Adams Zekeri

SOBOLEV SPACES, EMBEDDING THEOREMS AND APPLICATIONS TO PDEs

A Thesis Presented to the Department of

Pure and Applied Mathematics

African University of Science and Technology

In Partial Fulﬁlment of the Requirements for the Degree of

Master of Science

Adams, Zekeri Uni. No 40660

Supervised by

Prof. Khalli Ezzinbi

African University of Science and Technology

P.M.B 681, Garki, Abuja F.C.T

Nigeria.

http://www.aust.edu.ng

June, 2019. ACKNOWLEDGEMENTS

My heartfelt gratitude goes to my amiable thesis supervisor professor Khalli Ezzinbi whose guid- iance and wealth of experience has contributed immensely to the success of this work. You have truly inspired me to push the limits of my mathematical reasoning not just in this work, but also in the courses you taught me and for this, I will be eternally grateful to you.

Words cannot express the depth of my gratitude to my parent and my siblings for their prayers and encouragement and to Pastor Ezekiel Omolowo. God bless you all.

Furthermore, I owe my gratitude also to my Professors; Gane Samb Lo, Charles Chidume, Ngalla Djitt`e,Dr. Ma’aruf and Dr. Usman. for their doggedness, devotion and steadfastness in teaching their courses in details which contributed to prepare me for this research work.

I am further grateful to all my course mates and friends for their immense contribution and for been there for me when i need them. God bless you all. Special thanks to my friend and Brother Francis Obah for his care, support and encouragement in the course of this programme. Also, my heartfelt gratitude goes to Agaba Regina for all she did for me.

My sincere appreciation goes to my friends MamiGrace Omateyi, Henry, Martha, Lydia, Cyril, Julius, Arnold, Chiwendu, Mercy, Jonathan Auta, Moses, Friday Okpe, Emmanuel Omateyi and all my friends who stood for me, encouraged and supported me in divers way. Special thanks goes to Abdul, Joyce, Cecilia, Baike, Joshua, Sodiq, Ogwuche, Shedrach and Patience. God bless you all.

Above all, my greatest thanks goes to the Almighty God Jehovah for the priviledge of life and the wisdom to carry out this work, but most especially for surprising me positively in ways I never imagined.

i CERTIFICATION

This is to certify that the thesis titled ”SOBOLEV SPACES, EMBEDDING THEOREMS AND APPLICATIONS TO PDEs” submitted to the school of postgraduate studies, African University of Science and Technology (AUST), Abuja, Nigeria for the award of the Master’s degree is a record of original research work carried out by Adams, Zekeri in the department of Pure and Applied Mathematics.

ii APPROVAL

SOBOLEV SPACES, EMBEDDING THEOREMS AND APPLICATIONS TO PDEs

Adams, Zekeri

A Thesis Presented to the Department of Pure and Applied Mathematics

RECOMMENDED: ......

Supervisor, Prof. Khalli Ezzinbi

APPROVED BY: ......

Chief Academic Oﬃcer, Prof. Charles Ejike Chidume

......

Date

iii COPYRIGHT

iv INTRODUCTION

The space of weak derivatives called the Sobolev Spaces which was discovered in the 19300s by the Russian Mathematician Sergei Sobolev have great importance in the theory of Partial differential equations and calculus of Variations. Image denoising refers to the process of restoring a damaged image with missing information. Mathematical models often in the form of partial differential equations are often formulated to solve this problem. The boundary conditions on this models in the form of Diritchlet and Neumann conditions helps in solving the models formulated. For example, consider the image denoising model which requires to solve the minimization problem given by 1 min (u(x) − f(x))2dx + J(∇u(x))dx , (0.0.1) u∈X(Ω) 2 ˆΩ ˆΩ n where Ω is a bounded open domain of R . Here f :Ω −→ R is a gray value image such that f(x) is the light intensity of the noisy image at point x. u :Ω −→ R is the denoised image which is required to minimize the problem (0.0.1). n J : R −→ R defined by J(|∇u|) = |∇u|p, p > 1. is required to be a convex, lower semi-continuous function in order to smoothen the observation u. X(Ω) is a space of funtions on Ω and is taken to be the space W m,p(Ω), 1 < p < ∞. Solving problem(0.0.1) is equivalent to solving the diffusion equation given by u − div(β(∇u)) = f in Ω (0.0.2) β(∇u).v = 0 on ∂Ω n n Here β : R −→ R is the subdifferential of J and v is the normal to the boundary of Ω, ∂Ω. Problem (0.0.3) describe the filtering process of the original corrupted image f. In this work, we focus on seeking the existence of weak solutions to elliptic PDEs which are of the form of problem (0.0.3). In chapter one of this work, we focussed on the study of the Lp Spaces.Also, we emphasized on relevant theroems in Linear functional analysis related to the Lp Spaces. In chapter two of this work, we study the theory of weak derivatives and distributions which lead us to the study on Sobolev spaces. Also, relevant theorems and concepts in Sobolev spaces was discussed in the chapter. In chapter three, we show the existence of a weak solution to some linear and nonlinear elliptic PDEs. Some of the elliptic partial differential equations to be considered are of Dirichlet and Neumann boundary conditions given below. Example 0.0.1 −∆u + u = f in Ω (0.0.3) u = 0 on ∂Ω

v This is a Dirichlet problem with homogeneous boundary condition.

Example 0.0.2

−∆u + u = f in Ω ∂u (0.0.4) ∂n = 0 on ∂Ω This a Neumann problem with homogeneous boundary condition.

Example 0.0.3

−∆u + u = f in Ω, f ∈ L2(Ω) ∂u 2 (0.0.5) ∂n = g on ∂Ω, g ∈ L (∂Ω). This a Neumann problem with non homogeneous boundary condition.

Example 0.0.4

−∆u = f in Ω, f ∈ L2(Ω) ∂u (0.0.6) u = 0 on Γ0, ∂n = 0 on Γ1. This is a mixed problem with Dirichlet and Neumann homogeneous boundary conditions.

vi DEDICATION

To God almighty and to my late Father.

vii CONTENTS

Acknowledgmenti

Certiﬁcation ii

Approval iii

Introduction vi

Dedication vii

1 LP - spaces 2 1.1 Deﬁnition and basic properties...... 2 1.2 Main properties of LP - spaces...... 5 1.3 Convolution...... 10

2 Sobolev spaces and Embedding Theorems 13 2.1 Theory of distributions...... 13 2.2 Derivatives in distribution sense...... 15 2.3 Sobolev spaces...... 18 2.3.1 Basic deﬁnitions...... 18 m,p 2.3.2 The space W0 (Ω)...... 20 2.3.3 Extension operator...... 23 2.3.4 Trace Theorem...... 23 2.4 Embedding Theorems...... 24

3 Elliptic Partial Differential Equations 33 3.1 Definitions and existence Theorem...... 34 3.2 Applications to elliptic partial differential equations...... 35 3.3 Monotone operators...... 42 3.3.1 Yosida approximation...... 44 3.4 Application to nonlinear elliptic partial differential equations...... 44

1 CHAPTER 1

LP - SPACES

Introduction The Lp- spaces plays an important role in mathematics, especially in functional analysis. It is a space of functions that was discovered by Frigyes Riesz (1910). These spaces plays a vital role in measure and probability spaces. They also have great importance in the theoretical discussion of problems in physics, statistics, ﬁnance, engineering and other disciplines.

In statistics, problems on regression can be viewed as those of computing orthogonal projections on subspaces of L2 spaces.

In physics,the space L2 which is a Hilbert space is central to many applications especially in quan- tum mechanics in view of the fact that it helps in the interpretation of the wave functions which are been interpreted as a probability distribution functions.

The Lp- spaces are also vital in the theory of partial diﬀerential equations(PDEs) whose solutions are in the Sobolev spaces, W m,p. These spaces consists of functions, u, in Lp whose derivatives of order less than or equal to m are also in Lp.

The objective of this chapter is to acquint us with the study on Lp- spaces and discuss some important properties of the Lp- spaces needed in this work. Also relevant Theorems on Lp- spaces in linear functional analysis needed in this work will be discussed.

n In the following, Ω is an open subset of R with the lebesgue measure dx.

1.1 Deﬁnition and basic properties

p Deﬁnition 1.1.1 [6] Let p ∈ R, 1 ≤ p < ∞. We deﬁne the L -space by

p p L (Ω) = {f :Ω −→ R measurable such that |f| dx < ∞} ˆΩ provided with the following norm

1 p p kfkLp = |f| dx , for all f ∈ Lp(Ω), and 1 ≤ p < ∞. ˆΩ For p = ∞, we deﬁne L∞(Ω) by ∞ L (Ω) = {f :Ω −→ R measurable such that there exists a constant C > 0, satisfying |f(x)| ≤ C a.e in Ω} provided with the norm

2 kfkL∞ = ess sup |f(x)| = inf{C > 0 : |f(x)| < C a.e in Ω}. x∈Ω Remark 1.1.2 Let 1 ≤ p ≤ ∞. The conjugate exponent q of p is deﬁned by 1 1 p + q = 1. Lemma 1.1.3 [6] Let λ ∈ (0, 1). Then,

xλ ≤ (1 − λ) + λx, for x > 0.

Proof . Set f(x) = (1 − λ) + λx − xλ. Then, we have that f 0 (x) = λ − λxλ−1. We obtain that x = 1 is a critical point(in fact, a minimum pont). Hence, f(1) = 0 ≤ f(x). Then, 0 ≤ (1 − λ) + λx − xλ. Hence, we have that xλ ≤ (1 − λ) + λx.

Lemma 1.1.4 [6] Let a, b ≥ 0 and λ ∈ (0, 1), then the following inequality holds aλb1−λ ≤ λa + (1 − λ)b.

a Proof . For a = 0 or b = 0, the inequality holds. Now, a, b > 0 and from Lemma 1.1.3, set x = b . Then, we obtain that

a λ a b ≤ (1 − λ) + λ b . Thus, we obtain that aλb1−λ ≤ λa + (1 − λ)b.

Lemma 1.1.5 [6] (Cauchy Young’s Inequality) 1 1 Let a, b ≥ 0. Then, for p + q = 1, we have that ap bq ab ≤ p + q . Proof . The result is trivial if either a = 0 or b = 0. Let a, b > 0. Then,

ab = exp(logab) = exp(loga + logb) 1 1 = exp( logap + logbq). p q By convexity of the exponential funtion, we get that 1 1 ab ≤ exp(logap) + exp(logbq) p q 1 1 ab ≤ ap + bq. p q

3 Theorem 1.1.6 [6] (H¨older’sinequality) 1 1 p q Suppose that 1 ≤ p ≤ ∞ and 1 ≤ q ≤ ∞ with p + q = 1. If f ∈ L (Ω) and g ∈ L (Ω), then fg ∈ L1(Ω). Moreover,

kfgkL1 ≤ kfkLp kgkLq .

Proof . From Lemma (1.1.4), we have that |f|p |g|p 1 Set a = p , b = q , λ = p , f 6= 0, g 6= 0, for all x ∈ Ω. Then, kfkLp kgkLq

|f| |g| 1 |f|p 1 |g|q . ≤ p + q . kfk p kgk q p q L L kfkLp kgkLq Integrating both sides, we get that

1 1 p 1 q |f||g|dx ≤ p |f| dx + q |g| dx kfkLp kgkLq ˆΩ pkfkLp ˆΩ qkgkLq ˆΩ p q kfkLp kgkLq = p + q pkfkLp qkfkLq 1 1 = + p q = 1.

Hence, fg ∈ L1(Ω). Furthermore,

1 |fg|dx ≤ 1. kfkLp kgkLq ˆΩ Consequently, we have that

kfgkL1 ≤ kfkLp kgkLq .

Remark 1.1.7 [1](Generalized H¨older’sinequality) Assume f , f , ..., f are functions such that f ∈ Lpi , for 1 ≤ i ≤ k with 1 = 1 + 1 + .. + 1 = 1. 1 2 k i p p1 p2 p p k Then, the product f = f1.f2...fk belongs to L and

kfkLp ≤ kf1kLp1 kf2kLp2 ...kfkkLpk .

1 1 n−1 n−1 Let (n − 1) non-negative functions g1, g2, ..., gn ∈ L (Ω) be given. Then gi ∈ L (Ω) for each i. By the generalized Holder’s inequality, we obtain that,

n−1 1 1 1 Y 1 g n−1 g n−1 ...g n−1 dx ≤ kg k n−1 (Ω). ˆ 1 2 n−1 i L1 Ω i=1

Theorem 1.1.8 [6] (Minkowski’s inequality) Let 1 ≤ p ≤ ∞ and f, g ∈ Lp(Ω). Then,

kf + gkLp ≤ kfkLp + kgkLp . Proof . If f + g = 0 a.e on Ω, the inequality holds. Assume f + g 6= 0 a.e on Ω. Then,

|f + g|p = |f + g||f + g|p−1 ≤ (|f| + |g|) |f + g|p−1.

4 integrating, we have that

|f + g|pdx ≤ (|f| + |g|)|f + g|p−1dx. ˆΩ ˆΩ By H¨older’s inequality, we obtain that

p p q |f + g| dx ≤ (kfkLp + kgkLp ) kf + gkLp ˆΩ p p q kf + gkLp ≤ (kfkLp + kgkLp ) kf + gkLp 1 p(1− q ) kf + gkLp ≤ kfkLp + kgkLp . Consequently, we have that

kf + gkLp ≤ kfkLp + kgkLp .

Lemma 1.1.9 [6] (Lp interpolation inequality) r t 1 a 1−a Let 1 ≤ r ≤ s ≤ t ≤ ∞ and suppose that u ∈ L (Ω) ∩ L (Ω). Then, for s = r + t , we have that, a 1−a kukLs ≤ kukLr kukLt .

Proposition 1.1.10 [3] If µ(Ω) < ∞, then for 1 ≤ p < q ≤ ∞, Lq(Ω) ,→ Lp(Ω).

1.2 Main properties of LP - spaces

p p Deﬁnition 1.2.1 [6] Let (fn)n≥0 be a sequence in L (Ω). Then, (fn)n≥0 converges to f ∈ L (Ω) if

kfn − fkLp −→ 0 as n → ∞.

Theorem 1.2.2 [1] (Monotone convergence Theorem (MCT)) 1 Let (fn)n≥0 be a sequence of functions in L (Ω) that satisﬁes

f1 ≤ f2 ≤ ... ≤ fn ≤ fn+1 ≤ ... a.e on Ω. Moreover, if

kfnkL1 ≤ C < ∞, for all n ∈ N. Then,

1 lim fn(x) = f(x) exists with f ∈ L (Ω) and kfn − fkL1 −→ 0 as n → ∞. n→∞

Theorem 1.2.3 [1] (Dominated convergence Theorem(DCT)) p Let(fn)n≥0 be a sequence of functions in L (Ω) such that fn(x) −→ f(x) a.e on Ω as n → ∞. If p there exist a function g ∈ L (Ω) such that |fn(x)| ≤ |g(x)| a.e on Ω for all n ∈ N. Then fn −→ f p in L (Ω), that is, kfn − fk −→ 0 as n → ∞.

Theorem 1.2.4 [1] (Fatou-Lemma) 1 Let (fn)n≥0 be a sequence of functions in L (Ω) satisfying: (i) for all n, fn ≥ 0, a.e. (ii)kfnkL1 < ∞. 1 (iii) lim inf fn(x) = f(x) a.e. on Ω. Then, f ∈ L (Ω) and n→∞

5 f = lim inf fn(x) ≤ lim inf fn(x). ˆΩ ˆΩ n→∞ n→∞ ˆΩ Theorem 1.2.5 [6] Lp(Ω) is a Banach space for any p ∈ [1, ∞].

Proof . Let > 0 be given. Case I: for p = ∞, we show that L∞(Ω) is a Banach space. ∞ Let (un)n≥0 be a Cauchy sequence in L (Ω). Then, there exists N1 ∈ N such that,

kun − umkL∞ = supx∈Ω|un(x) − um(x)| < for all n, m ≥ N1 Hence, for each x ∈ Ω,

|un(x) − um(x)| < for all n, m ≥ N1.

Hence, (un(x))n≥0 is a Cauchy sequence in R a.e in Ω and converges to u a.e in Ω. Fix n and letting m → ∞, we get that,

|un(x) − u(x)| < for all n ≥ N1, which implies that,

kun − ukL∞ = supx∈Ω|un(x) − u(x)| < for all n ≥ N1. ∞ ∞ Hence, un −→ u in L (Ω). Thus, L (Ω) is Banach. p Case II: for 1 ≤ p < ∞. Let (un)n≥0 be a Cauchy sequence in L (Ω). Then, there exists N ∈ N such that,

kun − umkLp < , for all n, m ≥ N.

Then, by construction we get a subsequence (unk )k≥0 such that, 1 p kunk − unk+1 kL < 2k for k ≥ 0. Pn Deﬁne, gn = k=1 |unk (x) − unk+1 (x)|. p p 1 Now, gn is non-negative monotone increasing sequence in L (Ω). Thus, gn ∈ L (Ω) for all n ≥ 0. Also, we have that

p p kgnkL1 = kgnkLp n !p X ≤ kunk − unk+1 k k=1 n !p X 1 < 2k k=1 ∞ !p X 1 < 2k k=1 = 1.

p p 1 p p Hence, kgnkL1 ≤ 1. Then, by MCT 1.2.2, there exists g ∈ L (Ω) such that gn(x) −→ g (x) a.e in Ω. Hence, g ∈ Lp(Ω). Now, for r > m, we have that

|unm − unr | ≤ |unm − unm+1 | + |unm+1 − unm+2 | + ... + |unr−1 − unr | r−1 m−1 X X = |unk − unk+1| − |unk − unk+1| k=1 k=1

= gr−1 − gm−1 −→ 0 as r, m → ∞.

Hence, (unk )k≥0 is Cauchy in R a.e in Ω. Thus,

6 unk (x) −→ u(x) a.e in Ω. Furthermore,

|unm − unr | ≤ gr−1 − gm−1 ≤ gr−1 ≤ g. Fix m and letting r → ∞, we get that

|unm − u| ≤ g. Consequently, we get that

p p |unm − u| ≤ g . This implies that,

p 1 |unm − u| ∈ L (Ω).

p p Hence, unm − u ∈ L (Ω). Thus, u ∈ L (Ω). p p p 1 Since, |unm − u| −→ 0 a.e in Ω and |unm − u| ≤ g ∈ L (Ω). Then, DCT 1.2.3 , we get that,

p p kunk − ukLp = k(unk − u) kL1 −→ 0 as k → ∞. p That is, unk −→ u in L (Ω). p Since (unk )k≥0 is a subsequence of the Cauchy sequence (un)n≥0 and converges to u ∈ L (Ω). p Hence, un −→ u in L (Ω).

Theorem 1.2.6 [3] (Converse of DCT) p p Let (fn)n≥0 be a sequence of functions in L (Ω) such that fn −→ f in L (Ω) for 1 ≤ p < ∞. Then, p there exists a subsequence (fnk )k≥0 of (fn)n≥0 and g ∈ L (Ω) such that |fnk (x)| ≤ |g(x)| a.e. on

Ω and fnk (x) −→ f(x) a.e. on Ω.

p Theorem 1.2.7 [6] For 1 ≤ p < ∞, suppose that (fn)n≥0 ⊂ L (Ω) and fn(x) −→ f(x) a.e on Ω. p If kfnkLp −→ kfkLp as n → ∞, then kfn − fkLp −→ 0 as n → ∞, that is, fn −→ f in L (Ω).

Proof . We know that for a, b ≥ 0 and 1 ≤ p < ∞ , that

a+b p 1 p p 2 ≤ 2 (a + b ). Hence,

(a + b)p ≤ 2p−1(ap + bp).

Thus, we get that

|a − b|p ≤ |a + b|p ≤ 2p−1(|a|p + |b|p).

Let a = fn, b = f. Then,

p p−1 p p |fn − f| ≤ 2 (|fn| + |f| ), and

p−1 p p p 0 ≤ 2 (|fn| + |f| ) − |fn − f| .

7 Since fn(x) −→ f(x) a.e. on Ω. Then,

p p p−1 p p p 2 |f| = lim 2 (|fn| + |f| ) − |fn − f| ˆΩ ˆΩ n→∞ p−1 p p p = lim inf 2 (|fn| + |f| ) − |fn − f| . ˆΩ n→∞ By Fatou’s Lemma 1.2.4, we get that p−1 p p p ≤ lim inf 2 (|fn| + |f| ) − |fn − f| n→∞ ˆΩ p p−1 p p p p lim inf |fn − f| ≤ lim inf 2 |fn| + |f| − 2 |f| . n→∞ ˆΩ n→∞ ˆΩ ˆΩ

Since kfnkLp −→ kfkLp as n → ∞. Then, we have that

p p−1 p p p p lim inf kfn − fk p ≤ 2 lim inf kfnk p + kfk p − 2 kfk p . n→∞ L n→∞ L L L Hence, we have that lim inf kfn − fkLp = 0. n→∞ Furthermore, we get that p p p−1 p p p 2 |f| ≤ lim inf 2 (|fn| + |f| ) − |fn − f| . ˆΩ n→∞ ˆΩ p−1 p p p ≤ lim sup 2 (|fn| + |f| ) − |fn − f| . n→∞ ˆΩ p p−1 p p p p lim sup |fn − f| ≤ 2 lim sup |fn| + |f| − 2 |f| . n→∞ ˆΩ n→∞ ˆΩ ˆΩ ˆΩ

Since kfnkLp −→ kfkLp . Then, we have that p p−1 p p p p lim sup kfn − fkLp ≤ 2 lim sup kfnkLp + kfkLp − 2 kfkLp . n→∞ n→∞

Consequently, lim sup kfn − fkLp = 0. n→∞ Hence, we get

lim sup kfn − fkLp = lim inf kfn − fkLp = 0, n→∞ n→∞ which implies that,

lim kfn − fkLp = 0. n→∞

Theorem 1.2.8 [2] Let 1 ≤ p < ∞. Then, the dual of Lp(Ω) denoted (Lp(Ω))∗ = Lq(Ω), where 1 1 p + q = 1. Proof . Let g ∈ Lq(Ω) and for every f ∈ Lp(Ω). Consider the map

p Tg : L (Ω) −→ R deﬁned as p Tg(f) = fgdx for all f ∈ L (Ω). ˆΩ Clearly, Tg is linear. Furthermore,

|Tg(f)| ≤ |fg| ≤ kfkLp kgkLq , ˆΩ

8 p ∗ which shows that Tg is well-deﬁned and bounded. Hence, Tg ∈ (L (Ω)) and

kTgk(LP )∗ ≤ kgkLq .

Now, let h ∈ Lq(Ω) be arbitrary. Deﬁne

sgn(h)|h|q−1 g˜ = q−1 . khkLq Then, we have that |h|p(q−1) |h|q |g˜|p ≤ = = 1. p(q−1) q ˆΩ ˆΩ ˆΩ khk q khkLq L Hence,g ˜ ∈ Lp(Ω). Furthermore,

hsgn(h)|h|q−1 |h|q T (˜g) = = = khk q . h ˆ q−1 ˆ q−1 L Ω khkLq Ω khkLq Hence, we have that Th(˜g) = khkLq . Consequently, we have that kThk(Lp)∗ ≥ khkLq . Since h ∈ Lq(Ω) is arbitrary. Then, in particular for h = g ∈ Lq(Ω), we have that

kTgk(Lp)∗ ≥ kgkLq .

Consequently, we have that kTgk(Lp)∗ = kgkLq , which shows that (Lp(Ω))∗ is isometrically isomorphic to Lq(Ω). This complete the proof.

p p Proposition 1.2.9 [6] Let (un)n≥0 be a sequence in L (Ω) such that un −→ u in L (Ω). Then p un * u in L (Ω).

p Proof . Given that un −→ u in L (Ω). Then kun − ukLp −→ 0 as n → ∞. Let v ∈ Lq(Ω). Then,

unv − uv = (un − u)v . ˆΩ ˆΩ ˆΩ By Holder’s inequality, we obtain that

unv − uv ≤ kun − ukLp kvkLq −→ 0 as n → ∞ ˆΩ ˆΩ

Hence,

unv −→ uv as n → ∞ ˆΩ ˆΩ

p which implies that, un * u in L (Ω).

p Theorem 1.2.10 [6] Suppose 1 ≤ p < ∞ and un * u in L (Ω). Then, (un)n≥0 is bounded in Lp(Ω) and

9 kukLp(Ω) ≤ lim inf kunkLp . n→∞

p Theorem 1.2.11 [6] Supppose 1 ≤ p < ∞ and the sequence (un)n≥0 is bounded in L (Ω). Then, p there is a subsequence still denoted by (un)n≥0 and a function u ∈ L (Ω) such that p un * u in L (Ω).

Deﬁnition 1.2.12 [2] A Banach space E is called uniformly convex if for any ∈ (0, 2] there exists a δ = δ() ∈ (0, 1) such that for all x, y ∈ E with kxk ≤ 1, kyk ≤ 1 and kx − yk ≥ , then 1 k 2 (x + y)k ≤ 1 − δ. Theorem 1.2.13 [2] (Milman Pettis) Every uniformly convex real Banach space E is reﬂexive.

Theorem 1.2.14 [2] (Clarkson’s inequality) Let f, g ∈ Lp(Ω). Then, 1 f+g q f−g q 1 p 1 p p−1 (i)k 2 kLp + k 2 kLp ≤ 2 kfkLp + 2 kgkLp , 1 < p ≤ 2. f+g p f−g p 1 p 1 p (ii)k 2 kLp + k 2 kLp ≤ 2 kfkLp + 2 kgkLp , for 2 ≤ p < ∞. Theorem 1.2.15 [1] Let 1 < p < ∞. Then, Lp(Ω) is uniformly convex. Hence, it is reﬂexive.

p Proof . Let ∈ (0, 2], f, g ∈ L (Ω) with kfkLp ≤ 1, kgkLp ≤ 1 and kf − gkLp ≥ . Then, for 2 ≤ p < ∞, by Clarkson’s inequality, we get that

f+g p p k 2 kLp ≤ 1 − ( 2 ) 1 f+g p p k 2 kLp ≤ 1 − 1 − ( 2 )

1 p p We choose δ = 1 − ( 2 ) . Hence, for ∈ (0, 2], with kf − gk ≥ , then f+g k 2 kLp ≤ 1 − δ. Consequently, Lp(Ω) is uniformly convex for 2 ≤ p < ∞. Furthermore, for 1 < p ≤ 2. p Let ∈ (0, 2], f, g ∈ L (Ω) with kfkLp ≤ 1, kgkLp ≤ 1 and kf − gkLp ≥ . Then, by Clarkson’s inequality

f+g q q k 2 kLp ≤ 1 − ( 2 ) . 1 f+g q q k 2 kLp ≤ 1 − 1 − ( 2 ) .

1 q q We choose δ = 1 − ( 2 ) . Hence, for ∈ (0, 2], with kf − gk ≥ , then f+g k 2 kLp ≤ 1 − δ. Consequently, Lp(Ω) is uniformly convex for 1 < p ≤ 2. Thus, by Milman Pettis 1.2.13, Lp(Ω) is reﬂexive for 1 < p < ∞.

1.3 Convolution

Theorem 1.3.1 [1] (Tonelli) Let F :Ω1 × Ω2 −→ R be a measurable function such that (i)

|F (x, y)|dµ2 < ∞ for a.e x ∈ Ω. ˆΩ2 (ii)

dµ1 |F (x, y)|dµ2 < ∞. ˆΩ1 ˆΩ2

10 1 Then, F ∈ L (Ω1 × Ω2). Theorem 1.3.2 [1] (Fubini) Assume that F ∈ L1(Ω × Ω ). Then, for a.e x ∈ Ω ,F (x, y) ∈ L1(Ω ) and F (x, y)dµ ∈ 1 2 1 y 2 Ω2 2 1 1 Lx(Ω1). Similarly, for a.e. y ∈ Ω2, F (x, y) ∈ Lx(Ω1) and ´

1 F (x, y)dµ1 ∈ Ly(Ω2). ˆΩ1 Moreover, we have that

dµ1 F (x, y)dµ2 = dµ2 F (x, y)dµ1 = F (x, y)dµ1dµ2. ˆΩ1 ˆΩ2 ˆΩ2 ˆΩ1 ˆΩ1 ˆΩ2

1 n p n Deﬁnition 1.3.3 [1] Let f ∈ L (R ) and g ∈ L (R ), with 1 ≤ p ≤ ∞. we deﬁne the convolution of f and g denoted as f ∗ g as follows

(f ∗ g)(x) = f(x − y)g(y)dy. ˆRn

1 n p n p n Theorem 1.3.4 [1] Let f ∈ L (R ) and g ∈ L (R ). Then, f ∗ g ∈ L (R ). Moreover,

kf ∗ gkLp ≤ kfkL1 kgkLp .

Proof . Case I: for p = ∞.

(f ∗ g)(x) = f(x − y)g(y)dy ˆRn

≤ supy∈Rn |g(y)| |f(x − y)|dy ˆRn kf ∗ gkL∞ ≤ kfkL1 kgkL∞

Case II: for p = 1.

(f ∗ g)(x) = f(x − y)g(y)dy. ˆRn Then,

|(f ∗ g)(x)| ≤ |f(x − y)||g(y)|dy ˆRn n Set h(x, y) = f(x − y)g(y) for a.e y ∈ R . Thus,

|h(x, y)|dx = |g(y)| |f(x − y)|dx < ∞. ˆRn ˆRn Also,

dy |h(x, y)|dx = |g(y)|dy |f(x − y)|dx = kgkL1 kfkL1 < ∞. ˆRn ˆRn ˆRn ˆRn 1 n n Hence, by Fubini’s Theorem 1.3.2, h ∈ L (R × R ). We apply Tonelli’s Theorem 1.3.1 to obtain that,

|(f ∗ g)(x)|dx ≤ |g(y)|dy |f(x − y)|dx. ˆRn ˆRn ˆRn

11 That is,

kf ∗ gkL1 ≤ kfkL1 kgkL1 . Case III: 1 < p < ∞.

(f ∗ g)(x) = f(x − y)g(y)dy. ˆRn But,

1 1 1 1 |f(x − y)| = |f(x − y)| p |f(x − y)| q , where + = 1. p q 1 1 |f(x − y)||g(y)| = |f(x − y)| p |f(x − y)| q |g(y)|

1 1 |f(x − y)||g(y)|dy = |f(x − y)| p |f(x − y)| q |g(y)|dy. ˆRn ˆRn Thus, by Holder’s inequality we get that,

1 1 q p |(f ∗ g)(x)| ≤ |f(x − y)|dy |f(x − y)||g(y)|pdy ˆRn ˆRn 1 1 p q p = kfkL1 |f(x − y)||g(y)| dy ˆRn p p q p |(f ∗ g)(x)| ≤ kfkL1 |f(x − y)||g(y)| dy ˆRn p p q p |(f ∗ g)(x)| dx ≤ kfkL1 |f(x − y)||g(y)| dydx . ˆRn ˆRn ˆRn By Tonelli’s Theorem 1.3.1 we obtain that,

p p q p kf ∗ gkLp ≤ kfkL1 kfkL1 kgkLp Then,

1 1 q p p p kf ∗ gkL ≤ kfkL1 kfkL1 kgkL kf ∗ gkLp ≤ kfkL1 kgkLp .

12 CHAPTER 2

SOBOLEV SPACES AND EMBEDDING THEOREMS

Introduction The Sobolev space, W m,p is a vector space consisting of functions, u in Lp with derivatives of order less than or equal to m in Lp. These spaces is equiped with a norm that is a combination of the norm of the function, u in Lp and that of the derivatives up to order m.

The space which was discovered in the 1930’s by the Russian mathematician Sergei Sobolev was introduced mainly for the need of the theory of partial diﬀerential equations(PDEs). These spaces are special class of distribution which makes them closely related with the theory of distributions. Also, they spaces is well connected with the spaces of continuous and uniformly continuous functions.

The Sobolev spaces have great importance in the theory of partial differential equations and its applications in mathematical physics. Solutions to partial differential equations are well situated in the Sobolev spaces. Furthermore, the Sobolev spaces are vital tools in the theory of approximations, spectral theory and differential geometry.

This chapter is aimed at exposing us to the Sobolev spaces in details. Also, the theory of distributions will be discussed in this chapter. The sobolev inequalities mainly the Morrey’s inequality and the Gagliardo-Nirenberg inequality which are vital in the theory of the Sobolev spaces will be discussed . Furthermore, the theory of embeddings in Sobolev spaces are of great importance. We shall discuss the general sobolev embeddings and compact embeddings of Sobolev spaces such as the Rellich-Kondrachov compact embedding theorem will be discussed.

n In the following Ω is an open subset of R and ∂Ω is the boundary of Ω.

2.1 Theory of distributions

Definition 2.1.1 [3] We define a multi-index, α = (α1, α2, ..., αn) as an n-turple of non-negative integer numbers. Its length is defined as follows

|α| = α1 + α2 + ... + αn. Each multi index α determines a partial diﬀerential operator of order |α| namely

α ∂α1+α2+...+αn D u = ( α1 α2 αn )u. ∂x1 ∂x2 ...∂xn

Given two multi-indices, α = (α1, α2, ..., αn) and β = (β1, β2, ..., βn), for β ≤ α (means βi ≤ αi, ∀i = 1, 2, ..., n). Then,

13 α = α! = α1! ... αn! . β β!(α−β)! β1!(α1−β1)! βn!(αn−βn)!

Definition 2.1.2 [3] We define the support of a continuous function, φ, denoted by, supp(φ) as the closure of the set {x ∈ Ω: φ(x) 6= 0}. In general, we define the support of a function, f, as the complement of union of all open sets, θ, in Ω such that f = 0 a.e. in θ.

Deﬁnition 2.1.3 [3] We deﬁne D(Ω) as the space of C∞ functions with compact support in Ω . It is usually called test functions.

n Deﬁnition 2.1.4 [3] Let Ω ⊂ R and K be a compact subset of Ω. We deﬁne

DK (Ω) = {φ ∈ D(Ω) : supp(φ) ⊆ K}.

p Deﬁnition 2.1.5 [3] We deﬁne Lloc(Ω), 1 ≤ p < ∞ as the set of all measurable functions u in Ω such that |u(x)|pdx < ∞, ˆK for any K compact set contained in Ω.

Deﬁnition 2.1.6 [3] Let T : D(Ω) −→ R be linear. Then, T is a distribution on Ω if for every compact set K ⊆ Ω, there exists C > 0 and n ∈ N such that

|T (φ)| ≤ CkφkCn , for all φ ∈ DK (Ω),

α where kφkCn = sup |D φ(x)|. x∈k,|α|≤n

Deﬁnition 2.1.7 [3] We deﬁne D0 (Ω) the space of distributions on Ω.

1 Example 2.1.8 [3] Let f ∈ Lloc(Ω). Then, Tf : D(Ω) −→ R deﬁned as follows

Tf (φ) = f(x)φ(x)dx, for all φ ∈ D(Ω) ˆΩ is a distribution on Ω. Proof . Indeed, let λ ∈ R and φ, ψ ∈ D(Ω), then

Tf (λφ + ψ) = f(x)(λφ(x) + ψ(x))dx ˆΩ = λTf (φ) + Tf (ψ)

So Tf is linear. Furthermore, let K ⊆ Ω be compact such that supp(φ) ⊆ K. Then, |Tf (φ)| ≤ |f(x)|dx kφkc0 for allφ ∈ DK (Ω). ˆK

Example 2.1.9 [3](Dirac distribution) Let δ : D(R) −→ R deﬁned by hδ, ψi = φ(0) for all φ ∈ D(R). Then, δ is a distribution called the Dirac distribution. Clearly, δ is linear. Let φ ∈ D(R), and K ⊂ R such that supp(φ) ⊆ K. Then,

14 | hδ, ψi | = |φ(0)| ≤ kφ(x)kC0 , for φ ∈ Dk(R).

P 1 Example 2.1.10 [3] Let T : D(R) −→ R deﬁned by hT, ψi = n≥1 anψ( n ) and we assume that P 0 n≥1 |an| < ∞. Then, T ∈ D (Ω). Indeed, T is well-deﬁned and linear. Let K be a compact subset of R such that supp(ψ) ⊆ K, then

X 1 | hT, ψi | = anψ( ) n n≥1 X ≤ sup |ψ(x)| |an|. x∈K n∈N∩K P Set C = n∈ ∩K |an|. N0 Hence, T ∈ D (R).

0 0 Deﬁnition 2.1.11 [3] Let (Tn)n≥1 be a sequence of distributions in D (Ω) and T ∈ D (Ω). We 0 say that Tn −→ T in D (Ω) if and only if for all φ ∈ D(Ω),

hTn, φi −→ hT, φi as n → ∞.

0 Example 2.1.12 [3] Let Tn = cosnx ∈ D (R). Let φ ∈ D(R) and a > 0 such that supp(φ) ⊆ [−a, a]. Then,

hTn, φi = cos(nx)φ(x)dx ˆR a = cos(nx)φ(x)dx ˆ−a a a 1 1 0 = sin(nx)φ(x) − sin(nx)φ (x)dx n −a n ˆ−a Now, a a 1 0 1 0

0 ≤ sin(nx)φ (x)dx ≤ φ (x) dx −→ 0 as n → ∞ n ˆ−a n ˆ−a Also,

1 a n sin(nx)φ(x) −a −→ 0 as n → ∞ It follows that;

hTn, φi −→ 0 = h0, φi as n → ∞.

0 Hence, Tn −→ 0 in D (R) as n → ∞.

2.2 Derivatives in distribution sense

0 n α Deﬁnition 2.2.1 [3] Let T ∈ D (Ω) and α ∈ N . We deﬁne D T by; hDαT, φi = (−1)|α| hT,Dαφi for all φ ∈ D(Ω).

Theorem 2.2.2 [3] Let T ∈ D0 (Ω). Then, for all multi index, α we have that DαT ∈ D0 (Ω).

0 0 n Theorem 2.2.3 [7] Let (Tn)n≥0 ⊆ D (Ω) such that Tn −→ T in D (Ω). Then, for all α ∈ N ,

α α 0 D Tn −→ D T in D (Ω) as n → ∞.

15 0 Proof . Since Tn −→ T in D (Ω). Then, for all ψ ∈ D(Ω),

hTn, ψi −→ hT, ψi as n → ∞

n Let α ∈ N . Then, α |α| α hD Tn, ψi = (−1) hTn,D ψi But ψ ∈ D(Ω) implies that Dαψ ∈ D(Ω). Hence, we have that |α| α |α| α (−1) hTn,D ψi −→ (−1) hT,D ψi . Consequently, α |α| α α hD Tn, ψi −→ (−1) hT,D ψi = hD T, ψi . Hence, we obtain that, α α 0 D Tn −→ D T in D (Ω) as n → ∞.

1 Deﬁnition 2.2.4 [7] Let f ∈ Lloc(Ω) and α = (α1, α2, ..., αn) a multi-index. Then, the distribution Tf admits a distributional derivative deﬁned as

α |α| |α| α D Tf (φ) = (−1) TDαf (φ) = (−1) f(x)D φ(x)dx, for all φ ∈ D(Ω). ˆΩ

1 If there exists g ∈ Lloc(Ω) such that α Tg = D Tf , That is,

g(x)φ(x)dx = (−1)|α| f(x)Dαφ(x)dx ˆΩ ˆΩ

Then, g is called the αth derivative of f written as g = Dαf.

Example 2.2.5 [3] Consider the function ( 0, if x ≤ 0 f(x) = (2.2.1) x, if x > 0

It derivative is the Heaviside function deﬁned by, ( 0, if x ≤ 0 H(x) = (2.2.2) 1, if x > 0

1 1 th Lemma 2.2.6 [7] Let f ∈ Lloc(Ω) and suppose g, g˜ ∈ Lloc(Ω) are the weak α derivative of f such that

fDα(φ)dx = (−1)|α| gφdx = (−1)|α| gφdx˜ for all φ ∈ D(Ω) ˆΩ ˆΩ ˆΩ Then, we have that

g =g ˜ a.e in Ω.

Proof . We begin the proof by stating the following

16 1 n Lemma 2.2.7 [3] Let v ∈ Lloc(Ω) with Ω non-empty open set in R . If,

v(x)φ(x)dx = 0, for all φ ∈ D(Ω). ˆΩ Then, v = 0 a.e in Ω.

1 1 Since g, g˜ ∈ Lloc(Ω). Then g − g˜ ∈ Lloc(Ω). Let φ ∈ D(Ω). Consequently, we have that

(g − g˜)φ = 0. ˆΩ Hence, g =g ˜ a.e in Ω.

Definition 2.2.8 [6] We define Cm(Ω) the space of functions m times continuously differentiable on Ω.

n 0 Theorem 2.2.9 [7] Let Ω ⊆ R be open connected set and assume u ∈ D (Ω), such that ∂u = 0, for i = 1, 2, ..., n . Then, u is a constant function. ∂xi

n Deﬁnition 2.2.10 [6] Let w be a function with x ∈ R such that

w ∈ D(Ω), w(x) ≥ 0, w(x) = 0 if |x| ≥ 1 and w(x)dx = 1. ˆRn −n x n for ρ > 0, the function wρ(x) = ρ w( ρ ) , x ∈ R

n with wρ ∈ D(R ), wρ(x) ≥ 0, wρ(x) = 0 if |x| ≥ ρ and wρ(x)dx = 1 is called a mollifier. ˆRn

1 n Definition 2.2.11 [1] Let u ∈ Lloc(R ). Then, the mollification (or regularization) of u denoted as uρ is the convolution of u with the standard mollifier, that is,

uρ(x) = (wρ ∗ u)(x) = wρ(x − y)u(y)dy. ˆRn

k n Theorem 2.2.12 [3] Let u ∈ C (R ). Then, for all multi index α, with length |α| ≤ k we have k n α α that u ∗ wρ ∈ C (R ) and D (u ∗ wρ) = D u ∗ wρ.

Deﬁnition 2.2.13 (Absolute continuity) Let u :[a, b] −→ R. Then, u is said to be absolutely continuous if for any > 0, there exists δ > 0 such that for any ﬁnite set of disjoint intervals in [a, b]. If 0 0 Pn 0 (x1, x1), ..., (xn, xn) ⊆ [a, b] with i=1 |xi − xi| < δ, then Pn 0 i=1 |u(xi) − u(xi)| < .

Remark 2.2.14 The existence of deriavtive of a function is related to the absolute continuity of the function. Let u :[a, b] −→ R. Then, u is absolutely continuous if and only if there exists a function v ∈ L1(a, b) such that

x u(x) = u(a) + v(t)dt, x ∈ [a, b]. ˆa

17 1 0 1 Corollary 2.2.15 [7] Let u ∈ Lloc(a, b) have a distributional derivative u ∈ L (a, b). Then, there exists an absolutely continuous function u˜ such that

u˜(x) = u(x) for a.e x ∈ (a, b). u0 (x) = lim u˜(x+h)−u˜(x) . h→0 h ∞ n Corollary 2.2.16 [3](Leibnitz formula) Let u, v ∈ C (Ω) and α, β ∈ N a multi-index sets. Then, X α! Dα(uv) = DβuDα−βv. β!(α − β)! β≤α

Deﬁnition 2.2.17 [3] Let α ∈ C∞(Ω) and T ∈ D0 (Ω). Then, hαT, ψi = hT, αψi, for all ψ ∈ D(Ω)

Theorem 2.2.18 [3] Let T ∈ D0 (Ω). Then, αT ∈ D0 (Ω) for all α ∈ C∞(Ω).

Proof Let ψ ∈ D(Ω) and α ∈ C∞(Ω). Then, αψ ∈ D(Ω) (since supp(αψ) ⊆ supp(α) ∩ supp(ψ)). Let K be compact set such that supp(ψ) ⊆ K. Since T ∈ D0 (Ω), then by applying Leibnitz formula there exists C > 0 and n ∈ N such that

| hT, αψi | ≤ Ckψkcn .

2.3 Sobolev spaces

2.3.1 Basic deﬁnitions Deﬁnition 2.3.1 [1] The Sobolev space, W m,p(Ω) is the space of all functions u belonging to Lp(Ω) such that for every multi-index α, with |(α)| ≤ m, the derivative Dαu exists and belong to p m,p p α L (Ω). That is; W (Ω) = {u ∈ L (Ω) : D u ∈ Lp(Ω), |α| ≤ m}. The standard norm in W m,p(Ω) is given by

1   p X kuk = |Dαu(x)|pdx , 1 ≤ p < ∞. m,p ˆ  Ω |α|≤m X α kukm,∞ = esssupx∈Ω|D u(x)|, for p = ∞. |α|≤m

Theorem 2.3.2 [6] W m,p(Ω), is a Banach space, for p ∈ [1, ∞].

1 p α α p |D un − D uk| −→ 0 as n, k → ∞. ˆΩ α p Hence, (D un)n≥0 is a Cauchy sequence in L (Ω) for each α, such that |α| ≤ m. Since Lp(Ω) is Banach, there exist u, v ∈ Lp(Ω) such that

18 p α p un −→ u in L (Ω),D un −→ v in L (Ω). p 1 1 α 1 But L (Ω) ⊂ Lloc(Ω). Hence, un −→ u in Lloc(Ω) and D un −→ v in Lloc(Ω). Consequently, v = Dαu ∈ Lp(Ω). Hence, u ∈ W m,p(Ω). For p = ∞. Let > 0 be given. m,∞ Let (un)n≥0 be a Cauchy sequence in W (Ω). Then, there exists No ∈ N such that, P α α kun − ukkm,∞ = |α|≤m supx∈Ω|D un − D uk| < , for all n, k ≥ No. α ∞ Hence, for each α with |α| ≤ m,(D un)n≥0 is Cauchy in L (Ω). Consequently, there exist α ∞ α α v = D u ∈ L (Ω) such that D un −→ D u as n → ∞. Hence, ﬁx n and letting k → ∞, we get that P α α kun − ukm,∞ = |α|≤m supx∈Ω|D un − D u| < , for all n ≥ No Thus, u ∈ W m,∞(Ω). Hence, W m,p(Ω) is a Banach space for 1 ≤ p ≤ ∞.

P α m,p Proposition 2.3.3 [1] The norm |α|≤m kD ukLp is equivalent to the standard norm in W (Ω). Proof . Let u ∈ W m,p(Ω), 1 ≤ p < ∞. Then, we show that the norm

∗ X α kukW m,p = kD ukLp |α|≤m is equivalent to the standard norm given by

1   p X α p kukW m,p =  kD ukLp  . |α|≤m Now, we have that

1   p X α p kukW m,p =  kD ukLp  |α|≤m X α ≤ kD ukLp |α|≤m ∗ = kukW m,p It remain to show that there exists a constant k > 0 such that ∗ kukW m,p ≤ kkukW m,p (2.3.1) Suppose for contradiction that (2.3.1) does not hold. Then, for all n ∈ N, there exists a sequence m,p (un)n≥1 in W (Ω) such that ∗ kunkW m,p > nkunkW m,p . Hence, we have that kunkW m,p 1 ∗ < . kunkW m,p n un ∗ 1 Set vn = ∗ . Then, kvnkW m,p = 1 and kvnkW m,p < −→ 0 as n → ∞. kunkW m,p n Thus, for all multi index, α with length |α| ≤ m, we have that α kD vnkLp −→ 0 as n → ∞. Consequently, we have that ∗ kvnkW m,p −→ 0 as n → ∞. ∗ But, kvnkW m,p = 1. Hence, we have that 1 = 0 (condradiction). Thus, (2.3.1) holds. This complete the proof.

19 Deﬁnition 2.3.4 [1] We consider a special case of the Sobolev space for p = 2 denoted by Hm(Ω) = W m,2(Ω). It is a Hilbert space endowed with the following inner product,

X α α hu, vi m = D u.D vdx. H ˆ |α|≤m Ω

m m,2 Similarly, we deﬁne H0 (Ω) = W0 (Ω).

m,p Deﬁnition 2.3.5 [6] We deﬁne the space Wloc (Ω) as the space of functions which are locally in W m,p(Ω). That is,

m,p p α p Wloc (Ω) = {u ∈ Lloc(Ω) : D u ∈ Lloc(Ω), for all α such that |α| ≤ m}. Proposition 2.3.6 [7] Let (a, b) be an open interval and u ∈ W 1,p(a, b). Then, there exists an 0 p absolutely continuous function f :(a, b) −→ R having derivative f ∈ L (a, b) which coincides with u ∈ W 1,p(a, b).

1,p p 1 Proof . Let u ∈ W (a, b), then u ∈ L (a, b) ⊆ Lloc(a, b). Then, by Corollary 2.2.15, there is an absolutely continuous function f = u a.e on Ω and f 0 = u0 ∈ Lp(a, b).

m,p Proposition 2.3.7 [1] Let u ∈ W0 (Ω), for p ∈ [1, ∞] and let ( u(x) if x ∈ Ω u˜(x) = (2.3.2) n 0 if x ∈ R \Ω

m,p n Then, u˜ ∈ W (R ).

m,p Proof . Let u ∈ W0 (Ω). Then, there exists a sequence (un)n≥0 ⊆ D(Ω), such that un −→ u in W m,p(Ω) as n → ∞. Now, ( u (x) if x ∈ Ω u˜ (x) = n (2.3.3) n n 0 if x ∈ R \Ω n Then,u ˜n ∈ D(R ), for all n ≥ 0. Furthermore,

ku˜n − u˜kW m,p(Rn) = kun − ukW m,p(Ω) −→ 0 as n → ∞. m,p n Hence,u ˜ ∈ W (R ).

m,p 2.3.2 The space W0 (Ω) m,p Deﬁnition 2.3.8 [1] The subspace W0 (Ω) is deﬁned as the closure of D(Ω) with respect to the norm in W m,p(Ω).

m,p m,q 1 1 Theorem 2.3.9 [6] Let u ∈ W (Ω) and v ∈ W0 (Ω), where p + q = 1. Then,

(Dαu) vdx = (−1)|α| uDαvdx where |α| ≤ m. ˆΩ ˆΩ m,p m,p Proof . Since D(Ω) is dense in W0 (Ω). Then, for v ∈ W0 (Ω) there exists a sequence (vn)n≥0 ∈ D(Ω) such that

20 kvn − vkW m,p(Ω) −→ 0 as n → ∞. Now, by deﬁnition of the derivative of Dαu, we have that,

α |α| α (D u) vn = (−1) uD vn, for |α| ≤ m. ˆΩ ˆΩ

We need to show that,

α α (D u) vn −→ (D u) v, as n → ∞. ˆΩ ˆΩ Also,

|α| α |α| α (−1) uD vn −→ (−1) uD v, as n → ∞. ˆΩ ˆΩ Now,

α α α (D u) vn − (D u) v = D u(vn − v) ˆΩ ˆΩ ˆΩ α ≤ |D u||vn − v| ˆΩ α ≤ kD ukLp kvn − vkLq α ≤ kD ukW m,pkvn − vkW m,q −→ 0, as n → ∞.

Hence,

α α (D u) vn −→ (D u) v, as n → ∞ ˆΩ ˆΩ

Furthermore, |α| α α α (−1) uD vn − uD v = uD (vn − v) ˆΩ ˆΩ ˆΩ α ≤ |u||D (vn − v)| ˆΩ α ≤ kukLp kD (vn − v)kLq α ≤ kukW m,pkD (vn − v)kW m,q −→ 0, as n → ∞.

Which implies that,

|α| α |α| α (−1) uD vn −→ (−1) uD v, as n → ∞. ˆΩ ˆΩ Hence,

(Dαu) vdx = (−1)|α| uDαvdx where |α| ≤ m. ˆΩ ˆΩ

1,p Deﬁnition 2.3.10 [1] We deﬁne the space W0 (Ω) for p ∈ [1, ∞) as the closure of D(Ω) in W 1,p(Ω). It is associated with the norm in W 1,p(Ω). 1,2 1 For p = 2, the space W0 (Ω) = H0 (Ω) is a Hilbert space associated with the inner product in H1(Ω) = W 1,2(Ω).

21 n Theorem 2.3.11 [1] Let Ω be bounded, open subset of R . Then, 1,p 1,p W0 (Ω) = {u ∈ W (Ω) : u |∂Ω= 0}.

1 0 Proposition 2.3.12 [1] Let G ∈ C (R) be such that G(0) = 0 and |G (s)| ≤ M for all s ∈ R and for some constant M. Let u ∈ W 1,p(Ω) with 1 ≤ p ≤ ∞. Then, Gou ∈ W 1,p(Ω) and ∂ (Gou) = G0 (u(x)) ∂u for all i = 1, 2, ..., n. ∂xi ∂xi

Lemma 2.3.13 [1] Let u ∈ W 1,p(Ω) with 1 ≤ p < ∞ and assume that supp(u) is a compact subset 1,p of Ω. Then, u ∈ W0 (Ω).

n 1,p Theorem 2.3.14 [1] Let Ω ⊆ R be open and bounded. Let u ∈ W0 (Ω) for 1 ≤ p < ∞. Then, there exists a constant c such that

∂ψ u ≤ ckψkLq for all ψ ∈ D(Ω), i = 1, 2, ..., n. ˆΩ ∂xi

1,p Proof . Let u ∈ W0 (Ω). Then, there exists a sequence (un)n≥0 in D(Ω) such that un −→ u in W 1,p(Ω). Consequently, u −→ u in Lp(Ω) and ∂un −→ ∂u in Lp(Ω) for all i = 1, 2, ..., n. n ∂xi ∂xi Let ψ ∈ D(Ω). Then, ∂ψ ∂un un = − ψ. ˆΩ ∂xi ˆΩ ∂xi Hence, we have that

∂ψ ∂un un ≤ ψ . ˆΩ ∂xi ˆΩ ∂xi Then, by H¨older’sinequality we get that

∂ψ ∂un un ≤ kψkLq . ˆΩ ∂xi ∂xi Lp Taking limits as n → ∞, we have that

∂ψ ∂u u ≤ kψkLq . ˆΩ ∂xi ∂xi Lp

Set max ∂u = c < ∞. Consequently, 1≤i≤n ∂xi Lp

∂ψ u ≤ ckψkLq for all ψ ∈ D(Ω), i = 1, 2, ..., n. ˆΩ ∂xi

1,p −1,q Deﬁnition 2.3.15 [1] The dual space of W0 (Ω) for 1 ≤ p < ∞ is denoted by W (Ω). We 1 −1 deﬁne the dual of H0 (Ω) by H (Ω). If Ω is bounded, then

1,p −1,q W0 (Ω) ⊂ L2(Ω) ⊂ W (Ω), for 1 ≤ p < ∞. m,p −m,q In general, the dual space of W0 (Ω), for 1 ≤ p < ∞ is denoted by W (Ω).

1 1 where p + q = 1.

Theorem 2.3.16 [7] (Poinc´are’s inequality) n Suppose Ω ⊆ R is bounded. Then, there exists a constant C > 0 (depending on Ω) such that, 1,p p kukL ≤ Ck5ukLp , for all u ∈ W0 (Ω).

22 Proof . Let u ∈ D(Ω), thus support of u is compact. Then, for each i = 1, 2, ..., n and x = (x1, x2, ..., xn) with supp(u) ⊆ [xi, a], we have that

xi u(x) = ∇u(x)dsi, for all i = 1, 2, ..., n. ˆa Hence, we obtain that xi p p |u(x)| ≤ |∇u(x)|dsi . ˆa Consequently, by Holder’s inequality, there exists some constant C > 0 such that

|u(x)|pdx ≤ Cp |∇u(x)|pdx, for all u ∈ D(Ω). ˆΩ ˆΩ Hence, p kukLP ≤ C k∇ukLp , for all u ∈ D(Ω). 1,p 1,p 1,p Let u ∈ W0 (Ω). Since D(Ω) = W0 (Ω) in W (Ω) norm. Then, there exists a sequence (un)n≥0 ⊆ D(Ω) such that kun − ukW 1,p(Ω) −→ 0 as n → ∞. which implies that,

p p |un − u| dx + |∇(un − u)| dx −→ 0 as n → ∞. ˆΩ ˆΩ Hence, p p p p |un| dx −→ |u| dx and |∇un| dx ←→ |∇u| dx as n → ∞. ˆΩ ˆΩ ˆΩ ˆΩ Thus, for (un)n≥0 ⊆ D(Ω), we have that

kunkLp ≤ Ck∇unkLp Then, taking limits as n → ∞, we get that 1,p kukLp ≤ Ck∇ukLp , for all u ∈ W0 (Ω).

2.3.3 Extension operator Theorem 2.3.17 [7] n Let Ω ⊂⊂ Ω˜ ⊆ R be open set such that the closure of Ω is compact subset of Ω˜ and assume that the 1 1,p 1,p n boundary of Ω, ∂Ω is C .Then, there exists a bounded linear operator E : W (Ω) −→ W (R ) and a constant C such that, (i). Eu(x) = u(x) for a.e x ∈ Ω (ii). Eu(x) = 0, for x∈ / Ω˜

(iii). kEukW 1,p(Rn) ≤ CkukW 1,p(Ω)

2.3.4 Trace Theorem The trace theorem plays an important role in the existence and uniqueness of solution to boundary value problem to Partial Diﬀerential equations. It helps in extending the restriction of a function to its boundary. n 1 Theorem 2.3.18 [7] Let Ω ⊆ R , n ≥ 2 be a bounded open set with C boundary and let 1 ≤ p < ∞. Then, there exists a bounded linear operator 1,p T : W (Ω) −→ Lp(∂Ω) and a constant C > 0 depending on p and on the set Ω such that,

1,p kT ukLp(∂Ω) = kukLp(∂Ω) ≤ CkukW (Ω).

23 2.4 Embedding Theorems

Deﬁnition 2.4.1 [1] Let X1 and X2 be Banach spaces. We say that X1 is continuously embedded into X2(denoted as X1 ,→ X2) if for any u ∈ X1, we have u ∈ X2 and there exists a constant C > 0 such that,

kukX2 ≤ CkukX1 .

We deﬁne the embedding operator(Linear bounded operator) as J : X1 −→ X2 which takes u ∈ X1 into the same u considered as an element of X2. The embedding operator J is compact if any bounded set in X1 is a compact set in X2. If X1 ⊂ X2 and the embedding operator J : X1 −→ X2 is compact, then we say that X1 is compactly embedded in X2 (denoted as X1 ⊂⊂ X2). Furthermore, X1 ⊂⊂ X2 means that for any bounded sequence (un)n≥0 ⊆ X1, there exists a subsequence (unk )k≥0 which converges in X2.

Theorem 2.4.2 [1] The following embeddings hold: m ,p m ,p (a). W 1 (Ω) ,→ W 2 (Ω), if m1 > m2. m,p (b). W (Ω) ,→ Lp(Ω), if m > 0.

Proof . Consider the injection map

T : W m1,p(Ω) −→ W m2,p(Ω) deﬁned as T u = u for all u ∈ W m1,p(Ω).

Now, T is linear. We show that the graph of T , G(T ), is closed. m ,p m ,p m ,p Let (un)n≥0 ⊂ W 1 (Ω) such that un −→ u in W 1 (Ω) and T un −→ v in W 2 (Ω). Then, we show that T u = v. m ,p Then, T un = un −→ v in W 2 (Ω), which implies that, T u = u = v

Hence, T is closed. Then, by closed graph theorem we have that T is continuous. Thus,

W m1,p(Ω) ,→ W m2,p(Ω)

Similarly, (b) holds.

n Definition 2.4.3 (Hölder’sspace) Let Ω ⊆ R be open and bounded. Then, the function space defined as k,γ n k o C (Ω) = u ∈ C (Ω) : kukCk,γ < ∞ . is called the Hölder’sspace with exponent γ, where 0 ≤ γ ≤ 1. We define the Höldernorm as follows

X α X α kukCk,γ = kD ukC0 + [D u]γ. |α|≤k |α|=k where |u(x) − u(y)| [u]γ := sup γ . x,y∈Ω,x6=y |x − y|

Theorem 2.4.4 [7] (Morrey’s inequality) n 1 n 1,p n Assume n < p < ∞ and set ρ = 1 − p > 0 and let u ∈ C (R ) ∩ W (R ). Then, u is H¨older continuous. Moreover, there exists a constant C, depending only on p and n such that

kukC0,ρ(Rn) ≤ CkukW 1,p(Rn). Proof . We state the following Lemmas to help us in the prove of Morrey’s inequality.

n Lemma 2.4.6 [6] For n < p ≤ ∞, let B(x, r) ⊂ R and let y ∈ B(x, r). Then, n 1− p 1 n |u(x) − u(y)| ≤ C|x − y| kDukLp(Rn) for all u ∈ C (R ).

1 n We now proceed with the prove of Morrey’s inequality. Let u ∈ C (R ), then |u(x)| ≤ |u(x) − u(y)| + |u(y)| ρ ≤ C1|x − y| kDukLp(Rn) + |u(y)|. Taking the average of the Right hand side over the ball centered at x with radius 1, we get that,

|u(x)| ≤ |u(y)|dy + C1kDukLp(Rn) B(x,1)

≤ C2kukLp(Rn) + C1kukW 1,p(Rn). Hence, |u(x) − u(y)| kuk 0,ρ n = sup n |u(x)| + sup C (R ) x∈R x6=y |x − y|ρ 1 n ≤ CkukW 1,p(Rn), for all u ∈ C (R ). For some constant C. n 1,p n n Since D(R ) is dense in W (R ). Then, there exists a sequence (un)n≥1 ⊂ D(R ) such that 1,p n un −→ u in W (R ). Hence, for n, k ∈ N

kun − ukkC0,ρ(Rn) ≤ Ckun − ukkW 1,p(Rn). 0,ρ n 0,ρ n This implies that (un) is Cauchy in C (R ) which is Banach. Hence, there existsu ˜ ∈ C (R ) such that

0,ρ n un −→ u˜ in C (R ). n Thus, by uniqueness,u ˜ = u a.e in R . Consequently, we have that 1 n 1,p n kukC0,ρ(Rn) ≤ CkukW 1,p(Rn) for all u ∈ C (R ) ∩ W (R ).

Corollary 2.4.7 [7] (Embedding) n 1 n Let Ω ⊆ R be a bounded open set with C boundary. Assume n < p < ∞ and set ρ = 1 − p > 0. Then, every function u ∈ W 1,p(Ω) coincides a.e with a function u˜ ∈ C0,ρ(Ω). Moreover, there exists a constant C such that

ku˜kC0,ρ(Rn) ≤ CkukW 1,p(Rn).

n n Proof . Let Ω˜ = {x ∈ R : d(x, Ω) < 1}. Then Ω ⊂⊂ Ω˜ ⊆ R . Hence, by the extension Theorem 2.3.17 for each u ∈ W 1,p(Ω), there exists an extension map

1,p 1,p n E : W (Ω) −→ W (R ) such that Eu = u a.e in Ω. 1 n 1,p n 1 n Since C (R ) is dense in W (R ). Then, there exists a sequence (gn)n≥0 ⊆ C (R ) such that

25 1,p n gn −→ Eu in W (R ). By the Morrey’s inequality 2.4.4. Then, we get that

kgn − gmkC0,ρ(Rn) ≤ Ckgn − gmkW 1,p(Rn) 0,ρ n 0,ρ n n Hence, (gn)n≥0 is Cauchy in C (R ) and so converges to g ∈ C (R ) for a.e x ∈ R . n Then, by uniqueness g(x) = Eu(x) for a.e x ∈ R . So that, g(x) = u(x) a.e x ∈ Ω. By the Morrey’s inequality 2.4.4 and the extension Theorem 2.3.17. Thus, there exist C1,C2 > 0 such that

kgkC0,ρ(Ω) = kgkC0,ρ(Rn) ≤ C1kukW 1,p(Rn) ≤ C2kukW 1,p(Ω).

Deﬁnition 2.4.8 [7] Let 1 ≤ p < n, we deﬁne the Sobolev conjugate of p by

∗ np 1 1 1 p = n−p > p, where p∗ = p − n . Theorem 2.4.9 [7] (Gagliardo-Nirenberg inequality) Assume 1 ≤ p < n. Then, there exists a constant C, depending only on p and n such that

1 n kukLp∗(Rn) ≤ Ck 5 ukLp(Rn), for all u ∈ Cc (R )

∗ n Proof . Let p = 1. Then, p = n−1 . 1 n Let u ∈ Cc (R ), then u has a compact support. Then, for each i ∈ 1, 2, ..., n and x = (x1, x2, ..., xn).We have that,

u(x) = Dxi u(x)dsi ˆ−∞ Then,

|u(x)| ≤ |Dxi u(x)|dsi, 1 ≤ i ≤ n. ˆ−∞ Hence, n ∞ 1 n Y n−1 |u(x)| n−1 ≤ |D u(x)|ds (2.4.1) ˆ xi i i=1 −∞

We integrate (2.4.1) with respect to x1. Then,

∞ ∞ 1 ∞ n ∞ 1 n n−1 Y n−1 |u(x)| n−1 dx ≤ |D u(x)ds |D u(x)|ds dx ˆ 1 ˆ x1 1 ˆ ˆ xi i 1 −∞ −∞ −∞ i=2 −∞ Then, by the generalized Holder’s inequality, we get that

∞ ∞ 1 n ∞ ∞ 1 n n−1 Y n−1 |u(x)| n−1 dx ≤ |D u(x)ds |D u(x)|ds dx ˆ 1 ˆ x1 1 ˆ ˆ xi i 1 −∞ −∞ i=2 −∞ −∞

Following the same argument for x2, x3, ..., xn and after n integrations we get that

∞ ∞ n ∞ ∞ 1 n Y n−1 ... |u(x)| n−1 dx ...dx ≤ ... |D u(x)|dx ...dx ˆ ˆ 1 n ˆ ˆ xi 1 n −∞ −∞ i=1 −∞ −∞ Then,

n n n−1 |u(x)| n−1 dx ≤ | 5 u(x)|dx ˆRn ˆRn

26 Hence, n−1 n n |u(x)| n−1 dx ≤ | 5 u(x)|dx (2.4.2) ˆRn ˆRn Consequently,

kuk n ≤ k 5 ukL1( n) L n−1 (Rn) R Case 2: for 1 < p < n. β p(n−1) We set u = |g| , where β = n−p . Then, equation (2.4.2) becomes

n−1 n βn β−1 |g(x)| n−1 dx ≤ β|g(x)| | 5 g|dx ˆRn ˆRn By Holder’s inequality, we get that

n−1 1 1 n q p βn (β−1)q p |g(x)| n−1 dx ≤ β |g| | 5 g| (2.4.3) ˆRn ˆRn ˆRn p np ∗ But, (β − 1)q = (β − 1) p−1 = n−p = p . βn ∗ n−1 1 1 Also, n−1 = p and n − q = p∗ . Hence, equation (2.4.3) becomes,

n−1 1 1 ∗ n ∗ q p |g(x)|p dx ≤ β |g|p | 5 g|p ˆRn ˆRn ˆRn Consequently, we have that

kgk p∗ n ≤ Ck 5 gk p n . L (R ) L (R )

Corollary 2.4.10 [7] (Embedding) n 1 Let Ω ⊆ R be a bounded open domain with C boundary and assume 1 ≤ p < n. Then, for every q ∈ [1, p∗], there exists a constant C such that

1,p kukLq(Ω) ≤ CkukW 1,p(Ω), for all u ∈ W (Ω).

n n Proof . Let Ω˜ = {x ∈ R : d(x, Ω) < 1}. Then Ω ⊂⊂ Ω˜ ⊆ R . Hence, by the extension Theorem 2.3.17 for each u ∈ W 1,p(Ω), there exists an extension map

1,p 1,p n E : W (Ω) −→ W (R ) such that Eu = u a.e in Ω. Then, applying the Gagliardo-Nirenberg inequality 2.4.9 and the extension Theorem 2.3.17. Thus, 1,p there exists C1,C2,C3 and C4 such that for each u ∈ W (Ω).

∗ kukLq(Ω) ≤ C1kukLp (Ω) ≤ C2k 5 ukLp(Rn) ≤ C3kEukW 1,p(Rn) ≤ C4kukW 1,p(Ω).

1,p ∞ Theorem 2.4.11 [1] Let Ω ⊆ R be bounded. Then, W (Ω) ,→ L (Ω), for all 1 ≤ p ≤ ∞.

p−1 Proof . Let Ω = R and v ∈ D(R). Then, for 1 ≤ p ≤ ∞. Set G(v) = |v| v and w = G(v). 0 0 0 p−1 0 Thus, w ∈ D(R) and w = G (v).v = p|v| v . Consequently, for x ∈ R, we have that x G(v(x)) = w(x) = w0(t)dt. ˆ−∞

27 Consequently, we have that x 0 |v(x)|p ≤ p|v(t)|p−1|v (t)|dt. ˆ−∞ Thus, we get p p−1 0 kvkL∞ ≤ pkvkL∞ |v (t)|dt. ˆR Then, by H¨older’sinequality we get that

0 kvkL∞ ≤ ckv kLp . Consequently, we have that 0 ∞ kvkL ≤ ckv kW 1,p(R). 1,p 1,p Now, let u ∈ W (R). Since D(R) is dense in W (R). Then, there exists a sequence (un)n≥0 in 1,p D(R) such that un −→ u in W (R). Consequently, we have that

∞ kunkL ≤ ckunkW 1,p(R). Taking limits as n → ∞, we have that

1,p ∞ kukL ≤ ckukW 1,p(R) for all u ∈ W (R). Furthermore, let u ∈ W 1,p(Ω), for 1 ≤ p ≤ ∞. Then, by applying the extension Theorem 2.3.17 1,p and using the fact that D(R) is dense in W (R) we have that for some constants c1 and c2,

kukL∞(Ω) ≤ c1kEukW 1,p(R) ≤ c2kukW 1,p(Ω).

n Remark 2.4.12 Let Ω ⊆ R be bounded. Then, for n ≥ 2 we have the continuous embedding W 1,p(Ω) ,→ L∞(Ω) for all n < p < ∞.

Theorem 2.4.13 [6](Arzela-Ascoli’s Theorem) Let Ω be compact and suppose that (un)n≥0 ⊆ 0 C (Ω), kunkC0 ≤ M < ∞ and the family {un : n ≥ 0} is equicontinuous.Then, there exists a subsequence (unk )k≥0 such that unk −→ u uniformly on Ω.

n 1 Theorem 2.4.14 [7] Let Ω ⊆ R be a bounded open set with C boundary. Then, for n < p < ∞ and q ∈ [p, ∞], W 1,p(Ω) is compactly embedded in Lq(Ω).

1,p 1,p Proof . Since for every u ∈ W (Ω), for n < p < ∞ is H¨older continuous. Let (un)n≥1 ⊆ W (Ω) be a bounded sequence. Then, each un is H¨oldercontinuous and uniformly bounded. Consequently, each un is equicontinuous and uniformly bounded. Thus, by Ascela- Ascoli’s compactness theorem

2.4.13, there exists a subsequence (unk )k≥1 which converges uniformly to a continuous function u q on Ω. Since Ω is bounded, then by (DCT) 1.2.3 it implies that kunk − ukL (Ω) −→ 0 as k → ∞, for all q ∈ [p, ∞]. This implies that W 1,p(Ω) ,→ Lq(Ω) compactly, whenever n < p < ∞ and q ∈ [p, ∞].

Theorem 2.4.15 [7](Rellich’s Theorem) n 1 Let Ω ⊆ R be a bounded open domain with C boundary and assume 1 ≤ p < n. Then, for every ∗ ∗ np 1,p q ∈ [1, p ] (where p = n−p > p is called the Sobolev conjugate of p), then W (Ω) ⊂⊂ Lq(Ω) for q ∈ [1, p∗] .

1,p Proof . Let (un)n≥0 be a bounded sequence in W (Ω) and Ω˜ be a bounded, open domain such ˜ n ˜ n that Ω ⊂⊂ Ω ⊂ R and supp(un) ⊆ Ω. Then, by the extension Theorem (un)n≥0 is deﬁned on R and vanishes outside it compact support. Since q < p∗ and Ω˜ is bounded, then by Gagliardo-Nirenberg inequality (2.4.9) we have that,

28 kunkLq(Rn) = kunkLq(Ω)˜ ≤ C1kunkLp∗ (Ω)˜ ≤ C2kunkW 1,p(Ω)˜ q n For some constants C1 and C2. Hence, (un) is uniformly bounded in L (R ). Let wγ be a standard molliﬁer. Consider the molliﬁed function, γ un = wγ ∗ un γ with supp(un) ⊂ Ω˜ for all n ≥ 1.

γ We show that kun − unkLq(Ω)˜ −→ 0 as γ → 0. Now,

γ un(x) = wγ(x − y)un(y)dy ˆRn 1 y = n w( )un(x − y)dy ˆ|y|<γ γ γ

y Lety ˜ = γ , then y = γy˜ and dy = γdy˜. Hence,

γ un(x) = w(˜y)un(x − γy˜)dy˜ ˆ|y˜|<1 But,

1 d 1 un(x − γy˜) − un = (un(x − γty˜)) dt = −γ 5un(x − γty˜)˜ydt ˆ0 dt ˆ0 Then,

γ un(x) − un(x) = w(˜y)(un(x − γy˜) − un) dy˜ ˆ|y<˜ 1 1 = −γ w(˜y) 5un(x − γty˜)˜ydt dy˜ ˆ|y˜|<1 ˆ0 Consequently,

1 γ |un(x) − un(x)|dx ≤ γ | 5 un(x − γty˜)|dtdx ˆΩ˜ ˆΩ˜ ˆ0 Let z = x − γty˜. Then,

γ |un(x) − un(x)|dx ≤ γ | 5 un(z)|dz ˆΩ˜ ˆΩ˜ Hence, γ kun − unkL1(Ω)˜ ≤ γCkunkW 1,p(Ω)˜ for some constant C. Using the Lp interpolation Lemma 1.1.9 for 0 < a < 1, we get that

kkuγ − u k ≤ kuγ − u ka kuγ − u k1−a ≤ C γa n n Lq(Ω)˜ n n L1(Ω)˜ n n Lp∗ (Ω)˜ 0

For some constant C0. For ﬁxed δ > 0 choose γ > 0 small enough such that δ kkuγ − u k ≤ kuγ − u ka kuγ − u k1−a ≤ C γa ≤ n n Lq(Ω)˜ n n L1(Ω)˜ n n Lp∗ (Ω)˜ 0 2 γ Hence, kun − unkLq(Ω)˜ −→ 0 as γ → 0. γ But, un = wγ ∗ un. Then, for some constants C1 and C2 and for ﬁxed γ > 0, we have that

γ −n kunkL∞(Ω)˜ ≤ kwγkL∞(Ω)˜ kunkL1(Ω)˜ ≤ C1γ < ∞. (2.4.4)

29 Also, γ −n−1 k 5 unkL∞(Ω)˜ ≤ k 5 wγkL∞(Ω)˜ kunkL1(Ω)˜ ≤ C2γ < ∞. (2.4.5) γ Hence, from (2.4.4) and (2.4.5) we have that (un) is uniformly bounded and equicontinuous. Then, γ by Arzela-Ascoli’s compactness Theorem there exists a subsequence (unk )k≥0 which converges uniformly to uγ on Ω.˜ Thus, we have that

γ γ γ γ γ lim sup ku − u k q ≤ lim sup ku − u k q + ku − u k q + ku − u k q + ku − u k q nk nj L nk nk L nk L nj L nj nj L j,k→∞ j,k→∞ ≤ δ

1 1 1 Then, for δ = 2 , 4 , 8 ..., we construct a subsequence (unk )k≥0 such that 1 lim sup kunk − unj kLq(Ω)˜ ≤ k j,k→∞ 2 Thus,

lim sup kunk − unj kLq(Ω)˜ = 0 j,k→∞

q Hence, (unk )n≥0 is a Cauchy sequence. Thus, it converges to u ∈ L (Ω). The proof is complete.

n 1 Corollary 2.4.16 [7] Let Ω be an open subset of R of class C with compact boundary, ∂Ω. If n > 2, then

1 1 1 1 H (Ω) ,→ Lp∗ (Ω), for p∗ = 2 − n . where p∗ is the Sobolev conjugate of p.

Proof . We apply the Rellich Theorem for p = 2. Then, we have that H1(Ω) = W 1,2(Ω) ,→ Lp∗ (Ω).

n 1 Theorem 2.4.17 [7] Let Ω ⊆ R be a bounded open set with C boundary, then the following embeddings holds for the Sobolev Space, W m,p(Ω),

n m,p 1 1 m 1. If m − p < 0, then W (Ω) ,→ Lq(Ω), where q = p − n .

n m,p 2. If m − p = 0, then W (Ω) ,→ Lq(Ω), for every 1 ≤ q ≤ ∞. 3. If k ≥ 0 and γ > 0. Then, W m,p(Ω) ,→ Ck,γ(Ω).

n m,p q 1 1 m Proof (1) Let m − p < 0. Then, we show that W (Ω) ,→ L (Ω) with q = p − n . Let u ∈ W m,p(Ω). Then, we show that u ∈ Lq(Ω) and there exists a constant c > 0 such that

kukLq ≤ ckukW m,p .

m,p α 1,p n Since u ∈ W (Ω), then D u ∈ W (Ω) for all α ∈ N with |α| ≤ m − 1. Then, by (Gadliardo- Nirenberg) inequality (2.4.9) we have that

α 1 1 1 kD uk p∗ ≤ ckuk m,p , where = − . L W p∗ p n

m−1,p∗ ∗ Consequently, u ∈ W (Ω). Set p1 = p and repeating the same argument, we have that m−2,p∗ u ∈ W 1 and 1 1 1 kuk ∗ ≤ ckuk m−1,p , where = − . W m−2,p1 W 1 ∗ p1 p n

30 ∗ ∗ ∗ Following the same pattern and setting p2 = p1, p3 = p2, ..., pj = pj−1 we get that 1 1 j W m,p(Ω) ,→ W m−1,p1 (Ω) ,→ W m−2,p2 (Ω) ,→ ... ,→ W m−j,pj (Ω), where = − . pj p n

Doing this after m steps and set pm = q, we get that 1 1 m W m,p(Ω) ,→ Lq(Ω), where = − . q p n

(2). In the case when mp = n. Then, from (1), repeating the argument (m − 1) steps, we get that 1 1 m − 1 W m,p(Ω) ,→ W 1,n(Ω) ,→ W 1,n−(Ω), where = − and for any > 0. pm−1 p n Consequently, we have that

n(n − ) W 1,n−(Ω) ,→ Lq(Ω), where q = . n − (n − )

Since, > 0 is arbitrary, we have that

W m,p(Ω) ,→ Lq(Ω), for 1 ≤ q < ∞.

(3). Furthermore, let k ≥ 0 and γ > 0 and let u ∈ W m,p(Ω). From (1), we choose j to be the m−j,p smallest integer such that pj > n. Then, we have that u ∈ W j (Ω) and 1 j 1 1 1 j − 1 − = < < − . p n pj n p n

Hence, for every multi-index α with |α| ≤ m − j − 1. Then, by Morrey’s inequality (2.4.4) we get that n n Dαu ∈ W 1,pj (Ω) ,→ C0,γ(Ω), with γ = 1 − = 1 − + j. pj p Consequently, for all multi-index α with |α| ≤ m − j − 1 we have u ∈ Cm−j−1,γ(Ω). Set k = m − j − 1. Hence, we have that

W m,p(Ω) ,→ Ck,γ(Ω).

5 4,2 Example 2.4.18 [7] Let Ω be open unit ball in R and assume u ∈ W (Ω). Then, by applying the Gadliardo Nirenberg inequality (2.4.9) and Morrey’s inequality (2.4.4) twice, we get that

4,2 3, 10 2,10 1, 1 W (Ω) ,→ W 3 (Ω) ,→ W (Ω) ,→ C 2 (Ω).

4,2 n 5 3 where the net smoothness of u ∈ W (Ω) given by m − p = 4 − 2 = 2 .

Theorem 2.4.19 [7] (Poinc´aire Wittinger inequality) n 1 Let Ω ⊆ R be a bounded, connected open set with C boundary. Then, for 1 ≤ p < ∞, there exists a constant C depending only on p and Ω, such that

1,p u − udx ≤ C k5ukLp for all u ∈ W (Ω) (2.4.6) Ω Lp where 1 udx = udx. Ω mes(Ω) ˆΩ

31 1,p Proof . Suppose (2.4.6) is false. Then, for n ≥ 1 there exists (un)n≥1 ⊂ W (Ω) such that

kun − undxkLp(Ω) > nk 5 unkLp(Ω) Ω Then,

k 5 u k p 1 n L < . kun − Ω undxkLp n ffl u − u dx Let v = n Ω n . n ffl p kun− Ω undxkL ffl 1 Then, kvnkLp = 1, k 5 vnkLp(Ω) < n and

vndx = 0 Ω . Hence, k 5 vnkLp −→ 0 as n → ∞. 1,p But (vn)n≥1 is bounded in W (Ω) for 1 ≤ p < n. Then, by Rellich Theorem (2.4.15), there exists a subsequence (vnk )k≥1 ⊆ (vn)n≥1 such that p vnk −→ v in L (Ω)

Also, for n < p < ∞ by the embedding (2.4.7). Then (vn) is uniformly bounded and Holder’s continuous. Then, by Arzela- Ascoli’s compactness theorem, there exists a subsequence (vnk ) which converges to v in L∞. Then, p 5vnk −→ 5v in L (Ω). Furthermore,

p p k 5 vnk kL −→ k 5 vkL . Hence, 5v = 0 and since Ω is connected we have that v = k a.e in Ω. Also, p vnk −→ v in L (Ω) Then,

vnk dx −→ vdx Ω Ω

But,

vnk dx = 0. Ω Consequently, we have that vdx = 0. Ω Hence, k = v = 0 . But v = 0, implies that kvkLp = 0. Also,

vnk −→ v Then,

p p kvnk kL −→ kvkL

But, kvnkkLp = 1. Then, kvkLp = 1, which is a contradiction. Hence, inequality (2.4.6) holds.

32 CHAPTER 3

ELLIPTIC PARTIAL DIFFERENTIAL EQUATIONS

Introduction Partial differential equations are used majorly to formulate models involving functions of several variables. They are used to point out differs happenings in sound, heat, diffusion, electrostatics, electrodynamics and fluid dynamics.

A partial diﬀerential equation is a diﬀerential equation involving partial derivatives of a dependent variable with more than one independent variable. It is often denoted as PDE.

The order of a partial diﬀerential equation is the order of the highest derivative that appear in the equation.

A partial differential equation is said to be linear if the dependent variable and all its partial derivatives are expressed as a linear combination in which the coefficients are independent of the dependent variable and it derivatives. A partial differential equation which is not linear is said to be non-linear.

The following examples gives an illustration of a linear PDE and a non-linear PDE. Let u(x, y) be a function of two independent variables x and y. Then, the equation

∂2u ∂2u ∂u + u + x = x3 ∂x2 ∂x∂y ∂y is a second order non-linear PDE. The equation ∂2u ∂2u + = 0 ∂x2 ∂y2 is a second order linear PDE. Classification of partial differential equations. Consider the general linear partial differential equation of second order of n variables given by

n n X ∂2u X ∂u a + b + cu = d i,j ∂x ∂x i ∂x i,j=1 i j i=1 i

The classification of the PDE depends on the signs of the eigenvalues of the n×n coefficient matrix, A = [ai,j]1≤i,j≤n. Let p denotes the number of positive eigenvalues and let z denotes the number of zero eigenvalues. Then, the partial differential equation is classified as follows (a) It is elliptic if z = 0 and p = n or z = 0 and p = 0. (b) It is parabolic if z > 0.

33 (c) it is hyperbolic if z = 0 and p = 1 or z = 0 and p = n − 1. Example of an elliptic PDE is the Laplace equation given by

n X ∂2u ∆u = = 0. ∂x2 i=1 i Example of a parabolic PDE is the heat equation given by ∂u = c2∆u, where c is constant. ∂t Example of a hyperbolic PDE is the wave equation given by

∂2u = c2∆u, where c is constant. ∂t2 In this chapter, we focus on solving some linear and non-linear elliptic partial diﬀerential equations with homogeneous and non-homogeneous boundary conditions. We show the existence of a unique weak solution to some Dirichlet and Neumann problems with homogeneous and non-homogeneous boundary conditions. Also, some mixed problems will be discussed. Furthermore, we discuss the concept of monotone operators which aid us in solving the non-linear elliptic problems. We begin the chapter by discussing some important concepts needed to show the existence of the solution to the problems we shall solve.

3.1 Deﬁnitions and existence Theorem

Deﬁnition 3.1.1 We deﬁne the Laplacian,∆, of a function u by,

n X ∂2u ∆u = . ∂x2 i=1 i ∂u We deﬁne ∂n = ∇u.n as the outward normal vector of u, where n is the unit normal vector to ∂Ω, pointing outward.

Theorem 3.1.2 [3](Green’s formula) Let u ∈ H2(Ω) and v ∈ H1(Ω). Then,

∂u (∆u)v = vdσ − ∇u.∇v. ˆΩ ˆ∂Ω ∂n ˆΩ where dσ is the surface measure on ∂Ω.

Lemma 3.1.3 [3] Let A ⊆ H where H is a Hilbert space. Then, A = H if and only if A⊥ = {0}, where A⊥ = {y ∈ H : hx, yi = 0 for all x ∈ A} .

Proof Suppose A = H, then we show that A⊥ = {0}. Let y ∈ A⊥. Then, for all x ∈ A we have that

hx, yi = 0.

Consequently, y ∈ H implies that there exists a sequence (xn)n≥0 ⊆ A such that xn −→ y as n → ∞. Hence, we obtain that hxn, yi = 0. Taking limit as n → ∞, we get that

hy, yi = kyk2 = 0.

34 Thus, A⊥ = 0 ( since y ∈ A⊥ was arbitrarily).

Conversely, suppose that A⊥ = {0}, then we show that A = H. For contradiction, suppose A 6= H. Then, there existsy ˜ ∈ H,y ˜ 6= 0 such that hy,˜ xi = 0 for all x ∈ A. But, hx, yi = 0 for all x ∈ A, implies that y = 0. Hence,y ˜ = 0 which is a contradiction. Thus, A = H.

Remark 3.1.4 Let u ∈ H2(Ω), then ∂u ∈ H1(Ω). Then, by Trace Theorem, we get that ∂u | ∈ ∂xi ∂xi ∂Ω L2(∂Ω).

Theorem 3.1.5 [3](Density of traces) n Let Ω ⊆ R be smooth. Then,

1 2 {u |∂Ω: u ∈ H (Ω)} = L (∂Ω).

Theorem 3.1.6 [1](Lax-Milgram Theorem) Let H be a real Hilbert space and a, a bilinear continuous form on H. Suppose a is coercive, that is, there exists α > 0 such that

a(u, u) ≥ αkuk2, for all u ∈ H.

Then, for all f ∈ H∗, there exists a unique v ∈ H such that

a(v, u) = f(u), for all u ∈ H.

3.2 Applications to elliptic partial diﬀerential equations

In this section, we solve some linear elliptic PDEs with homogeneous and non-homogeneous boundary conditions. Consider the Dirichlet problem with homogeneous boundary condition given by ( −∆u + u = f in Ω, f ∈ L2(Ω) (3.2.1) u = 0 on ∂Ω.

Due to the Diritchlet boundary condition given, we look for the solution

1 1 u ∈ H0 (Ω) = u ∈ H (Ω) : u |∂Ω= 0 , which satisﬁes (3.2.1). 1 1 Let u ∈ H0 (Ω) be a solution of (3.2.1). Then, for all v ∈ H0 (Ω), we have that

1 −∆u.vdx + uvdx = fvdx, for all v ∈ H0 (Ω). ˆΩ ˆΩ ˆΩ Then, by Green’s formula, we get that ∂u ∇u.∇vdx − .vdσ + uvdx = fvdx (3.2.2) ˆΩ ˆ∂Ω ∂n ˆΩ ˆΩ 1 Since v ∈ H0 (Ω), then (3.2.2) becomes

∇u.∇vdx + uvdx = fvdx. ˆΩ ˆΩ ˆΩ 1 Set H = H0 (Ω). Let T1 : H × H −→ R deﬁned by

T1(u, v) = ∇u.∇vdx + uvdx, for all u, v ∈ H. ˆΩ ˆΩ

35 and let T2 : H −→ R deﬁned by

T2(v) = fvdx, for all v ∈ H. ˆΩ

Now, T1 is bilinear. Furthermore,

|T1(u, v)| ≤ |∇u||∇v|dx + |u||v|dx ˆΩ ˆΩ ≤ k∇ukL2 k∇vkL2 + kukL2 kvkL2

By Poincaire’s inequality, for some C > 0, we get that

|T1(u, v)| ≤ Ckuk 1 . H0

Hence, T1 is continuous. Also,

2 2 T1(u, u) = |∇u| + |u| ≥ kukH1 ˆΩ ˆΩ 0 ∗ Consequently, T1 is coercive. Also, we have that T2 ∈ H . 1 Then, by the Theorem 3.1.6, there exists a unique u ∈ H0 (Ω) which satisﬁes (3.2.1). 1 We show that u ∈ H0 (Ω) satisﬁes (3.2.1). Now, 1 ∇u.∇v + uv = fv, for all v ∈ H0 (Ω). (3.2.3) ˆΩ ˆΩ ˆΩ But,

n X ∂u ∂v ∇u.∇v = . ˆ ˆ ∂x ∂x Ω i=1 Ω i i n X ∂u ∂v = , ∂x ∂x i=1 i i n X ∂2u = − , v ∂x2 i=1 i = h−∆u, vi .

Hence, (3.2.3) becomes

1 h−∆u, vi + hu, vi = hf, vi , for all v ∈ H0 (Ω).

Consequently, we have that −∆u + u = f in Ω 1 Since u ∈ H0 (Ω), then u = 0 on ∂Ω.

Consider the Neumann problem with a homogeneous boundary condition given by ( −∆u + u = f in Ω, f ∈ L2(Ω) ∂u (3.2.4) ∂n = 0 on ∂Ω.

We seek a unique weak solution u ∈ H1(Ω) which satisﬁes (3.2.4). Let u ∈ H1(Ω) be a solution to (3.2.4). Then, for all v ∈ H1(Ω), we get that

−∆u.vdx + uvdx = fvdx. ˆΩ ˆΩ ˆΩ

36 Then, by Green’s formula we get that ∂u ∇u.∇vdx − vdσ + uvdx = fvdx. ˆΩ ˆ∂Ω ∂n ˆΩ ˆΩ

∂u But, ∂n = 0 on ∂Ω. Thus, we have that

∇u.∇vdx + uvdx = fvdx. ˆΩ ˆΩ ˆΩ

1 Set H = H (Ω). Let T1 : H × H −→ R deﬁned by

T1(u, v) = ∇u.∇v + uv, for all u, v ∈ H. ˆΩ ˆΩ

Now, T1 is bilinear, continuous and coercive. furthermore, let T2 : H −→ R deﬁned by

T2(v) = fv, for all v ∈ H. ˆΩ

∗ 1 Then, T2 ∈ H . Consequently, by Theorem (3.1.6), there exists a unique weak solution u ∈ H (Ω) which satisﬁes (3.2.4). We show that u ∈ H1(Ω) satisﬁes (3.2.4). Now, ∇u.∇v + uv = fv. (3.2.5) ˆΩ ˆΩ ˆΩ But, ∇u.∇v = h−∆u, vi , for all v ∈ H1(Ω) ˆΩ Hence, (3.2.5) becomes

h−∆u, vi + hu, vi = hf, vi , for all v ∈ H1(Ω).

Thus, we get that −∆u + u = f in Ω. ∂u It remain to show that ∂n = 0 on ∂Ω. Since u ∈ H1(Ω) satisﬁes −∆u + u = f. Then, for all v ∈ H1(Ω) we get that

− ∆u.v + uv = fv. ˆΩ ˆΩ ˆΩ Hence, ∂u ∇u.∇v − vdσ + uv = fv. (3.2.6) ˆΩ ˆΩ ∂n ˆΩ ˆΩ from (3.2.5) and (3.2.6), we get that ∂u vdσ = 0, for all v ∈ H1(Ω). ˆΩ ∂n

∂u 1 ⊥ Consequently, ∂n ∈ u |∂Ω: u ∈ H (Ω) . Then by the lemma(3.1.3), we have that ∂u = 0 on ∂Ω. ∂n We consider a mixed problem of Diritchlet and Neumann homogeneous boundary conditions given by ( −∆u = f in Ω, f ∈ L2(Ω) ∂u (3.2.7) u = 0 on Γ0, ∂n = 0 on Γ1.

37 n where ∂Ω = Γ0 ∪ Γ1, mes(Γ0) > 0, mes(Γ1) > 0 and Ω is connected and smooth in R . 1 1 Let H = u ∈ H (Ω) : u = 0 on Γ0 . Then, we show that H is a closed subset of H (Ω). 1 Let (un)n≥0 ⊆ H such that un −→ u in H (Ω). Then, by Trace Theorem (2.3.18)

2 un −→ u, in L (∂Ω).

By converse of DCT (1.2.6), there exists a subsequence (unk )k≥0 ⊆ (un)n≥0 such that

unk (x) −→ u(x) a.e on Γ0.

But, unk = 0 on Γ0. Hence, u = 0 on Γ0. Consequently, H is closed . Thus, H is a Hilbert space. We seek a unique solution u ∈ H which satisﬁes (3.2.7). Let u ∈ H be a solution of (3.2.7). Then, for all v ∈ H, we have that

− ∆u.v = fv. ˆΩ ˆΩ Then, by Green’s formular, we get that ∂u ∇u.∇v − .v = fv ˆΩ ˆΩ ∂n ˆΩ Hence, ∂u ∂u ∇u.∇v − .v − .v = fv ˆΩ ˆΓ0 ∂n ˆΓ1 ∂n ˆΩ

∂u But, v = 0 on Γ0 and ∂n = 0 on Γ1. Hence, we have that

∇u.∇v = fv. ˆΩ ˆΩ

Let T1 : H × H −→ R deﬁned by

T1(u, v) = ∇u.∇v, for all u, v ∈ H. ˆΩ

Also, let T2 : H −→ R deﬁned by

T2(v) = fv, for all v ∈ H. ˆΩ ∗ Now, T2 is a bounded linear map. Thus, T2 ∈ H . Furthermore, T1 is bilinear and continuous. We show that T1 is coercive. That is, there exists α > 0 such that 2 T1(u, u) ≥ αkukH1 .

For contradiction, suppose T1 is not coercive. Then, for all n ≥ 1, there exists (un)n≥1 ⊂ H such that 1 T (u , u ) < ku k2 . 1 n n n n H1 Then,

2 1 2 |∇un| < kunkH1 ˆΩ n

un Set vn = . Then, kunkH1 2 1 |∇vn| < . ˆΩ n

38 Hence,

2 ∇vn −→ 0 in L (Ω)

1 But, kvnkH1 = 1. Since (vn)n≥1 is bounded in H (Ω). Then, by Rellich’s Theorem 2.4.15 there 2 exists a subsequence (vnk )k≥1 ⊆ (vn)n≥1 which converges to v in L (Ω). Consequently, 2 ∇vnk −→ ∇v in L (Ω). Then, by uniqueness of limits, we have that

∇v = 0 a.e on Ω.

Since Ω is connected, then v = c, where c is constant. 1 Thus, vnk −→ k in H (Ω). Then, by Trace Theorem 2.3.18 we have that

2 vnk −→ c in L (∂Ω).

Consequently, we get that 2 vnk −→ c in L (Γ0). That is, 2 2 |vnk | dσ −→ c mes(Γ0). ˆΓ0 2 But, vnk = 0 on Γ0. Hence, c mes(Γ0) = 0 Which implies that, c = 0. Thus, we have that

1 vnk −→ 0 in H (Ω).

But, kvnk kH1 = 1 which implies that 1 = 0 (Impossible). Consequently, T1 is coercive. Then, by Theorem 3.1.6 there exist a unique weak solution u ∈ H which satieﬁes (3.2.7). We show that u ∈ H satisﬁes (3.2.7). Now,

∇u.∇v = fv, for all v ∈ H. ˆΩ ˆΩ But, ∇u.∇v = h−∆u, vi . ˆΩ Hence, we get that h−∆u, vi = hf, vi for all v ∈ H, which implies that −∆u = f in Ω.

Also, u ∈ H implies that u = 0 on Γ0. ∂u It remain to show that ∂n = 0 on Γ1. Since u ∈ H satiesﬁes −∆u = f. Then, for all v ∈ H, we have that

− ∆u.v = fv. ˆΩ ˆΩ Thus, ∂u ∇u.∇v − vdσ = fv. ˆΩ ˆΩ ∂n ˆΩ Consequently, ∂u vdσ = 0 for all v ∈ H. ˆΓ1 ∂n

39 ∂u ⊥ Hence, ∂n ∈ {u |Γ1 : u ∈ H} . 2 By the Theorem 3.1.5 {u |Γ1 : u ∈ H} is dense in L (Γ1). Consequently, by lemma 3.1.3 ∂u = 0 on Γ . ∂n 1

We consider the non-homogeneous Neumann problem given by ( −∆u + u = f in Ω, f ∈ L2(Ω) (3.2.8) ∂u 2 ∂n = g on ∂Ω, g ∈ L (∂Ω).

We seek a unique solution u ∈ H1(Ω) that satisﬁes (3.2.8). Let u ∈ H1(Ω) be a solution of (3.2.8). Then, for all v ∈ H1(Ω) we have that

−∆u.vdx + u.vdx = f.vdx. ˆΩ ˆΩ ˆΩ Then, by Green’s formula we get that ∂u ∇u.∇vdx − .vdσ + u.vdx = fvdx. ˆΩ ˆ∂Ω ∂n ˆΩ ˆΩ

∂u But, ∂n = g on ∂Ω. Hence, we have that

∇u.∇v + u.v = fv + g.vdσ. ˆΩ ˆΩ ˆΩ ˆ∂Ω

1 Set H = H (Ω) and let a : H × H −→ R deﬁned by

a(u, v) = ∇u.∇v + u.v for all u, v ∈ H. ˆΩ ˆΩ Now, a is bilinear, continuous and coercive. Let L : H −→ R deﬁned by

L(v) = fv + g.vdσ, for all v ∈ H. ˆΩ ˆ∂Ω Now, L is linear. Also,

|L(v)| ≤ |f.v| + |g.v|dσ ˆΩ ˆ∂Ω ≤ kfkL2(Ω)kvkL2(Ω) + kgkL2(∂Ω)kvkL2(∂Ω) ≤ kfkL2(Ω) + kgkL2(∂Ω) kvkH1 Hence, L is bounded. Consequently, we have that L ∈ H∗. Then, by the Theorem 3.1.6, there exists a unique u ∈ H1(Ω) that satisﬁes (3.2.8). We show that u ∈ H1(Ω) satisﬁes (3.2.8). Now, ∇u.∇v + u.v = fv + g.vdσ (3.2.9) ˆΩ ˆΩ ˆΩ ˆ∂Ω For v ∼ ψ ∈ D(Ω). Then, we have that

g.v = 0. ˆ∂Ω Hence, we have ∇u.∇v + u.v = fv. ˆΩ ˆΩ ˆΩ

40 Which implies that h−∆u, vi + hu, vi = hf, vi , for all v ∈ H1(Ω). Consequently, −∆u + u = f in Ω, f ∈ L2(Ω). Since u satisﬁes −∆u + u = f. Then, for all v ∈ H1(Ω) we get that

− ∆u.v + u.v = fv. ˆΩ ˆΩ ˆΩ Hence, ∂u ∇u.∇v − .vdσ + u.v = f.v (3.2.10) ˆΩ ˆ∂Ω ∂n ˆΩ ˆΩ From equation (3.2.9) and equation (3.2.10), we get that ∂u − g .v = 0, for all v ∈ H1(Ω). ˆ∂Ω ∂n

∂u 1 ⊥ Consequently, ∂n − g ∈ u |∂Ω: u ∈ H (Ω) . Thus, we apply Theorem 3.1.5 and lemma 3.1.3 to obtain that ∂u − g = 0, on ∂Ω. ∂n Hence, we have that ∂u = g on ∂Ω. ∂n n Let Ω be smooth and connected in R . We consider the mixed problem with a non-homogeneous boundary condition given by ( −∆u = f in Ω, f ∈ L2(Ω) ∂u (3.2.11) u = 0 on Γ0, ∂n = g on Γ1.

2 where mes(Γ0) > 0, mes(Γ1) > 0 and g ∈ L (Γ1). 1 1 Let H = u ∈ H (Ω) : u = 0 on Γ0 which is a closed subspace of the Hilbert space H (Ω). Thus, we seek a unique weak solution u ∈ H which satisﬁes (3.2.11). Let u ∈ H be a solution of (3.2.11). Then, for all v ∈ H

−∆u.vdx = f.vdx. ˆΩ ˆΩ Then, by Green’s formula we get that ∂u ∇u.∇v − vdσ = fvdx. ˆΩ ˆ∂Ω ∂n ˆΩ Consequently, we get that ∇u.∇v − g.vdσ = fvdx. ˆΩ ˆΓ1 ˆΩ Then, ∇u.∇v = fvdx + g.vdσ. ˆΩ ˆΩ ˆΓ1 Now, let T1 : H × H −→ R deﬁned by

T1(u, v) = ∇u.∇v, for all u, v ∈ H. ˆΩ

T1 is bilinear, continuous and coercive. Also, let T2 : H −→ R deﬁned by

T2(v) = fvdx + g.vdσ, for all v ∈ H. ˆΩ ˆΓ1

41 ∗ Indeed, T2 ∈ H since T2 is linear and

|T2(v)| ≤ |f.v|dx + |g.v|dσ ˆΩ ˆΓ1

2 2 2 2 ≤ kfkL (Ω)kvkL (Ω) + kgkL (Γ1)kvkL (Γ1) 2 2 1 ≤ kfkL (Ω) + kgkL (Γ1) kvkH ∗ Which implies that T2 is bounded. Consequently, T2 ∈ H . Then, by the Theorem 3.1.6 there exists a unique u ∈ H which satisﬁes (3.2.11). We show that u ∈ H satisﬁes (3.2.11). Now, for all v ∈ H we get that

∇u.∇v = fvdx + g.vdσ. (3.2.12) ˆΩ ˆΩ ˆΓ1

For v ∼ ψ ∈ D(Γ1). Then, we get that

g.vdσ = 0. ˆΓ1 Hence, ∇u.∇v = fvdx, ˆΩ ˆΩ which implies that h−∆u, vi = hf, vi , for all v ∈ H. Thus, −∆u = f in Ω, f ∈ L2(Ω)

Also, u ∈ H implies that u = 0 on Γ0. Since u ∈ H satisﬁes −∆u = f Then, we get that ∂u ∇u.∇vdx − .vdσ = fvdx ˆΩ ˆ∂Ω ∂n ˆΩ Hence, we get that ∂u ∇u.∇vdx − .vdσ = fvdx, for all v ∈ H (3.2.13) ˆΩ ˆΓ1 ∂n ˆΩ Consequently, from (3.2.12) and (3.2.13) we get that ∂u − g dσ = 0. ˆΓ1 ∂n ∂u ⊥ Hence, ∂n − g ∈ {u |Γ1 : u ∈ H} . Then, we apply the lemma (3.1.3) and we obtain that ∂u = g on Γ . ∂n 1 3.3 Monotone operators

Notations. Let X and Y be real Banach spaces and T : X −→ Y be a multi-valued operator. We define the graph of T , G(T ) as follows; G(T ) = {(x, y) ∈ X × Y : y ∈ T x} . We define the effective domain of T , D(T ) as follows; D(T ) = {x ∈ X : T x 6= ∅} . Let X be a Banach space. We denote it dual by X∗.

42 Deﬁnition 3.3.1 [9] A subset A ⊂ X × X∗ is said to be a monotone set provided that

hx∗ − y∗, x − yi ≥ 0, whenever (x, x∗), (y, y∗) ∈ A.

Deﬁnition 3.3.2 [9] Let T : X −→ X∗ be a multivalued operator. Then, T is a monotone operator provided that for all x, y ∈ X,

hx∗ − y∗, x − yi ≥ 0, x∗ ∈ T x and y∗ ∈ T y.

Furthermore, T is a monotone operator if and only if it graph, G(T ) is a monotone set.

Deﬁnition 3.3.3 [9] A monotone set B ⊂ X × X∗ is said to be maximal monotone if it is the maximal monotone set of all monotone subsets of X × X∗. Thus, a monotone operator T is said to be maximal monotone if it graph, G(T ) is a maximal monotone set.

Example 3.3.4 [9] Let H be a Hilbert space and T : H −→ H∗ a linear map and single valued. Then, T is a monotone operator provided that

hx, T xi ≥ 0, for all x ∈ H.

Example 3.3.5 [9] Let T : X −→ 2X∗ a multivalued map called the duality map and deﬁned by

T x = x∗ ∈ X∗ : hx∗, xi = kxk2, with kxk = kx∗k .

Then, T is a monotone operator. Indeed, for all x, y ∈ X with x∗ ∈ T x and y∗ ∈ T y. Then,

hx∗ − y∗, x − yi = hx∗, xi − hx∗, yi − hy∗, xi + hy∗, yi = kxk2 − kxkkyk − kykkxk + kyk2 = (kxk − kyk)2

Hence, we obtain that hx∗ − y∗, x − yi ≥ 0.

Example 3.3.6 [9] The sub-diﬀerential operator T : X −→ X∗ deﬁned by

∂T (x) = {x∗ ∈ X∗ : hx∗, y − xi ≤ T (y) − T (x), for all y ∈ X} . is a monotone operator (in fact it is a maxiamal monotone operator).

Indeed, for all x, y ∈ X and x∗ ∈ T x, y∗ ∈ T y. Then,

hx∗ − y∗, x − yi = hx∗, x − yi − hy∗, x − yi ≥ T (x) − T (y) − T (x) + T (y) = 0

Hence, we obtain that hx∗ − y∗, x − yi ≥ 0.

Deﬁnition 3.3.7 [8] Let T : X −→ X∗ be a single valued operator from D(T ) = X. Then, T is said to be demi-continuous if (un)n≥0 ⊆ X such that un −→ u in X as n → ∞, then T un * T u in X∗ as n → ∞.

43 3.3.1 Yosida approximation The Yosida approximations(or regularization) plays an important role in the smooth approximation of a maximal monotone operator, A. Definition 3.3.8 [9] Let X be a reflexive strictly convex Banach space with strictly convex dual X∗ and let A be maximal monotone operator. Then, we define the Yosida approximation of A as ∗ the operator Aλ : X −→ X such that 1 A x = J(x − J x), λ > 0, x ∈ X. λ λ λ ∗ Where J : X −→ X is the duality mapping. The resolvent Jλx := xλ is defined to be the unique solution of the resolvent equation 0 ∈ J(xλ − x) + λAxλ.

Proposition 3.3.9 [8] The Yosida approximation Aλ is single-valued, maximal monotone, bounded on bounded subsets and demi-continuous operator from X to X∗.

Proposition 3.3.10 [8] Let X = H a Hilbert space and A a maximal monotone operator. Then, −1 (i). Jλ = (I + λA) 1 (ii). kAλx − Aλyk ≤ λ kx − yk, for all x, y ∈ D(A), λ > 0. Deﬁnition 3.3.11 [2] Let D be a subset of a real vector space X and f : D −→ R be a map. Then, f is said to be convex if D is a convex set and for each t ∈ [0, 1] and for each x1, x2 ∈ D, we have that f(tx1 + (1 − t)x2) ≤ tf(x1) + (1 − t)f(x2).

Furthermore, f is proper if there exists at least an x0 ∈ D such that f(x0) 6= ∞

Deﬁnition 3.3.12 [2] A function f : X −→ R is called lower semi continuous at x0 ∈ X if and only if for all λ ∈ R such that λ < f(x0), there exists a neighbourhood of x0, V , such that f(x) > λ for all x ∈ V .

Consider the elliptic variational problem given by T u + ∂ψ(u) = f, f ∈ X (3.3.1)

∗ Here, ψ : X −→ R is a proper, lower semi continuous and convex function on X. Also, ∂ψ ⊂ X×X is the sub-diﬀerential of ψ. Theorem 3.3.13 [8] Let T : X −→ X∗ be a monotone, demi-continuous operator and ψ : X −→ R a proper, lower semi continuous and convex function on X. Assume that there exists u0 ∈ D(ψ) such that (u − u , T u) + ψ(u) lim 0 = +∞. kuk→∞ kuk Then, problem (3.3.1) has atleast a solution. Moreover, the set of solutions is bounded, convex and closed in X and if the operator T is strictly monotone(That is, hT u − T v, u − vi = 0 if and only if u = v ), then the solution is unique.

3.4 Application to nonlinear elliptic partial diﬀerential equations

In this section, we propose to solve some nonlinear elliptic PDEs with homogeneous boundary conditions.

Consider the nonlinear elliptic partial diﬀerential equation with a homogeneous Dirichlet boundary condition given by λu − div β(∇u) = f in Ω, f ∈ L2(Ω) x (3.4.1) u = 0 on ∂Ω n n n where Ω is smooth and open in R . Here, β : R −→ R is a maximal monotone operator.

44 Theorem 3.4.1 [8] Suppose β is continuous and satisﬁes the following conditions: p−1 n (i) |β(r)| ≤ c1 1 + |r| , for all r ∈ R .

p n (ii) β(r).r ≥ ω|r| − c2, for all r ∈ R . 1,p ∗ where ω, c1, c2 are positive constants. Then, for each f ∈ (W0 (Ω)) and λ > 0, there exists 1,p u ∈ W0 (Ω) that satisﬁes (3.4.1).

We solve problem (3.4.1) with the assumptions given in Theorem 3.4.1.

45 CONCLUSION

n This thesis was devoted to the study of a space of functions on a subset Ω of R and its derivative called the Sobolev space with it applications to some elliptic partial differential equations. In chapter one of this thesis, we discussed the Lp spaces and some of it important concepts relevant in this thesis. Chapter two was devoted to the study of the Sobolev spaces. We also discussed the concept of embeddings(continuous and compact embeddings) of the Sobolev spaces with other spaces of functions like the Hölderspace. In chapter three, we were concerned with the applications to solve elliptic partial differential equations. We considered some Dirichlet and Neumann problems of linear elliptic partial differential equations. By applying the Lax-Milgram Theorem, we established the existence of a unique solution to these problems. Also, we solve some nonlinear partial differential equations with Dirichlet and Neumann (respec- tively) boundary conditions as follows:

λu − div β(∇u) = f in Ω, f ∈ L2(Ω) x (3.4.2) u = 0 on ∂Ω

λu − div β(∇u) = f in Ω x (3.4.3) β(∇u) = 0 on ∂Ω

We were able to establish the existence of a solution to these nonlinear problems by applying an appropriate theorem. We note that further research can be done on the nonlinear problems.

46 BIBLIOGRAPHY

[1] H. Brezis; Functional Analysis, Sobolev Spaces and Partial Differential Equation, Springer, New york, (2010). [2] C.E. Chidume; Applicable Functional Analysis, International centre for Theoretical Physics Trieste, Italy, July 2006. [3] K. Ezzinbi; Lecture Notes on Sobolev Spaces and Elliptic PDE, AUST-Abuja, Nigeria, 2019. [4] K. Ezzinbi; Lecture Notes on Metric Spaces and Differential Calculus, AUST-Abuja, Nigeria, 2018. [5] K. Ezzinbi; Lecture Notes on Ordinary Differential Equations, AUST-Abuja, Nigeria, 2018. [6] S. Shkoller; Introduction to Sobolev Spaces, May 2011.

[7] A. Bressan; Lecture Notes on Sobolev Spaces, February, 2012. [8] V. Barbu; Non-linear Diﬀerential Equations of Monotone types in Banach spaces, Springer New york, 2010. [9] R.R. Phelps; Lectures Notes on Maximal Monotone Operator, given at Prague/Paseky Summer School, Czech Republic, August, 1993. [10] K.H. Karlsen; Note on Weak Convergence February, 2006.