VARIATIONAL INEQUALITY IN HILBERT SPACES AND THEIR APPLICATIONS

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

By

Udeani Cyril Izuchukwu 40680

June, 2019 Acknowledgements

Firstly, all the glory, thanks, praise, adoration be given to the Almighty God, who made this thesis and my MSc. programme a success. I would like to express my deepest gratitude to my supervisor Professor Khalil Ezzinbi, who inspired me and gave me the motivation and impetus to start this work, and guided me all the way; for his patience and indeed all his invaluable support. I also wish to express my profound gratitude to Professor Charles Ejike Chidume, for his management of the Mathematics Institute in a diligent and patient manner and for the thorough academic guidance. I am as well grateful to my classmates, for their advice and encouragement through out my stay at AUST-Abuja. I will not also forget to ap- preciate all the Faculties that taught me at AUST-Abuja. I wish to thank the Board, President, faculty and staff of AUST for all forms of support and encouragements they gave me which enabled me undergo the MSc. programme. I would also like to thank specially Pan African Material Institute(PAMI) for their sponsorship. A spe- cial appreciation to my family and friends for their words of encouragement during times of despair.

i Certification

This is to certify that the thesis titled “ VARIATIONAL INEQUALITY IN HILBERT SPACES AND THEIR APPLICATIONS” 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 Udeani Cyril Izuchukwu in the department of Pure and Applied Mathematics.

ii VARIATIONAL INEQUALITY IN HILBERT SPACES AND THEIR APPLICATIONS

By Udeani Cyril Izuchukwu A Thesis Presented to the Department of Pure and Applied Mathematics

RECOMMENDED: ...... Supervisor, Prof. K. Ezzinbi

APPROVED BY: ...... Chief Academic Officer, Prof. Charles Ejike Chidume

...... Date

iii ©2019 Udeani Cyril Izuchukwu

ALL RIGHTS RESERVED

iv Dedication

This work is dedicated to God Almighty and my mother for her great support towards my success.

v Abstract

The study of variational inequalities frequently deals with a mapping F from a vector space X or a convex subset of X into its dual X 0 . Let H be a real Hilbert space and a(u, v) be a real bilinear form on H. Assume that the linear and continuous mapping A : H −→ H 0 determines a bilinear form via the pairing a(u, v) = hAu, vi. Given K ⊂ H and f ∈ H 0 . Then, Variational inequality(VI) is the problem of finding u ∈ K such that a(u, v − u) ≥ hf, v − ui, for all v ∈ K. In this work, we outline some results in theory of variational inequalities. Their relationships with other problems of Nonlinear Analysis and some applications are also discussed

Keywords

Sobolev spaces, Variational inequalities,Hilbert Spaces, Elliptic variational inequal- ities

vi Contents

0 Acknowledgmenti

Certification ii

Approval iii

Dedicationv

Abstract vi

Introduction viii

CHAPTER ONE1

1 Linear Functional Analysis1 1.1 Hilbert Spaces...... 1 1.2 Function spaces...... 7 1.3 Sobolev spaces...... 9

2 Variational Inequalities in RN 20 2.1 Basic Theorems and Definition about Fixed point...... 20 2.2 First Theorem about variational inequalities...... 21 2.3 Some problems leading to variational inequality...... 24

3 Variational Inequality in Hilbert Spaces 30 3.1 Problem...... 30 3.2 Application...... 34

4 CONCLUSION 38

vii Introduction

In the study of variational inequalities, we are frequently concern with a mapping F from a vector space X or a convex subset of X into its dual X 0 . Variational inequal- ities and Complementary problems are of fundamental importance in a wide range of mathematical and applied problems, such as programming, traffic engineering, economics and equilibrium problems. The idea and techniques of the variational inequalities are being applied in a variety of diverse areas in sciences and proved to be productive and innovative. It has been shown that this theory provides a sim- ple, natural and unified framework for a general treatment of unrelated problems. The fixed point theory has played an important role in the development of various algorithms for solving variational inequalities. Using the projection operator tech- nique, one usually establishes an equivalence between the variational inequalities and the fixed point problem. The alternative equivalent formulation was used by Lions and Stampacchia [8] to study the existence of a solution of the variational inequalities. Projction methods and its variant forms represent important tools for finding the approximate solution of variational inequalities. In this work, we intend to present the element of variational inequalities and free boundary problems with several examples and their applications. The usual setting of the scalar variational inequality is the following: Let K be a nonempty subset of Rn and h., .i denote the scalar product in Rn. Let an operator F : K → Rn be given. The Stampacchia variational inequality problem [8] consist of finding y ∈ K such that hF (y), x − yi ≥ 0, for all x ∈ K while in the case of the Minty variational inequality [10], it is to find y ∈ K such that hF (y), y − xi ≥ 0, for all x ∈ K. Historically, Stampacchia variational inequality(SVI) was introduced by P.Hartman and G. Stampacchia, and was subsequently expanded by Stampacchia in several paper[8]. The study of the Minty variational inequality problem goes back to Minty, who studied the relationships of Stampachia variational inequality (SVI) and Minty variational inquality (MVI) in the case when F is a monotone operator, but without using the ’Variational Inequality’ terminology[10]. New impetus has been given to the field by the recent paper of Giannessi. Readers interested in more details on the

viii Stampacchia Variational Inequality in finite dimension may consult the nice survey of Harker and Pang[7] for more motivations, examples, results and comprehensive bibliography.

Definition 0.0.1 Let H be a real Hilbert space and a(u, v) be a real bilinear form on H. Assume that the linear and continuous mapping A : H −→ H 0 determines a bilinear form via the pairing

a(u, v) = hAu, vi.

Given K ⊂ H and f ∈ H 0 . Then, Variational inequality(VI) is the problem of finding u ∈ K such that

a(u, v − u) ≥ hf, v − ui, for all v ∈ K.

The aim of this work is to obtain the solution of the differential equation ( −u00 + u = f on I = (0, 1), (0.0.1) u(0) = α, u(1) = β. with α, β ∈ R given and f ∈ L2(I) given using variational approach. Suppose we multiply (0.0.1) by ϕ ∈ C1(I) and integrate by part; we obtain Z Z Z u0 ϕ0 + uϕ = fϕ, for all ϕ ∈ C1(I), ϕ(0) = ϕ(1) = 0. (0.0.2) I I I We obtain the solution equation (0.0.2) using Stampacchia and Lax-Milgram Theorem[2] The first chapter is divided into three sections. The first section introduces Hilbert space, some examples and some of its properties. The second section briefly review the main function space used is this work and the final section discussed the gen- eral Sobolev spaces. Chapter two introduces the concept of variational inequality in RN , we stated the theorem about variational inequalities, some problems lead- ing to variational inequality were discussed in connection with convex function. In chapter three, we started Stammpacchia and Lax-Milgram Theorems. Variational inequalities in Hilbert spaces were disscussed and their applications

ix CHAPTER 1

Linear Functional Analysis

The aim of this Chapter is to recall basic results from functional analysis and Distri- bution theory. The chapter is divided into three sections. The first section introduces Hilbert spaces and some basic properties of Hilbert spaces. The second section in- troduces basic concept of Distribution theory and the last section deals with basic results about Sobolev spaces that are of important in the remaining chapters.

1.1 Hilbert Spaces

Let us recall some definitions, theorems, and elementary properties on Hilbert spaces.

Definition 1.1.1 Let E be a linear space over K,(K = R or C). An inner product on E is a function h., .i : E × E −→ K such that the following are satisfied, for x, y, z ∈ E; λ, µ ∈ K: (i) hx, xi ≥ 0 and hx, xi = 0 if and only if x = 0

(ii) hx, yi = hx, yi

(iii) hλx + µy, zi = λhx, zi + µhy, zi. The pair (E, h., .i) is called an inner product space.

Remark 1.1.2 A complete inner product space is called a Hilbert space.

Examples 1 Euclidean space: The space RN is a Hilbert space with the inner product defined by PN hx, yi = i=1 xiyi where, x = (x1, x2, ..., xN ) and y = (y1, y2, ..., yN ). We obtain that

1 N 1 p P 2 2 kxk = hx, xi = ( i=1 xi ) .

2 Space L2(Ω). Z 2 L2(Ω) := {f :Ω −→ R : f is measurablbe and f dx < ∞} Ω where Ω is an open set in RN , is a Hilbert space with the inner product defined by

Z hf, gi = f(x)g(x)dx, Ω and

Z 2 1 kfk = ( |f(x)| dx) 2 . Ω Proposition 1.1.3 [2](Cauchy-Schwart’s Inequality) Let E be an inner prod- uct space. For arbitary x, y ∈ E we have

|hx, yi|2 ≤ hx, xihy, yi

Proof. Let x, y ∈ E be arbitrary. Take z ∈ C with |z| = 1 and let t ∈ R. Then, 0 ≤ htzx + y, tzx + yi = htzx, tzxi + htzx, yi + hy, tzxi + hy, yi = t2zzhx, xi + tzhx, yi + tzhy, xi + hy, yi = t2|z|hx, xi + tzhx, yi + tzhx, yi + hy, yi = t2hx, xi + 2tRe(zhx, yi) + hy, yi ≤ t2hx, xi + 2t|zhx, yi| + hy, yi = t2hx, xi + 2t|z||hx, yi| + hy, yi = t2hx, xi + 2t|hx, yi| + hy, yi. (1.1.1) t2hx, xi + 2t|hx, yi| + hx, xi is a quadratic function with variable t ∈ R. Since,

t2hx, xi + 2t|hx, yi| + hy, yi ≥ 0, for arbitrary t ∈ R, Hence,

|hx, yi|2 ≤ hx, xihy, yi, for all x, y ∈ E.

Theorem 1.1.4 [2] Let h., .i be an inner product on E, then the mapping

x 7−→ kxk = phx, xi is a norm on E.

Proof. Let x, y ∈ E be arbitrary. From the definition of the inner product, we have

2 hx, xi = 0 ⇔ x = 0, hence

kxk2 = hx, xi = 0 ⇔ x = 0.

Also,

hλx, λxi = |λ|hx, xi.

And

kλxk = phx, xi = p|λ|2hx, xi = |λ|phx, xi = kxk.

Let x, y ∈ E. Then,

kx + yk = hx + y, x + yi = hx, xi + hx, yi + hy, xi + hy, yi = kxk2 + 2Re(hx, yi) + kyk2 ≤ kxk2 + 2|hx, yi| + kyk2 ≤ kxk2 + 2p(hx, xihy, yi) + kyk2 = kxk2 + 2(kxk + kyk) + kyk2 = (kxk + kyk)2.

Definition 1.1.5 Let E be a linear space and F ⊂ E is said to be convex if for each x, y ∈ F and λ ∈ [0, 1] we have

λx + (1 − λ)y ∈ E.

Proposition 1.1.6 [4](Parallelogram Law) Let E be an inner product space, then for x, y ∈ E

kx + yk2 + kx − yk2 = 2(kxk2 + kyk2).

Theorem 1.1.7 [4] Let H denote a real Hilbert space and let K be a closed, convex subset of H. Then for each x ∈ H there exists unique y ∈ K such that

kx − yk = inf{kx − ηk : η ∈ K}. (1.1.2)

Proof. Let ηk ∈ K be a minimizing sequence such that

lim kηk − xk = d = infη∈K kη − xk. k→∞ Since H is a Hilbert Space, then by the Parallelogram Law, we have

kx + yk2 + kx − yk2 = 2(kxk2 + kyk2), for all x, y ∈ H.

3 Thus,

2 2 2 2 kηk − ηpk + kηk + ηpk = 2(kηkk + kηpk ); for ηk, ηp ∈ K. By convexity of K, we have

1 1 1 2 ηk + (1 − 2 )ηp = 2 (ηk + ηp) ∈ K.

But d = infη∈K kx − ηk ≤ kx − ηk, for all η ∈ K. 1 Take η = 2 (ηk + ηp). Then, 1 d ≤ kx − (η + η )k 2 k p 1 ⇒ d2 ≤ kx − (η + η )k2 2 k p 1 ⇒ −d2 ≥ −kx − (η + η )k2. 2 k p

Now, using the Parallelogram Law and setting y = x − ηk ∈ H and x = x − ηp ∈ H, we obtain that

2 0 ≤ kηk − ηpk 2 2 2 = 2kx − ηkk + 2kx − ηpk − k2x − (ηk + ηp)k 1 = 2kx − η k2 + 2kx − η k2 − 4kx − (η + η )k2 k p 2 k p 2 2 2 ≤ 2kx − ηkk + 2kx − ηpk − 4d .

But lim kx − ηkk = d. Then, k→∞

2 2 2 2kx − ηkk + 2kx − ηpk − 4d −→ 0 as n −→ ∞. Consequently,

2 kηk − ηpk −→ 0 as n −→ ∞ ⇒ kηk − ηpk −→ 0 as n −→ ∞

Hence, (ηk)k≥0 is Cauchy sequence in K. Since H is a Hilbert space, then there existsx ˆ ∈ H such that

ηk −→ xˆ.

But K is a closed subset of H and ηk ∈ K, thusx ˆ ∈ K. Therefore,

kx − xˆk = lim kx − ηkk = d. k→∞ For uniqueness: Letx, ˆ yˆ ∈ K such that

kx − xˆk = infη∈K kx − ηk and

kx − yˆk = infη∈K kx − ηk.

4 By the Parallelogram law and convexity of K, we obtain

0 ≤ kxˆ − yˆk 1 = 2kx − xˆk2 + 2kx − yˆk2 − 4kx − (ˆx − yˆ)k2 2 ≤ 2d2 + 2d2 − 4d2 = 0.

Then kxˆ − yˆk = 0 ⇔ xˆ =y ˆ. Hence, there is a unique y ∈ K such that

kx − yk = infη∈K kx − ηk.

Remark 1.1.8 The point y ∈ H satisfying (1.1.2) is called a projection of x on K and y = PK x.

Corollary 1.1.9 [2] Let K be a closed, convex subset of a Hilbert space H. Then the operator PK is nonexpansive, that is

0 0 0 kPK x − PK x k ≤ kx − x k, for all x, x ∈ H.

Proof. 0 0 0 0 Let x, x ∈ H such that y = PK x, y = PK x . Then, for y, y ∈ K we have hy, η − yi ≥ hx, η − yi, for all η ∈ K, and

hy0 , η − y0 i ≥ hx0 , η − y0 i , for all η ∈ K.

Setting η = y0 and η = y in the first and second inequality respectively we obtain

hy, y0 − yi ≥ hx, y0 − yi and hy0 , y − y0 i ≥ hx0 , y − y0 i.

Adding we obtain that,

hy, y0 − yi + hy0 , y − y0 i ≥ hx, y0 − yi + hx0 , y − y0 i, hence

hy, y0 − yi − hy0 , y0 − yi ≥ hx, y0 − yi − hx0 , y0 − yi, and

hy − y0 , y0 − yi ≥ hx − x0 , y0 − yi.

Consequently,

−hy − y0 , y − y0 i ≥ hx − x0 , y0 − yi.

Then,

hy − y0 , y − y0 i ≤ hx − x0 , y − y0 i

5 ky − y0 k2 = hy − y0 , y − y0 i ≤ hx − x0 , y − y0 i ≤ |hx − x0 , y − y0 i| ≤ kx − x0 kky − y0 k, and thus

ky − y0 k ≤ kx − x0 k.

Therefore,

0 0 0 kPK x − PK x k ≤ kx − x k, for all x, x ∈ H.

Theorem 1.1.10 [8] Let K be a closed convex subset of a real Hilbert space H. Then y = PK x, the projection of x on K, if and only if y ∈ K such that hy, η − yi ≥ hx, η − yi, for all η ∈ K.

Proof. Let x ∈ H and y = PK x. Since K is convex, then tη + (1 − t)y = y + t(η − y) ∈ K, for all η ∈ K, 0 ≤ t ≤ 1.

Set φ(t) = kx − (y + t(η − y))k2, 0 ≤ t ≤ 1.

φ(t) = kx − y − t(η − y)k2 = kx − yk2 − 2tRehx − y, η − yi + t2kη − yk2 = kx − yk2 − 2thx − y, η − yi + t2kη − yk2.

Then,

φ0 (t) = −2hx − y, η − yi + 2tkη − yk2, thus, φ0 (0) = −2hx − y, η − yi. Therefore, the function φ attains its minimum at t = 0. Thus,

φ0 (0) ≤ 0 ⇔ hx − y, η − yi ≤ 0, η ∈ K ⇔ hx, η − yi − hy, η − y >≤ 0 ⇔ hy, η − yi ≥ hx, η − yi, η ∈ K.

Let y ∈ K. Then,

hy, η − yi ≥ hx, η − yi, η ∈ K.

Thus,

hy, η − yi − hx, η − yi ≥ 0 and

hy − x, η − yi ≥ 0

6 0 ≤ hy − x, η − yi = hy − x, (η − x) + (x − y)i ≤ −kx − yk2 + hy − x, η − xi.

Therefore,

ky − xk2 ≤ hy − x, η − xi ≤ |hy − x, η − xi| ≤ ky − xkkη − xk.

Thus, ky − xk ≤ kη − xk. Hence for each x ∈ H there exists y ∈ K such that

ky − xk = infη∈K kη − xk. Corollary 1.1.11 [2] Let H be a real Hilbert space and K a closed subspace of H. Then, for arbitrary vector x ∈ H, there exist a unique vector y˜ ∈ K such that kx − y˜k ≤ kx − yk, for all y ∈ K.

Theorem 1.1.12 [2](Reiesz Theorem) Let H be a Hilbert space. Then H0 = H, where H0 denote the dual of H.

Theorem 1.1.13 [4](Riesz Representation Theorem) Let H be a Hilbert space and let f be a bounded linear functional on H. Then, there exists a unique vector of x0 ∈ H such that

f(x) = hx, x0i, for each x ∈ H.

Moreover, kfk = kx0k.

1.2 Function spaces

We recall some definitions of function spaces used in this thesis

Definition 1.2.1 An open connected set Ω ⊂ RN is called a domain. By Ω , we denote the closure of Ω; ∂Ω is the boundary and Ωo is the interior of Ω.

N x = (x1, x2, ..., xN ) ∈ R and α = (α1, α2, ..., αN ) ∈ N is a multi-index.

|α| = α1 + α2 + ... + αN

|α| α ∂ u D u := α1 α2 αN ∂x1 ∂x2 ...∂xN

∇u = (∂1u, ∂2u, ..., ∂N u) N X 2 1 |∇u| = ( |∂ju| ) 2 j=1

7 Definition 1.2.2 Let f :Ω −→ R be continuous. We define support of f by supp(f) = {x ∈ Ω: f(x) 6= 0}. The function is said to be of compact support on Ω if the support is a compact set contained inside Ω. The space of test functions in Ω is denoted by D(Ω) and defined

∞ D(Ω) := {f :Ω −→ R,C : support(f) is compact} = {f ∈ C∞(Ω) : supp(f) is compact}

Definition 1.2.3 Let (ψn)n≥1 be a sequence in D(Ω) and ψ ∈ D(Ω). Then, ψn → ψ in D(Ω) if

(i) there exists a compact set K ⊂ Ω such that, supp(ψn) ⊂ K, for all n ≥ 1.

α α n (ii) D ψn → D ψ uniformly on K as n → ∞ and for all α ∈ N .

Definition 1.2.4 A distribution on Ω is any continuous linear mapping T : D(Ω) → 0 R. The set of all distribution on Ω is denoted by D (Ω).

Means that if

ψn → 0 in D(Ω) , then (T, ψn) → 0 in R. Example.

The map δ : D(R) → R defined by hδ, ψi = δ(ψ) = ψ(0) is a distribution. It is usually called Dirac distribution. To see this, we have that δ is linear, since for ψ1, ψ2 ∈ D(Ω) and α ∈ R,

δ(ψ1 + αψ2) = hδ, ψ1 + αψ2i

= (ψ1 + αψ2)(0)

= ψ1(0) + αψ2(0)

= hδ, ψ1i + hδ, ψ2i

= hδ, ψ1i + αhδ, ψ2i

= δ(ψ1) + δ(ψ2).

Hence, δ is linear. Let {ψn}n≥1 ⊂ D(R) such that ψn → 0 as n → ∞ on D(R). But ψn → 0 on D(R) implies that there exists a compact set K ⊂ R such that (j) supp(ψn) ⊆ K and for all j ∈ N, ψn → 0 uniformly on R. Thus, 0 ≤ |ψn(0)| ≤ supx∈K |ψn(x)|→ 0 as n → ∞. And then hδ, ψni = ψn(0) → 0 as n → ∞. Therefore, δ is continuous and hence δ is a distribution.

Definition 1.2.5 A funtion f :Ω → R is said to be locally integrable if for any compact set, K ⊂ Ω, we have that

Z |f(x)|dx < ∞. K

8 1 The collection of all locally integrable funtionals is denoted by Lloc(Ω). For any 1 f ∈ Lloc(Ω), f gives a distribution Tf defined by

Z (Tf , ψ) = f(x)ψ(x)dx, for all ψ ∈ D(Ω). Ω

1 0 Remark 1.2.6 Lloc(Ω) ⊆ D (Ω).

0 Theorem 1.2.7 [6] Let T ∈ D (Ω) be a distribution on an open set Ω in RN , and α, a multi index. Then, for all α ∈ Nn, n ≥ 1 (DαT, ψ) = (−1)|α|(T,Dαψ), Dαψ ∈ D0(Ω), for all ψ ∈ D(Ω).

Definition 1.2.8 By Ck(Ω), we denote the space of k times differentiable (real valued) functions on Ω.

Definition 1.2.9 [2] By Ck,λ(Ω), 0 < λ < 1, we indicate the functions k times continuously differentiable in Ω whose derivative of order k are continuous , 0 < λ < 1.

1.3 Sobolev spaces

Definition 1.3.1 Let Ω be open set in RN . Let p ∈ R with 1 ≤ p < +∞. Z p LP (Ω) := {f :Ω −→ R; measurable : |f| dλ < +∞} Ω where λ is a measure on Ω and Z p 1 kfkp := ( |f| dλ) p Ω

L∞(Ω) := {f :Ω −→ R : f is essentially bounded}, i.e f ∈ L∞(Ω) ⇔ there exists c > 0 such that |f(x) ≤ c a.e on Ω and kfk∞ = inf{c > 0: |f(x)| ≤ c a.e on Ω}.

Theorem 1.3.2 [6] LP (Ω)is a Banach space for 1 ≤ p ≤ ∞.

Proposition 1.3.3 [6](Holder’s Inequality) Let f ∈ Lp(Ω), g ∈ Lq(Ω) such that 1 1 p + q = 1. Then fg ∈ L1(Ω). Moreover, Z Z Z p 1 q 1 |fg| ≤ ( |f| ) p ( |g| ) q = kfkpkkgkq. (1.3.1) Ω Ω Ω Definition 1.3.4 The space Hm(Ω) is called Sobolev space of order m and it is defined as

m α H (Ω) := {f ∈ L2(Ω) : D f ∈ L2(Ω), |α| ≤ m}, endowed with the inner product

α α m < f, g >H (Ω)=< f, g >L2(Ω) +Σα|≤m < D f, D g >L2(Ω) . (1.3.2)

9 and

α 2 1 m m 2 kfkH (Ω) = (kfkL2(Ω) + Σ|α|≤m|D f| ) , for all f, g ∈ H (Ω).

Theorem 1.3.5 [6] The spaces Hm(Ω) , m ≥ 0 endowed with the inner product (1.1.1) are Hilbert spaces.

Proof. m Let (fn)n≥1 be a Cauchy sequence in H (Ω). Let  > 0 be given. Then, there exists no ∈ N such that

kfn − fmkHm(Ω) for all n, m ≥ no.

Thus

kf − f k2 + P kDαf − Dαf k2 < 2, for all n, m ≥ n n m L2(Ω) |α|≤m n m o which implies

kf − f k2 < 2 and P kDαf − Dαf k2 < 2, for all n, m ≥ n . n m L2(Ω) |α|≤m m m o then

P α α 2 kfn − fmk <  and |α|≤mkD fn − D fmk ≤  , for all n, m ≥ no.

α Thus, we obtain that (fn)n≥1 is a Cauchy sequence in L2(Ω) and (D (fn))n is a Cauchy sequence in L2(Ω). Since L2(Ω) is complete, then there exist f, fi ∈ L2(Ω) such that

α fn −→ f in L2(Ω) as n −→ ∞ and D fn −→ fi in L2(Ω) as n −→ ∞.

0 Since L2(Ω) ⊂ D (Ω) we obtain that

0 α fn −→ f in D (Ω) and D fn −→ fi. But Dαf −→ Dαf in D0 (Ω) as n −→ ∞. By uniqueness of limit we obtain that α 0 D f = fi in D (Ω). Thus,

α α fn −→ f in L2(Ω) and D fn −→ D f in L2(Ω), |α| ≤ m. Thus f ∈ Hm(Ω) with

P α α kfn − fkL2(Ω) −→ 0 and |α|≤mkD fn − D fkL2(Ω) −→ 0 an n −→ ∞. Hence,

m f ∈ H (Ω) and kfn − fkHm(Ω) −→ 0 as n −→ ∞. Therefore, Hm(Ω) is a Hilbert space. For m = 1 we obtain

1 ∂f H (Ω) := {f ∈ L2(Ω) : ∈ L2(Ω), i = 1, 2, ..., N} ∂xi

10 and on H1(Ω) we have the following inner product

N X ∂f ∂g hf, gi 1 = hf, gi + h , i H (Ω) L2(Ω) ∂x ∂x L2(Ω) i=1 i i N Z X Z ∂f ∂g = fg + ∂x ∂x Ω i=1 Ω i i and

q 1 kfkH1(Ω) = hf, giH1(Ω), for all f ∈ H (Ω) v u N u 2 X ∂f = tkfk + k kL (Ω). L2(Ω) ∂x 2 i=1 i

1 Definition 1.3.6 We define H0 (Ω) := D(Ω) |H1(Ω).

1 1 H0 (Ω) is a Hilbert space with the norm k.kH1(Ω) and we define the norm on H0 (Ω) by

sZ 2 kuk 1 := |∇u| . H0 (Ω) Ω

Theorem 1.3.7 [6](Poincare’s inequality) Suppose Ω is bounded. Then, there exists c > 0 such that

Z Z 2 2 2 1 u dx ≤ c |∇u| dx, for all u ∈ H0 (Ω). Ω Ω Proof. N Since Ω is bounded. Then Ω ⊆ Πi=1[ai, bi]. We proceed this way, we first prove it in D(Ω). Let ϕ ∈ D(Ω) such that ϕ(t, x) = ϕ(t, x2, x3, ..., xN ). R t ∂ But ϕ(s, x)ds = ϕ(t, x) − ϕ(a1, x) = ϕ(t, x). Then using Cauchy Schwartz a1 ∂s inequality, we obtain that

Z t ∂ ϕ2(t, x) = ( ϕ(s, x)ds)2 a1 ∂s Z t ∂ 2 ≤ (t − a1) ( ϕ(s, x)) ds. a1 ∂s

11 Integrating, we obtain that Z Z Z t 2 ∂ 2 ϕ (t, x)dtdx ≤ ((t − a1) ( ϕ(s, x)) )ds)dtdx Ω Ω a1 ∂s Z Z b1 ∂ 2 ≤ (t − a1)( ϕ(s, x)) )dsdtdx Ω a1 ∂s Z b1 Z ∂ 2 = (t − a1)dt ( ϕ(s, x)) dsdx a1 Ω ∂s Z 1 2 ∂ 2 = (b1 − a1) ( ϕ(s, x)) dsdx 2 Ω ∂s Z 1 2 2 ≤ (b1 − a1) |∇ϕ| . 2 Ω

2 1 2 Choose c = 2 (b1 − a1) > 0. Then Z Z ϕ2 ≤ c2 |∇ϕ|2, for all ϕ ∈ D(Ω). (1.3.3) Ω Ω

1 Now, let u ∈ H0 (Ω) = D(Ω) |H1(Ω). Then, there exist (ϕp)p≥1 ⊂ D(Ω) such that

kϕp − uk → 0 as p → ∞.

We have Z Z 2 2 2 ϕp ≤ c |∇ϕp| . (1.3.4) Ω Ω

Z Z 2 2 |ϕp − u| + |∇(ϕp − u)| → 0, as n → ∞. Ω Ω And thus,

Z Z Z Z 2 2 2 2 ϕp → u and |∇ϕp| → |∇u| . Ω Ω Ω Ω Letting p → ∞ in equation (1.3.4), we obtain that

Z Z 2 2 2 1 u ≤ c |∇u| , for all u ∈ H0 (Ω). Ω Ω Theorem 1.3.8 [6](Poincare-Wirtinger’s inequality) Suppose Ω is smooth and connected, then for any u ∈ H1(Ω) there exists c > 0 such that

Z Z |u − uˆ|2 ≤ c2 |∇u|2, for all u ∈ H1(Ω), Ω Ω where 1 Z uˆ = u. mes(Ω) Ω

12 Corollary 1.3.9 The norm k.k 1 is equivalent to k.k 1 . H0 (Ω) H (Ω) Proof of Corollary. 1 Let u ∈ H0 (Ω). Then, Z 2 2 2 1 kukH (Ω) = kukL2(Ω) + |∇u| Ω Z ≥ |∇u|2 Ω 2 = kuk 1 . H0 (Ω) Thus,

1 1 kukHo (Ω) ≤ kukH (Ω). (1.3.5) Now using Poincare’s inequality we obtain that Z Z 2 2 2 kukH1(Ω) = u dx + |∇u| dx Ω Ω Z Z ≤ c2 |∇u|2dx + |∇u|2 Ω Ω Z = (c2 + 1) |∇u|2 Ω 2 2 = (c + 1)kuk 1 , H0 (Ω) which implies 2 kuk 1 ≤ (c + 1)kuk 1 and thus H (Ω) H0 (Ω) 1 βkukH1(Ω) ≤ kukH1(Ω), β = . (1.3.6) 0 (c2 + 1)

Therefore, from equation (1.3.5) and (1.3.6) we obtain

βkuk 1 ≤ kuk 1 ≤ kuk 1 . H (Ω) H0 (Ω) H (Ω)

1 Therefore, the two norms are equivalent on H0 (Ω).

N Theorem 1.3.10 [6] Let Ω be smooth in R ,N ≥ 2 and D(Ω) = {u |Ω: u ∈ D(RN )}. D(Ω) = H1(Ω).

1 Then, γ : D(Ω) → L2(∂Ω) is continuous with the H (Ω) norm. Hence, γ is exten- sible by continuity over H1(Ω). i.e

1 γ : H (Ω) → L2(∂Ω) is continuous u → ∂u = u |∂Ω. Moreover, there exists α > 0 such that

Z 2 2 1 u dσ ≤ α kukH1(Ω), for all u ∈ H (Ω). ∂Ω

13 Application Let Ω be smooth and connected in RN . Define

Z Vˆ = {u ∈ H1(Ω) : udσ = 0}. ∂Ω Then Vˆ is closed in H1(Ω). ˆ 1 To see this, let (un)n≥1 be a sequence in V such that un → u in H (Ω). Since (u ) ⊂ Vˆ , then n n≥1 Z undσ = 0. ∂Ω Thus, we obtain that

Z 2 2 |un − u| dσ ≤ α kun − ukH1(Ω). ∂Ω

1 But un → u in H (Ω), thus kun − ukH1(Ω) → 0 as n → ∞. Which implies

Z 2 |un − u| dσ → 0 as n → ∞. ∂Ω But

Z Z 2 1 1 |un − u|dσ ≤ ( |un − u| dσ) 2 (mes(∂Ω)) 2 . ∂Ω ∂Ω Hence, since mes(∂Ω) > 0 we have

Z |un − u|dσ → 0 in L1(Ω). ∂Ω And we obtan that

Z Z undσ → udσ. ∂Ω ∂Ω

But Z undσ = 0. ∂Ω Then by uniqueness of limit we obtain that

Z udσ = 0. ∂Ω

14 Hence u ∈ Vˆ and therefore Vˆ is closed.

1 Theorem 1.3.11 [6](Rellich Theorem) If Ω is smooth, then H (Ω) ,→ L2(Ω) is 1 compact. Moreover, if (un)n≥1 is a bounded sequence in H (Ω), then there exists a subsequence (unk )k≥1 of (un)n≥1 such that (unk )k≥1 converges in L2(Ω). Application Let Ω be smooth and connected in RN . Define

Z V = {u ∈ H1(Ω) : u = 0}. Ω Then, Poincare’s inequality is true on V . Thus, there exists c > 0 such that

Z Z u2 ≤ |∇u|2, for all u ∈ V. Ω Ω

We proceed by contradiction. Suppose for all n ∈ N there exists (un) ∈ V such that

Z Z 2 √ 2 2 un > ( n) |∇un| , Ω Ω thus

Z Z 2 2 un > n |∇un| . Ω Ω But

Z Z Z Z 2 2 2 2 un + |∇un| ≥ un > n |∇un| . Ω Ω Ω Ω Hence, Z 2 kunkH1(Ω) > n |∇un| . (1.3.7) Ω un Let vn = . Then kvnkH1(Ω) = 1, for all n ≥ 1. kunkH1(Ω) 1 Since (vn)n≥1 is bounded in H (Ω). Then by Rellich Theorem, there exists a subse- quence (vnk )k≥1 of (vn)n≥1 and f ∈ L2(Ω) such that vnk → f in L2(Ω). 1 Multiplying equation (1.3.7) by 2 , we obtain that kunkH1(Ω)

Z 1 2 n 2 |∇un| < 1, for all n ≥ 1. kunkH1(Ω) Ω

15 Which implies that

Z 2 1 |∇vn| < , for all n ≥ 1. Ω n

Thus Z 2 |∇vn| → 0 asn → ∞. Ω And hence Z 2 |∇vnk | → 0 as k → ∞. Ω

∂vnk Then, for all i = 1, 2, ..., N → 0 in L2(Ω). ∂xi 0 But vnk → f in L2(Ω) as k → ∞ and since L2(Ω) ⊆ D (Ω), we obtain that

0 ∂vnk 0 vnk → f in D (Ω) and → 0 in D (Ω). ∂xi ∂v ∂f And by convergence in D0(Ω), we have that nk → → in D0(Ω). Thus, by ∂xi ∂xi ∂f uniqueness of limits = 0, for all i = 1, 2, ..., N. And therefore f is constant. ∂xi 1 Thus f =c ˜ and by the above argument, we have that vnk → c˜ in H (Ω).

But V is closed and vnk ∈ V , thenc ˜ ∈ V . It implies that Z cdx˜ = 0 Ω

1 and thusc ˜ = 0. Hence, vnk → 0 in H (Ω) as k → ∞. 1 But kvnk kH (Ω) = 1, a contracdiction. Therefore, the claim is true.

Definition 1.3.12 More generally, we define for every 1 ≤ p < ∞ and for m ≥ 0, the Sobolev spaces

m,p α W (Ω) = {f ∈ Lp(Ω) : D f ∈ Lp(Ω), |α| ≤ m} endowed with the following norm

1 p α p p kfk m,p = kfk p + (Σ kD fk ) . W (Ω) L (Ω) |α|≤m LP (Ω) We define

m,q W0 = D(Ω) |W m,q(Ω) . m,q Thus, W0 is the closure of D(Ω) with respect to the norm k.kW m,q(Ω). m m,2 m m,2 When q = 2, we write H (Ω) = W (Ω) and H0 (Ω) = W0 (Ω). For m = 0 we have that

0,q W (Ω) = Lq(Ω).

Theorem 1.3.13 [2] Suppose Ω is smooth, then

m,q m,q m−1 W0 (Ω) := {f ∈ W (Ω) : f = Df = ... = ...D f = 0 on ∂Ω}.

16 For p = 2, we obtain that

m,2 m,2 m−1 W0 (Ω) := {f ∈ W (Ω) : f = Df = ... = D f = 0 on ∂Ω}.

Theorem 1.3.14 [6] W m,p(Ω) is Banach space.

Proof. m,q Let (fn)n≥1 be a Cauchy in W (Ω). Let  > 0 be given, then there exists n0 ∈ N such that

kfn − fkkW m,q(Ω) < , for all n, k ≥ n0. Then,

q q 1 (kf − f k + P kDαf − Dαf k ) q < , for all n, k ≥ n . n k Lq(Ω) |α|≤m n k Lq(Ω) 0 And kf − f kq + P kDαf − Dαf kq < q, for all n, k ≥ n . n k Lq(Ω) |α|≤m n k Lq(Ω) 0 Consequently, kf − f kq < q and P kDαf − Dαf kq < q, for all n, k ≥ n , n k Lq(Ω) |α|≤m n k Lq(Ω) 0 thus

α α kfn − fkkLq(Ω) <  and kD fn − D fkkLq(Ω) < , for all n, k ≥ n0.

α Hence, (fn)n≥1 and (D fn)n are Cauchy sequences in Lq(Ω) and since Lq(Ω) is complete, then there exists f, fi ∈ Lq(Ω) such that

α fn −→ f in Lq(Ω) as n −→ ∞ and D fn −→ fi in Lq(Ω) as n −→ ∞.

0 But Lq(Ω) ⊂ D (Ω), we obtain that

0 α α 0 fn −→ f in D (Ω) as n −→ ∞ and D fn −→ D f as n −→ ∞ in D (Ω)

α 0 By uniqueness of limit we obtain that D f = fi in D (Ω). Thus α α fn −→ f as n −→ ∞ in Lq(Ω) and D fn −→ D f as n −→ ∞ in Lq(Ω), |α| ≤ m. Hence

P α α kfn − fkkLq(Ω) −→ 0 as n −→ ∞ and |α|≤mkD fn − D fkLq(Ω) −→ 0 as n −→ ∞ which implies that kfn − fkW m,q(Ω) −→ 0 as n −→ ∞. Thus,

m,q m,q f ∈ W (Ω) and kfn − fkW m,q(Ω) −→ 0 as n −→ ∞ in W (Ω). Therefore, W m,q(Ω) is a Banach space.

Theorem 1.3.15 [6](Green’s Formula) Let Ω be smooth in Rn, u ∈ H2(Ω) and v ∈ H1(Ω). Then,

Z Z Z ∂u ∇u∇v = − ∆uv + vdσ, n ≥ 2 Ω Ω ∂(Ω) ∂n

17 ∂u ∂u where denotes the normal derivatives defined by = ∇u.~n and ~n denote the ∂n ∂n normal vector.

Definition 1.3.16 The bilinear form a : H × H → R is coercive on H if there exists α > 0 such that

a(v, v) ≥ αkvk2, for all v ∈ H

Example Let Ω be smooth and connected with ∂Ω = Γ0 ∪ Γ1. Define

1 H = {u ∈ H (Ω) : u |Γ0 = 0}. Then, the bilinear form Z a(u, v) = ∇u∇v Ω is coercive on H. To see this, we proceed by contradiction. suppose it is not coercive then for all n ≥ 1 there exists (un)n ∈ H such that

1 2 a(un, un) < n kunkH1(Ω). Thus, Z 2 1 2 |∇un| < kunkH1(Ω). (1.3.8) Ω n

un 1 Let vn = . Then, kvnkH1(Ω) = 1 Multiplying equation (1.3.8) by , we kunkH1(Ω) kunk obtain that

Z 2 1 |∇vn| < . Ω n Which implies that Z 2 |∇vn| → 0 Ω in L2(Ω). But |vn| = 1, hence (vn)n≥1 is bounded and by Rellich theorem there exists a subsequence (vnk )k≥1 ⊆ (vn)n≥1 such that (vnk ) → g in L2(Ω). Thus, (vnk ) → g in ∂v ∂g 0 0 ∂g D0(Ω) and nk → in D (Ω). By uniquness of limit in D (Ω). Thus, = 0. ∂xi ∂xi ∂xi 1 Since Ω is connected we have that g =c ˆ, a constant. Thus, vnk → cˆ in H (Ω). By Trace theorem we obtain that

vnk |∂Ω→ cˆ in L2(∂Ω). Thus

vnk |Γ0 → cˆ in L2(Γ0).

18 Hence Z 2 2 |vnk | dσ → (ˆc) mes(Γ0). Γ0 But Z 2 |vnk | dσ → 0. Γ0 2 Therefore, (ˆc) mes(Γ0) = 0. Since mes(Γ0) > 0, thenc ˆ = 0. And hence vnk → 0 in H1(Ω).

1 But kvnk kH (Ω) = 1, a contradiction. Therefore the bilinear form is coercive on H.

Definition 1.3.17 A bilinear form a : H × H −→ R is said to be continuous if there exists a constant c > 0 such that

|a(u, v)| ≤ ckukkvk, for u, v ∈ H

Example The bilinear form a : H1(Ω) × H1(Ω) → R defined by

Z Z a(u, v) = ∇u∇v + λ(x)u(x)v(x)dσ is continuous, λ ∈ L∞(∂Ω). Ω ∂Ω To see this we apply Cauchy schwartz inequality. Let u, v ∈ H1(Ω), thus Z Z |a(u, v)| = | ∇u∇v + λ(x)u(x)v(x)dσ| Ω ∂Ω Z Z ≤ |∇u∇v| + |λ(x)u(x)v(x)|dσ Ω ∂Ω Z Z Z 2 1 2 1 ≤ ( |∇u| ) 2 ( |∇v| ) 2 + |λ| |u(x)v(x)|dσ Ω Ω ∂Ω

≤ kukL2(Ω)kvkL2(Ω) + kλk∞kukL2(∂Ω)kvkL2(∂Ω) 2 ≤ kukH1(Ω)kvkH1(Ω) + α kλk∞kukH1(Ω)kvkH1(Ω) 2 = (1 + α kλk∞)kukH1(Ω)kvkH1(Ω).

2 Take c = (1 + α kλk∞), then

|a(u, v)| ≤ ckukH1(Ω)kvkH1(Ω). Therefore, a is continuous.

19 CHAPTER 2

Variational Inequalities in RN

Given K ⊂ RN and F : K −→ RN , a continuous mapping. Then, the Variational inequalities(VI) is the problem of finding a point u ∈ K such that

hF (u), v − ui ≥ 0, v ∈ K. (2.0.1)

Variational inequalities(VI) are closely related with many general problems of non- linear Analysis such as complementary, fixed point and optimization problem. The simplest examples of variational inequalites is the problem of solving a system of equation. Here, we intend to discuss variational inequalities in RN , fixed point and some elementary problem that are associated to variational inequality. In particu- lar,we discuss the connection between variational inequalities and convex funtions.

2.1 Basic Theorems and Definition about Fixed point

Definition 2.1.1 Let S be a metric space with metric d. A mapping F : S −→ S is said to be a strictly contraction map if there exists α ∈ [0, 1[

d(F (x),F (y)) ≤ αd(x, y) , for all x, y ∈ S.

Remark 2.1.2 if α = 1, then F is nonexpansive.

Theorem 2.1.3 [3](Banach’s fixed point Theorem) Let S be a complete met- ric space and let F : S −→ S be a strict contraction mapping. Then, there exist a unique fixed point of F .

Theorem 2.1.4 [3](Brouwer’s fixed point Theorem) Let F be a continuous mapping from a closed ball G ⊂ RN into itself. Then, F admit at least one fixed point in G.

Theorem 2.1.5 [3](Schauder’s fixed point Theorem) Let G be a compact convex subset of RN and F be a continuous mapping from G into itself. Then, F admits a fixed point in G.

20 2.2 First Theorem about variational inequalities

Theorem 2.2.1 [8] Let K be compact and convex set in RN and let F : K −→ RN be continuous. Then, there exists x ∈ K such that

hF (x), y − xi ≥ 0, for all y ∈ K.

Proof. Let Π : RN −→ RN be the identication and (., .) be the scalar product on N R . Let PK (I − ΠF ): K −→ K be continuous, where Ix = x. Then by Schauder fixed point Theorem, PK (I − ΠF ) admits a fixed point. Thus there exists x ∈ K such that

PK (I − ΠF )x = x. By the characterisation of projection Theorem we obtain that

(x, y − x) ≥ ((I − ΠF )x, y − x), for allx, y ∈ k = (x − ΠF (x), y − x) = (x, y − x) − Π(F (x), y − x), for allx, y ∈ K.

Then,

Π(F (x), y − x) ≥ (x, y − x) − (x, y − x) = 0, for all x, y ∈ K, namely

(F (x), y − x) ≥ 0, for all y ∈ K.

Therefore, there exists x ∈ K such that

hF (x), y − xi ≥ 0, for all y ∈ K.

Applications Variational Inequality theory provides us with a tool for: formulating a variety of equilibrium problems; qualitatively analysing the problem in terms of existence and uniquness of solutions and stability. Many of the applications explored to date that have been formulated, studied and solved as variational inequality problems are in fact, network problems. Indeed, many mathematical problems can be formulated as variational inequality problems and several examples applicable to equilibrium analysis follows thus Systems Equations Many classical economic equilibrium problems have been formulated as systems of equation, since market clearing conditions necessarily equate the total supply with the total demand. In terms of variational inequality problem, the formulation of a system of equation is as follow.

Proposition 2.2.2 [9] Let F : RN −→ RN be a mapping. Then for any x ∈ RN we have that

hF (x), y − xi ≥ 0, for all y ∈ RN if and only if F (x) = 0.

21 Proof. Suppose x ∈ RN such that F (x) = 0, then x is a solution to the variational inequality. Let x ∈ RN . Then for any ν ∈ RN , there exists  ≥ 0 and y ∈ RN such that ν = (y − x).

Then, for all ν ∈ RN hF (x), νi = hF (x), (y − x)i ≥ 0 = hF (x), y − xi ≥ 0.

Therefore hF (x), νi ≥ 0, for all ν ∈ RN . Hence F (x) = 0, for all x ∈ RN . An example (market equilibrium with equalities) As an illustration, we present an example of a system of equation. Consider m consumers, with a typical consumer denoted by j, and n commodities, with a typical commodities i. Let p denote the N-dimensional vector of the commodity price with commodities {p1, p2, ..., pN }. Assume that the demand for commodity i, di, may in general depend on the price of all the commodities, that is Pm j di(p) = j=1 di (p)

j where di (p) denotes the demand for commodity i by consumer j at the price vector p. Similarly, assume that the supply of a commodity i, si, may in general depend on the price of all the commodities ,that is

m si(p) = Σj=i

j where si (p) denote the supply of commodityi of consumer j at the price vector p. We then argue the aggregate demands for the commodities into the n-dimensional col- umn vector d with component: {d1, d2,...,dN } and the aggregate supplies of the com- modities into the N-dimensional column vectors with component: {s1, s2, ..., sN }. But the market equilibrium conditions requires that the supply of each commodity must be equal to the demands for each commodity at the equilibrium price p∗, thus we obtain the following system of equation s(p∗) − d(p∗) = 0. Now, suppose we define the vectors x ≡ p and F (x) = s(p) − d(p), we obtain a nonlinear equation. Problem 1 Given K closed and convex subset of RN and F : K −→ RN continuous, find x ∈ K such that hF (x), y − xi ≥ 0, for all y ∈ K. The problem (1) does not always have a solution. If K is bounded, then the problem admits a solution guaranteed by Theorem 2.2.1. If K is not bounded, then the 0 problem does not admits a solution. For instance, take K = R and f : K −→ R defined by f(x) = exp(x). Then,

22 f(x)(y − x) ≥ 0, for all y ∈ R, has no solution. To see this, take y = −1 + x, then

0 ≤ f(x)(y − x) = exp(x)(−1 + x − x) = −1 exp(x).

⇒ −1 exp(x) ≥ 0, impossible. N Now, given a convex set K, we set KR = K ∩B(0,R). Returning to F : K → R ,we have that there exists at least one xR ∈ KR such that

hF (xR), y − xRi ≥ 0, for all y ∈ KR (2.2.1) whenever KR 6= ∅ by the previous theorem.

Theorem 2.2.3 [8] Let K ⊂ RN be closed and convex. Let F : K −→ RN be continuous. Set KR = K ∩ B(0,R). A necessary and sufficient condition that there exists a solution to problem (1) is that there exists xR ∈ KR such that

hF (xR), y − xRi ≥ 0, for all y ∈ K and

|xR| < R.

Proof. N Let KR = K ∩ B(0,R), 0 ∈ R and x ∈ K. Then, hF (x), y − xi ≥ 0, y ∈ K.

|x| < R ⇒ x ∈ B(0,R) ⊂ B(0,R), thus x ∈ KR such that Hence x ∈ KR such that

hF (xR), y − xRi ≥ 0, for all y ∈ KR

Suppose that xR ∈ KR such that |xR| < R. Since |xR| < R then given y ∈ K, take  ≥ 0 sufficiently small such that ω = xR + (y − xR) ∈ KR. Thus xR ∈ KR:

0 ≤ hF (xR), ω − xRi

= hF (xR), xR + (y − xR) − xRi, for ally ∈ K

hF (xR), (y − xR)i

= hF (xR), y − xRi, which implies hF (xR), y − xRi ≥ 0, for all y ∈ K Hence,

hF (xR), y − xRi ≥ 0, for all y ∈ K.

Corollary 2.2.4 [8] N Let F : K −→ R be a mapping. Suppose there exists x0 ∈ K such that hF (x) − F (x ), x − x i 0 0 −→ +∞ as |x| −→ +∞. |x − x0|

23 Then, problem (1) has a solution in K.

Proof. Let x ∈ K. Choose M > |F (x0)| and R > |x0 such that

hF (x) − F (x0), x − x0i ≥ M|x − x0|, for |x| ≥ R and x ∈ K. Then,

M|x − x0| ≤ hF (x) − F (x0), x − x0i

= hF (x), x − x0i − hF (x0), x − x0i, which implies hF (x), x − x0i ≥ M|x − x0| + hF (x0), x − x0i. But

hF (x0), x − x0i ≤ |hF (x0), x − x0i|

≤ |F (x)||x − x0| and −|F (x0)||x − x0| ≤ −hF (x0), x − x0i. Then,

hF (x), x − x0i ≥ M|x − x0| + hF (x0), x − x0i

= M|x − x0| − hF (x0), x0 − xi

≥ M|x − x0| − |F (x0)||x − x0|

= (M − |F (x0)|)|x − x0

≥ (M − |F (x0)|)|x| − |x0|, |x − x0| ≥ |x| − |x0|.

Since M > |F (x0)|, then

(M − |F (x0)|)(|x| − |x0|) > 0, for |x| = R therefore hF (x), x − x0i ≤ 0.

2.3 Some problems leading to variational inequal- ity

Theorem 2.3.1 [8] Let K ⊂ RN be a closed convex set and f ∈ C1(RN ). Set F (x) = gradf(x). Suppose there exists x ∈ K such that

f(x) = miny∈K f(y). Then, x is a solution of the following variational inequality; (F (x), y − x) ≥ 0, for all y ∈ K.

Proof. Let x ∈ K be such that f(x) = miny∈K f(y). Since K is convex, then x + t(y − x) = ty + (1 − t)x ∈ K , for y ∈ K and 0 ≤ t ≤ 1

Set φ(t) = f(x + t(y − x)), 0 ≤ t ≤ 1. If t = 0, then φ(0) = f(x) = miny∈K f(y). Hence φ attains its minimum when t = 0 and φ0 (t) = f 0 (x + t(y − x))(y − x).

24 Then,

0 ≤ φ0 (0) = f 0 (x)(y − x) = (f 0 (x), y − x) = (gradf(x), y − x) = (F (x), y − x).

Therefore, (F (x), y − x) ≥ 0, for all y ∈ K.

Remark 2.3.2 In general, the converse does not hold.

Definition 2.3.3 (Convex and Concave function) Let K be a subset of a real vector space and f : K → R ∪ {+∞} be a map. Then, (a) f is said to be convex if

(i) K is a convex set, and

(ii) For each t ∈ [0, 1] and for each x1, x2 ∈ K we have that

f(tx1 + (1 − t)x2) ≤ tf(x1) + (1 − t)f(x2). (b) f is concave if

(i) K is a convex set, and

(ii) For each t ∈ [0, 1] and for each x1, x2 ∈ K, we have that

f(tx1 + (1 − t)x2) ≥ tf(x1) + (1 − t)f(x2).

Remark 2.3.4 f is convex if and only if (−f) is concave,and f is concave if and only if (−f) is convex.

Properties of convex functions

Definition 2.3.5 Let X be a normed space and K ⊂ X. Let f : K → R ∪ +∞ be a map. The epigragraph of f is the set defined by

epi(f):= {(x, α) ∈ X × R : x ∈ D(f) and f(x) ≤ α}.

Theorem 2.3.6 [4] Let X be a real normed space. A mapping f : X → R ∪ +∞ is convex if and only if epi(f) is convex.

Proof. (⇒) Suppose that f is convex. Let (x1, α1), (x1, α2) ∈ epi(f) and t ∈ [0, 1]. Then, f(x1) ≤ α1 and f(x2) ≤ α2. Now, since f is convex,

f(tx1 + (1 − t)x2) ≤ tf(x1) + (1 − t)f(x2)

≤ tα1 + (1 − t)α2.

Thus,

25 (tx1 + (1 − t)x2, tα1 + (1 − t)α2) ∈ epi(f). But

(tx1 + (1 − t)x2, tα1 + (1 − t)α2) = t(x1, α1) + (1 − t)(x2, α2). Hence, epi(f) is convex. (⇐) Suppose epi(f) is convex. Let x1, x2 ∈ D(f) and t ∈ [0, 1]. Then x1, x2 ∈ D(f) implies that f(x1), f(x2) ∈ (R). Thus, (x1, f(x1)), (x2, f(x2)) ∈ epi(f). But epi(f) is convex. Hence,

t(x1, f(x1)) + (1 − t)(x2, f(x2)) ∈ epi(f) and

(tx1 + (1 − t)x2, tf(x1) + (1 − t)f(x2)) ∈ epi(f). Consequently,

f(tx1 + (1 − t)x2) ≤ tf(x1) + (1 − t)f(x2). Therefore, f is convex.

Definition 2.3.7 A mapping f : X → R ∪ +∞ is called strictly convex if for each x1, x2 ∈ D(f), x1 6= x2, and for each t ∈ (0, 1) we have

f(tx1 + (1 − t)x2) < tf(x1) + (1 − t)f(x2). Moreover, f is strictly concave if (−f) is strictly convex.

Definition 2.3.8 [4] Let X be normed vector space, λ ∈ R and f : X → R be a map. The section of f; Sf,λ is defined by

Sf,λ := {x ∈ D(f): f(x) ≤ λ}.

Theorem 2.3.9 [4] Suppose f is a convex function, then the sections of f are convex

Proof. Let x1, x2 ∈ Sf,λ and t ∈ [0, 1]. Then, f(x1) ≤ λ and f(x2) ≤ λ. By the convexity of f we obtain

f(tx1 + (1 − t)x2) ≤ tf(x1) + (1 − t)f(x2) ≤ tλ + (1 − t)λ = λ.

Thus,

f(tx1 + (1 − t)x2) ≤ λ. And consequently,

tx1 + (1 − t)x2 ∈ Sf,λ.

Therefore, Sf,λ is convex.

Theorem 2.3.10 Let f1 and f2 be convex functions. Then f1 + f2 is convex.

26 Examples of convex functions. Example 1 In R, the following functions are convex f(x) = x2, f(x) = x4 and f(x) = |x| Example 2 0 Let g ∈ Rn and define f : Rn → R by f(x) = hx, g0 i. Then, f is convex. n To see this, let x1, x2 ∈ R and t ∈ [0, 1]. Then,

0 f(tx1 + (1 − t)x2) = htx1 + (1 − t)x2, g i 0 0 = htx1, g i + h(1 − t)x2, g i 0 0 = thx1, g i + (1 − t)hx2, g i

= tf(x1) + (1 − t)f(x2). Therefore, f is convex. Example 3 Let a : Rn × Rn → R be a bilinear map. Suppose for each x ∈ Rn, a(x, x) ≥ 0. Then f : Rn → R defined by f(x) = a(x, x)

n is convex. To see this, let x1, x2 ∈ R and t ∈ [0, 1]. Then, f(tx1 + (1 − t)x2) − tf(x1) − (1 − t)f(x2) = a(tx1 + (1 − t)x2, tx1 + (1 − t)x2) − ta(x1, x1) − (1 − t)a(x2, x2)

= a(tx1, tx1) + a(tx1, (1 − t)x2) + a((1 − t)x2, tx1) + a((1 − t)x2, (1 − t)x2) − ta(x1, x1) − (1 − t)a(x2, x2) 2 2 = t a(x1, x1) + t(1 − t)a(x1, x2) + t(1 − t)a(x2, x1) + (1 − t) a(x2, x2) − ta(x1, x1) − (1 − t)a(x2, x2) 2 2 = (t − t)a(x1, x1) + t(1 − t)[a(x1, x2) + a(x2, x1)] + ((1 − t) − (1 − t))a(x2, x2) 2 2 = (t − t)a(x1, x1) + t(1 − t)[a(x1, x1) + a(x2, x2) − a(x1 − x2, x1 − x2)] + ((1 − t) − (1 − t))a(x2, x2)

= −t(1 − t)a(x1 − x2, x1 − x2) ≤ 0. Thus,

f(tx1 + (1 − t)x2) − tf(x1) − (1 − t)f(x2) ≤ 0, and consequently,

f(tx1 + (1 − t)x2) ≤ tf(x1) − (1 − t)f(x2). Therefore, f is convex.

Theorem 2.3.11 [1] Suppose f : Rn → R is convex and x ∈ K satisfying (F (x), y − x) ≥ 0, for all y ∈ K. Then,

f(x) = miny∈K f(y).

Proof. Let x, y ∈ K such that (F (x), y − x) ≥ 0. Since K is convex, then yt + (1 − t)x = x + (y − x)t ∈ K. By convexity of f, we have

27 f(x + (y − x)t) ≤ f(x) + tf(y − x).

Then,

f 0 (x + t(y − x))(y − x) ≤ f(y − x) = f(x − y).

Thus, f(y) ≥ f(x) + f 0 (x + t(y − x))(y − x). If t = 0, we obtain that

f(y) ≥ f(x) + f 0 (y − x) = f(x) + (F (x), y − x) ≥ f(x).

Then, f(x) ≤ f(y), for all y ∈ K. Therefore,f(x) = miny∈K f(y)

0 Theorem 2.3.12 [5] Let f : K −→ R ,K ⊂ RN , be a continuously differentiable convex function. Then, F (x) = gradf(x) is monotone.

Proof. Let x, x0 ∈ K. Then,

hF (x), y − xi ≥ 0, for all y ∈ K and

hF (x0 ), y − x0 i ≥ 0, for all y ∈ K.

Since f is convex, then we have that for all y ∈ K

f(y) ≥ f(x) + (F (x), y − x), (2.3.1) and f(y) ≥ f(x0 ) + (F (x0 ), y − x0 ). (2.3.2) Set x = y and x0 in the variational inequality of x and x0 respectively and adding, we obtain that

f(x0 ) + f(x) ≥ f(x) + f(x0 ) + (F (x0 ), x − x0 ) + (F (x), x0 − x).

Then,

⇒(F (x), x0 − x) + (F (x0 ), x − x0 ) ≤ 0 ⇒ (F (x), x0 − x) − (F (x0 ), x0 − x) ≤ 0 ⇒ (F (x) − F (x0 ), x0 − x) ≤ 0 ⇒ −(F (x0 ) − F (x), x0 − x) ≤ 0 ⇒ (F (x0 ) − F (x), x0 − x) ≥ 0.

Hence, F (x) = gradf(x) is monotone. Problem 2 N N Let R+ = {x = (x1, x2, ..., xN ) ∈ R : xi ≥ 0, i = 1, 2, ..., N} be closed and convex N N N N subset of R . Let F : R+ −→ R be a mapping. Find x0 ∈ R+ such that

28 N F (x0) ∈ R+ and (F (x0), x0) = 0.

N Theorem 2.3.13 [11] The point x0 ∈ R+ is a solution to problem (2) if and only N if x0 ∈ R+ such that

N (F (x0), y − x0) ≥ 0, for all y ∈ R+.

Proof. N Suppose x0 ∈ R+ is a solution to problem (2), then

N F (x0) ∈ R+ and (F (x0), x0) = 0.

N Thus (F (x0), y) ≥ 0, for all y ∈ R+

(F (x0), y − x0) = (F (x0), y) − (F (x0), x0)

= (F (x0), y) ≥ 0.

N N N Therefore, x0 ∈ R+:(F (x0), y − x0) ≥ 0, for all y ∈ R+. Now suppose x0 ∈ R+:

N (F (x0), y − x0) ≥ 0, for all y ∈ R+.

Let y = x0 + ei, where ei = (0, 0, ..., 0, 1, 0, ..., 0)

0 ≤ (F (x0), y − x0)

= (F (x0), x0 + ei − x0)

= (F (x0), ei)

= Fi(x0)

= F (x0), thus,F (x ) ∈ N . Since x ∈ N such that (F (x ), y − x ) ≥ 0, for all y ∈ N , then 0 R+ 0 R0 0 0 R+ for y = 0

(F (x0), 0 − x0) ≥ 0

⇒ (F (x0), −x0) ≥ 0

⇒ −(F (x0), x0) ≥ 0

⇒ (F (x0), x0) ≤ 0.

N N But (F (x0), x0) ≥ 0, for any x0 ∈ R+ and F (x0) ∈ R+. Therefore (F (x0), x0) = 0.

29 CHAPTER 3

Variational Inequality in Hilbert Spaces

Here, we study variational inequalities in Hilbert space. Some basic theorems and proofs are presented in this chapter. This will be used in obtaining our main exis- tence and uniqueness theorem. The study of variational inequalites started being considered around nintheenth century. Many differential equations that arise from different kind of application are solved by a very simple calculation. This approach does not give the existence and uniqueness of classical and weak solutions. Hence, the concept of Variational approach is paramount. Let H be a real Hilbert space and a(u, v) be a real bilinear form on H. Assume that the linear and continuous mapping A : H −→ H 0 determines a bilinear form via the pairing

a(u, v) = hAu, vi.

3.1 Problem

Let H be a real Hilbert space and f ∈ H 0 . Let K ⊂ H be closed and convex. Find u ∈ K such that a(u, v − u) ≥ hf, v − ui, for all v ∈ K. (3.1.1)

Theorem 3.1.1 [2](Stampacchia Theorem) Let a(u, v) be a continuous coercive bilinear form on H. Let K ⊂ H be a nonempty closed and convex with f ∈ H 0 . Then there exists a unique solution to problem (3.1.1). 0 Moreover, if u1, u2 ∈ K are solutions to problem (3.1.1) corresponding to f1, f2 ∈ H , then 1 ku − u k ≤ kf − f k 0 , α > 0. (3.1.2) 1 2 H α 1 2 H Proof. Let u1, u2 ∈ H be solutions of the variational inequality ui ∈ K such that

a(ui, v − ui) ≥ hfi, v − uii, v ∈ K, i = 1, 2. Thus,

30 a(u1, v − u1) ≥ hf1, v − u1i ,for all v ∈ K and

a(u2, v − u2) ≥ hf2, v − u2i, for all v ∈ K.

Setting v = u1 and v = u2 in the variational inequality of u1 and u2 respectively, we obtain

a(u1, u2 − u1) ≥ hf1, u2 − u1i and a(u2, u1 − u2) ≥ hf2, u1 − u2i. Adding gives

a(u1, u2 − u1) + a(u2, u1 − u2) ≥ hf1, u2 − u1i + hf2, u1 − u2i

⇒ a(u2, u1 − u2) − a(u1, u1 − u2) ≥ h2, u1 − u2i − hf1, u1 − u2i

⇒ a(u2 − u1, u1 − u2) ≥ hf2 − f1, u1 − u2i

⇒ −a(u1 − u2, u1 − u2) ≥ hf2 − f1, u1 − u2i

⇒ a(u1 − u2, u1 − u2) ≤ hf2 − f1, u1 − u2i.

But a is coercive, then there exists α > 0 such that

2 αku1 − u2k ≤ a(u1 − u2, u1 − u2)

≤ hf1 − f2, u1 − u2i

≤ |hf1 − f2, u1 − u2i|

≤ kf1 − f2kku1 − u2k.

Then,

αku1 − u2k ≤ kf1 − f2kku1 − u2k,

1 and therefore ku1 − u2k ≤ α kf1 − f2k. Let a be symmetric and define I : H −→ R by I(u) = a(u, u) − 2hf, ui, for all u ∈ H.

0 Let d = infu∈ K I(u). Then d > −∞. To see this we use the fact that f ∈ H and a is corecive.

I(u) = a(u, u) − 2hf, ui 2 ≥ αkukH − 2kfkH0 kukH

2 1 2 2 ≥ αkuk − kfk 0 − αkuk H α H H 1 2 = − kfk 0 . α H

1 2 thus, d ≥ − kfk 0 > −∞ , α > 0. α H Now, d = infu∈K I(u) ⇒ for all n ∈ N, there exists un ∈ K such that

1 d ≤ I(un) < d + n .

31 Since a is coercive, then there exists α > 0 such that

2 αkun − umk ≤ a(un − um, un − um)

= a(un, um) − 2a(un, um) + a(um, um)

= a(un, un) + a(um, um) + a(un, un) + a(um, um) − a(un, un) − a(um, um) − 2a(un, um)

= 2a(un, un) + 2a(um, um) − {a(un, un) + a(um, um) + 2a(un, um)}

= 2a(un, un) + 2a(um, um) − {a(un + um, un + um)} 1 1 = 2a(u , u ) + 2a(u , u ) − 4a( (u + u ), (u + u ). (3.1.1) n n m m 2 n m 2 n m

But I(un) = a(un, un) − 2hf, uni.Thus

0 2I(un) = 2a(un, un) − 4hf, uni, for all f ∈ H 1 1 1 1 ⇒ 4I( 2 (un + um) = 4a( 2 (un + um), 2 (un + um)) − 8hf, 2 (un + um)i. Then, 1 1 1 1 2I(u ) + 2(u ) − 4I( (u + u )) = 2a(u , u ) + 2a(u , u ) − 4hf, u i − 4hf, u i − 4a( (u + u ), (u + u )) + 8 < f, (u + u ) > n m 2 n m n n m m n m 2 n m 2 n m 2 n m 1 1 1 = 2a(u , u ) + 2a(u , u ) − 4a( (u + u ), (u + u )) − 4{hf, u i+, hf, u i} + 4hf, (u + u )i n n m m 2 n m 2 n m n m 2 n m 1 1 = 2a(u + u ) + 2a(u + u ) − 4a( (u + u ), (u + u )) − 4hf, u + u i + hf, u + u i n n m m 2 n m 2 n m n m n m 1 1 = 2a(u , u ) + 2a(u , u ) − 4a( (u + u ), (u + u )). n n m m 2 n m 2 n m Using characterisation of infimum we obtain 1 αku − u k2 ≤ 2I(u ) + 2I(u ) − 4I( (u + u )) n m n m 2 n m 1 1 ≤ 2(d + ) + 2(d + ) − 4d 2 m 1 1 = 2( + ) −→ 0, as n,m −→ ∞. n m

Hence (un)n≥1 is a Cauchy sequence and since H is complete,then there exists u ∈ H such that un −→ u. But K is closed, then u ∈ K. By continuity of I we obtain un −→ u ⇒ I(un) −→ I(u) as n −→ ∞, then I(u) = d. Now, let v ∈ K. Since K is convex we obtain u + t(v − u) ∈ K, 0 ≤ t ≤ 1. d 0 Thus, d = I(u) ≤ I(u + t(v − u)) ⇒ dt (I(u + t(v − u)))|t=o ≥ 0 i.e (v − u)I (u) ≥ 0 I(u) ≤ I(u + t(v − u)) = a(u + t(v − u), u + t(v − u)) − 2hf, u + t(v − u)i = a(u, u) + 2a(u, t(v − u)) + t2a(v − u, v − u) − 2thf, v − ui − 2hf, ui = a(u, u) + 2ta(u, v − u) + t2a(v − u, v − u) − 2thf, v − ui − 2hf, ui. = I(u) + 2ta(u, v − u) + t2a(v − u, v − u) − 2thf, v − ui.

Thus, 2ta(u, v − u) + t2a(v − u, v − u) − 2thf, v − ui ≥ 0 and t hence a(u, v − u) + 2 a(v − u, v − u) ≥ hf, v − ui, for all v ∈ K and for all t ∈ [0, 1]. Letting t = 0 we obtain

32 a(u, v − u) ≥ hf, v − ui, for all v ∈ K. Hence, u ∈ K is a solution to the variational inequality. Corollary 3.1.2 [2](Lax-Milgram Theorem) Assume that a(u, v) is a continu- ous coercive bilinear form on H. Then, for any f ∈ H 0 , there exists a unique u ∈ H such that a(u, v) = hf, vi, for all v ∈ H. Moreover, if a is symmetric, then u is characterised by the property

1 1 u ∈ H and 2 a(u, u) − hf, ui = minv∈H { 2 a(v, v) − hf, vi}.

N Example 3.1.3 Let Ω ⊂ R be measurable and choose ϕ ∈ L2(Ω). Define

K = {v ∈ L2(Ω) : v ≥ 0, a.e inΩ}

Let Z a(u, v) = u(x)v(x)dx Ω be the scalar product on L2(Ω). Then, for any f ∈ L2(Ω), there exists a unique u ∈ K:

Z Z u(v − u)dx ≥ f(v − u)dx, for all v ∈ K Ω Ω .

Claim 1: K is closed and convex. To see this, let (vn)n≥1 ⊂ K such that vn −→ v as n −→ ∞. But

vn ∈ K ⇒ vn ≥ ϕ, a.e in Ω, for some ϕ ∈ L2(Ω)

⇒ v = lim vn ≥ ϕ, a.e in Ω n→∞ ⇒ v ∈ K. Hence, K is closed. For Converxity: Let v, u ∈ K and 0 ≤ t ≤ 1.

v ∈ K ⇒ v ≥ ϕ1 a.e in Ω for some ϕ1 ∈ L2(Ω), thus tv ≥ tϕ1 a.e in Ω, 0 ≤ t ≤ 1. Also u ∈ K ⇒ u ≥ ϕ2 a.e in Ω, for some ϕ2 ∈ L2(Ω), then (1 − t)u ≥ (1 − t)ϕ2 a.e in Ω, 0 ≤ t ≤ 1. Now,

tv + (1 − t)u ≥ tϕ1 + (1 − t)ϕ2

= ϕ2 + (ϕ1 − ϕ2)t

=: ϕ3 ∈ L2(Ω). then tv + (1 − t)u ∈ K. Therefore, K is convex. We have that a(u, v) is coercive since it is an inner produvt on L2(Ω). And therefore, by Theorem 3.1.1 there exists u ∈ K such

33 Z Z u(v − u)dx ≥ f(v − u)dx, for all v ∈ K. Ω Ω Claim 2: u = max{ϕ, f}

then, max{ϕ, f} = f(x), if ϕ(x) ≤ f(x) and maxϕ, f = ϕ(x), if f(x) ≤ ϕ(x).

Now Z Z Z u(v − u)dx = ϕ(v − ϕ)dx + f(v − f)dx Ω (f<ϕ) (ϕ≤f) Z Z ≥ f(v − ϕ) + f(v − f)dx (f≤ϕ) (ϕ≤f) Z ≥ f(v − f)dx, v − ϕ ≥ 0, for all v ∈ K. Ω Hence, u is a solution to the variational inequality.

3.2 Application

Example 3.2.1 Consider the following Problem ( −u00 + u = f on I = (0, 1), (3.2.1) u(0) = α, u(1) = β. with α, β ∈ R given and f ∈ L2(I) given. We proceed as follows: Defined in the space H1(I) the set K by

K := {v ∈ H1(I): v(0) = α and v(1) = β}.

Claim. K is nonempty, closed and convex. To see this, consider the mapping v : I −→ R defined by v(x) = α, if ∈ [0, 1) and v(x) = β, if x = 1.

We have that v ∈ K since v ∈ H1(I), v(0) = α and v(1) = β. Let (vn)n≥1 ⊂ K such that vn −→ v as n −→ ∞. But vn ∈ K ⇒ vn(0) = α and vn(1) = β. Thus,

v(0) = lim vn(0) = α and v(1) = lim vn(1) = β. n→∞ n→∞ Hence, v ∈ K since v(0) = α and v(1) = β. Therefore, K is closed. For convexity. Let v, u ∈ K and 0 ≤ t ≤ 1. But u ∈ K ⇒ u(0) = α and u(1) = β. Thus,

tu(0) = tα and tu(1) = tβ.

34 Also, v ∈ K ⇒ v(0) = α and v(1) = β. Thus, (1 − t)v(0) = (1 − t)α and (1 − t)v(1) = (1 − t)β. Now, (tu + (1 − t)v)(0) = tu(0) + (1 − t)v(0) = tα + (1 − t)α = α. Consequently,

(tu + (1 − t)v)(1) = tu(1) + (1 − t)v(1) = tβ + (1 − t)β = β. Hence, tu + (1 − t)v ∈ K. Therefore, K is convex. We assume that u is a classical solution of (3.2.1), then multiplying (3.2.1) by v ∈ D(I) and integrating by part, we obtain Z Z Z u0 (v − u)0 + u(v − u) = f(v − u), for all v ∈ K. (3.2.2) I I I Hence, Z Z Z u0 (v − u)0 + u(v − u) ≥ f(v − u), for all v ∈ K. (3.2.3) I I I Set Z Z a(u, v − u) = u0 (v − u)0 + u(v − u). I I But Z Z a(u, v) = u0 v0 + uv I I is continuous and coercive. Therefore by Stampacchia Theorem there exist a unique function u ∈ K satisfying(3.2.3). 1 We proceed to recover the classical solution of (3.2.1). Let w ∈ H0 (I), setting v = u + w in the equation (3.2.3) we obtain Z Z Z u0 w0 + uw = fw, for all w ∈ K. (3.2.4) I I I 1 Since f ∈ L2(I) and u ∈ H0 (I) is a weak solution of the problem, then u ∈ H2(I).Suppose f ∈ C(I), then the weak solution u belongs to C2(I). To see this, we obtain from the assumption that f − u ∈ C(I). Thus (u0 )0 ∈ C(I) and hence u ∈ C2(I). Therefore, u is a classical solution of the problem. Example 3.2.2 ( −∆u + u = f on Ω, f ∈ L (Ω) 2 (3.2.5) u = g on Γ. We proceed thus: Suppose that there exist a function g ∈ H1(Ω) ∩ C(Ω) such that g = g on Γ. Define in the space H1(Ω) the set K by

35 K := {v ∈ H1(Ω) : v − g = 0 on Γ}.

K is nonempty, closed and convex. Assume that u ∈ K is a weak solution of the problem, then Z Z Z ∇u(∇v − ∇u) + u(v − u) ≥ f(v − u), for all v ∈ K. (3.2.6) Ω Ω Ω

Set Z Z a(u, v − u) = ∇u(∇v − ∇u) + u(v − u). Ω Ω But Z Z a(u, v) = ∇u∇v + uv Ω Ω is continuous and coercive. Therefore by Stampacchia Theorem there exist a unique function u ∈ K satisfying the equation (3.2.6). The classical solution is recovered following the same approach in the previous ex- ample. 1 Let w ∈ H0 (Ω), setting v = u ± w we obtain Z Z Z ∇u∇w + uw = fw, for all w ∈ K. (3.2.7) Ω Ω Ω 1 2 Since f ∈ L2(Ω) and u ∈ H0 (Ω) is a weak solution of (3.2.5), then u ∈ H (Ω).Suppose f ∈ C(Ω), then the weak solution u belongs to C2(Ω). To see this, we obtain from the assumption that f − u ∈ C(Ω). Thus (u0 )0 ∈ C(Ω) and hence u ∈ C2(Ω). Therefore, u is a classical solution of (3.2.5) .

Example 3.2.3 (Application Problem) Let Ω be smooth on Rn, n ≥ 2. Con- sider the problem ( −∆u + u = f on Ω, f ∈ L (Ω) 2 (3.2.8) u |∂Ω= 0. We proceed thus: 1 Assume that the problem has a solution. Define the space H0 (Ω) by

1 1 H0 (Ω) := {u ∈ H (Ω) : u |∂Ω= 0}

1 Multiplying the equation (3.2.8) by v ∈ H0 (Ω) and integrating on Ω we have Z Z Z − ∆uv + uv = fv (3.2.9) Ω Ω Ω We then use Green Formula to obtain Z Z Z 1 ∇u∇v + uv = fv, for all H0 (Ω). (3.2.10) Ω Ω Ω

The bilinear form a(u, v) = hu, viH1(Ω) is coercive and continuous and f is in the 1 1 dual of H0 (Ω). Therefore by Lax-Milgram theorem, there exist a unique u ∈ H0 (Ω) such that

36 1 a(u, v) = hf, vi, for all v ∈ H0 (Ω).

Thus Z Z Z 1 ∇u∇v + uv = fv, for all H0 (Ω). (3.2.11) Ω Ω Ω Let v = ϕ ∈ D(Ω), we have

Z Z Z ∇u∇ϕ + uϕ = fϕ. Ω Ω Ω

Thus Z ∇u∇ϕ + hu, ϕi = hf, ϕi, Ω and consequently

−h∆u, ϕi + hu, ϕi = hf, ϕi. for all ϕ ∈ D(Ω).

Hence, −h∆u + u, ϕi = hf, ϕi. for all ϕ ∈ D(Ω). We have

−∆u + u = f on D0 (Ω).

2 But by regularity property we have ∆u = u − f ∈ L2(Ω),then u ∈ H (Ω). Therefore 1 2 the problem has a unique solution on H0 (Ω) ∩ H (Ω).

37 CHAPTER 4

CONCLUSION

In this work, we studied variational inequalities in Hilbert space. Some basic theo- rems and proofs were presented. We studied and obtained existence and uniqueness theorems for variational inequalities. Many differential equations that arise from different kind of application were solved by a very simple calculation. We discov- ered that this approach does not give the existence and uniqueness of classical and weak solutions. Hence, the concept of Variational approach is paramount. we es- tablished the existence and uniqueness of solutions of variational inequalities. This was achieved through the use of Stampacchia theorem and Lax-Milgram theorem. And its applications. We Considered the following Problem ( −u00 + u = f on I = (0, 1), (4.0.1) u(0) = α, u(1) = β. with α, β ∈ R given and f ∈ L2(I) given. And obtained its solution using variational approach via Stammpacchia Theorem. We also looked at its application in Rn and more generally in Hilbert Space. We also considered the problem of the form ( −∆u + u = f on Ω, f ∈ L (Ω) 2 (4.0.2) u = g on Γ. We obtained its solution using variational approach by applying Stammpacchia theorem.

38 Bibliography

[1] Blum, E. From Optimization and Variational Inequalities to Equilibrium Prob- lems Math.Student, pp. 123-145 Vol.63, 1994.

[2] Brezis, H. Functional Analysis, Sobolev Spaces and Partial Differential Equa- tion Spring Science and Business Media, 2010.

[3] Browder,F. E. Fixed Point Theory and Nonlinear Problems Proc. Sym. Pure. Math, pp. 49-88, Vol.39, 1983.

[4] Chidume, C. E. Applicable Functional Analysis University of Ibadan, Press, 2014.

[5] Cottle, R. W., Giannessi, F., and Lions, J. L. Variational Inequalities and Complementarity Problems: Theory and Applications John Wiley and Sons, 1980.

[6] Ezzinbi, K. Lecture Notes on Distribution Theory, Sobolev Spaces and Elliptic Partial Differential Equation African University of Science and Technology, Abuja, 2018.

[7] Harker, P. T. and Pang, J. S. Finite-Dimensional Variational Inequalities and Nonlinear Complementarity Problems: a Survey of Theory, Algorithms and Applications Mathematical Programming, Vol.48(1-3), pp. 161-220, 1990.

[8] Kinderlehrer, D. and Stampachia, G. An Introduction to Variational Inequal- ities and their Applications SIAM, Vol.31, 1980.

[9] Konnov, I. V. and Laitinen, E. Theory and Applications of Variational In- equalities University of Oulu, Department of Mathematical Sciences, 2002.

[10] Minty, G. J. On the Generalization of a Direct Method of the Calculus of Variations Bulletin of American Mathematical Society, Vol.73(3), pp. 315-321, 1967.

[11] Patriksson, M. Nonlinear Programming and Variational Inequalities Problems: a Unified Apploach Spring Science and Business Media, Vol.23, 2013.

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