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Page 1: Copyright by Lan Tang 2011

Copyright

by

Lan Tang

2011

Page 2: Copyright by Lan Tang 2011

The Dissertation Committee for Lan Tangcertifies that this is the approved version of the following dissertation:

Random Homogenization of p-Laplacian with Obstacles

on Perforated Domain and Related Topics

Committee:

Luis Caffarelli, Supervisor

Mikhail Vishik

Alexis Vasseur

Lexing Ying

Natasa Pavlovic

Aristotle Arapostathis

Page 3: Copyright by Lan Tang 2011

Random Homogenization of p-Laplacian with Obstacles

on Perforated Domain and Related Topics

by

Lan Tang, B.S.; M.S.

DISSERTATION

Presented to the Faculty of the Graduate School of

The University of Texas at Austin

in Partial Fulfillment

of the Requirements

for the Degree of

DOCTOR OF PHILOSOPHY

THE UNIVERSITY OF TEXAS AT AUSTIN

May 2011

Page 4: Copyright by Lan Tang 2011

Dedicated to my wife Lina.

Page 5: Copyright by Lan Tang 2011

Acknowledgments

Firstly, I would like to thank my advisor Luis Caffarelli for his guidance

and support during my Ph.D. study in Austin.

Also, I need to give thanks to Department of Mathematics, the Uni-

versity of Texas at Austin to provide me a perfect academic environment to

study. Here I want to express my great appreciation to Nancy Lamm, the

former secretary of Mathematics Department, for her much help in my life.

And I would like to thank my classmates and friends Nestor Guillen and Ray

Yang, with whom I made many important discussions in research.

I would also like to thank my former professor Lihe Wang for his guid-

ance and help to my early steps in the area of Partial Differential Equations.

Finally, but by far not least, I am indebted to my wife, Lina for her

support and understanding through many years of hard work.

v

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Random Homogenization of p-Laplacian with Obstacles

on Perforated Domain and Related Topics

Publication No.

Lan Tang, Ph.D.

The University of Texas at Austin, 2011

Supervisor: Luis Caffarelli

Let D ⊂ Rn be a bounded domain and (Ω,F,P) be a given probability

space. For each ω ∈ Ω and ε > 0, we denote by Tε(ω) the set of holes on D and

Dε = D \ Tε(ω). We assume that Tε(ω) =⋃k∈Zn

Baε(k,ω)(εk) and the p-capacity

(1 < p ≤ n) of each ball Baε(k,ω)(εk) is γ(k, ω)εn, where the random process

γ : Zn×Ω 7→ [0,+∞) is bounded and stationary ergodic. Also we let f and ψ

be bounded measurable functions. We define ψε = ψ on Dε and ψε = 0 on Tε.

We prove the asymptotic behaviour of the solutions to the following

problems: min∫

D

1

p|∇u|pdx−

∫D

fudx : u ∈ W 1,p0 (D), u ≥ 0 a. e. in Tε(ω)

and min

∫D

1

p|∇v|p − fvdx : v ∈ W 1,p

0 (D), v ≥ ψε a. e. in D .

Finally, we consider the parabolic fractional p-Laplacian and we prove

that any weak solution of parabolic fractional p-Laplacian is uniformly bounded

if the initial data is bounded in Lp(Rn), where p > 2, 0 < s < 1 and ps ≤ n.

vi

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Table of Contents

Acknowledgments v

Abstract vi

Chapter 1. Introduction 1

Chapter 2. Main Results for Random Homogenization 5

2.1 Main Assumptions and Settings . . . . . . . . . . . . . . . . . 5

2.2 Statement of Main Results . . . . . . . . . . . . . . . . . . . . 6

Chapter 3. Proof of Main Results for Random Homogenization 9

3.1 Classical Case : p = 2 . . . . . . . . . . . . . . . . . . . . . . . 9

3.2 p-Laplacian for 1 < p ≤ n . . . . . . . . . . . . . . . . . . . . . 11

Chapter 4. Construction of the Correctors wε 18

4.1 The Auxiliary Obstacle Problem . . . . . . . . . . . . . . . . . 19

4.2 Properties of wε . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.3 Proof of Lemma 3.2.1 . . . . . . . . . . . . . . . . . . . . . . . 32

Chapter 5. Lower Semicontinuous Property 37

5.1 Review of the Classical Case . . . . . . . . . . . . . . . . . . . 37

5.2 Revised Version: Proof of Lemma 3.2.2 . . . . . . . . . . . . . 38

Chapter 6. Other Topics: Parabolic Fractional p-Laplacian 42

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

6.2 Some Preliminary Lemmas . . . . . . . . . . . . . . . . . . . . 43

6.3 Proof of Theorem 6.1.1. . . . . . . . . . . . . . . . . . . . . . . 44

vii

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Chapter 7. Future Direction 50

7.1 Random Homogenization . . . . . . . . . . . . . . . . . . . . . 50

7.2 Parabolic Fractional p-Laplacian . . . . . . . . . . . . . . . . . 50

Bibliography 51

Vita 56

viii

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Chapter 1

Introduction

Let D ⊂ Rn be a bounded domain and (Ω,F,P) be a given probability

space. For each ω ∈ Ω and ε > 0, we denote by Tε(ω) the set of holes on D

and Dε = D \ Tε(ω).

Our main purpose is to study the following variational problem:

min∫

D

1

p|∇u|pdx−

∫D

fudx : u ∈ W 1,p0 (D), u ≥ 0 a. e. in Tε(ω)

where f is some measurable and bounded function.

From the variational problem above, we have the following questions:

Does the solution (minimizer) uε converge to a limit function as ε→ 0? If the

limit exists, how can we characterize it?

Roughly speaking, three cases can occur in this situation:

(1) The holes, in spite of their number, are too small and uε converges

to the limit function u which is a solution of the variational problem without

obstacles.

(2) The holes are too big and uε converges to the solution of the varia-

tional problem with obstacles almost everywhere.

1

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(3) Between the two cases, there is a third one where the holes have

a critical size depending on their number and distribution and where uε con-

verges to the solution of the variational problem with an additional term added

in the energy functional and this strange term comes from the holes.

In the following, we mainly consider the third case, i.e. Tε(ω) has a

critical size. This type of problems were first studied by L. Carbone and F.

Colombini [12] in periodic settings and then in more general frameworks by

E. De Giorgi, G. Dal Maso and P. Longo [20], G. Dal Maso and P. Longo [17]

and G. Dal Maso [16].

And in the early 1980’s, the special case p = 2 ( i. e. Laplacian ) in

the periodic settings was studied by D. Cioranescu and F. Murat in [14] and

[15]. More precisely, they let

Tε =⋃k∈Zn

Baε(εk)

and

aε =

r1ε

nn−2 if n > 2

exp(−r−12 ε−2) if n = 2

where r1 and r2 are both positive constants.

Let uε be the solution of the variational problem above for p = 2, then

uεε>0 is bounded in H1 and there is a subsequence of uε (we still denote

by uε) such that uε converges to u0 weakly in H1(D).

They proved that the weak limit u0 is the solution of the following

2

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problem:

min∫

D

1

2|∇u|2dx+

1

2Cu2− −

∫D

fudx : u ∈ H10 (D)

where the constant C =

∑k∈Zn

cap2(Baε(εk)), i. e. the sum of the 2-capacity of

Baε(εk) and the capacity of the set A ⊂ Rn is defined as follows (see [28]): if

1 < p < n , then

capp(A) = inf∫Rn

|∇h|p : h ∈ W 1,p(Rn), h ≥ 1 on A, lim|x|→∞

h(x) = 0

and if p = n and A ⊂ B1(0),

capn(A) = inf∫Rn

|∇h|n : h ∈ W 1,n0 (B1(0)), h ≥ 1 on A.

And later N. Ansini and A. Braides studied the periodic homogenization

of more general divergence structure in [3] with Γ-Convergence method. They

considered the following problem ( 1 < p ≤ n ):

min∫D

1

p|∇u|p − fudx : u ∈ W 1,p

0 (D) and u = 0 on Tε

where

Tε =⋃k∈Zn

Baε(εk)

and

aε =

r1ε

nn−p if 1 < p < n

exp(−r2ε−nn−1 ) if p = n

where r1 and r2 are both positive constants.

They proved that uε converges to some u0 weakly in W 1,p(D) and u0

solves the follwoing:

3

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min∫D

1

p|∇u|p +

1

pC|u|p − fudx : u ∈ W 1,p

0 (D)

where C is the sum of p-capacity of Baε(εk).

For the random settings, L. Caffarelli and A. Mellet recently studied

the special case p = 2 ( i. e. Laplacian ) in [8]. They assume that

Tε(ω) =⋃k∈Zn

Baε(k,ω)(εk)

and the capacity (p = 2) of each ball Baε(k,ω)(εk) satisfies the following:

cap2(Baε(k,ω)(εk)) = γ(k, ω)εn

where γ : Zn × Ω 7→ [0,+∞) is bounded and stationary ergodic , i. e. there

exists a family of measure-preserving transformations τk : Ω 7→ Ω satisfying

γ(k + k′, ω) = γ(k, τk′ω), ∀ k, k′ ∈ Zn and ω ∈ Ω,

and such that if A ⊂ Ω and τkA = A for all k ∈ Zn, then P (A) = 1 or

P (A) = 0.

They proved that there exists some nonnegative constant A0 ≥ 0 such

that uε(x, ω) converges weakly in H1(D) and almost surely ω ∈ Ω to the

solution u(x) of the following problem:

min∫D

1

2|∇v|2 +

1

2A0v

2− − fvdx : v ∈ H1

0 (D).

4

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Chapter 2

Main Results for Random Homogenization

2.1 Main Assumptions and Settings

Now we consider the case for the 1 < p ≤ n in the random settings.

We assume that the union of holes Tε(ω) satisfies the following:

Tε(ω) =⋃k∈Zn

Baε(k,ω)(εk)

and the p-capacity of each ball Baε(k,ω)(εk) satisfies :

capp(Baε(k,ω)(εk)) = γ(k, ω)εn

where γ : Zn×Ω 7→ [0,+∞) is stationary ergodic i. e. there exists a family of

measure-preserving transformations τk : Ω 7→ Ω satisfying

γ(k + k′, ω) = γ(k, τk′ω), ∀ k, k′ ∈ Zn and ω ∈ Ω,

and such that if A ⊂ Ω and τkA = A for all k ∈ Zn, then P (A) = 1 or

P (A) = 0. And we also assume that γ : Zn × Ω 7→ [0,+∞) is bounded.

Thus the radius of the ball

aε(k, ω) =

(γ(k,ω)nωn

)1

n−p (n−pp−1 )

1−pn−p ε

nn−p if 1 < p < n

exp(−(γ(k,ω)nωn

)−1n−1 ε−

nn−1 ) if p = n

Obviously, aε : Zn × Ω 7→ [0,+∞) is also stationary ergodic and bounded.

5

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2.2 Statement of Main Results

In the following, we assume that the union of all the holes Tε satisfies

the conditions in section 2.1.

Firstly we consider the following variational problem :

infv∈Kε

F(v)

where

F(v) =

∫D

1

p|∇v|p − fvdx

and

Kε = v ∈ W 1,p0 (D) : v ≥ 0 a. e. on Tε.

Let uε be the solution of such a variational problem, i.e.

F(uε) = infv∈Kε

F(v)

.

Obviously, uε is bounded in W 1,p0 (D), then we can choose a subse-

quence of uε (we still denote by uε) such that

uε u0 in W 1,p0 (D).

Our main purpose is to determine the variational functional F0 such

that for almost surely ω ∈ Ω,

F0(u0) = inf

v∈W 1,p0 (D)

F0(v).

In fact, F0 has the following form:

6

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Theorem 2.2.1. Let 1 < p ≤ n and Tε satisfies the assumptions above. Then

there exits a nonnegative real number α0 such that when ε goes to zero, the

solution uε(x, ω) of

min∫Rn

1

p|∇v|p − fvdx : u ∈ W 1,p

0 (D), u ≥ 0 a. e. in Tε(ω)

converges weakly in W 1,p(D) and almost surely ω ∈ Ω to the solution u0 of the

following minimization problem:

min∫D

1

p|∇v|p +

1

pα0v

p− − fvdx : ∀ v ∈ W 1,p

0 (D).

From Theorem 2.2.1, there is also a strange term in F0.

In the following we consider the following variational inequality with

oscillating obstacles:

min∫D

1

p|∇v|p − fvdx : v ∈ W 1,p

0 (D) and v ≥ ψε

where ψ be a measurable function in D and

ψε =

ψ in D \ Tε

0 on Tε .

We have the following result:

Theorem 2.2.2. For 1 < p ≤ n, if when ε goes to zero, the solution hε(x, ω)

of

min∫Rn

1

p|∇v|p − fvdx : v ∈ W 1,p

0 (D), v ≥ ψε a. e. in D

7

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converges weakly in W 1,p(D) and almost surely ω ∈ Ω to the solution h0 , then

h0 is the solution to the following variational problem:

min∫D

1

p|∇v|p +

1

pα0v

p− − fvdx : v ∈ W 1,p

0 (D) and v ≥ ψ a. e. in D

where the constant α0 is the same constant as in Theorem 2.2.1.

8

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Chapter 3

Proof of Main Results for Random

Homogenization

In this part, we will present the proofs for main results for random

homogenization. Firstly, we will review the classical proof for the linear case

p = 2 given by L. Caffarelli and A. Mellet in [8]. And motivated by this, we

will prove Theorem 2.2.1 and 2.2.2 by using two key lemmas.

3.1 Classical Case : p = 2

For the case p = 2, by [8] we know that the proof of main results

strongly depends on two lemmas. The first one is about the construction of

some suitable correctors:

Lemma 3.1.1. Assume that p = 2 and Tε(ω) satisfies the assumptions listed

above. Then there exist a nonnegative real number α0 and a function wε such

that 4wε = α0 in Dε(ω)wε(x, ω) = 1 for x ∈ Tε(ω)wε(x, ω) = 0 for x ∈ ∂D \ Tε(ω)wε(·, ω) → 0 weakly in H1

for a. s. ω ∈ Ω and wε also satisfies the following properties:

9

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(a) for any φ ∈ D(D),

limε→0

∫D

|∇wε|2φdx =

∫D

α0φdx.

(b) for any sequence vε ⊂ H10 (D) with the property: vε → v weakly in

H10 (D) and vε = 0 on Tε and any φ ∈ D(D), we have that

limε→0

∫D

∇wε · ∇vεφdx = −α0

∫D

vφdx.

Proof. See Proposition 2.2 in [8].

And another important lemma is the revised lower semicontinuity prop-

erty:

Lemma 3.1.2. If the Lemma 3.1.1. holds and uε is the solution of

min∫Rn

1

2|∇v|2 − fvdx : u ∈ H1

0 (D), u ≥ 0 a. e. in Tε(ω)

then

lim infε→0

∫D

|∇uε|2dx ≥∫D

|∇(u0)|2 + α0|u0−|2dx,

where u0− is the negative part of u0: u0−(x) = max−u0(x), 0.

Proof. See Proposition 3.1 in [14] and [15].

10

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In [8], L. Caffarelli and A. Mellet proved the random homogenization

results of Laplacian ( p = 2 ) with obstacles with Lemma 3.1.1 and 3.1.2 in

the following way:

∀v ∈ D(D) such that v− ∈ D(D), then v + v−wε is nonnegative on Tε.

Then

F(uε) ≤ F(v + v−wε).

And they expanded F(v+ v−wε) term by term and by Lemma 3.1.1., they got

F0(v) = limε→0

F(v + v−wε)

=

∫D

1

2|∇(v)|2 +

1

2α0|v−|2dx−

∫D

fvdx

Hence

F0(v) ≥ lim supε→0

F(uε).

And by Lemma 3.1.2.,they got

F0(v) ≥ F0(u0),

which concludes Theorem 2.2.1. The proof for Theorem 2.2.2 is similar.

3.2 p-Laplacian for 1 < p ≤ n

For the p-Laplacian, we expect similar results as the linear case:

(1) we need to construct the correctors wε with a similar way as [8] ,

however, we need to give more delicate estimates for the correctors due to the

nonlinearity of p-Laplacian.

11

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(2) since the proof of Lemma 3.1.2 (i.e. Proposition 3.1 in [14] and [15])

strongly depends on the condition p = 2, hence we cannot apply the method

there to the nonlinear case of p-Laplacian and we need to use some properties

of the correctors to overcome this .

In fact, we also have the following key lemmas for p-Laplacian:

Lemma 3.2.1. Assume that Tε(ω) satisfies the assumptions listed above. Then

there exist a nonnegative real number α0 and a function wε such that

4pwε = α0 in Dε(ω)

wε(x, ω) = 1 for x ∈ Tε(ω)

wε(x, ω) = 0 for x ∈ ∂D \ Tε(ω)

wε(·, ω) → 0 weakly in W 1,p0 (D)

for a. s. ω ∈ Ω and wε also satisfies the following properties:

(a) for any φ ∈ D(D) and 0 < p′ < p,

limε→0

∫D

|∇wε|p′φdx = 0

(b) for any φ ∈ D(D),

limε→0

∫D

|∇wε|pφdx =

∫D

α0φdx.

(c) for any sequence vε ⊂ W 1,p0 (D) with the property: vε → v weakly in

W 1,p0 (D) and vε = 0 on Tε and any φ ∈ D(D), we have that

limε→0

∫D

|∇wε|p−2∇wε · ∇vεφdx = −α0

∫D

vφdx.

12

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And for uε, we have the following lower semicontinuous property:

Lemma 3.2.2. If Lemma 3.2.1 holds and uε is the solution of

min∫Rn

1

p|∇v|p − fvdx : u ∈ W 1,p

0 (D), u ≥ 0 a. e. in Tε(ω).

We assume that u0 is the weak limit of uε in W 1,p0 (D). Then

lim infε→0

F(uε) ≥ F0(u0),

where F0 is defined as follows:

F0(v) =

∫D

1

p|∇v|p +

1

pα0v

p− − fvdx, ∀ v ∈ W

1,p0 (D)

and α0 is the same constant as the one in Lemma 3.2.1.

The proofs of these two lemmas will be given in Chapter 4 and 5. In

the following, we will give the proof for Theorem 2.2.1 and 2.2.2.

Proof. Let φ ∈ D(D) such that φ− ∈ D(D). Then

F(uε) ≤ F(φ+ φ−wε).

Next, we can estimate F(φ+ φ−wε) as follows:

F(φ+ φ−wε) =

∫D

1

p[ |∇φ+∇φ−wε +∇wεφ−|p ]dx

−∫D

[fφ+ fφ−wε]dx.

13

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If p is an integer, then

|∇φ+∇φ−wε +∇wεφ−|p ≤ |∇φ+∇φ−wε| + |∇wεφ−|p

=

p∑k=0

Ckp |∇φ+∇φ−wε|k · |∇wεφ−|p−k

= |∇φ+∇φ−wε|p + |∇wεφ−|p

+

p−1∑k=1

Ckp |∇φ+∇φ−wε|k · |∇wεφ−|p−k

By Lemma 3.2.1,

wε −→ 0 weakly in W 1,p(D).

Then wε → 0 strongly in Lp(D) as ε→ 0. Thus

limε→0

∫D

|∇φ+∇φ−wε|pdx =

∫D

|∇φ|pdx

By the property (b) in Lemma 3.2.1 , we have

limε→0

∫D

|∇wεφ−|pdxdx =

∫D

α0φp−dx

For any k: 1 ≤ k ≤ p− 1, we have that

Ckp |∇φ+∇φ−wε|k ≤ C + C|wε|k

where C is a constant (not depending on k).

From (a) in Lemma 3.2.1, we know that (1 ≤ k ≤ p− 1)

limε→0

∫D

|∇wεφ−|p−kdx = 0

And by Holder inequality,

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∫D

|wε|k|∇wεφ−|p−kdx ≤ ∫D

|wε|pdxkp ·

∫D

|∇wεφ−|pdxp−kp

Hence

limε→0

∫D

|wε|k|∇wεφ−|p−kdx = 0.

Therefore (for p is an integer)

lim supε→0

∫D

|∇φ+∇φ−wε +∇wεφ−|pdx ≤∫D

|∇φ|pdx+

∫D

α0φp−dx

If p is not an integer, then we let m be the integer part of p ( Thus

0 < p−m < 1). Hence

|∇φ+∇φ−wε +∇wεφ−|p ≤ |∇φ+∇φ−wε| + |∇wεφ−|p−m

× |∇φ+∇φ−wε| + |∇wεφ−|m

≤ |∇φ+∇φ−wε|p−m + |∇wεφ−|p−m

× |∇φ+∇φ−wε| + |∇wεφ−|m

= |∇φ+∇φ−wε|p + |∇wεφ−|p

+m∑k=1

Ckm|∇φ+∇φ−wε|p−k|∇wεφ−|k

+m∑k=1

Ckm|∇φ+∇φ−wε|k|∇wεφ−|p−k.

If we use the same argument as above and we can get the same conclu-

sion for p is not an integer.

15

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Thus for any p : 1 < p ≤ n, we have

lim supε→0

∫D

|∇φ+∇φ−wε +∇wεφ−|pdx ≤∫D

|∇φ|pdx+

∫D

α0φp−dx

Hence

F0(φ) ≥ lim supε→0

F(φ+ φ−wε)

≥ lim infε→0

F(uε)

By Lemma 3.2.2, we have

F0(φ) ≥ F0(u0).

And the set φ ∈ D(D) : φ− ∈ D(D) is dense in W 1,p0 (D), then u0 is

the solution of

min∫D

1

p|∇v|p +

1

pα0v

p− − fvdx : ∀ v ∈ W 1,p

0 (D).

Now we go to the proof of Theorem 2.2.2:

Proof. Let φ ∈ D(D) such that φ− ∈ D(D) and φ ≥ ψ a. e. in D. Then

φ+ + (wε+ − 1)φ− ≥ ψε in D

Obviously, for wε+, we have the following property:

lim supε→0

∫D

|∇wε+|p|φ−|pdx ≤ α0

∫D

φp−dx

16

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Thus from the proof for Theorem 2.2.1, we have that

lim supε→0

∫D

1

p|∇φ+ + (wε+ − 1)φ−|pdx

= lim supε→0

∫D

1

p|∇φ+ +∇wε+φ− + (wε+ − 1)∇φ−|pdx

≤∫D

1

p|∇φ|pdx+

α0

p

∫D

φp−dx

And since φ+ + (wε+ − 1)φ− ≥ ψε in D, then∫D

1

p|∇φ+ + (wε+ − 1)φ−|pdx ≥

∫D

f · (φ+ + (wε+ − 1)φ−)dx

+

∫D

1

p|∇hε|p − fhεdx

And wε+ converges to 0 weakly in W 1,p0 (D) and hε converges to h0 weakly

in W 1,p0 (D), hence

lim infε→0

∫D

1

p|∇hε|p − fhεdx

≤∫D

1

p|∇φ|pdx+

α0

p

∫D

φp−dx− fφdx

By Lemma 3.2.2, we have that∫D

1

p|∇h0|pdx+

∫D

1

pα0(h0)

p−dx− fh0dx

≤∫D

1

p|∇φ|pdx+

α0

p

∫D

φp−dx−∫D

fφdx

And φ ∈ D(D) : φ− ∈ D(D) and φ ≥ ψ a.e. in D is dense in

v ∈ W 1,p0 (D) : v ≥ ψ a.e. in D, therefore h0 is the solution to the following

variational problem: min∫D

1

p|∇v|p +

1

pα0v

p− − fvdx : v ∈ W 1,p

0 (D) and v ≥

ψ a. e. in D.

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Chapter 4

Construction of the Correctors wε

In this chaper, we will prove Lemma 3.2.1. In order to prove it, we

need to follow the following :

(1) Find the crital value α0 and then define the correctors wε.

(2) Show that wε is bounded in W 1,p(D), where the correctors wε

are defined by

wε(x, ω) = infv(x) : 4pv ≤ α0 in Dε, v ≥ 1 on Tε and v = 0 on ∂D

(3) Show that wε −→ 0 in Lp(D) as ε→ 0.

(4) Prove the property (a)-(c) of Lemma 3.2.1, i. e.

(a) for any φ ∈ D(D) and 0 < p′ < p,

limε→0

∫D

|∇wε|p′φdx = 0

(b) for any φ ∈ D(D),

limε→0

∫D

|∇wε|pφdx =

∫D

α0φdx.

(c) for any sequence vε ⊂ W 1,p0 (D) with the property : vε → v weakly

in W 1,p0 (D) and vε = 0 on Tε and any φ ∈ D(D), we have that

limε→0

∫D

|∇wε|p−2∇wε · ∇vεφdx = −α0

∫D

vφdx.

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4.1 The Auxiliary Obstacle Problem

Here we will use the method from [8]. Firstly we consider the following

obstacle problem: for every open set A ⊂ Rn, α ∈ R and ∀x ∈ A and ω ∈ Ω ,

we define

vεα,A(x, ω) = inf v(x) : 4pv(·) ≤ α−∑

k∈Zn∩ε−1A

γ(k, ω)εnδ(· − εk) in A,

v ≥ 0 in A, v = 0 on ∂A

where 1 < p ≤ n and 4pw = ∇ · (| ∇w |p−2∇w) is the p-Laplacian operator.

And we set

mεα(A, ω) = |x ∈ A : vεα,A = 0|.

From [8], we can find that for any given ε > 0, mεα(·, ω) is subadditive

for each ω ∈ Ω and the process mεα(A, ω) is stationary ergodic.

Hence by [8] and [10], for any real number α, there is a constant l(α) ≥ 0

such that

limε→0

mεα(B1(x0), ω)

|B1(x0)|= l(α) ,

i.e.

limε→0

|x ∈ B1(x0) : vεα, B1(x0)= 0|

|B1(x0)|= l(α),

for any B1(x0) ⊂ Rn.

About the function l(α) , we have the following :

Proposition 4.1.1.

(i) l(α) a nondecreasing function of α ;

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(ii) l(α) = 0 for α < 0;

(iii) l(α) > 0 for α is large enough.

Proof. (i) For its monotonicity, we consider two parameters α ≤ α′ and we

will compare l(α) and l(α′). By comparison principle, for any A ⊆ Rn,

vεα′,A(x, ω) ≤ vεα,A(x, ω), a. e. x ∈ A.

Hence

x ∈ A : vεα,A(x, ω) = 0 ⊆ x ∈ A : vεα′,A(x, ω) = 0,

which implies l(α) ≤ l(α′) for α ≤ α′.

(ii) If α < 0, we let β = |α|(2−p)/(p−1)α then

4pβ

c(n, p)|x− x0|

pp−1 − β

c(n, p) = |β|p−2β = α

and βc(n,p)|x − x0|

pp−1 − β

c(n,p)is positive in B1(x0) and vanishes on ∂(B1(x0)).

Then we deduce that:

vεα,B1≥ β

c(n, p)|x− x0|

pp−1 − β

c(n, p)> 0 in B1(x0).

Hence

mεα(B1(x0), ω) = 0.

Therefore l(α) = 0.

(iii) Let a = a(k, ω) = n

√ncγ(k,ω)

α, where the constant c depends on p

and n. More precisely, it should be determined by the following: for 1 < p < n,

4pc1p−1 |x|

p−np−1 = −δ(x); and for p = n, 4pc

1n−1 log

1

|x| = −δ(x).

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We define the function gεα,k(x, ω) for any α ∈ Rn as following:

gεα,k(x, ω) =

∫ aε

r

(cγ(k, ω)εns1−n − α

ns)

1p−1ds, if 0 ≤ r = |x− εk| ≤ aε;

0, if x ∈ B1 \Baε(εk).

Obviously if the parameter α is large enough, then 12≥ n

√ncγ(k,ω)

α, which

implies the function gεα,k(x, ω) is only concentrated on the cell ball B ε2(εk) for

α very large. From the definition of gεα,k(x, ω), we know that (for α is large)

4p gεα,k(x, ω) ≤ α− γ(k, ω)εnδ(x− εk) in B1,

and gεα,k(x, ω) = 0 if x ∈ B1 \Baε(εk).

Now we consider the sum of all gεα,k:∑k∈ ε−1B1∩Zn

gεα,k

By the definition, we know that for any two different k, k′ ∈ ε−1B1∩Zn,

gεα,k and gεα,k′ have disjoint support. Hence if we let gεα =∑

k∈ ε−1B1∩Zngεα,k, then

4p gεα(x, ω) ≤ α−

∑k∈ ε−1B1∩Zn

γ(k, ω)δ(x− εk)

And gεα(x, ω) ≥ 0 for x ∈ B1 and gεα(x, ω) = 0 on ∂B1.

Therefore

0 ≤ vεα,B1(x, ω) ≤ gεα(x, ω), for x ∈ B1.

Thus ⋃k∈ ε−1B1∩Zn

(B1 \Baε(εk)) ⊂ x ∈ B1 : vεα,B1= 0,

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which implies

mεα(B1, ω) ≥ ωn − Cε−n(aε)n = 1− Can,

where ωn is the volume of the unit ball B1.

And when α is large enough, then a will be small enough such that

ωn − Can ≥ 12ωn. Therefore

mεα(B1, ω) ≥ 1

2ωn > 0, if α is large enough

Then l(α) > 0 if α is large enough.

Next We choose the critical value α0 by the following way:

α0 = supα : l(α) = 0.

Then by Proposition 4.1.1, α0 is finite and nonnegative.

In the following, we will define wε as follows:

wε(x, ω) = infv(x) : 4pv ≤ α0 in Dε, v ≥ 1 on Tε and v = 0 on ∂D

Therefore wε satisfies the following conditions:4pw

ε(x, ω) = α0 for x ∈ Dε(ω)

wε(x, ω) = 1 for x ∈ Tε(ω)

wε(x, ω) = 0 for x ∈ ∂D \ Tε(ω).

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4.2 Properties of wε

Firstly, we will show

Proposition 4.2.1. wε is uniformly bounded in W 1,p0 (D).

Proof. We will split the proof into two parts: wε is uniformly bounded in

Lp(D) and ∇wε is also uniformly bounded in Lp(D).

To prove the first part: wε is uniformly bounded in Lp(D), we need

to introduce an auxiliary function v(x): let v be the solution to the following

problem: 4pv = α0 in D

v = 0 on ∂D

By Comparison Principle for almost surely ω ∈ Ω,

v(x) ≤ wε(x, ω) ≤ 1 for a. e. x ∈ D.

Hence ∫D

|wε|pdx ≤ C

which implies that wε is uniformly bounded in Lp(D)

To show that ∇wε is also uniformly bounded in Lp(D), we define the

function hε(x, ω) as following: if 1 < p < n, then

hε(x, ω) =

1, x ∈ Tε

( ε2)p−np−1 − |x− εk|

p−np−1

( ε2)p−np−1 − (aε)

p−np−1

, x ∈ B ε2\Baε(εk)

0, otherwise

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and if p = n, then

hε(x, ω) =

1, x ∈ Tε

log |x− εk| − log ε2

log aε − log ε2

, x ∈ B ε2\Baε(εk)

0, otherwise

Obviously, wε − hε = 0 on Tε and ∂D. Hence

α0

∫D

(hε − wε)dx =

∫D

4pwε(hε − wε)dx

=

∫D

|∇wε|pdx−∫D

|∇wε|p−2∇wε · ∇hεdx

Then by Holder inequality and Young’s Inequality∫D

|∇wε|pdx ≤ (

∫D

|∇wε|p)p−1p · (

∫D

|∇hε|pdx)1p

+ α0

∫D

(hε − wε)dx

≤ p− 1

p

∫D

|∇wε|pdx+1

p

∫D

|∇hε|pdx+ Cα0

Thus ∫D

|∇wε|pdx ≤∫D

|∇hε|pdx+ pα0

∫D

|hε − wε|dx

≤ C

where C is a universal constant depending on n and α0.

Therefore

wε is uniformly bounded in W 1,p0 (D).

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Another important property of wε is the following:

Proposition 4.2.2. wε −→ 0 in Lp(D) as ε→ 0

To prove this fact: limε→0

∫D

|wε|pdx = 0, we need to compare wε with

vεα0,D. Roughly speaking, we will show that, near the singular points (the

holes), their liming behaviour should be very close.

First of all , we will consider some asymptotic properties of vεα0,Das

ε → 0. Since D is bounded, then without loss of generality we can assume

that D ⊂ B1. In the following we will use vε0 and vε0 to denote vεα0,B1and

min(vεα0,B1, 1). And about vε0, we have the following facts:

Lemma 4.2.3.

(i) we have that vεα0,D(x, ω) ≥ hεk(x, ω)−o(1) for a. e. x ∈ B ε

2(εk) and a. s. ω ∈

Ω, where

hεk(x, ω) =

c

1p−1γ(k, ω)

1p−1 ε

np−1 |x− εk|

p−np−1 , if 1 < p < n,

−c1

n−1γ(k, ω)1

n−1 εnn−1 log |x− εk|, if p = n.

and the constant c is the same constant as Proposition 4.1.1 (iii).

(ii) For any τ > 0, vετ (x, ω)→ 0 in Lp(B1) as ε→ 0 for a. s. ω ∈ Ω, where

vετ is defined as follows:

vετ (x, ω) = infv(x) : 4pv ≤ α0 + τ −∑

k∈Zn∩ε−1B1

γ(k, ω)εnδ(· − εk)

in B1, v ≥ 0 on B1, v = 0 on ∂B1

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and let vετ = min(vετ , 1). Hence

vεα0+τ,D= min(vεα0+τ,D

, 1) converges to 0 in Lp(D)

Proof. ( i) Let b(k, ω) = n

√ncγ(k, ω)

α0

. Then we define the function hεα0,k(x, ω)

as follows: if b ≥ 12, then

hεα0,k(x, ω) =

∫ ε

2

r

(cγ(k, ω)εns1−n − α0

ns)

1p−1ds, 0 ≤ r ≤ 1

0, r ≥ ε2

and if b ≤ 12, then

hεα0,k(x, ω) =

∫ bε

r

(cγ(k, ω)εns1−n − α0

ns)

1p−1ds, 0 ≤ r ≤ bε

0, r ≥ bε

where r = |x− εk| and x ∈ B 12ε(εk).

If b ≥ 12

, we have that ∀ x ∈ Bbε(εk) and ω ∈ Ω,

4phεα0,k

(x, ω) = α0 − γ(k, ω)εnδ(x− εk) in B ε2(εk)

and hεα,k(x, ω) = 0 if |x− εk| = ε2. Hence

4phεα0,k

(x, ω) ≥ 4pvεα0,D

(x, ω), ∀ x ∈ B ε2(εk)

and hεα0,k(x, ω) = 0 ≤ vε0(x, ω) when |x − εk| = ε

2. By comparison principle,

we have

hεα0,k(x, ω) ≤ vεα0,D

(x, ω), a. e. x ∈ B ε2(εk) and a. s. ω ∈ Ω

For the case b ≤ 12, the proof is similar as above. Thus

hεα0,k(x, ω) ≤ vεα0,D

(x, ω), a. e. x ∈ B ε2(εk) and a. s. ω ∈ Ω.

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And by direct simple computation, we know that

hεα0,k(x, ω) ≥ hεk(x, ω)− o(1), a. e. x ∈ B ε

2(εk) and a. s. ω ∈ Ω

Therefore

vεα0,D(x, ω) ≥ hεk(x, ω)− o(1), a. e. x ∈ B ε

2(εk) and a. s. ω ∈ Ω

which concludes Lemma 4.2.3 (i).

(ii) From the definition of vετ, we know that for a. e. ω ∈ Ω

−∑

k∈Zn∩ε−1B1

γ(k, ω)εnδ(· − εk) ≤ 4pvετ ≤ α0 + τ in B1

Hence

〈4pvετ , v

ετ 〉 ≥ −

∑k∈Zn∩ε−1B1

γ(k, ω)εnvετ (εk)

By (i), we have that

vετ ≥ hεk(x, ω)− o(1), a. e. x ∈ B ε2(εk) and a. s. ω ∈ Ω

which concludes that vετ (εk) = 1

Thus from integration by parts∫D

|∇vετ |pdx ≤ C

where C is a universal constant. Therefore vετ are bounded in W 1,p(B1).

From [8] and [10], we have that for a. s. ω ∈ Ω,

limε→0

|vετ = 0 ∩Br(x0)||Br(x0)|

= l(α0 + τ) > 0, for any Br(x0) ⊆ B1.

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By the Poincare-Sobolev inequality (see Lemma 4.8 in [25]), there exists

a constant C = C(α0 + τ, n) such that∫Br(x0)

|vετ |pdx ≤ Crp∫Br(x0)

|∇vετ |pdx

for any Br(x0) ⊆ B1

Since vετ are bounded in W 1,p(B1), hence∫B1

|vετ |pdx ≤ Crp

which implies that

limε→0

∫B1

|vετ |pdx = 0.

And vετ (x, ω) ≥ vεα0+τ, D(x, ω) ≥ 0 for a. e. x ∈ D and a. s. ω ∈ Ω ,

thus

limε→0

∫D

|vεα0+τ,D|pdx = 0

To finish the proof of Proposition 4.2.2 , we need to pass the limiting

property of vεα0+τ,Dto wε. First of all, we introduce a new auxiliary function

wετ as follows:

wετ (x, ω) = infv(x) : 4pv ≤ α0 + τ in Dε, v ≥ 1 on Tε and v = 0 on ∂D

Obviously, for a. s. ω ∈ Ω, wε(x, ω) ≥ wετ (x, ω) for a. e. x ∈ D and

wετ is also bounded in W 1,p(D) by Proposition 4.2.1.

More precisely, wετ satisfies the following property:

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Lemma 4.2.4.

(i) For any τ > 0 and a. s. ω ∈ Ω,

‖wε − wετ‖W 1,p(D) ≤Cδ1/(p−1), 2 ≤ p ≤ nCδ, 1 < p ≤ 2

where C depends only on p, n and α0.

(ii) For a. s. ω ∈ Ω, wετ has the following asymptotic behaviour:

limε→0

∫D

|wετ |pdx = 0.

Proof. (i) Next we apply the well-known inequality

(|ξ|p−2ξ − |η|p−2η) · (ξ − η) ≥ γ

|ξ − η|2(|ξ|+ |η|)p−2, 1 < p ≤ 2|ξ − η|p, 2 ≤ p ≤ n

for any nonzero ξ, η ∈ Rn and a constant γ = γ(n, p) > 0.

If 2 ≤ p ≤ n, then we have the following:∫D

4pwετ (x, ω)−4pw

ε(x, ω) · wε(x, ω)− wετ (x, ω)dx

=

∫D

|∇wετ |p−2∇wετ − |∇wε|p−2∇wε · ∇wετ −∇wεdx

≥ γ

∫D

|∇wετ −∇wε|pdx.

And by Holder inequality and Poincare inequality, we have∫D

τwε(x, ω)− wεδ(x, ω) dx ≤ Cτ∫D

|wε(x, ω)− wετ (x, ω)|pdx1p

≤ Cτ∫D

|∇wε(x, ω)−∇wετ (x, ω)|pdx1p

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Therefore ∫D

|∇wετ −∇wε|pdx ≤ Cτpp−1

which implies that

‖wετ − wε‖W 1,p(D) ≤ Cτ1p−1

where C depends only on p, n.

If 1 < p ≤ 2, then by Holder inequality∫D

4pwετ −4pw

ε · wε − wετdx

=

∫D

τwε(x, ω)− wετ (x, ω)dx

≥ γ

∫D

|∇(wε − wετ )|2(|∇wε|+ |∇wετ |)p−2dx

≥ γ(

∫D

|∇(wε − wετ )|pdx)2/p × (

∫D

(|∇wε|+ |∇wετ |)pdx)1−2/p

And by Holder inequality and Poincare inequality,∫D

τwε(x, ω)− wετ (x, ω)dx ≤ Cτ∫D

|∇wε(x, ω)−∇wετ (x, ω)|pdx1/p .

Then

∫D

|∇wε(x, ω)−∇wετ (x, ω)|pdx1/p

≤ Cτ(

∫D

(|∇wε|+ |∇wετ |)pdx)2/p−1

≤ Cτ(

∫D

|∇wε|pdx+

∫D

|∇wετ |pdx)2/p−1

≤ Cτ .

Therefore

‖wε − wετ‖W 1,p(D) ≤Cτ 1/(p−1), 2 ≤ p ≤ nCτ, 1 < p ≤ 2

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where C depends only on p, n and α0.

(ii) And by Lemma 4.2.3 (i) and comparison principle, we know that

0 ≤ (wετ )+ ≤ vεα0+τ, D+ o(1)

hence

limε→0

∫D

(wετ )p+dx = 0.

Next we will consider (wετ )−. In Bε/2(εk), we suppose that

supBε/2(εk)

(wετ )− > 0

Since 4pwεδ = α0 +τ in Dε, then wετ is continuous in D and so is (wετ )−.

Then if we apply Harnack inequality (see [28]) to (wετ )−, we will have that for

a. s. ω ∈ Ω,

supBε/2(εk)

(wετ )− = o(1), for ε is small

which implies that

limε→0

∫D

(wετ )p−dx = 0

Thus

limε→0

∫D

|wετ |pdx = 0

which concludes Lemma 4.2.4.

Hence by Lemma 4.2.4, we can prove Proposition 4.2.2:

limε→0

∫D

|wε|pdx = 0.

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Remark 4.2.1. By we can select a subsequence from wε such that this sub-

sequence (we still use wε to denote it) converges weakly to zero in W 1,p(D).

4.3 Proof of Lemma 3.2.1

To finish the proof, we only need to prove the Property (a)-(c) in Lemma

3.2.1.

Proof. (a) Without loss of the generality, we assume that φ ∈ D(D) and

φ ≥ 0 on D. Let θ be an any small positive number (0 < θ < 1).To prove

property (a), we need to prove the two facts:

lim supε→0

∫D∩wε≤θ

|∇wε|p′φdx ≤ C(α0, φ)θ,

and

lim supε→0

∫D∩wε>θ

|∇wε|p′φdx = 0

Now we let wεθ = (θ − wε)+, then wεθ ∈ W 1,p(D) and

wεθ converges to θ weakly in W 1,p0 (D).

And since θ < 1, then

wεθ = 0 on Tε.

From integration by parts,

limε→0

∫D

|∇wε|p−2∇wε · ∇(wεθφ)dx = −α0θ

∫D

φdx

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which implies

limε→0∫D∩wε≤θ

|∇wε|pφdx −∫D∩wε≤θ

|∇wε|p−2∇wε · ∇φ wεθdx

= α0θ

∫D

φdx

Since wε is bounded in W 1,p, then by Holder inequality,

|∫D∩wε≤θ

|∇wε|p−2∇wε · ∇φ wεθdx| ≤ C∫D∩wε≤θ

(wεθ)pdx1/p

≤ C∫D

(wεθ)pdx1/p

And wε converges to 0 in W 1,p weakly, then

limε→0∫D

(wεθ)pdx

1p = θ

Thus

lim supε→0

∫D∩wε≤θ

|∇wε|pφdx ≤ Cθ

Now let 0 < p′ < p, then by Holder inequality, we have that∫D∩wε≤θ

|∇wε|p′φdx ≤ ∫D∩wε≤θ

|∇wε|pφdxp′p ·

∫D

φdxp−p′p

Thus

lim supε→0

∫D∩wε≤θ

|∇wε|p′φdx ≤ C(α0, φ)θp′p ,

And for the integral

∫D∩wε>θ

|∇wε|p′φdx, we still apply Holder in-

equality, then∫D∩wε>θ

|∇wε|p′φdx ≤ ∫D

|∇wε|pφdxp′p ·

∫D∩wε>θ

φdxp−p′p

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And wε 0 weakly in W 1,p0 (D), then

limε→0

∫D∩wε>θ

φdx = 0

Then

lim supε→0

∫D∩wε>θ

|∇wε|p′φdx = 0

Therefore

lim supε→0

∫D

|∇wε|p′φdx ≤ C(α0, φ)θp′p

And θ is an arbitrary small positive number, so

limε→0

∫D

|∇wε|p′φdx = 0.

(b) By integration by parts, we have:

α0

∫D

φ(1− wε)dx =

∫D

∇ · (| ∇wε |p−2∇wε)φ(1− wε)dx

=

∫D

∇φ · ∇wε|∇wε|p−2(wε − 1)dx

+

∫D

φ|∇wε|pdx

Since wε goes to 0 weakly in W 1,p(D), hence

limε→0

α0

∫D

φ(1− wε)dx = α0

∫D

φdx

And wε converges to 0 strongly in Lp(D) and ∇wε is bounded in Lp(D) , hence

by Holder inequality, we have that

limε→0

∫D

∇φ · ∇wε|∇wε|p−2wεdx = 0

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Finally, by (a), we know that

limε→0

∫D

∇φ · ∇wε|∇wε|p−2dx = 0

Therefore

limε→0

∫D

φ|∇wε|pdx =

∫D

α0φdx.

(c) From integration by parts, we have that∫D

∇ · (|∇wε|p−2∇wε)vεφdx = −∫D

φ|∇wε|p−2∇wε · ∇vεdx

−∫D

vε|∇wε|p−2∇wε · ∇φdx

which concludes that

−∫D

α0vεφ =

∫D

φ|∇wε|p−2∇wε · ∇vεdx+

∫D

vε|∇wε|p−2∇wε · ∇φdx

Since vε is bounded in W 1,p0 (D) (1 < p ≤ n), then by Sobolev imbedding

theorem ( Corollary 1.58 in [28]), vε is bounded in Lq for some p < q <∞ and

‖vε‖q ≤ C‖vε‖W 1,p(D).

Hence by Holder inequality,

|∫D

vε|∇wε|p−2∇wε · ∇φdx| ≤ ∫D

|vε|q1q ∫D

|∇wε|(p−1)q′ |∇φ|q′dx1q′

where q′ = qq−1 . And q > p, then (p− 1)q′ < p, which implies that (by (a))

limε→0∫D

|∇wε|(p−1)q′|∇φ|q′dx1q′ = 0

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Hence

limε→0

∫D

vε|∇wε|p−2∇wε · ∇φdx = 0

Therefore

limε→0

∫D

|∇wε|p−2∇wε · ∇vεφdx = −α0

∫D

vφdx.

36

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Chapter 5

Lower Semicontinuous Property

5.1 Review of the Classical Case

If there is a sequence uε in W 1,p(D) ( p > 1 ) and uε −→ u weakly in

W 1,p(D), then we have the following classical lower semicontinuity property

(see [23]):

Proposition 5.1.1.

lim infε→0

∫D

|∇uε|pdx ≥∫D

|∇u|pdx.

The proof of this property is simple ( also see the proof in [23] for

more general energy functional than p-Laplacian): since uε −→ u weakly in

W 1,p(D), then

limε→0

∫D

|∇u|p−2∇u · ∇uεdx =

∫D

|∇u|pdx.

And by Young’s inequality, we have∫D

|∇u|p−2∇u · ∇uεdx ≤ (p− 1)/p

∫D

|∇u|pdx+ 1/p

∫D

|∇uε|pdx.

Now if we combine the two estimates above, we will get

lim infε→0

∫D

|∇uε|pdx ≥∫D

|∇u|pdx.

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5.2 Revised Version: Proof of Lemma 3.2.2

In this chapter, we will prove Lemma 3.2.2. And the method we used

here also can generalize the result for the case p = 2 ( see Proposition 3.1 in

[14] and [15]) .

Lemma 3.2.2. If Lemma 3.2.1 holds and uε is the solution of

min∫Rn

1

p|∇v|p − fvdx : u ∈ W 1,p

0 (D), u ≥ 0 a. e. in Tε(ω).

We assume that u0 is the weak limit of uε in W 1,p0 (D). Then

lim infε→0

F(uε) ≥ F0(u0),

where F0 is defined as follows:

F0(v) =

∫D

1

p|∇v|p +

1

pα0v

p− − fvdx, ∀ v ∈ W

1,p0 (D).

and α0 is the same constant as the one in Lemma 3.2.1.

Proof. Let us decompose uε = uε+ − uε−, where uε+ = maxuε, 0 and uε− =

max−uε, 0.

Obviously, there is a subsequence uε′ of uε such that

lim infε→0

F(uε) = limε′→0

F(uε′)

For uε′ , there is a subsequence uε′′ then uε′′

+ → u0+ weakly in W 1,p

( uε′′

− → u0− weakly in W 1,p, respectively). For simplicity, in the following we

will use uε to denote uε′

and uε′′

i. e.

lim infε→0

F(uε) = limε→0

F(uε)

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and uε+ → u0+ weakly in W 1,p ( uε− → u0− weakly in W 1,p, respectively). It is

obvious that

∫D

|∇uε|pdx =

∫D

|∇uε+|pdx +

∫D

|∇uε−|pdx and

∫D

|∇u0|pdx =∫D

|∇u0+|pdx+

∫D

|∇u0−|pdx.

For uε+, we apply the classical lower semicontinuity property:

lim infε→0

∫D

|∇uε+|pdx ≥∫D

|∇u0+|pdx

In order to prove Lemma 3.2.2, we need to prove the following revised

lower semicontinuity property:

lim infε→0

∫D

|∇uε−|pdx ≥∫D

|∇u0−|pdx+

∫D

α0(u0−)pdx

Let θ be an any (small) positive number and φ is a test function ( which

is in D(D) ). Firstly we will show that

lim infε→0

1

p

∫wε≤θ∩D

|∇uε−|pdx ≥∫D

|∇φ|p−2∇φ · ∇u0−dx

− p− 1

p

∫D

|∇φ|pdx (5.1)

In fact, by Young’s inequality, we have the following∫wε≤θ∩D

|∇φ|p−2∇φ · ∇uε−dx ≤1

p

∫wε≤θ|∇uε−|pdx+

∫wε≤θ

p− 1

p|∇φ|pdx.

Since wε converges to 0 in Lp(D), then |wε > θ| → 0 as ε goes to 0.

Hence

limε→0

∫wε>θ∩D

|∇φ|pdx = 0

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which implies that ( by Holder inequality )

limε→0

∫wε>θ∩D

|∇φ|p−2∇φ · ∇uε−dx = limε→0

∫wε>θ∩D

|∇φ|pdx = 0

Since uε− converges to u0− weakly in W 1,p(D), then we have the estimate (5.1):

lim infε→0

1

p

∫wε≤θ∩D

|∇uε−|pdx ≥∫D

|∇φ|p−2∇φ · ∇u0−dx

− p− 1

p

∫D

|∇φ|pdx

Next we will prove that

1

p

∫wε>θ∩D

|∇uε−|pdx ≥ −p− 1

p

∫D

|∇wεφ|pdx−∫D

|∇wεφ|p−2∇wε · ∇uε−φdx

− Cθ − Cθp−1p (5.2)

From Young’s inequality , we have that

−∫wε>θ∩D

|∇wεφ|p−2∇wε · ∇uε−φdx ≤p− 1

p

∫wε>θ∩D

|∇wεφ|pdx

+1

p

∫wε>θ∩D

|∇uε−|pdx

Then by the proof of Lemma 3.2.1 (a)∫wε<θ∩D

|∇wεφ|pdx ≤ Cθ

which implies that ( by Holder inequality)

|∫wε<θ∩D

|∇wεφ|p−2∇wε · ∇uε−φdx| ≤ ∫wε<θ∩D

|∇wεφ|pdxp−1p

× ∫D

|∇uε−|pdx1p

≤ Cθp−1p .

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Thus

1

p

∫wε>θ∩D

|∇uε−|pdx ≥ −p− 1

p

∫D

|∇wεφ|pdx

−∫D

|∇wεφ|p−2∇wε · ∇uε−φdx− Cθ − Cθp−1p

which concludes (5.2).

Now we combine the two estimates: (5.1) and (5.2) and apply Lemma

3.2.1, then we have the following:

lim infε→0

1

p

∫D

|∇uε−|p ≥ lim infε→0

1

p

∫wε≤θ∩D

|∇uε−|pdx+1

p

∫wε>θ∩D

|∇uε−|pdx

≥∫D

|∇φ|p−2∇φ · ∇u0−dx−p− 1

p

∫D

|∇φ|pdx

− p− 1

pα0

∫D

|φ|pdx+ α0

∫D

|φ|p−2φ · u0−dx− Cθ − Cθp−1p

Since θ is arbitary small, then

lim infε→0

1

p

∫D

|∇uε−|pdx ≥∫D

|∇φ|p−2∇φ · ∇u0−dx−p− 1

p

∫D

|∇φ|pdx

− p− 1

pα0

∫D

|φ|pdx+ α0

∫D

|φ|p−2φ · u0−dx

Then if we let φ = u0− (since the test functions are dense in W 1,p0 (D)), we have

lim infε→0

∫D

|∇uε−|pdx ≥∫D

|∇u0−|pdx+ α0

∫D

(u0−)pdx.

41

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Chapter 6

Other Topics: Parabolic Fractional p-Laplacian

6.1 Introduction

In this chapter, we will discuss a class of non local operators : parabolic

fractional p-Laplacian.

We assume that n ≥ 2, 0 < s < 1, 0 < ps ≤ n and p > 2. The

measurable function K : Rn ×Rn −→ R is symmetric in x, y for any x 6= y

and satisfies the following estimate: there are two positive constants λ and Λ

such that

λ

|x− y|n+ps≤ K(x, y) ≤ Λ

|x− y|n+ps, ∀ x, y ∈ Rn

Usually, the parabolic fractional p-Laplacian is like the following form:

ut(x, t) +

∫Rn

K(x, y)|u(x)− u(y)|p−2(u(x)− u(y))dy = 0 (6.1)

In the following we will introduce the weak solution of (6.1).

Definition 6.1.1. We say w ∈ C([a, b];L2(Rn))∩Lp([a, b];W s,p(Rn))

to be a weak solution to the parabolic fractional p-Laplacian (6.1) if and only

if ∀ η ∈ C∞0 (Rn) and t ∈ [a, b], the following holds:∫Rn

wt(x, t)η(x)dx+

∫∫Rn×Rn

K(x, y)|w(x)−w(y)|p−2(w(x)−w(y))η(x)dxdy = 0.

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Our main purpose is to prove the following regularity property for the

weak solution :

Theorem 6.1.1. Any weak solution w ∈ C([a, b];L2(Rn))∩Lp([a, b];W s,p(Rn))

to (6.1) is uniformly bounded on [t0, b] ×Rn for any a < t0 < b if the initial

data for w is bounded in Lp(Rn) .

6.2 Some Preliminary Lemmas

In order to prove Theorem 6.1.1, we need some preliminary tools. The

first one concerns the geometric convergence of sequences :

Lemma 6.2.1. Let Xnn≥0 be a sequence of positive numbers stisfying the

recurrence relation

Xn+1 ≤ CBnXn1+α

where C,B > 1 and α > 0 are given. Then there is a positive number δ0 which

depends only on C,B and α such that if X0 ≤ δ0,

limn→∞

Xn = 0.

Proof. See [29].

The second important tool is the interpolation inequality:

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Lemma 6.2.2. Let u : [a, b]×Rn −→ Rn satisfy

u ∈ Lp0([a, b];Lq0(Rn)) ∩ Lp1([a, b];Lq1(Rn))

where 1 ≤ p0, q0, p1, q1 ≤ ∞. Then there exists a constant 0 ≤ θ ≤ 1 such that

u ∈ Lp([a, b];Lq(Rn)) and

‖u‖Lp([a,b];Lq(Rn)) ≤ ‖u‖θLp0 ([a,b];Lq0 (Rn))‖u‖1−θLp1 ([a,b];Lq1 (Rn))

where1

p=

θ

p0+

1− θp1

and1

q=

θ

q0+

1− θq1

.

Proof. See [4].

6.3 Proof of Theorem 6.1.1.

In this part, we will prove that if the initial data for w is bounded in

Lp(Rn), then it is essentially bounded. Here we will use the method from [11]

and [6].

Lemma 6.3.1. There is a constant ε0 ∈ (0, 1) denpending only on n, p, s, λ

and Λ such that any weak solution w : [−2, 0]×Rn → R of (7.1), the following

is true:

If ∫ 0

−2

∫Rn

|w(x, t)|pdxdt ≤ ε0,

then

w(t, x) ≤ 1/2 on [−1, 0]×Rn.

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Proof. ∀λ > 0 and let wλ = supw − λ, 0 be the test function. Then

∂t

∫Rn

w2λ(x, t)dx+

∫Rn

∫Rn

K(x, y)|w(x)−w(y)|p−2(w(x)−w(y))(wλ(x)−wλ(y))dxdy = 0.

By simple computation, we have

∂t

∫Rn

w2λ(x, t)dx+

∫Rn

∫Rn

K(x, y)|wλ(x)− wλ(y)|pdxdy ≤ 0.

For any −2 ≤ t1 ≤ t2 ≤ 0, then∫Rn

w2λ(x, t2)dx+

∫ t2

t1

∫ ∫Rn×Rn

K(x, y)|wλ(t, x)−wλ(t, y)|pdtdxdy ≤∫Rn

w2λ(x, t1)dx.

Let Tk = (−1− 2−k), λk = 1/2− 2−k−1 and wk = (w − λk)+. For Tk−1 ≤ s ≤

Tk ≤ t, then∫Rn

w2k(x, t)dx+

∫ t

Tk

∫ ∫Rn×Rn

K(x, y)|wk(t, x)−wk(t, y)|pdtdxdy ≤∫Rn

w2k(x, s)dx.

We define

Uk4= sup

0≥t≥Tk

∫Rn

w2k(x, t)dx+

∫ 0

Tk

∫ ∫Rn×Rn

K(x, y)|wk(t, x)−wk(t, y)|pdtdxdy.

Then

Uk ≤∫Rn

w2k(x, s)dx for Tk−1 ≤ s ≤ Tk (6.2)

which implies that

Uk ≤ 2k∫ 0

Tk−1

∫Rn

w2k(x, t)dxdt. (6.3)

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Since ps ≤ n, then by Sobolev’s inequality (see [1] and [29] for fractional

Sobolev spaces), there is some positive constant q such that q > p and∫ 0

Tk−1

(

∫Rn

|wk−1(x, t)|qdx)pq dt ≤ C(n, p, s)Uk−1.

Let θ be a nonnegative number from 0 to 1 such that

θ

p=

1− θ2

q=

1

r.

Since q > p > 2, then

0 < θ < 1 and r > p > 2.

Then by Lemma 6.2.2, for any v ∈ C(Tk−1, 0;L2(Rn))∩Lp(Tk−1, 0;Lq(Rn)),

‖v‖Lr(Tk−1,0;Lr(Rn)) ≤ ( supt≥Tk−1

‖v‖L2)1−θ · ‖v‖θLp(Tk−1,0;Lq(Rn))

which concludes

‖wk−1‖Lr(Tk−1,0;Lr(Rn)) ≤ U1−θ2

k−1Uθp

k−1

i. e. ∫ ∞Tk−1

∫Rn

|wk−1(x, t)|rdx ≤ U1+r(1−θ)/2k−1 (6.4)

By (6.3), we have

Uk ≤ 2k∫ 0

Tk−1

∫Rn

w2k(x, t)dxdt ≤ 2k

∫ 0

Tk−1

∫Rn

w2k−1(x, t)χwk>0dxdt.

Hence for any a > 0, by Chebychev’s inequality we have

Uk ≤ 2k∫ 0

Tk−1

∫Rn

w2k−1(x, t)χwk−1>2−k−1dxdt

≤ 2k∫ 0

Tk−1

∫Rn

w2k−1(x, t)(

wk−12−k−1

)adxdt

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Thus

Uk ≤ 2(1+2a)k

∫ 0

Tk−1

∫Rn

w2+ak−1(x, t)dxdt. (6.5)

Since r > 2, we can let r be 2 + a for some a > 0. Using (6.4) and (6.5),

Uk ≤ 2(1+2a)kU1+αk−1

where 0 < α = r(1− θ)/2 < 1 since1− θ

2+θ

q=

1

r.

Therefore by Lemma 6.2.1, there exists some positive constant δ0, such

that if U1 ≤ δ0, then

limk→0

Uk = 0.

From the inequality (6.5) (here we choose a = p− 2), we have that

U1 ≤ 22p−3∫ 0

−2

∫Rn

|w(x, t)|pdxdt

Thus if we let ε0 = δ0/22p−3 and

∫ 0

−2

∫Rn

|w(x, t)|pdxdt ≤ ε0, then

w(t, x) ≤ 1/2 on [−1, 0]×Rn.

By Lemma 6.3.1, we have

Corollary 6.3.2. Let w : [−2, 0] × Rn → R be a weak solution to (6.1). If∫Rn

|w(−2, x)|pdxdt ≤ 2ε0 ( the constant ε0 is the same as the one in Lemma

6.3.1.), then

w(t, x) ≤ 1/2 on [−1, 0]×Rn.

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Proof. By (6.2), we have

U1 ≤∫Rn

w21(−2, x)dx.

And by Chebychev’s inequality,∫Rn

w21(−2, x)dx ≤

∫Rn

w20(−2, x)χw1(x)>0dx

≤∫Rn

|w(−2, x)|2χw0(x)>1/4dx

≤ 4p−2∫Rn

|w(−2, x)|pdx.

Thus

U1 ≤ 4p−2∫Rn

|w(−2, x)|pdx.

Hence if ∫Rn

|w(−2, x)|pdx ≤ 2ε0,

then

U1 ≤ δ0.

Therefore by Lemma 6.2.1, we have

Uk −→ 0 as k →∞

which implies the following:

w(t, x) ≤ 1/2 on [−1, 0]×Rn.

And from Corollary 6.2.2., we know that

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Corollary 6.3.3. Any weak solution w : [−2, 0]×Rn → R of (6.1) with initial

data in Lp(Rn) is uniformly bounded in [−1, 0]×Rn.

Remark 6.3.1. From Corollary 6.2.3, we can conclude Theorem 6.1.1.

49

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Chapter 7

Future Direction

7.1 Random Homogenization

For the random homogenization of p-Laplacian on perforated domain,

there are still some open problems:

Open Problem 1: We assume that the sum of the capacity of each

hole is bounded. If the holes are not periodically distributed and they can be

in any shape, then can we still have the homogenization result ?

Open Problem 2: We know that there is a very beautiful result about

periodic homogenization for more general structure than p-Laplacian (see [3]),

then can we generalize it to the random case?

7.2 Parabolic Fractional p-Laplacian

For the linear case p = 2, Caffarelli , Chan and Vasseur [6] proved that

any weak solution is locally Holder continuous. Hence for Parabolic Fractional

p-Laplacian (p > 2), we still expect the Holder continuity for the weak solution.

The key to prove the Holder continuity is to develop the De Giorgi oscillation

lemma for the bounded weak solution.

50

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Vita

Lan Tang was born in Wuhan, China on January 30th, 1980. He went

to Huazhong University of Science and Technology where he received his degree

as Master of Science in June 2004. In September 2005, he started his Ph.D

study in the University of Texas at Austin.

Permanent address: [email protected]

This dissertation was typeset with LATEX† by the author.

†LATEX is a document preparation system developed by Leslie Lamport as a specialversion of Donald Knuth’s TEX Program.

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