 9, 2015


• There is a subsequence of Uω,ǫ (same notation for subsequence) such that

Uω,ǫ|Ωǫf

converges weakly to U in W 1,p(Ωǫf ) as ω → 0.

• The limit function U satisfies (2.2) and (2.3).

So Theorem 2.2 is proved for G,F ∈ Lp(Ωǫf ), p ∈ (1,∞). For general case, Theorem

2.2 can be proved by a limiting argument.

Based on the above uniform results (that is, Theorem 2.1, Theorem 2.2), we then

apply semigroup theory to obtain the uniform estimates for parabolic equations.

Proof of Theorem 2.3: By A1–A2, A5 as well as by tracing the proof of

Theorem 2.1 [23], we know the solution of (1.2) exists uniquely and satisfies, for

p ∈ (n,∞),

‖U‖C1([0,T ];Lp(Ω)) + ‖U‖C([0,T ];Bp) ≤ c(‖U0‖Bp + ‖F‖Cσ([0,T ];Lp(Ω))

),

where c is a constant independent of ω, ǫ. (1.2) can be written as, for fixed t ∈ (0, T ],

−∇ · (Kω2,ǫ∇U(·, t)) = F (·, t) − ∂tU(·, t) in Ω,

U(·, t) = 0 on ∂Ω.

By Theorem 2.1, we see, for p ∈ (n,∞),

‖Kω,ǫ∇U(·, t)‖Lp(Ω) ≤ c(‖U0‖Bp + ‖F‖Cσ([0,T ];Lp(Ω))

),

where c is a constant independent of ω, ǫ. So Theorem 2.3 is proved.

Proof of Theorem 2.4: By A1–A2, A6 as well as by tracing the proof of

Theorem 2.1 [23], we know that the solution of (2.4) exists uniquely and satisfies,

for p ∈ (n,∞),

‖U‖C1([0,T ];Lp(Ωǫf )) + ‖∆U‖C([0,T ];Lp(Ωǫ

f )) ≤ c(‖U0‖Dp + ‖F‖Cσ([0,T ];Lp(Ωǫ

f ))

),

where c is a constant independent of ǫ. (2.4) can be written as, for fixed t ∈ (0, T ],

−∆U(·, t) = F (·, t) − ∂tU(·, t) in Ωǫf ,

∇U(·, t) · ~nǫ = 0 on ∂Ωǫm,

U(·, t) = 0 on ∂Ω.

By Theorem 2.2, we see, for p ∈ (n,∞),

‖∇U‖C([0,T ];Lp(Ωǫf )) ≤ c

(‖U0‖Dp + ‖F‖Cσ([0,T ];Lp(Ωǫ

f ))

),

where c is a constant independent of ǫ. So Theorem 2.4 is proved.

5. Local uniform Lipschitz estimate

We prove a local Lipschitz estimate, that is, Lemma 3.7. The idea of proof follows

from the arguments in [4]. We first derive local Holder estimate in subsection 5.1

and then derive local Lipschitz estimate in subsection 5.2.

July 9, 2015


5.1. Holder estimate

An open set O ⊂ Rn with boundary ∂O is said to satisfy a uniform exterior ball

condition, if there exists a r > 0 with the following property: For each x ∈ ∂O,

there exists a point y = y(x) ∈ Rn such that Br(y) \ x ⊂ Rn \O and x ∈ ∂Br(y).

If O ⊂ Rn is a nonempty open bounded Lipschitz set and satisfy a uniform exterior

ball condition, then O is called a semiconvex domain.

In this subsection, we assume (1) A2 holds and (2) O is a semiconvex domain.

If 0 ∈ ∂O, by rotation, there is a Lipschitz function Ψ : Rn−1 → R such that

Ψ(0) = 0,

B1(0) ∩ O/r = B1(0) ∩ (x′, xn) ∈ Rn : rxn > Ψ(rx′) if r ∈ (0, 1].(5.1)

If r = 0, we define B1(0) ∩ O/r ≡ B1(0) ∩ (x′, xn) ∈ Rn : xn > 0. Similar to Ωǫf

and Ωǫm, one can also define analogous Oǫ

f and Oǫm. Define Kν,ǫ,r as

Kν,ǫ,r ≡

1 in Oǫ

f/r

ν in Oǫm/r

for ν, ǫ, r ∈ (0, 1].

Lemma 5.1. Let ω, ǫ, r ∈ (0, 1], ǫ ≤ r, either B2(0) ⊂ O/r or 0 ∈ ∂O/r, and ϕ be

a solution of−∇ · (Kω2,ǫ,r∇ϕ) = 0 in B2(0) ∩ O/r,

ϕ = 0 on B2(0) ∩ ∂O/r.

There is a constant c independent of ω, ǫ, r such that

‖ϕ‖H1(B1/2(0)∩O/r) ≤ c‖Kω,ǫ,rϕ‖L2(B2(0)∩O/r).

Proof. Let c denote a constant independent of ω, ǫ, r. By energy method,

‖Kω,ǫ,r∇ϕ‖L2(B1(0)∩O/r) ≤ c‖Kω,ǫ,rϕ‖L2(B2(0)∩O/r). (5.2)

For any z ∈ B1(0) ∩ O/r, we move z to the origin of the coordinate system by

translation and define

K(x) ≡ Kω2,ǫ,r(ǫrx)

ϕ(x) ≡ ϕ( ǫrx)

for x ∈ B1(z) ∩O/ǫ.

Then ϕ satisfies−∇ · (K∇ϕ) = 0 in B1(z) ∩ O/ǫ,

ϕ = 0 on B1(z) ∩ ∂O/ǫ.

By (3.5)1,

‖∇ϕ‖L2(B1/2(z)∩O/ǫ) ≤ c‖ϕ‖L2(B1(z)∩Oǫf /ǫ). (5.3)

By Poincare inequality [12], (5.3) implies

‖∇ϕ‖2L2(Bǫ/2r(z)∩O/r) ≤ c‖∇ϕ‖2

L2(Bǫ/r(z)∩Oǫf /r). (5.4)

July 9, 2015


By covering B1(0) ∩ O/r with a finite number of balls of radius ǫ/2r, (5.4) implies

‖∇ϕ‖L2(B1/2(0)∩O/r) ≤ c‖∇ϕ‖L2(B1(0)∩Oǫf/r).

Together with (5.2) and Poincare inequality [12], we prove the lemma.

The rest of this subsection is to prove the following lemma.

Lemma 5.2. Suppose δ > 0 and ω, ǫ ∈ (0, 1], any solution of

−∇ · (Kω2,ǫ,1∇Φ) = 0 in B1(0) ∩ O

Φ = 0 on B1(0) ∩ ∂O

satisfies

‖Φ‖C0,µ(B1/2(0)∩O) ≤ c‖Kω,ǫ,1Φ‖L2(B1(0)∩O), (5.5)

where µ ≡ δn+δ and c is a constant independent of ω, ǫ.

The interior estimate of (5.5) is given in subsection 5.1.1 and the boundary

estimate of (5.5) is in subsection 5.1.2.

5.1.1. Interior Holder estimate

Assume B1(0) ⊂ O.

Lemma 5.3. For any δ > 0, there are constants θ1, θ2 ∈ (0, 1) with θ1 < θ22 and a

constant ǫ0 ∈ (0, 1) (depending on δ, θ2, Yf ) such that if

−∇ · (Kω2,ν,1∇ϕ) = 0 in B1(0),

‖Kω,ν,1ϕ‖L2(B1(0)) ≤ 1,(5.6)

then, for any ω ∈ (0, 1], ν ∈ (0, ǫ0], and θ ∈ [θ1, θ2],

−

∫

Bθ(0)

∣∣ϕ− (ϕ)Bθ(0)

∣∣2 dx ≤ θ2µ, (5.7)

where µ ≡ δn+δ .

Proof. Consider the following problem

−∇ · (Kω∗∇ϕ∗) = 0 in B2/3(0), (5.8)

where Kω∗for ω∗ ∈ [0, 1] is from (3.24). By Theorem 1.2 in page 70 [11] and (3.25),

there is a small θ ∈ (0, 2/3) such that

−

∫

Bθ(0)

∣∣ϕ∗ − (ϕ∗)Bθ(0)

∣∣2dx ≤ θ2µ′

−

∫

B2/3(0)

|ϕ∗|2dx, (5.9)

July 9, 2015


for some µ′ ∈ (µ, 1). We choose θ1, θ2 ∈ (0, 2/3) such that θ1 < θ22 and (5.9) holds

if θ ∈ [θ1, θ2]. Now we claim (5.7). If not, there is a sequence ων , θν , ϕν satisfying

(5.6) and

ων → ω∗ ∈ [0, 1]

θν → θ∗ ∈ [θ1, θ2]

−

∫

Bθν (0)

∣∣ϕν − (ϕν)Bθν (0)

∣∣2 dx > θ2µν

as ν → 0. (5.10)

By Lemma 5.1 and tracing the proof of Theorem 2.3 [3], there is a subsequence

(same notation for subsequence) such thatϕν → ϕ∗ in L2(B2/3(0)) strongly

Kω2ν ,ν,1∇ϕν → Kω∗

∇ϕ∗ in L2(B2/3(0)) weaklyas ν → 0. (5.11)

Also the ϕ∗ above is a solution of (5.8). By (5.9)–(5.11), we conclude

θ2µ∗ = lim

ν→0θ2µ

ν ≤ limν→0

−

∫

Bθν (0)

∣∣ϕν − (ϕν)Bθν (0)

∣∣2 dx

= −

∫

Bθ∗(0)

|ϕ∗ − (ϕ∗)Bθ∗ (0)|2dx ≤ θ2µ′

∗ −

∫

B2/3(0)

|ϕ∗|2dx. (5.12)

But (5.12) is impossible if θ2 is small enough. Therefore, there is a ǫ0 such that

(5.7) holds for ν ≤ ǫ0.

Lemma 5.4. For any δ > 0, there are constants θ1, θ2 ∈ (0, 1) with θ1 < θ22 and a

constant ǫ0 ∈ (0, 1) (depending on δ, θ2, Yf ) such that if

−∇ · (Kω2,ǫ,1∇Φ) = 0 in B1(0), (5.13)

then, for any ω ∈ (0, 1], ǫ ∈ (0, ǫ0], θ ∈ [θ1, θ2], and k satisfying ǫ/θk ≤ ǫ0,

−

∫

Bθk (0)

∣∣∣Φ − (Φ)Bθk (0)

∣∣∣2

dx ≤ θ2kµ|Jω,ǫ|2, (5.14)

where µ ≡ δn+δ and Jω,ǫ ≡ ‖Kω,ǫ,1Φ‖L2(B1(0)).

Proof. The proof is done by induction on k. For k = 1, we define ϕ ≡ Φ/Jω,ǫ

and use Lemma 5.3 with ν = ǫ to obtain (5.14). Suppose (5.14) holds for some k

satisfying ǫ/θk ≤ ǫ0, then we define

ϕ(x) ≡ J−1ω,ǫ θ

−kµ(Φ(θkx) − (Φ)B

θk (0)

)in B1(0).

Then ϕ satisfies (5.6) with ν = ǫ/θk. By changing variable and employing Lemma

5.3, we obtain (5.14) with k + 1 in place of k.

Lemma 5.5. For any δ > 0, there is a constant ǫ∗ ∈ (0, 1) (depending on δ, Yf)

such that if ω ∈ (0, 1] and ǫ ∈ (0, ǫ∗], then any solution of (5.13) satisfies

[Φ]C0,µ(B1/2(0))≤ c‖Kω,ǫ,1Φ‖L2(B1(0)), (5.15)

July 9, 2015


where c is a constant independent of ω, ǫ. See Lemma 5.4 for µ.

Proof. Let θ1, θ2, ǫ0, Jω,ǫ be same as those in Lemma 5.4 and define ǫ∗ ≡ ǫ0θ2/2.

Denote by c a constant independent of ω, ǫ. Because of θ1 < θ22, for any r ∈ [ǫ/ǫ0, θ2],

there are θ ∈ [θ1, θ2] and k ∈ N satisfying r = θk. Lemma 5.4 implies

−

∫

Br(0)

∣∣Φ − (Φ)Br(0)

∣∣2 dy ≤ cr2µ|Jω,ǫ|2 for r ∈ [ǫ/ǫ0, θ2]. (5.16)

Take r = 2ǫǫ0

and define

ϕ(y) ≡ J−1ω,ǫǫ

−µ(Φ(ǫy) − (Φ)B2ǫ/ǫ0

(0)

)in B2/ǫ0(0).

Then ϕ satisfies−∇ · (Kω2,1,1∇ϕ) = 0 in B2/ǫ0(0),

‖ϕ‖L2(B2/ǫ0(0)) ≤ c.

(3.5)1 of Lemma 3.5 implies [ϕ]C0,µ(B1/ǫ0(0)) ≤ c. Together with (5.16), then (5.16)

holds for r ∈ (0, θ2). Next we shift the origin of the coordinate system to any point

z ∈ B1/2(0) and repeat above argument to see that (5.16) with 0 replaced by any

z ∈ B1/2(0) also holds for r ∈ (0, θ2). By Theorem 1.2 in page 70 [11], we prove the

lemma.

Remark 5.1. Let ǫ∗ be same as that in Lemma 5.5. By (3.5)1 of Lemma 3.5,

we know that if ω ∈ (0, 1] and ǫ ∈ [ǫ∗, 1], any solution of (5.13) satisfies (5.15).

Together with Lemma 5.5, we conclude that any solution of (5.13) satisfies (5.15)

if ω, ǫ ∈ (0, 1].

5.1.2. Boundary Holder estimate

We assume 0 ∈ ∂O.

Lemma 5.6. For any δ > 0, there are constants θ1, θ2 ∈ (0, 1) with θ1 < θ22, and a

constant ǫ0 ∈ (0, 1) (depending on δ, θ2, Yf , ‖∇Ψ‖L∞(Rn−1)) such that if

−∇ · (Kω2,ǫ,r∇ϕ) = 0 in B1(0) ∩ O/r,

ϕ = 0 on B1(0) ∩ ∂O/r,

‖Kω,ǫ,rϕ‖L2(B1(0)∩O/r) ≤ 1,

(5.17)

then, for any ω, ǫ, r ∈ (0, 1], ǫ/r ≤ ǫ0, and θ ∈ [θ1, θ2],

−

∫

Bθ(0)∩O/r

|ϕ|2dx ≤ θ2µ, (5.18)

where µ ≡ δn+δ . See (5.1) for Ψ.

July 9, 2015


Proof. Consider the following problem−∇ · (Kω∗

∇ϕ∗) = 0 in B2/3(0) ∩ O/r∗,

ϕ∗ = 0 on B2/3(0) ∩ ∂O/r∗,(5.19)

where Kω∗is from (3.24) and ω∗, r∗ ∈ [0, 1]. Note O/r∗ is a semiconvex domain

(see the beginning of subsection 5.1) as well as Kω∗is a symmetric positive definite

matrix and can be diagonalizable. By Theorem 4.5 [20], Theorem 7.26 [12], and

(3.25), there is a small θ ∈ (0, 2/3) such that

−

∫

Bθ(0)∩O/r∗

|ϕ∗|2dx ≤ θ2µ′

−

∫

B2/3(0)∩O/r∗

|ϕ∗|2dx, (5.20)

where µ′ ∈ (µ, 1). We choose θ1, θ2 ∈ (0, 2/3) such that θ1 < θ22 and (5.20) holds if

θ ∈ [θ1, θ2]. Now we claim (5.18). If not, there is a sequence ωǫ, rǫ, θǫ, ϕǫ satisfying

(5.17) and

ωǫ, rǫ → ω∗, r∗ ∈ [0, 1]

θǫ → θ∗ ∈ [θ1, θ2]

−

∫

Bθǫ(0)∩O/rǫ

|ϕǫ|2dx > θ2µ

ǫ

as ǫ/rǫ → 0. (5.21)

By Lemma 5.1 and tracing the proof of Theorem 2.3 [3], there is a subsequence

(same notation for subsequence) such thatϕǫ → ϕ∗ in L2(B2/3(0) ∩ O/r∗) strongly

Kω2ǫ ,ǫ,rǫ

∇ϕǫ → Kω∗∇ϕ∗ in L2(B2/3(0) ∩ O/r∗) weakly

as ǫ/rǫ → 0. (5.22)

Also the ϕ∗ above is a solution of (5.19). By (5.20)–(5.22), we conclude

θ2µ∗ = lim

ǫ/rǫ→0θ2µ

ǫ ≤ limǫ/rǫ→0

−

∫

Bθǫ(0)∩O/rǫ

|ϕǫ|2dx

= −

∫

Bθ∗(0)∩O/r∗

|ϕ∗|2dx ≤ θ2µ′

∗ −

∫

B2/3(0)∩O/r∗

|ϕ∗|2dx. (5.23)

But (5.23) is impossible if θ2 is small enough. So there is a ǫ0 such that (5.18) holds

for ǫ/r < ǫ0.

Lemma 5.7. For any δ > 0, there are constants θ1, θ2 ∈ (0, 1) with θ1 < θ22, and a

constant ǫ0 ∈ (0, 1) (depending on δ, θ2, Yf , ‖∇Ψ‖L∞(Rn−1)) such that if−∇ · (Kω2,ǫ,1∇Φ) = 0 in B1(0) ∩ O,

Φ = 0 on B1(0) ∩ ∂O,(5.24)

then, for any ω ∈ (0, 1], ǫ ∈ (0, ǫ0], θ ∈ [θ1, θ2], and k satisfying ǫ/θk ≤ ǫ0,

−

∫

Bθk (0)∩O

|Φ|2dx ≤ θ2kµ|Jω,ǫ|2, (5.25)

July 9, 2015


where µ ≡ δn+δ and Jω,ǫ ≡ ‖Kω,ǫ,1Φ‖L2(B1(0)∩O).

Proof. The proof is done by induction on k. For k = 1, we set ϕ ≡ Φ/Jω,ǫ. Then

(5.25) is deduced from Lemma 5.6 with r = 1. Suppose (5.25) holds for some k

satisfying ǫ/θk ≤ ǫ0, then we define

ϕ(x) ≡ J−1ω,ǫθ

−kµΦ(θkx) in B1(0) ∩ O/θk.

Then ϕ satisfies (5.17) with r = θk. By changing variable and employing Lemma

5.6 with r = θk, we obtain (5.25) with k + 1 in place of k.

Lemma 5.8. For any δ > 0, there is a constant ǫ∗ ∈ (0, 1) (depending on δ, Yf ,

‖∇Ψ‖L∞(Rn−1)) such that if ω ∈ (0, 1] and ǫ ∈ (0, ǫ∗], then any solution of (5.24)

satisfies

[Φ]C0,µ(B1/2(0)∩O) ≤ c‖Kω,ǫ,1Φ‖L2(B1(0)∩O), (5.26)

where c is a constant independent of ω, ǫ. See Lemma 5.7 for µ.

Proof. Let θ1, θ2, ǫ0, Jω,ǫ be those in Lemma 5.7 and define ǫ∗ ≡ minǫ0θ2/3, ǫ∗

where ǫ∗ is the one in Lemma 5.5. Denote by c a constant independent of ω, ǫ. For

any x ∈ Bθ2/3(0) ∩ O, define ξ(x) ≡ |x − x0| where x0 ∈ ∂O satisfying |x − x0| =

miny∈∂O |x− y|. Then we have either case (1) ξ(x) > 2ǫ3ǫ0

or case (2) ξ(x) ≤ 2ǫ3ǫ0

.

Let us consider case (1). Because of θ1 < θ22, for any r ∈ [ǫ/ǫ0, θ2], there are

θ ∈ [θ1, θ2] and k ∈ N satisfying r = θk. Since ξ(x) ∈ [ 2ǫ3ǫ0, θ2

3 ], by Lemma 5.7,

−

∫

Br(x0)∩O

|Φ|2dy ≤ r2µ|Jω,ǫ|2 for r ∈ [32ξ(x), θ2].

So, for s ∈ [ ξ(x)2 , θ2

3 ],

−

∫

Bs(x)∩O

∣∣Φ − (Φ)Bs(x)∩O

∣∣2 dy ≤ cs2µ|Jω,ǫ|2. (5.27)

Next we move the origin of the coordinate system to x and define

ϕ(y) ≡ J−1ω,ǫξ

−µ(x)(Φ(ξ(x)y) − (Φ)Bξ(x)(x)

)in B1(x).

Then ϕ satisfies

−∇ · (Kω2,ǫ/ξ(x),1∇ϕ) = 0 in B1(x). (5.28)

Take s = ξ(x) < 1 in (5.27) to see ‖ϕ‖L2(B1(x)) ≤ c. Apply Remark 5.1 to (5.28) to

obtain [ϕ]C0,µ(B1/2(x)) ≤ c. Which implies

−

∫

Bs(x)

∣∣Φ − (Φ)Bs(x)

∣∣2 dy ≤ cs2µ|Jω,ǫ|2 for s < ξ(x)

2 . (5.29)

July 9, 2015


Next we consider case (2). Because of θ1 < θ22 , for any r ∈ [ǫ/ǫ0, θ2], there are

θ ∈ [θ1, θ2] and k ∈ N satisfying r = θk. By Lemma 5.7,

−

∫

Br(x0)∩O

|Φ|2dy ≤ r2µ|Jω,ǫ|2 for r ∈ [ǫ/ǫ0, θ2]. (5.30)

This implies, for s ∈ [ ǫ3ǫ0, θ2

3 ],

−

∫

Bs(x)∩O

∣∣Φ − (Φ)Bs(x)∩O

∣∣2 dy ≤ cs2µ|Jω,ǫ|2. (5.31)

Again we move the origin of the coordinate system to x and define

ϕ(y) ≡ J−1ω,ǫǫ

−µ(Φ(ǫy) − (Φ)Bǫ/ǫ0

(x)∩O

)in B1/ǫ0(x) ∩ O/ǫ.

Then ϕ satisfies−∇ · (Kω2,ǫ,ǫ∇ϕ) = 0 in B1/ǫ0(x) ∩ O/ǫ,

ϕ = −J−1ω,ǫǫ

−µ(Φ)Bǫ/ǫ0(x)∩O on B1/ǫ0(x) ∩ ∂O/ǫ.

Let us take s = ǫ/ǫ0 in (5.31) to see ‖ϕ‖L2(B1/ǫ0(x)∩O/ǫ) ≤ c and take s = ǫ/ǫ0 in

(5.30) to see |J−1ω,ǫǫ

−µ(Φ)Bǫ/ǫ0(x)∩O| ≤ c. By (3.5)1 of Lemma 3.5,

[ϕ]C0,µ(B1/2ǫ0(x)∩O/ǫ) ≤ c. (5.32)

(5.32) implies that (5.31) holds for s ≤ ǫ2ǫ0

.

The Holder estimate of Φ follows from (5.27), (5.29), (5.31), (5.32), and Theorem

1.2 in page 70 [11].

Remark 5.2. Let ǫ∗ be same as that in Lemma 5.8. By (3.5)1 of Lemma 3.5, we

know that if ω ∈ (0, 1] and ǫ ∈ [ǫ∗, 1], any solution of (5.24) satisfies (5.26). Together

with Lemma 5.8, any solution of (5.24) satisfies (5.26) if ω, ǫ ∈ (0, 1].

Remark 5.1, Remark 5.2, and maximal principle imply Lemma 5.2.

5.2. Lipschitz estimate

A1–A2 are assumed and we show Lemma 3.7. The interior estimate of (3.26) is in

subsection 5.2.1 and the boundary estimate of (3.26) is in subsection 5.2.2.

5.2.1. Interior gradient estimate

We assume B1(0) ⊂ Ω.

Lemma 5.9. There exist θ, ǫ0 ∈ (0, 1) such that if−∇ · (Kω2,ν∇ϕ) = 0 in B1(0),

‖Kω,νϕ‖L2(B1(0)) ≤ 1,(5.33)

July 9, 2015


then, for any ω ∈ (0, 1] and ν ∈ (0, ǫ0),

supBθ(0)

|ϕ(x) − ϕ(0) − (x+ Xω,ν(x))bω,ν | ≤ θ4/3, (5.34)

where bω,ν ≡ K−1ω −

∫

Bθ(0)

Kω2,ν∇ϕ dx and K−1ω is the inverse matrix of Kω. See

(3.23)–(3.24) for Xω,ν ,Kω.

Proof. Assume −∇ · (Kω∗∇ϕ∗) = 0 in B2/3(0), where Kω∗

for ω∗ ∈ [0, 1] is from

(3.24). If θ is small enough, by (3.25) and Taylor expansion,

supBθ(0)

∣∣ϕ∗(x) − ϕ∗(0) − x(∇ϕ∗)Bθ(0)

∣∣ ≤ θ3/2‖ϕ∗‖L2(B2/3(0)). (5.35)

Fix a small θ so that (5.35) holds and we prove (5.34) by contradiction. If not, there

is a sequence ων, ϕν satisfying (5.33) and, as ν → 0,ων → ω∗ ∈ [0, 1],

supBθ(0)

|ϕν(x) − ϕν(0) − (x+ Xω,ν(x))bω,ν | > θ4/3. (5.36)

By Lemma 5.1 and Lemma 5.2 and by tracing the proof of Theorem 2.3 [3], there

is a subsequence (same notation for subsequence) such thatϕν → ϕ∗ in C(B2/3(0))

Kω2ν ,ν∇ϕν → Kω∗

∇ϕ∗ in L2(B2/3(0)) weaklyas ν → 0. (5.37)

(5.37) implies that ϕ∗ satisfies −∇·(Kω∗∇ϕ∗) = 0 in B2/3(0). Together with (5.35),

(5.36), and (5.37), we get contradiction if θ is small. So we prove (5.34).

Lemma 5.10. There exist θ, ǫ0 ∈ (0, 1) such that if Φ satisfies

−∇ · (Kω2,ǫ∇Φ) = 0 in B1(0), (5.38)

then, for any ω ∈ (0, 1], ǫ ∈ (0, ǫ0), and k satisfying ǫ/θk ≤ ǫ0, there are constants

aω,ǫk , b

ω,ǫk satisfying|aω,ǫ

k | + |bω,ǫk | ≤ cJω,ǫ,

supB

θk (0)

|Φ(x) − Φ(0) − ǫaω,ǫk − (x + Xω,ǫ(x))b

ω,ǫk | ≤ θ4k/3Jω,ǫ,

(5.39)

where Jω,ǫ ≡ ‖Kω,ǫΦ‖L2(B1(0)) and c is independent of ω, ǫ.

Proof. If ϕ ≡ Φ/Jω,ǫ, then it satisfies (5.33) with ν = ǫ. By Lemma 5.9, we obtain

(5.39) for k = 1 where aω,ǫ1 = 0, b

ω,ǫ1 = K−1

ω −

∫

Bθ(0)

Kω2,ǫ∇Φ dx. If (5.39) holds for

some k satisfying ǫ/θk ≤ ǫ0, we define

ϕ(x) ≡Φ(θkx) − Φ(0) − ǫaω,ǫ

k −(θkx+ Xω,ǫ(θ

kx))b

ω,ǫk

θ4k/3Jω,ǫ

in B1(0).

July 9, 2015


By induction and (3.22), we see ϕ satisfies (5.33) with ν = ǫ/θk. Apply Lemma 5.9

to obtain

supBθ(0)

∣∣∣ϕ(x) − ϕ(0) −(x+ Xω,ǫ/θk(x)

)bω,ǫ/θk

∣∣∣ ≤ θ4/3, (5.40)

where bω,ǫ/θk ≡ K−1ω −

∫

Bθ(0)

Kω2,ǫ/θk∇ϕ dx. (5.40) can be written as

supBθ(0)

∣∣∣Φ(θkx) − Φ(0) + ǫXω,1(0)bω,ǫk −

(θkx+ Xω,ǫ(θ

kx))b

ω,ǫk

−Jω,ǫθ4k/3

(x+ θ−kXω,ǫ(θ

kx))bω,ǫ/θk

∣∣∣ ≤ Jω,ǫθ4(k+1)/3. (5.41)

Define

aω,ǫk+1 ≡ −Xω,1(0)bω,ǫ

k and bω,ǫk+1 ≡ b

ω,ǫk + Jω,ǫθ

k/3bω,ǫ/θk . (5.42)

By energy method, we know that |bω,ǫ/θk | is bounded uniformly in ω, ǫ, k. So (5.39)1holds. Substituting (5.42) into (5.41) and changing variables, we obtain (5.39)2 for

k + 1 case.

Lemma 5.11. There exists ǫ0 ∈ (0, 1) such that if ω ∈ (0, 1] and ǫ ∈ (0, ǫ0), any

solution of (5.38) satisfies

‖∇Φ‖L∞(B1/2(0)) ≤ c‖Kω,ǫΦ‖L2(B1(0)), (5.43)


Proof. Let Jω,ǫ be that in Lemma 5.10; c denote a constant independent of ω, ǫ;

and k ∈ N satisfying ǫ/θk ≤ ǫ0 < ǫ/θk+1. By Lemma 5.10,

supB ǫ

ǫ0(0)

|Φ(x) − Φ(0) − ǫaω,ǫk − (x+ Xω,ǫ(x))b

ω,ǫk | ≤ c

∣∣ ǫǫ0

∣∣4/3Jω,ǫ.

Define

ϕ(x) ≡Φ(ǫx) − Φ(0) − ǫaω,ǫ

k −(ǫx+ Xω,ǫ(ǫx)

)b

ω,ǫk

ǫ4/3Jω,ǫ

in B 1ǫ0

(0).

Then ϕ satisfies−∇ · (Kω2,1∇ϕ) = 0 in B1/ǫ0(0),

‖ϕ‖L∞(B1/ǫ0(0)) ≤ c.

(3.5)2 of Lemma 3.5 implies

‖ϕ‖C1,0(B1/2ǫ0(0)∩Ωǫ

f /ǫ)∩C1,0(B1/2ǫ0(0)∩Ωǫ

m/ǫ) ≤ c. (5.44)

Since ∇ϕ(x) =∇Φ(ǫx)−(I+∇Xω,1(x))bω,ǫ

k

ǫ1/3Jω,ǫ, |∇Φ(ǫx)| ≤ cJω,ǫ for x ∈ B1/2ǫ0(0) by

(3.23), (5.44), and Lemma 5.10. We prove (5.43).

Remark 5.3. Let ǫ0 be same as that in Lemma 5.11. By (3.5)2 of Lemma 3.5, we

know that if ω ∈ (0, 1] and ǫ ∈ [ǫ∗, 1], any solution of (5.38) satisfies (5.43). Together

with Lemma 5.11, any solution of (5.38) satisfies (5.43) if ω, ǫ ∈ (0, 1].

July 9, 2015


5.2.2. Boundary gradient estimate

Assume 0 ∈ ∂Ω and (3.2). Let Qd(0) ≡ Πni=1[−di, di] denote a cube, where d =

(d1, · · · , dn), di ∈ [3, 4]. Find a smooth cut-off function ρ ∈ C∞0 (Qd(0)) such that

ρ ∈ [0, 1] and ρ = 1 in Q2(0). For any ω, ǫ, r ∈ (0, 1] and ǫ ≤ r, we find W(n)ω,ǫ,r ∈

H1(Qd(0) ∩ Ω/r) by solving−∇ ·

(Kω2,ǫ,r

(∇W

(n)ω,ǫ,r + ~en

))= 0 in Qd(0) ∩ Ω/r,

W(n)ω,ǫ,r =

(1 − ρ

)X

(n)ω,ǫ/r on ∂(Qd(0) ∩ Ω/r),

(5.45)

where ~en is the unit vector in the n-th coordinate direction. See (3.22) for X(n)ω,ǫ/r

and (3.3) for Kω2,ǫ,r. We adjust the d of Qd(0) so that if ǫ(Ym + j) ⊂ Ωǫm for any

j ∈ Zn, then |Qd(0) ∩ ǫr (Y + j)| is either 0 or | ǫ

r |n. Define

D ≡ Qd(0) ∩ Ω/r,

D∗ ≡⋃

j∈Znǫr

(Ym+j)⊂Qd(0)∩Ωǫm/r

ǫr (Y + j). (5.46)

From (5.46)1, the definition of Ωǫm, and A1, we see

ρ = 0 on ∂D ∩ ∂D∗,

D \ D∗ ⊂ Ωǫf/r,

D \ D∗ ⊂ x ∈ D : dist(x, ∂Ω/r) ≤ c ǫr,

D is a simply-connected semiconvex domain,

(5.47)

where c is a constant independent of ǫ, r. See the beginning of subsection 5.1 for a

semiconvex domain. Let Gǫ,r(x, y) denote the Green’s function of−∇y ·

(Kω2,ǫ,r∇yGǫ,r(x, ·)

)= δ(x, ·) in D,

Gǫ,r(x, ·) = 0 on ∂D.(5.48)

By [16], Gǫ,r(x, ·) ∈W 1,1(D) exists uniquely. Next we give a local L∞ estimate.

Lemma 5.12. If x∗ ∈ D, ω, ǫ, r ∈ (0, 1], ǫ ≤ r, and t > 0, any solution of−∇ · (Kω2,ǫ,r∇ϕ) = 0 in Bt(x

∗) ∩ D

ϕ = 0 on Bt(x∗) ∩ ∂D

(5.49)

satisfies

∣∣∣Kω,ǫ,rϕ∣∣∣ (x∗) ≤ c

∣∣∣∣ −∫

Bt(x∗)∩D

|Kω,ǫ,rϕ(y)|2dy

∣∣∣∣1/2

, (5.50)

for some constant c independent of ω, ǫ, r, x∗, t.

Proof. First we assume x∗ = 0 ∈ D and define ϕ(y) = ϕ(ty). Then (5.49) implies−∇ · (Kω2,ǫ,rt∇ϕ) = 0 in B1(0) ∩ D/t,

ϕ = 0 on B1(0) ∩ ∂D/t.

July 9, 2015


Note ǫrt ≤ 1 or 1 < ǫ

rt . If ǫrt ≤ 1 (resp. 1 < ǫ

rt), Lemma 5.2 and (5.47)4 (resp.

Theorem 7.26 [12] and Lemma 3.6) imply

‖Kω,ǫ,rtϕ‖L∞(B1/4(0)∩D/t) ≤ c‖Kω,ǫ,rtϕ‖L2(B1(0)∩D/t), (5.51)

where c is a constant independent of ǫ, ω, r, t. By (5.51),

|Kω,ǫ,rϕ(0)| ≤ c

∣∣∣∣∫

B1(0)∩D/t

|Kω,ǫ,rtϕ(y)|2dy

∣∣∣∣12

≤ c

∣∣∣∣ −∫

Bt(0)∩D

|Kω,ǫ,rϕ(y)|2dy

∣∣∣∣12

.

So (5.50) holds for x∗ = 0 case. If x∗ 6= 0, we shift x∗ to the origin of the coordinate

system and repeat the above argument to obtain (5.50).

Lemma 5.13. Let ω, ǫ, r ∈ (0, 1], s ∈ (0, 1), ǫ ≤ r, and n ≥ 3. There is a constant

c independent of ω, ǫ, r, s such that, for any x, y ∈ D,

∣∣Gǫ,r(x, y)∣∣ ≤ c|x− y|2−nK1/ω,ǫ,r(x)K1/ω,ǫ,r(y),

|Gǫ,r(x, y)| ≤ c|ξr(x)|s|x− y|2−n−sK1/ω,ǫ,r(x)K1/ω,ǫ,r(y),

|Gǫ,r(x, y)| ≤ c|ξr(x)|s|ξr(y)|

s|x− y|2−n−2sK1/ω,ǫ,r(x)K1/ω,ǫ,r(y),

(5.52)

where ξr(x) (resp. ξr(y)) denotes the distance from x (resp. y) to the boundary

∂Ω/r.

Proof. Let c be a constant independent of ω, ǫ, r, s and set t ≡ |x− y|.

Proof of (5.52)1. Take F ∈ C∞0 (Bt/3(y) ∩D), and find ϕ ∈ H1(D) satisfying

−∇ · (Kω2,ǫ,r∇ϕ) = Kω,ǫ,rF in D,

ϕ = 0 on ∂D.

Note ϕ is solvable uniquely in H1(D). By Theorem 4.31 [2] and Theorem 2.1 [1],

‖Kω,ǫ,rϕ‖L

2nn−2 (D)

≤ c‖Kω,ǫ,r∇ϕ‖L2(D). (5.53)

By [16] and Lemma 5.12,

ϕ(x) =

∫

Bt/3(y)∩D

Gǫ,r(x, z)Kω,ǫ,r(z)F (z)dz,

|Kω,ǫ,rϕ(x)| ≤ c

∣∣∣∣ −∫

Bt/3(x)∩D

|Kω,ǫ,rϕ|2dz

∣∣∣∣12

≤ c

∣∣∣∣ −∫

Bt/3(x)∩D

∣∣Kω,ǫ,rϕ∣∣ 2n

n−2 dz

∣∣∣∣n−22n

.

(5.54)

(5.53)–(5.54) imply∣∣∣∣∫

Bt/3(y)∩D

Gǫ,r(x, z)Kω,ǫ,r(z)F (z)dz

∣∣∣∣

≤ cK1/ω,ǫ,r(x)

∣∣∣∣ −∫

Bt/3(x)∩D

∣∣Kω,ǫ,rϕ∣∣ 2n

n−2 dz

∣∣∣∣n−22n

≤ ct2−n

2 K1/ω,ǫ,r(x)‖Kω,ǫ,r∇ϕ‖L2(D) ≤ ct4−n

2 K1/ω,ǫ,r(x)‖F‖L2(Bt/3(y)∩D). (5.55)

July 9, 2015


(5.55) and Lemma 5.12 imply

∣∣Gǫ,r(x, y)∣∣ ≤ cK1/ω,ǫ,r(y)

∣∣∣∣ −∫

Bt/3(y)∩D

|Kω,ǫ,r(z)Gǫ,r(x, z)|2dz

∣∣∣∣12

≤ ct2−nK1/ω,ǫ,r(x)K1/ω,ǫ,r(y).

So (5.52)1 is proved.

Proof of (5.52)2. By (5.52)1, it is enough to show (5.52)2 for the case ξr(x) < t/6.

By (5.52)1,

|Gǫ,r(x, y)| ≤ c|x− y|2−nK1/ω,ǫ,r(x)K1/ω,ǫ,r(y)

for all x satisfying |x− x| < t/3. Applying Lemma 3.6 and Lemma 5.2 to Gǫ,r(·, y)

in Bt/3(x) ∩ D, we obtain

|Gǫ,r(x, y)| ≤c|ξr(x)|

s

|x− y|n−2+sK1/ω,ǫ,r(x)K1/ω,ǫ,r(y) for x ∈ Bt/6(x) ∩ D.

(5.52)2 follows by setting x = x. (5.52)3 is obtained by (5.48), (5.52)2, Lemma 5.2,

and a similar argument as that for (5.52)2.

Lemma 5.14. Solution of (5.45) exists uniquely in H1(D). For any ω, ǫ, r ∈ (0, 1]

and ǫ ≤ r, there is a constant c independent of ω, ǫ, r such that

|W(n)ω,ǫ,r(x)| ≤ cǫ/r for x ∈ D.

Proof. Let c denote a constant independent of ω, ǫ, r.

Step 1. Unique existence of a solution of (5.45) in H1(D) is clear. If we define

Y(n)ω,ǫ,r ≡ W

(n)ω,ǫ,r − X

(n)ω,ǫ/r in D∗ (see (5.46)), then

−∇ ·

(Kω2,ǫ,r∇Y

(n)ω,ǫ,r

)= 0 in D∗,

Y(n)ω,ǫ,r = W

(n)ω,ǫ,r − X

(n)ω,ǫ/r on ∂D∗.

By Theorem 8.1 [12], (3.23), and (5.47)1,

supD∗

|W(n)ω,ǫ,r| ≤ cǫ/r + sup

∂D∗\∂D

|W(n)ω,ǫ,r|. (5.56)

Next we want to derive

|W(n)ω,ǫ,r(x)| ≤ cǫ/r for x ∈ D \ D∗. (5.57)

If so, together with (5.56), the lemma is proved.

Step 2. Define ν ≡ ǫ/r. Suppose G(x, y) is the Green’s function of−∇y ·

(Kω2,ǫ,ǫ∇G(x, ·)

)= δ(x, ·) in D/ν,

G(x, ·) = 0 on ∂D/ν,

it is easy to see

G(x, y) = νn−2Gǫ,r(νx, νy). (5.58)

July 9, 2015


We claim, for any s ∈ (0, 1), there is a constant c independent of ǫ, ω, r, s such that

G(x, y) ≤ c|ξǫ(x)|s

|x−y|n−2+s K1/ω,ǫ,ǫ(x)K1/ω,ǫ,ǫ(y),

G(x, y) ≤ c|ξǫ(x)|s|ξǫ(y)|s

|x−y|n−2+2s K1/ω,ǫ,ǫ(x)K1/ω,ǫ,ǫ(y),

|∇yG(x, y)| ≤ c |ξǫ(x)|s

|x−y|n−1+s K1/ω,ǫ,ǫ(x)K1/ω,ǫ,ǫ(y) for |x− y| ≤ 1,

|∇yG(x, y)| ≤ c |ξǫ(x)|s|ξǫ(y)|s

|x−y|n−2+2s K1/ω,ǫ,ǫ(x)K1/ω,ǫ,ǫ(y) for |x− y| > 1,

(5.59)

where ξǫ(x) is the distance from x to the boundary ∂Ω/ǫ. By (5.58) and (5.52)2,

G(x, y) = νn−2Gǫ,r(νx, νy) ≤cνn−2|ξr(νx)|

s

νn−2+s|x− y|n−2+sK1/ω,ǫ,r(νx)K1/ω,ǫ,r(νy)

≤c|ξǫ(x)|

s

|x− y|n−2+sK1/ω,ǫ,ǫ(x)K1/ω,ǫ,ǫ(y). (5.60)

So we obtain (5.59)1. Similarly, by (5.52)3, we have (5.59)2. If t = |x − y| ≤ 1,

Lemma 3.6 and (5.60) imply

‖∇yG(x, ·)‖L∞(Bt/2(y)∩D/ν) ≤c

t‖G(x, ·)‖L∞(B3t/4(y)∩D/ν)

≤c|ξǫ(x)|

s

|x− y|n−1+sK1/ω,ǫ,ǫ(x)K1/ω,ǫ,ǫ(y).

So (5.59)3 holds. (5.59)4 follows from (3.5)2 and (5.59)2.

Step 3. We claim (5.57). The solution of (5.45) can be written as W(n)ω,ǫ,r =

X(n)ω,ǫ/r + U1 + U2, where U1 is the solution of

−∇ ·

(Kω2,ǫ,r∇U1

)= 0 in D,

U1 = −ρX(n)ω,ǫ/r on ∂D,

and U2 is the solution of

−∇ ·

(Kω2,ǫ,r(∇U2 + ∇X

(n)ω,ǫ/r + ~en)

)= 0 in D,

U2 = 0 on ∂D.

By (3.23) and maximal principle [12], we see

‖X(n)ω,ǫ/r‖L∞(D) + ‖U1‖L∞(D) ≤ cǫ/r. (5.61)

Set ν ≡ ǫ/r and D/ν(ℓ) ≡ x ∈ D/ν : dist(x, ∂Ω/ǫ) ≤ ℓ for ℓ > 2. By (5.47)3,

there is a ℓ∗ so that (D \ D∗)/ν ⊂ D/ν(ℓ∗). Find η ∈ C∞(D/ν) with support in

˜D/ν(ℓ∗ + 1) so that η ∈ [0, 1], η = 1 in D/ν(ℓ∗), and ‖∇η‖L∞(D/ν) ≤ c. By (3.23),

July 9, 2015


(5.47)2, and (5.59), for any x ∈ (D \ D∗)/ν and s ∈ (12 , 1),

U2(νx) = −

∫

D/ν

∇yG(x, y)Kω2,ǫ,ǫν(∇X(n)ω,1(y) + ~en)dy

= −

∫

D/ν

∇yG(x, y)(1 − η(y))Kω2,ǫ,ǫν(∇X(n)ω,1(y) + ~en)dy

−

∫

D/ν

∇yG(x, y)η(y)Kω2,ǫ,ǫν(∇X(n)ω,1(y) + ~en)dy

= −

∫

˜D/ν(ℓ∗+1)

G(x, y)∇η(y)Kω2,ǫ,ǫν(∇X(n)ω,1(y) + ~en)dy

−

∫

˜D/ν(ℓ∗+1)

∇yG(x, y)η(y)Kω2,ǫ,ǫν(∇X(n)ω,1(y) + ~en)dy

≤ c

∫

˜D/ν(ℓ∗+1)∩|x−y|≤1

ν|ξǫ(x)|s

|x− y|n−1+sdy

+c

∫

˜D/ν(ℓ∗+1)∩|x−y|>1

ν|ξǫ(x)|s

|x− y|n−2+2sdy ≤ cǫ/r. (5.62)

(5.62) implies ‖U2‖L∞(D\D∗) ≤ cǫ/r. Together with (5.61), we obtain (5.57). So we

prove the lemma.

Lemma 5.15. Let θ, ǫ0 be those in Lemma 5.9. There exist constants θ, ǫ0 ∈ (0, 1)

satisfying θ < θ, ǫ0 < ǫ0 such that if−∇ · (Kω2,ǫ,r∇ϕ) = 0 in B1(0) ∩ Ω/r,

ϕ = ϕb on B1(0) ∩ ∂Ω/r,(5.63)

and ifϕb(0) = ∂Tϕb(0) = 0,

‖Kω,ǫ,rϕ‖L2(B1(0)∩Ω/r), [ϕb]C1,α(B1(0)∩Ω/r) ≤ 1,

then, for any ω, ǫ, r ∈ (0, 1] and ǫ/r ≤ ǫ0,

supBθ(0)∩Ω/r

∣∣∣ϕ(x) −(xn + W(n)

ω,ǫ,r(x))dω,ǫ,r

∣∣∣ ≤ θ1+τ ,

where α ∈ (0, 1), τ = α2 , ∂Tϕb(0) is the tangential derivative of ϕb at 0, dω,ǫ,r is

the n-th component of K−1ω −

∫

Bθ(0)∩Ω/r

Kω2,ǫ,r∇ϕ dx, and K−1ω is the inverse matrix

of Kω (see (3.24)).

Proof. The proof is similar to that of Lemma 5.9. Let r∗, ω∗ ∈ [0, 1], (ϕ∗, ϕb,∗)

satisfy−∇ · (Kω∗

∇ϕ∗) = 0 in B2/3(0) ∩ Ω/r∗,

ϕ∗ = ϕb,∗ on B2/3(0) ∩ ∂Ω/r∗,

July 9, 2015


and ϕb,∗ is smooth with ϕb,∗(0) = ∂Tϕb,∗(0) = 0. By (3.25) and Taylor expansion,

there exist θ ∈ (0, 23 ) and τ ′ satisfying τ < τ ′ < α such that

supBθ(0)∩Ω/r∗

∣∣ϕ∗ − xn(∂nϕ∗)Bθ(0)∩Ω/r∗

∣∣

≤ θ1+τ ′(‖ϕ∗‖L∞(B2/3(0)∩Ω/r∗) + [ϕb,∗]C1,α(B2/3(0)∩Ω/r∗)

). (5.64)

If we fix a small θ ∈ (0, 1) so that (5.64) holds, the conclusion will follow by con-

tradiction provided we prove limǫ/r→0 ‖W(n)ω,ǫ,r‖L∞(B1(0)∩Ω/r) = 0. But that is the

result of Lemma 5.14. So we prove this lemma.

Lemma 5.16. θ, ǫ0, τ are same as those in Lemma 5.15. If−∇ · (Kω2,ǫ∇Φ) = 0 in B1(0) ∩ Ω,

Φ = 0 on B1(0) ∩ ∂Ω,(5.65)

then, for any ω ∈ (0, 1], ǫ ∈ (0, ǫ0), and k satisfying ǫ/θk ≤ ǫ0, there is a constant

dω,ǫk satisfying

|dω,ǫk | ≤ cJω,ǫ,

supB

θk (0)∩Ω

∣∣∣∣Φ −k−1∑

j=0

θτj(xn + θjW

(n)

ω,ǫ,θj(θ−jx)

)d

ω,ǫj

∣∣∣∣ ≤ θk(1+τ)Jω,ǫ,(5.66)

where Jω,ǫ ≡ ‖Kω,ǫΦ‖L2(B1(0)∩Ω) and c is a constant independent of ω, ǫ.

Proof. This is done by induction on k. When k = 1, (5.66) holds by Lemma 5.15

with r = 1. dω,ǫ0 is the n-th component of K−1

ω −

∫

Bθ(0)∩Ω

Kω2,ǫ∇Φdx. Suppose (5.66)

holds for some k satisfying ǫ/θk ≤ ǫ0, we define, in B1(0) ∩ Ω/θk,

ϕ(x) ≡ J−1ω,ǫθ

−k(1+τ)

(Φ(θkx) −

k−1∑

j=0

θτj(θkxn + θjW

(n)

ω,ǫ,θj(θk−jx)

)d

ω,ǫj

),

ϕb(x) ≡ −J−1ω,ǫθ

−k(1+τ)k−1∑

j=0

θτj θkdω,ǫj xn.

Then those functions satisfy (5.63) with r = θk. Note [ϕb]C1,α(B1(0)∩Ω/θk)= 0. By

induction,

max‖ϕ‖L∞(B1(0)∩Ω/θk), [ϕb]C1,α(B1(0)∩Ω/θk)

≤ 1. (5.67)

Apply Lemma 5.15,

supBθ(0)∩Ω/θk

∣∣∣∣ϕ(x) −(xn + W

(n)

ω,ǫ,θk(x)

)dω,ǫ,θk

∣∣∣∣ ≤ θ1+τ , (5.68)

July 9, 2015


where dω,ǫ,θk is the n-th component of K−1ω −

∫

Bθ(0)∩Ω/θk

Kω2,ǫ,θk∇ϕ dx. By energy

method and (5.67), |dω,ǫ,θk | is bounded uniformly in ω, ǫ, θk. Rewrite (5.68) in terms

of Φ in Bθk+1(0) to obtain

supB

θk+1 (0)∩Ω

∣∣∣∣Φ(x) −

k−1∑

j=0

θτj(xn + θjW

(n)

ω,ǫ,θj(θ−jx)

)d

ω,ǫj

−θkτ Jω,ǫ

(xn + θkW

(n)

ω,ǫ,θk(θ−kx)

)dω,ǫ,θk

∣∣∣∣ ≤ θ(k+1)(1+τ)Jω,ǫ.

If dω,ǫk ≡ Jω,ǫdω,ǫ,θk, we conclude that (5.66) holds for k + 1.

Lemma 5.17. Let ǫ0 be same as that in Lemma 5.16. Suppose ω ∈ (0, 1] and

ǫ ∈ (0, ǫ0), any solution of (5.65) satisfies

‖∇Φ‖L∞(B1/2(0)∩Ω) ≤ c‖Kω,ǫΦ‖L2(B1(0)∩Ω), (5.69)


Proof. By (3.2), we have a local coordinate x = (x′, xn) so that

B1(0) ∩ Ω =(x′, xn) ∈ Ω : |x′|2 + |xn|

2 < 1,Ψ(x′) < xn

.

To obtain the Lipschitz estimate in (5.69), it is suffice to show

sup(0,xn)∈B1/2(0)∩Ω

|∇Φ(0, xn)| ≤ c‖Kω,ǫΦ‖L2(B1(0)∩Ω). (5.70)

The reason is that one can repeat the same argument by varying the origin along

the boundary B1(0) ∩ ∂Ω and by adjusting the constant c to obtain the estimate.

Let θ, Jω,ǫ, τ are same as those in Lemma 5.16, let c be a constant independent

of ω, ǫ, and let k satisfy ǫ/θk ≤ ǫ0 < ǫ/θk+1. For any x ≡ (0, xn) ∈ B1/2(0) ∩ Ω, we

have either case (1): 12 θ

ℓ < xn ≤ 12 θ

ℓ−1 for 1 ≤ ℓ ≤ k or case (2): 0 ≤ xn ≤ 12 θ

k.

For case (1): By Lemma 5.16, we have

supB

θℓ−1 (0)∩Ω

∣∣∣∣Φ(y) −

ℓ−2∑

j=0

θτj(yn + θjW

(n)

ω,ǫ,θj(θ−jy)

)d

ω,ǫj

∣∣∣∣ ≤ cθℓ(1+τ)Jω,ǫ. (5.71)

Hence, by Lemma 5.14 and (5.71),

supB

θℓ−1 (0)∩Ω

|Φ| ≤ cJω,ǫ

(θℓ(1+τ) + (ξ(x) + ǫ)

ℓ−2∑

j=0

θτj

)≤ cξ(x)Jω,ǫ. (5.72)

Here we use ξ(x) ≡ |x−x0| where x0 ∈ ∂Ω so that |x−x0| = miny∈∂Ω |x− y|. Note

ǫ ≤ ǫ0θk ≤ 2ǫ0

12 θ

ℓ ≤ 2ǫ0xn ≤ cǫ0ξ(x). By (5.72),

supBξ(x)/2(x)

|Φ| ≤ cξ(x)Jω,ǫ. (5.73)

July 9, 2015


Then we move the origin of the coordinate system to x and define

ϕ(y) ≡ ξ(x)−1J−1ω,ǫΦ(ξ(x)y).

By (5.73),

‖ϕ‖L∞(B1/2(x)) ≤ c.

Function ϕ satisfies

−∇ · (Kω2,ǫ/ξ(x)∇ϕ) = 0 in B1/2(x).

By Lemma 5.11,

‖∇Φ‖L∞(Bξ(x)/4(x)) ≤ cJω,ǫ.

This proves (5.70) for case (1).

For case (2): Apply Lemma 5.16 to obtain

supB

θk (0)∩Ω

∣∣∣∣Φ(y) −

k−1∑

j=0

θτj(yn + θjW

(n)

ω,ǫ,θj(θ−jy)

)d

ω,ǫj

∣∣∣∣ ≤ cJω,ǫθk(1+τ).

By Lemma 5.14,

supB

θk (0)∩Ω

|Φ| ≤ cǫJω,ǫ. (5.74)

Define ϕ(y) ≡ ǫ−1J−1ω,ǫΦ(ǫy). By (5.74),

‖ϕ‖L∞(B1(0)∩Ω/ǫ) ≤ c.

Function ϕ satisfies−∇ · (Kω2,ǫ,ǫ∇ϕ) = 0 in B1(0) ∩ Ω/ǫ,

ϕ = 0 on B1(0) ∩ ∂Ω/ǫ.

(3.5)2 of Lemma 3.5 implies ‖ϕ‖C1,0(B1(0)∩Ωǫf /ǫ)∩C1,0(B1(0)∩Ωǫ

m/ǫ) ≤ c. This gives the

proof of (5.70) for case (2).

Remark 5.4. By Lemma 3.5 and Lemma 5.17, we know that if ǫ, ω ∈ (0, 1], any

solution of (5.65) satisfies ‖∇Φ‖L∞(B1/2(0)∩Ω) ≤ c‖Kω,ǫΦ‖L2(B1(0)∩Ω), where c is a

constant independent of ω, ǫ.

Lemma 3.7 is direct result of Remark 5.3 and Remark 5.4.

Acknowledgement

The author would like to thank the anonymous referee’s valuable suggestions for

improving the presentation of this paper. This research is supported by the grant

number NSC 102-2115-M-009 -014 from the research program of National Science

Council of Taiwan.

July 9, 2015


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