ﺭﺎﺘﺨﻤ ﻲﺠﺎﺒ ﺔﻌﻤﺎﺠbiblio.univ-annaba.dz/wp-content/uploads/2014/06/thesis.pdf · Systèmes Dynamiques et Analyse Fonctionnelle Devant le jury PRÉSIDENT

وزارة التعليم العايل والبحث العلمي

BADJI MOKHTAR UNIVERSITY OF ANNABA UNIVERSITÉ BADJI MOKHTAR DE ANNABA

جامعة باجي مختار -عنابة -

Faculté des Sciences

Département de Mathématiques

MÉMOIRE

Présenté en vue de l’obtention du diplôme de

MAGISTER EN MATHÉMATIQUES ÉCOLE DOCTORALE EN MATHÉMATIQUES

Par

Khaled ZENNIR

Intitulé

Existence et Comportement Asymptotique des

Solutions d’une Equation de Viscoélasticité

Non Linéaire de type Hyperbolique

Dirigé par

Prof. Hocine SISSAOUI

Option Systèmes Dynamiques et Analyse Fonctionnelle

Devant le jury

PRÉSIDENT

B. KHODJA

PROF.

U. B. M. ANNABA

RAPPORTEUR

H. SISSAOUI

PROF.

U. B. M. ANNABA

ÉXAMINATEUR

L. NISSE

M. C.

U. B. M. ANNABA

ÉXAMINATEUR

Y. LASKRI

M. C.

U. B. M. ANNABA

INVITE

B. SAID-HOUARI

Dr.

U. B. M. ANNABA

Année 2008

وزارة التعليم العايل والبحث العلمي

BADJI MOKHTAR UNIVERSITY OF ANNABA UNIVERSITÉ BADJI MOKHTAR DE ANNABA

جامعة باجي مختار -عنابة -

Sciences Faculty

Department of Mathematics

THESIS

Submitted for the obtention of

MAGISTER DIPLOMA IN MATHEMATICS DOCTORAL SCHOOL IN MATHEMATICS

By

Khaled ZENNIR

Existence et Comportement Asymptotique des

Solutions d’une Equation de Viscoélasticité

Non Linéaire de type Hyperbolique

Directed by

First Advisor: Prof. Hocine SISSAOUI Second Advisor : Dr. Belkacem SAÏD-HOUARI

Option

Dynamic Systems and Functional Analysis

PRESIDENT

B. KHODJA

PROF.

U. B. M. ANNABA

SUPERVISOR

H. SISSAOUI

PROF.

U. B. M. ANNABA

EXAMINER

L. NISSE

M. C.

U. B. M. ANNABA

EXAMINER

Y. LASKRI

M. C.

U. B. M. ANNABA

2008

Existence and Asymptotic Behavior of

Solutions of a Non Linear Viscoelastic

Hyperbolic Equation

ThanksThanksThanksThanks

In the first place at all, my gratitude goes to Prof. H. SISSAOUI, my supervisor. He always listens, advices, encourages and guides me, he was behind this work. I shall always be impressed by his availability and the care with which he comes to read and correct my manuscript.

My warm thanks go to Dr.B.SAID-HOUARI, who supervised this thesis with great patience and kindness. He was able to motivate each step of my work by relevant remarks and let me able to advance in my research, beginning by the DES study in university of Skikda. I thank him very sincerely for his availability even at a distance, with him I have a lot of fun to work.

I wish to thank Prof B. KHODJA, who makes me the honour of chairing the jury of my defense.

I am very grateful to Dr. L. NISSE and Dr. Y. LASKRI to have agreed to rescind this thesis and to be part of my defense panel.

Of course, this thesis is the culmination of many years of study. I am indebted to all those who guided me on the road of mathematics.

I thank from my heart, my teachers whose encouraged me during my studies and have distracted me the mathematics.

I think for my parents, whose the work could not succeed without their endless support and encouragement. They find in the realization of this work, the culmination of their efforts and express my most affectionate gratitude.

Khaled ZENNIR

Contents

1 Preliminary 1

1.1 Banach Spaces - De�nition and Properties . . . . . . . . . . . . . . . . . . . . 2

1.1.1 The weak and weak star topologies . . . . . . . . . . . . . . . . . . . . 3

1.1.2 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Functional Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.1 The Lp () spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.2 The Sobolev space Wm;p() . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.3 The Lp (0; T;X) spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.2.4 Some Algebraic inequalities . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.3 Existence Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.3.1 The Contraction Mapping Theorem . . . . . . . . . . . . . . . . . . . . 14

1.3.2 Gronwell�s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.3.3 The mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 Local Existence 17

2.1 Local Existence Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Global Existence and Energy Decay 39

3.1 Global Existence Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

3.2 Decay of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4 Exponential Growth 59

4.1 Growth result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

i

ii

Abstract

Our work, in this thesis, lies in the study, under some conditions on p, m and the functional g, the

existence and asymptotic behavior of solutions of a nonlinear viscoelastic hyperbolic problem

of the form 8>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>:

utt ��u� !�ut +tR0

g(t� s)�u(s; x)ds

+a jutjm�2 ut = b jujp�2 u; x 2 ; t > 0

u (0; x) = u0 (x) ; x 2 ut (0; x) = u1 (x) ; x 2

u (t; x) = 0; x 2 �; t > 0

; (P)

where is a bounded domain in RN (N � 1), with smooth boundary �; a; b; w are positiveconstants, m � 2; p � 2; and the function g satisfying some appropriate conditions.

Our results contain and generalize some existing results in literature. To prove our results many

theorems were introduced.

Keywords: Nonlinear damping, strong damping, viscoelasticity, nonlinear source, local solutions,

global solutions, exponential decay, polynomial decay, growth.

iii

Résumé

Notre travail, dans ce memoire consiste à étudier l�éxistence et le comportement asymptotique des

solutions d�un problème de viscoelasticité non lineaire de type hyperbolique suivant:8>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>:




u (0; x) = u0 (x) ; x 2 ut (0; x) = u1 (x) ; x 2

u (t; x) = 0; x 2 �; t > 0

; (P)

où, est un domaine borné de RN (N � 1), avec frontière assez regulière �; a; b; w sont desconstantes positives, m � 2; p � 2; et la fonction g satisfaite quelques conditions.

Nos résultats contiennent et généralisent certains résultats d�existences dans la littérature.

Pour la preuve, beaucoup théorèmes ont été présentés

Mots clés: Dissipation nonlinéaire, viscoelasticité, source nonlinéaire, solutions locale, solutions

globale, décroissance exponentielle de l�énergie, décroissance polynomiale, croissement.

iv

Notations

: bounded domain in RN :� : topological boundary of :

x = (x1; x2; :::; xN) : generic point of RN :dx = dx1dx2:::dxN : Lebesgue measuring on :

ru : gradient of u:�u : Laplacien of u:

f+; f� : max(f; 0); max(�f; 0):a:e : almost everywhere.

p0 : conjugate of p; i:e1

p+1

p0= 1:

D() : space of di¤erentiable functions with compact support in :

D0() : distribution space.

Ck () : space of functions k�times continuously di¤erentiable in :C0 () : space of continuous functions null board in :

Lp () : Space of functions p�th power integrated on with measure of dx:

kfkp =�R

jf(x)jp�1p

:

W 1;p () =nu 2 Lp () ; ru 2 (Lp ())N

o:

kuk1;p =�kukpp + kruk

pp

�1p:

W 1;p0 () : the closure of D () in W 1;p () :

W�1;p0 () : the dual space of W 1;p0 () :

H : Hilbert space.

H10 = W 1;2

0 :

If X is a Banach space

Lp (0; T ;X) =

�f : (0; T )! X is measurable ;

TR0

kf(t)kpX dt <1�:

L1 (0; T ;X) =

(f : (0; T )! X is measurable ;ess� sup

t2(0;T )kf(t)kpX <1

):

Ck ([0; T ] ;X) : Space of functions k�times continuously di¤erentiable for [0; T ]! X:

D ([0; T ] ;X) : space of functions continuously di¤erentiable with compact support in [0; T ] :

BX = fx 2 X; kxk � 1g : unit ball.

v

Introduction

In this thesis we consider the following nonlinear viscoelastic hyperbolic problem8>>>>>>>>>>>>>><>>>>>>>>>>>>>>:




u (0; x) = u0 (x) ; ut (0; x) = u1 (x) ; x 2

u (t; x) = 0; x 2 �; t > 0

; (1)

where is a bounded domain inRN (N � 1), with smooth boundary �; a; b; w are positive constants,and m � 2; p � 2: The function g(t) is assumed to be a positive nonincreasing function de�ned onR+ and satis�es the following conditions:(G1): g : R+ �! R+ is a bounded C1-function such that

g(0) > 0; 1�1Z0

g(s)ds = l > 0:

(G2): g(t) � 0; g0(t) � 0; g(t) � ��g0(t); 8 t � 0; � > 0:

In the physical point of view, this type of problems arise usually in viscoelasticity. This type

of problems have been considered �rst by Dafermos [12], in 1970, where the general decay was

discussed. A related problems to (1) have attracted a great deal of attention in the last two

decades, and many results have been appeared on the existence and long time behavior of solutions.

See in this directions [2;3;5� 8;18;29;33;34;38] and references therein.

In the absence of the strong damping �ut; that is for w = 0; and when the function g vanishes

identically ( i.e: g = 0); then problem (1) reduced to the following initial boundary damped wave

equation with nonlinear damping and nonlinear sources terms.

utt ��u+ a jutjm�2 ut = b jujp�2 u: (2)

Some special cases of equation (2) arise in quantum �eld theory which describe the motion of

charged mesons in an electromagnetic �eld.

Equation (2) together with initial and boundary conditions of Dirichlet type, has been extensively

studied and results concerning existence, blow up and asymptotic behavior of smooth, as well

vi

as weak solutions have been established by several authors over the past three decades. Some

interesting results have been summarized by Said-Houari in his master thesis [42]:

For b = 0, that is in the absence of the source term, it is well known that the damping term

a jutjm�2 ut assures global existence and decay of the solution energy for arbitrary initial data ( seefor instance [17] and [21]):

For a = 0; the source term causes �nite-time blow-up of solutions with a large initial data ( negative

initial energy). That is to say, the norm of our solution u(t; x) in the energy space reaches +1when the time t approaches certain value T � called " the blow up time", ( see [1] and [20] for more

details).

The interaction between the damping term a jutjm�2 ut and the source term b jujp�2 u makes theproblem more interesting. This situation was �rst considered by Levine [23;24] in the linear

damping case (m = 2), where he showed that solutions with negative initial energy blow up in �nite

time T �. The main ingredient used in [23] and [24] is the " concavity method" where the basic

idea of this method is to construct a positive function L(t) of the solution and show that for some

> 0; the function L� (t) is a positive concave function of t: In order to �nd such ; it su¢ ces to

verify that:d2L� (t)

dt2= � L� �2(t)

�LL00 � (1 + )L02(t)

�� 0; 8t � 0:

This is equivalent to prove that L(t) satis�es the di¤erential inequality

LL00 � (1 + )L02(t) � 0; 8t � 0:

Unfortunately, this method fails in the case of nonlinear damping term (m > 2):

Georgiev and Todorova in their famous paper [14], extended Levine�s result to the nonlinear damp-

ing case (m > 2): More precisely, in [14] and by combining the Galerkin approximation with the

contraction mapping theorem, the authors showed that problem (2) in a bounded domain with

initial and boundary conditions of Dirichlet type has a unique solution in the interval [0; T ) provided

that T is small enough. Also, they proved that the obtained solutions continue to exist globally in

time if m � p and the initial data are small enough. Whereas for p > m the unique solution of

problem (2) blows up in �nite time provided that the initial data are large enough. ( i.e: the initial

energy is su¢ ciently negative).

This later result has been pushed by Messaoudi in [35] to the situation where the initial energy

E(0) < 0: For more general result in this direction, we refer the interested reader to the works of

Vitillaro [47]; Levine [25] and Serrin and Messaoudi and Said-Houari [32]:

vii

In the presence of the viscoelastic term (g 6= 0) and for w = 0; our problem (1) becomes8>>>>>>>>>>>>>><>>>>>>>>>>>>>>:

utt ��u+tR0



u (0; x) = u0 (x) ; ut (0; x) = u1 (x) ; x 2

u (t; x) = 0; x 2 �; t > 0

: (3)

For a = 0; problem (3) has been investigated by Berrimi and Messaoudi [3]: They established the

local existence result by using the Galerkin method together with the contraction mapping theorem.

Also, they showed that for a suitable initial data, then the local solution is global in time and in

addition, they showed that the dissipation given by the viscoelastic integral term is strong enough to

stabilize the oscillations of the solution with the same rate of decaying ( exponential or polynomial)

of the kernel g: Also their result has been obtained under weaker conditions than those used by

Cavalcanti et al [7]; in which a similar problem has been addressed.

Messaoudi in [29]; showed that under appropriate conditions between m; p and g a blow up and

global existence result, of course his work generalizes the results by Georgiev and Todorova [14]

and Messaoudi [29]:

One of the main direction of the research in this �eld seems to �nd the minimal dissipation such

that the solutions of problems similar to (3) decay uniformly to zero, as time goes to in�nity.

Consequently, several authors introduced di¤erent types of dissipative mechanisms to stabilize these

problems. For example, a localized frictional linear damping of the form a(x)ut acting in sub-domain

w � has been considered by Cavalcanti et al [7]: More precisely the authors in [6] looked into

the following problem

utt ��u+tZ0

g(t� s)�u(s; x)ds+ a(x)ut + juj u = 0: (4)

for > 0; g a positive function and a : ! R+ a function, which may be null on a part of thedomain :

By assuming a(x) � a0 > 0 on the sub-domain w � ; the authors showed a decay result of an

exponential rate, provided that the kernel g satis�es

� �1g(t) � g0(t) � ��2g(t); t � 0; (5)

viii

and kgkL1(0;1) is small enough.

This later result has been improved by Berrimi and Messaoudi [2], in which they showed that the

viscoelastic dissipation alone is strong enough to stabilize the problem even with an exponential

rate.

In many existing works on this �eld, the following conditions on the kernel

g0(t) � ��gp(t); t � 0; p � 1; (6)

is crucial in the proof of the stability.

For a viscoelastic systems with oscillating kernels, we mention the work by Rivera et al [36]. In

that work the authors proved that if the kernel satis�es g(0) > 0 and decays exponentially to zero,

that is for p = 1 in (6), then the solution also decays exponentially to zero. On the other hand,

if the kernel decays polynomially, i.e. (p > 1) in the inequality (6), then the solution also decays

polynomially with the same rate of decay.

In the presence of the strong damping (w > 0) and in the absence of the viscoelastic term (g = 0),

the problem (1) takes the following form8>>>>>>><>>>>>>>:

utt ��u� !�ut + a jutjm�2 ut = b jujp�2 u; x 2 ; t > 0

u (0; x) = u0 (x) ; ut (0; x) = u1 (x) ; x 2

u (t; x) = 0; x 2 �; t > 0

: (7)

Problem (7) represents the wave equation with a strong damping �ut. When m = 2, this problem

has been studied by Gazzola and Squassina [13]: In their work, the authors proved some results

on well posedness and asymptotic behavior of solutions. They showed the global existence and

polynomial decay property of solutions provided that the initial data is in the potential well.

The proof in [13] is based on a method used in [19]. Unfortunately their decay rate is not opti-

mal, and their result has been improved by Gerbi and Said-Houari [16], by using an appropriate

modi�cation of the energy method and some di¤erential and integral inequalities.

Introducing a strong damping term �ut makes the problem from that considered in [42] and [14],

for this reason less results where known for the wave equation with strong damping and many

problem remain unsolved. ( See [13] and the recent work by Gerbi and Said-Houari [15]):

In this thesis, we investigated problem (1), in which all the damping mechanism have been considered

in the same time ( i.e. w > 0; g 6= 0; and m � 2); these assumptions make our problem di¤erent

ix

form those studied in the literature, specially the blow up result / exponential growth of solutions

(chapter4).

This thesis is organized as follows:

Chapter1:

In this chapter we introduce some notation and prepare some material needed for our work. The

main results of this chapter such as: the Lp� spaces, the Sobolev spaces, di¤erential and integral

inequalities and other theorems of functional analysis, can found in the books [4] and [43]:

Chapter2:

This chapter is devoted to the study of the local existence result, the main ingredient used in this

chapter is the Galerkin approximations ( the compactness method) introduced in the book of Lions

[26]; together with the �x point method.

Indeed, we consider �rst for u 2 C ([0; T ] ; H10 ) given, the following problem

vtt ��v � !�vt +

tZ0

g(t� s)�v(s; x)ds+ a jvtjm�2 vt = b jujp�2 u; x 2 ; t > 0 (8)

with the initial data

v (0; x) = u0 (x) ; vt (0; x) = u1 (x) ; x 2 (9)

and boundary conditions of the form

v (t; x) = 0; x 2 �; t > 0; (10)

and we will show that problem (8) � (10) has a unique local solutions v by the Faedo-Galerkinmethod, which consists in constructing approximations of the solution, then we obtain a priori

estimates necessary to guarantee the convergence of these approximations. We recall here that the

presence of nonlinearity on the damping term a jvtjm�2 vt forces us to go until the second a prioriestimate. We point out that the contraction semigroup method fails here, because of the presence

of the nonlinear terms.

Once the local solution v exists, we will use the contraction mapping theorem to show the local

existence of our problem (1). This will be done under the assumption that T is required to be small

enough (see formula (2:77)):

Chapter 3:

Our main purpose in this chapter is tow-fold:

First, we introduce a set W de�ned in (3:5) called " the potential well" or " stable set" and we

show that if we restrict our initial data in this set, then our solution obtained in chapter 2 is global

in time, that is to say, the norm

kutk2 + kruk2 ;

x

in the energy space L2() � H10 () of our solution is bounded by a constant independent of the

time t:

Second, We show that, if our solution is global in time, ( i.e: by assuming that the initial data

u0 2 W ) and if our function g satis�es the condition (6) ( for p = 1); then our solution decays timeasymptotically to zero. More precisely we prove that the decay rate is of the form (1 + t)2=(2�m)

if m > 2; whereas for m = 2; we obtain an exponential decay rate. (See Theorem 3.2.1). The

main tool used in our proof is an inequality due to Nakao [37]; in which this inequality has been

introduced in order to study the stability of the wave equation, but it is still works in our problem.

Chapter 4:

In this chapter we will prove that if the initial energy E(0) of our solution is negative ( this means

that our initial data are large enough), then our local solutions in bounded and

kutk2 + kruk2 !1

as t tends to +1: In fact it will be proved that the Lp�norm of the solution grows as an exponentialfunction. An essential tool of the proof is an idea used by Gerbi and Said-Houari [15], which based

on an auxiliary function ( which is a small perturbation of the total energy), in order to obtain a

di¤erential inequality leads to the exponential growth result provided that the following conditions

1Z0

g(s)ds <p� 2p� 1 ;

holds.

Chapter 1

Preliminary

Abstract

In this chapter we shall introduce and state some necessary materials needed in the proof of our

results, and shortly the basic results which concerning the Banach spaces, the weak and weak star

topologies, the Lp space, Sobolev spaces and other theorems. The knowledge of all this notations

and results are important for our study.

1

Chapter 1. Preliminary 2

1.1 Banach Spaces - De�nition and Properties

We �rst review some basic facts from calculus in the most important class of linear spaces "Banach

spaces".

De�nition 1.1.1 A Banach space is a complete normed linear space X: Its dual space X 0 is the

linear space of all continuous linear functional f : X ! R:

Proposition 1.1.1 ([43])

X 0 equipped with the norm k:kX0 de�ned by

kfkX0 = sup fjf(u)j : kuk � 1g ; (1.1)

is also a Banach space.

We shall denote the value of f 2 X 0 at u 2 X by either f(u) or hf; uiX0; X :

Remark 1.1.1 ([43]) From X 0 we construct the bidual or second dual X 00 = (X 0)0: Furthermore,

with each u 2 X we can de�ne '(u) 2 X 00 by '(u)(f) = f(u); f 2 X 0, this satis�es clearly

k'(x)k � kuk : Moreover, for each u 2 X there is an f 2 X 0 with f(u) = kuk and kfk = 1, so itfollows that k'(x)k = kuk :

De�nition 1.1.2 Since ' is linear we see that

' : X ! X 00;

is a linear isometry of X onto a closed subspace of X 00, we denote this by

X ,! X 00:

De�nition 1.1.3 If ' ( in the above de�nition) is onto X 00 we say X is re�exive, X u X 00:

Theorem 1.1.1 ([4]; Theorem III.16)

Let X be Banach space. Then, X is re�exive, if and only if,

BX = fx 2 X : kxk � 1g ;

is compact with the weak topology � (X;X 0) : (See the next subsection for the de�nition of � (X;X 0))

De�nition 1.1.4 Let X be a Banach space, and let (un)n2N be a sequence in X. Then un converges

strongly to u in X if and only if

limn!1

kun � ukX = 0;

and this is denoted by un ! u; or limn!1

un = u:

Section 1.1. Banach Spaces - De�nition and Properties


1.1.1 The weak and weak star topologies

Let X be a Banach space and f 2 X 0: Denote by

'f : X ! Rx 7! 'f (x);

(1.2)

when f cover X 0, we obtain a family�'f�f2X0 of applications to X in R:

De�nition 1.1.5 The weak topology on X; denoted by � (X;X 0) ; is the weakest topology on X for

which every�'f�f2X0 is continuous.

We will de�ne the third topology on X 0, the weak star topology, denoted by � (X 0; X) : For all

x 2 X: Denote by'x : X 0 ! R

f 7! 'x(f) = hf; xiX0; X ;(1.3)

when x cover X, we obtain a family ('x)x2X0 of applications to X 0 in R:

De�nition 1.1.6 The weak star topology on X 0 is the weakest topology on X 0 for which every

('x)x2X0 is continuous.

Remark 1.1.2 ([4]) Since X � X 00; it is clear that, the weak star topology � (X 0; X) is weakest

then the topology � (X 0; X 00), and this later is weakest then the strong topology.

De�nition 1.1.7 A sequence (un) in X is weakly convergent to x if and only if

limn!1

f(un) = f(u);

for every f 2 X 0; and this is denoted by un * u.

Remark 1.1.3 ([42]; Remark 1.1.1)

1. If the weak limit exist, it is unique.

2. If un ! u 2 X (strongly), then un * u (weakly):

3. If dimX < +1; then the weak convergent implies the strong convergent.


On the compactness in the three topologies in the Banach space X :

1- First, the unit ball

B � fx 2 X : kxk � 1g ; (1.4)



in X is compact if and only if dim(X) <1:

2- Second, the unit ball B0 in X 0 (The closed subspace of a product of compact spaces) is weakly

compact in X 0 if and only if X is re�exive.

3- Third, B0 is always weakly star compact in the weak star topology of X 0.

Proposition 1.1.3 ([4]; proposition III.12)

Let (fn) be a sequence in X 0. We have:

1.hfn

�* f in � (X 0; X)

i, [fn(x)! f(x); 8x 2 X] :

2. If fn ! f (strongly), then fn * f , in � (X 0; X 00),

If fn * f in � (X 0; X 00), then fn�* f , in � (X 0; X) :

3. If fn�* f , in � (X 0; X), then kfnk is bounded and kfk � lim inf kfnk :

4. If fn�* f , in � (X 0; X) and xn ! x (strongly) in X, then fn(xn)! f(x):

1.1.2 Hilbert spaces

The proper setting for the rigorous theory of partial di¤erential equation turns out to be the most

important function space in modern physics and modern analysis, known as Hilbert spaces. Then,

we must give some important results on these spaces here.

De�nition 1.1.8 A Hilbert space H is a vectorial space supplied with inner product hu; vi such thatkuk =

phu; ui is the norm which let H complete.

Theorem 1.1.2 ([42], Theorem 1.1.1)

Let (un)n2N is a bounded sequence in the Hilbert space H, then it possess a subsequence which

converges in the weak topology of H:


In the Hilbert space, all sequence which converges in the weak topology is bounded.

Theorem 1.1.4 ([42], Corollary 1.1.1)

Let (un)n2N be a sequence which converges to u; in the weak topology and (vn)n2N is an other sequence

which converge weakly to v; then

limn!1

hvn; uni = hv; ui: (1.5)


Let X be a normed space, then the unit ball

B0 =nl 2 X 0

: klk � 1o; (1.6)

of X0is compact in � (X 0; X).



Proposition 1.1.4 ([42]; Proposition 1.1.1)

Let X and Y be two Hilbert spaces, let (un)n2N 2 X be a sequence which converges weakly to u 2 X;let A 2 L(X;Y ): Then, the sequence (A (un))n2N converges to A(u) in the weak topology of Y:

Proof. For all u 2 X, the functionu 7! hA(u); vi

is linear and continuous, because

jhA(u); vij � kAkL(X; Y ) kukX kvkY ; 8u 2 X; v 2 Y:

So, according to Riesz theorem, there exists w 2 X such that

hA(u); vi = hu;wi, 8u 2 X:

Then,

limn!1

hA (un) ; vi = limn!1

hun; wi

= hu;wi = hA(u); vi: (1.7)

This completes the proof.



1.2 Functional Spaces

1.2.1 The Lp () spaces

De�nition 1.2.1 Let 1 � p � 1; and let be an open domain in Rn; n 2 N: De�ne the standardLebesgue space Lp(); by

Lp() =

8<:f : ! R : f is measurable andZ

jf(x)jp dx <1

9=; : (1.8)

Notation 1.2.1 For p 2 R and 1 � p <1, denote by

kfkp =

0@Z

jf(x)jp dx

1A1p

: (1.9)

If p =1; we have

L1() = ff : ! R : f is measurable and there exists a constant C

such that, jf(x)j � C a.e in g:(1.10)

Also, we denote by

kfk1 = Inf fC; jf(x)j � C a.e in g : (1.11)

Notation 1.2.2 Let 1 � p � 1; we denote by q the conjugate of p i.e.1

p+1

q= 1:

Theorem 1.2.1 ([48])

It is well known that Lp() supplied with the norm k:kp is a Banach space, for all 1 � p � 1:

Remark 1.2.1 In particularly, when p = 2, L2 () equipped with the inner product

hf; giL2() =Z

f(x)g(x)dx; (1.12)

is a Hilbert space.

Theorem 1.2.2 ([43], Corollary 3.2)

For 1 < p <1; Lp() is re�exive space.

Section 1.2. Functional Spaces


Some integral inequalities

We will give here some important integral inequalities. These inequalities play an important role

in applied mathematics and also, it is very useful in our next chapters.

Theorem 1.2.3 ([48], Hölder�s inequality )

Let 1 � p � 1. Assume that f 2 Lp() and g 2 Lq(); then, fg 2 L1() andZ

jfgj dx � kfkp kgkq : (1.13)

Corollary 1.2.1 (Hölder�s inequality - general form)

Lemma 1.2.1 Let f1; f2; :::fk be k functions such that, fi 2 Lpi(); 1 � i � k, and

1

p=1

p1+1

p2+ :::+

1

pk� 1:

Then, the product f1f2:::fk 2 Lp() and kf1f2:::fkkp � kf1kp1 ::: kfkkpk :

Lemma 1.2.2 ( [48], Young�s inequality)

Let f 2 Lp(R) and g 2 Lq(R) with 1 < p < 1; 1 < q < 1 and1

r=1

p+1

q� 1 � 0. Then

f � g 2 Lr(R) andkf � gkLr(R) � kfkLp(R) kgkLq(R) : (1.14)

Lemma 1.2.3 ([43]; Minkowski inequality)

For 1 � p � 1; we have

ku+ vkLp � kukLp + kvkLp : (1.15)

Lemma 1.2.4 ([43])

Let 1 � p � r � q;1

r=�

p+1� �

q; and 1 � � � 1: Then

kukLr � kuk�Lp kuk

1��Lq : (1.16)

Lemma 1.2.5 ([43])

If � () <1; 1 � p � q � 1; then Lq ,! Lp; and

kukLp � �()1p�1q kukLq :



1.2.2 The Sobolev space Wm;p()


Let be an open domain in RN ; Then the distribution T 2 D0() is in Lp() if there exists a

function f 2 Lp() such that

hT; 'i =Z

f(x)'(x)dx; for all ' 2 D();

where 1 � p � 1; and it�s well-known that f is unique.

De�nition 1.2.2 Let m 2 N and p 2 [0;1] : The Wm;p() is the space of all f 2 Lp(), de�ned

asWm;p() = ff 2 Lp(); such that @�f 2 Lp() for all � 2 Nm such that

j�j =Pn

j=1 �j � m; where, @� = @�11 @�22 ::@

�nn g:

(1.17)

Theorem 1.2.4 ([9])

Wm;p() is a Banach space with their usual norm

kfkWm;p() =Xj�j�m

k@�fkLp ; 1 � p <1, for all f 2 Wm;p(): (1.18)

De�nition 1.2.3 Denote by Wm;p0 () the closure of D() in Wm;p():

De�nition 1.2.4 When p = 2, we prefer to denote by Wm;2() = Hm () and Wm;20 () = Hm

0 ()

supplied with the norm

kfkHm() =

0@Xj�j�m

(k@�fkL2)2

1A 12

; (1.19)

which do at Hm () a real Hilbert space with their usual scalar product

hu; viHm() =Xj�j�m

Z

@�u@�vdx (1.20)

Theorem 1.2.5 ([42]; Proposition 1.2.1)

1) Hm () supplied with inner product h:; :iHm() is a Hilbert space.

2) If m � m0, Hm () ,! Hm0(), with continuous imbedding :

Lemma 1.2.6 ([26])

Since D() is dense in Hm0 () ; we identify a dual H

�m () of Hm0 () in a weak subspace on ;

and we have

D() ,! Hm0 () ,! L2 () ,! H�m () ,! D0();



Lemma 1.2.7 (Sobolev-Poincaré�s inequality)

If

2 � q � 2n

n� 2 ; n � 3

q � 2; n = 1; 2;

then

kukq � C(q; ) kruk2 ; (1.21)

for all u 2 H10 () :

The next results are fundamental in the study of partial di¤erential equations

Theorem 1.2.6 ([9] Theorem 1.3.1)

Assume that is an open domain in RN (N � 1), with smooth boundary �. Then,(i) if 1 � p � n; we have W 1;p � Lq(); for every q 2 [p; p�] ; where p� = np

n� p:

(ii) if p = n we have W 1;p � Lq(); for every q 2 [p; 1) :(iii) if p > n we have W 1;p � L1() \ C0;�(); where � = p� n

p:

Theorem 1.2.7 ([9] Theorem 1.3.2)

If is a bounded, the embedding (ii) and (iii) of theorem 1.1.4 are compacts. The embedding (i) is

compact for all q 2 [p; p�) :

Remark 1.2.2 ([26])

For all ' 2 H2(); �' 2 L2() and for � su¢ ciently smooth, we have

k'(t)kH2() � C k�'(t)kL2() : (1.22)

Proposition 1.2.2 ([43], Green�s formula)

For all u 2 H2(); v 2 H1() we have

�Z

�uvdx =

Z

rurvdx�Z@

@u

@�vd�; (1.23)

where@u

@�is a normal derivation of u at �.



1.2.3 The Lp (0; T;X) spaces

De�nition 1.2.5 Let X be a Banach space, denote by Lp(0; T;X) the space of measurable functions

f : ]0; T [! X

t 7�! f(t)

such that 0@ TZ0

kf(t)kpX dt

1A1p

= kfkLp(0;T;X) <1; for 1 � p <1: (1.24)

If p =1,kfkL1(0;T;X) = sup

t2]0;T [ess kf(t)kX : (1.25)

Theorem 1.2.8 ([42])

The space Lp(0; T;X) is complete.

We denote by D0 (0; T;X) the space of distributions in ]0; T [ which take its values in X, and let usde�ne

D0 (0; T;X) = L (D ]0; T [ ; X) ;

where L (�; ') is the space of the linear continuous applications of � to ': Since u 2 D0 (0; T; X) ;we de�ne the distribution derivation as

@u

@t(') = �u

�d'

dt

�; 8' 2 D (]0; T [) ; (1.26)

and since u 2 Lp (0; T; X) ; we have

u(') =

TZ0

u(t)'(t)dt; 8' 2 D (]0; T [) : (1.27)

We will introduce some basic results on the Lp(0; T; X) space. These results, will be very useful in

the other chapters of this thesis.

Lemma 1.2.8 ([26] Lemma 1.2 )

Let f 2 Lp(0; T; X) and@f

@t2 Lp(0; T; X); (1 � p � 1) ; then, the function f is continuous from

[0; T ] to X:i.e: f 2 C1(0; T; X):



Lemma 1.2.9 ([26])

Let ' = ]0; T [� an open bounded domain in R�Rn; and let g�; g are two functions in Lq (]0; T [ ; Lq()) ;1 < q <1 such that

kg�kLq(0;T;Lq()) � C; 8� 2 N (1.28)

and

g� ! g in ';

then

g� * g in Lq (') :

Theorem 1.2.9 ([9]; Proposition 1.4.17)

Lp(0; T; X) equipped with the norm k:kLp(0;T;X), 1 � p � 1 is a Banach space.


Let X be a re�exive Banach space, X 0 it�s dual, and 1 � p <1; 1 � q <1;1

p+1

q= 1: Then the

dual of Lp(0; T; X) is identify algebraically and topologically with Lq(0; T;X 0):


Let X; Y be to Banach space, X � Y with continuous embedding, then we have Lp(0; T; X) � Lp(0;

T; Y ) with continuous embedding.

The following compactness criterion will be useful for nonlinear evolution problems, especially in

the limit of the non linear terms.

Proposition 1.2.5 ([26]).

Let B0; B; B1 be Banach spaces with B0 � B � B1; assume that the embedding B0 ,! B is compact

and B ,! B1 are continuous. Let 1 < p < 1; 1 < q < 1; assume further that B0 and B1 are

re�exive.

De�ne

W � fu 2 Lp (0; T; B0) : u0 2 Lq (0; T; B1)g : (1.29)

Then, the embedding W ,! Lp (0; T; B) is compact.



1.2.4 Some Algebraic inequalities

Since our study based on some known algebraic inequalities, we want to recall few of them here.

Lemma 1.2.10 ([48];The Cauchy-Schwarz inequality)

Every inner product satis�es the Cauchy-Schwarz inequality

hx1 ; x2i � kx1k kx2k : (1.30)

The equality sign holds if and only if x1 and x2 are dependent.

Young�s inequalities :

Lemma 1.2.11 For all a; b 2 R+, we have

ab � �a2 +b2

4�; (1.31)

where � is any positive constant.

Proof. Taking the well-known result

(2�a� b)2 � 0 8a; b 2 R

for all � > 0, we have

4�2a2 + b2 � 4�ab � 0:

This implies

4�ab � 4�2a2 + b2

consequently,

ab � �a2 +1

4�b2:


Lemma 1.2.12 ([43])

For all a; b � 0; the following inequality holds

ab � ap

p+bq

q;

where,1

p+1

q= 1:



Proof. Let G = (0; 1), and f : G! R is integrable, such that

f(x) =

8><>:p log a, 0 � x � 1

p

q log b,1

p� x � 1

;

for all a; b � 0 and 1p+1

q= 1:

Since '(t) = et is convex, and using Jensen�s inequality

'

0@ 1

� (G)

ZG

f(x)dx

1A � 1

� (G)

ZG

' (f(x)) dx: (1.32)

Consequently, we have

1

� (G)

ZG

' (f(x)) dx =

1Z0

ef(x)dx =

1=pZ0

ep log adx+

1Z1=p

eq log bdx

=

1=pZ0

apdx+

1Z1=p

bqdx

=1

p(ap) +

�1� 1

p

�bq:

(1.33)

=ap

p+bq

q;

where, � (G) = 1 and

'

0@ 1

� (G)

ZG

f(x)dx

1A = e(R 10 f(x)dx) = e(

R 1=p0 p log adx+

R 11=p q log bdx)

= e(log a+log b) = elog ab

= ab: (1.34)

Using (1:32), (1:33) and (1:34) to conclude the result.



1.3 Existence Methods

1.3.1 The Contraction Mapping Theorem

Here we prove a very useful �xed point theorem called the contraction mapping theorem. We will

apply this theorem to prove the existence and uniqueness of solutions of our nonlinear problem.

De�nition 1.3.1 Let f : X ! X be a map of a metric space to itself. A point x 2 X is called a

�xed point of f if f(x) = x.

De�nition 1.3.2 Let (X; dX) and (Y; dY ) be metric spaces. A map ' : X ! Y is called a contrac-

tion if there exists a positive number C < 1 such that

dY ('(x); '(y)) � CdX(x; y); (1.35)

for all x; y 2 X.

Theorem 1.3.1 (Contraction mapping theorem [45] )

Let (X; d) be a complete metric space. If ' : X ! X is a contraction, then ' has a unique �xed

point.

1.3.2 Gronwell�s lemma

Theorem 1.3.2 ( In integral form)

Let T > 0; and let ' be a function such that, ' 2 L1(0; T ); ' � 0; almost everywhere and � be

a function such that, � 2 L1(0; T ); � � 0; almost everywhere and �' 2 L1 [0; T ] ; C1; C2 � 0.

Suppose that

�(t) � C1 + C2

tZ0

'(s)�(s)ds; for a.e t 2 ]0; T [ ; (1.36)

then,

�(t) � C1 exp

0@C2 tZ0

'(s)ds

1A ; for a.e t 2 ]0; T [ : (1.37)

Proof. Let

F (t) = C1 + C2

tZ0

'(s)�(s)ds; for t 2 [0; T ] ; (1.38)

we have,

�(t) � F (t);

Section 1.3. Existence Methods


From (1:38) we have

F 0(t) = C2'(t)�(t)

� C2'(t)�(t); for a.e t 2 ]0; T [ : (1.39)

Consequently,

d

dt

8<:F (t) exp0@� tZ

0

C2'(s)ds

1A9=; � 0; (1.40)

then,

F (t) � C1 exp

0@C2 tZ0

'(s)ds

1A ; for a.e t 2 ]0; T [ : (1.41)

Since � � F; then our result holds.

In particle, if C1 = 0; we have � = 0 for almost everywhere t 2 ]0; T [ :

1.3.3 The mean value theorem

Theorem 1.3.3 Let G : [a; b]! R be a continues function and ' : [a; b]! R is an integral positivefunction, then there exists a number x in (a; b) such that

bZa

G(t)'(t)dt = G(x)

bZa

'(t)dt: (1.42)

In particular for '(t) = 1; there exists x 2 (a; b) such that

bZa

G(t)dt = G(x) (b� a) : (1.43)

Proof. Let

m = inf fG(x); x 2 [a; b]g (1.44)

and

M = sup fG(x); x 2 [a; b]g (1.45)

of course m and M exist since [a; b] is compact.

Then, it follows that

m

bZa

'(t)dt �bZa

G(t)'(t)dt �M

bZa

'(t)dt: (1.46)



By monotonicity of the integral. dividing through byR ba'(t)dt; we have that

m �R baG(t)'(t)dtR ba'(t)dt

�M: (1.47)

Since G(t) is continues, the intermediate value theorem implies that there exists x 2 [a; b] such that

G(x) =

R baG(t)'(t)dtR ba'(t)dt

: (1.48)

Which completes the proof.


Chapter 2

Local Existence

Abstract

Our goal in this chapter is to study the local existence ( local well-possedness) of the problem (P );

for u in C ([0; T ] ; H10 ()). For this purpose we consider, �rst the related problem for u �xed in

C ([0; T ] ; H10 ()) 8>>>>>>>>>>>><>>>>>>>>>>>>:

vtt ��vt ��v +R t0g(t� s)�v(s; x)ds

+ jvtjm�2 vt = jujp�2 u; x 2 ; t > 0

v (0; x) = u0 (x) ; vt (0; x) = u1 (x) ; x 2

v (t; x) = 0; x 2 �; t > 0

; (2.1)

and we will prove the local existence of this problem by using the Faedo-Galerkin method. Then,

by using the well-known contraction mapping theorem, we can show the local existence of (P ):

Our techniques of proof follows carefully the techniques due to Georgiev and Todorova [14]; with

necessary modi�cations imposed by the nature of our problem. The �rst step of our proof is the

choice of the space where the local solution exists. The minimal requirement for this space is that

u(t; x) be time continuous. The weak space satisfying the above requirement is C ([0; T ] ; H), where

H = H10 ()�H1

0 () is the natural energy space for (P ).

17

Chapter 2. Local Existence 18

2.1 Local Existence Result

In order to prove our local existence results, let us introduce the following space

YT =

8>><>>:u : u 2 C ([0; T ] ; H1

0 ()) ;

ut 2 C([0; T ] ; H10 ()) \ Lm([0; T ]� )

9>>=>>; : (2.2)

Our main result in this chapter reads as follows:

Theorem 2.1.1 Let (u0; u1) 2 (H10 ())

2 be given. Suppose that m � 2; p � 2 be such that

max fm; pg � 2 (n� 1)n� 2 ; n � 3: (2.3)

Then, under the conditions (G1) and (G2), the problem (P ) has a unique local solution u(t; x) 2 YT ,for T small enough.

The proof of theorem 2.1.1 will be established through several lemmas. The presence of the term

jujP�2 u in the right hand side of our problem (P ) ; gives us negative values of the energy. For this

purpose we �xed u 2 C ([0; T ] ; H10 ()) in the right hand side of (P ) and we will prove that our

problem (2:1) ; admits a solution.

Lemma 2.1.1 ([14]; Theorem 2.1) Let (u0; u1) 2 (H10 ())

2, assume that m � 2; p � 2 and (2:3)holds. Then, under the conditions (G1) and (G2), there exists a unique weak solution v 2 YT to theproblem (2:1); for any u 2 C ([0; T ] ; H1

0 ()) given.

The proof of the above Lemma follows the techniques due to Lions [26], in order to deal with the

convergence of the non linear terms in our problem, we must take �rst our initial data (u0; u1) in a

high regularity (that is, u0 2 H2() \H10 (); and u1 2 H1

0 () \ L2(m�1)()).

Lemma 2.1.2 ([26]; Theorem 3.1) Let u 2 C ([0; T ] ; H10 ()). Suppose that

u0 2 H2() \H10 (); (2.4)

u1 2 H10 () \ L2(m�1)(); (2.5)

assume further that m � 2; p � 2. Then, under the conditions (G1) and (G2) there exists a uniquesolution v of the problem (2:1) such that

v 2 L1�[0; T ] ; H2() \H1

0 ()�; (2.6)

Section 2.1. Local Existence Result


vt 2 L1�[0; T ] ; H1

0 ()�; (2.7)

vtt 2 L1�[0; T ] ; L2()

�; (2.8)

vt 2 Lm ([0; T ]� ()) : (2.9)

The following technical Lemma will play an important role in the sequel.

Lemma 2.1.3 For any v 2 C1 (0; T;H2()) we have

Z

tZ0

g(t� s)�v(s):v0(t)dsdx =1

2

d

dt(g � rv) (t)� 1

2

d

dt

24 tZ0

g(s)

Z

jrv(t)j2 dxds

35�12(g0 � rv) (t) + 1

2g(t)

Z

jrv(t)j2 dxds:

where (g � u)(t) =tZ0

g(t� s)

Z

ju(s)� u(t)j2 dxds:

The proof of this result is given in [31], for the reader�s convenience we repeat the steps here.

Proof. It�s not hard to see

Z

tZ0

g(t� s)�v(s):v0(t)dsdx = �tZ0

g(t� s)

Z

rv0(t):rv(s)dxds

= �tZ0

g(t� s)

Z

rv0(t): [rv(s)�rv(t)] dxds

�tZ0

g(t� s)

Z

rv0(t):rv(t)dxds:

Consequently,

Z

tZ0


2

tZ0

g(t� s)d

dt

Z

jrv(s)�rv(t)j2 dxds

�tZ0

g(s)

0@ d

dt

1

2

Z

jrv(t)j2 dx

1A ds



which implies,Z

tZ0


2

d

dt

24 tZ0

g(t� s)

Z


35�12

d

dt

24 tZ0

g(s)

Z

jrv(t)j2 dxds

35�12

tZ0

g0(t� s)

Z


+1

2g(t)

Z

jrv(t)j2 dxds:


Proof of Lemma 2.1.2.

Existence:

Our main tool is the Faedo-Galerkin�s method, which consist to construct approximations of the so-

lutions, then we obtain a prior estimates necessary to guarantee the convergence of approximations.

Our proof is organized as follows. In the �rst step, we de�ne an approach problem in bounded

dimension space Vn which having unique solution vn and in the second step we derive the various

a priori estimates. In the third step we will pass to the limit of the approximations by using the

compactness of some embedding in the Sobolev spaces.

1. Approach solution:

Let V = H10 ()\H2() the separable Hilbert space. Then there exists a family of subspaces fVng

such that

i) Vn � V (dimVn <1); 8n 2 N:ii) Vn ! V; such that, there exist a dense subspace # in V and for all v 2 #; we can get sequencefvngn2N 2 Vn; and vn ! v in V:

iii) Vn � Vn+1 and [n2N�Vn = H10 () \H2():

For every n � 1; let Vn = Span fw1; :::; wng ; where fwig, 1 � i � n; is the orthogonal complete

system of eigenfunctions of �� such that kwjk2 = 1, wj 2 H2() \ L2(m�1)() for all j = 1; :::; n.Denote by f�jg the related eigenvalues, where wj are solutions of the following initial boundaryvalue problem 8>><>>:

��wj = �jwj

wj = 0 on �

j = 1; :::; in : (2.10)



According to (iii) ; we can choose vn0; vn1 2 [w1; :::; wn] such that

vn0 �nXj=1

�jnwj �! u0 in H10 () \H2(), (2.11)

vn1 �nXj=1

�jnwj �! u1 in H10 () \ L2(m�1)(); (2.12)

where�jn =

Ru0wjdx

�jn =Ru1wjdx:

We seek n functions 'n1 ; :::; 'nn 2 C2 [0; T ] ; such that

vn(t) =nXj=1

'nj (t)wj(x); (2.13)

solves the problem 8>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>:

R

((v00n(t)��v0n(t)��vn(t)) �dx

+R

�tR0

g(t� s)�vn(s)ds+ jv0n(t)jm�2 v0n(t)

��dx

=R

ju(t)jp�2 u(t)�dx

vn (0) = vn0; v0n (0) = vn1

; (2.14)

where the prime "0" denotes the derivative with respect to t. For every � 2 Vn and t � 0: Taking� = wj; in (2:14) yields the following Cauchy problem for a ordinary di¤erential equation with

unknown 'nj : 8>>>>>>>>><>>>>>>>>>:

'00nj (t) + �j'0nj (t) + �j'

nj (t) + �j

tR0

g(t� s)'nj (s)ds

+��'0hj (t)��m�2 '0nj (t) = j(t)

'nj (0) =R

u0wj; '0nj (0) =

R

u1wj; j = 1; :::; n

; (2.15)

for all j; where

j(t) =

Z

ju(t)jp�2 u(t)wj 2 C [0; T ] :



By using the Caratheodory theorem for an ordinary di¤erential equation, we deduce that, the

above Cauchy problem yields a unique global solution 'nj 2 H3 [0; T ] ; and by using the embedding

Hm [0; T ] ,! Cm�1 [0; T ], we deduce that the solution 'nj 2 C2 [0; T ]. In turn, this gives a unique

vn de�ned by (2:13) and satisfying (2:14):

2. The a priori estimates:

The next estimate prove that the energy of the problem (2:1) is bounded and by using a result in

[9]; we conclude that; the maximal time tn of existence of (2:15) can be extended to T .

The �rst a priori estimate:

Substituting � = v0n(t) into (2:14), we obtainZ

v00n(t)v0n(t)dx�

Z

�v0n(t)v0n(t)dx�

Z

�vn(t)v0n(t)dx

+

Z

tZ0

g(t� s)�vn(s)dsv0n(t)dx+

Z

jv0n(t)jm�2

v0n(t)v0n(t)dx (2.16)

=

Z

f(u)v0n(t)dx:

for every n � 1; where f(u) = ju(t)jp�2 u(t); Since the following mapping

L2() �! L2()

u 7! jujp�2 u

is continues, we deduce that,

jujp�2 u 2 L1�[0; T ] ; L2()

�:

So, f 2 H1 ([0; T ] ; H1()). Consequently, by using the Lemma 2.1.3 and (G1) we get easily

1

2

d

dtkv0n(t)k

22 + krv0n(t)k

22 +

1

2

d

dtkrvn(t)k22

+1

2

d

dt

24 tZ0

g(t� s)

Z

jrvn(s)�rvn(t)j2 dxds

35� 12

d

dt

0@krvn(t)k22 tZ0

g(s):ds

1A�12

tZ0

g0(t� s)

Z

jrvn(s)�rvn(t)j2 dxds+1

2g(t) krvn(t)k22 + kv0n(t)k

mm

=

Z

f(u):v0n(t)dx:



Therefore, we obtain

d

dtEn(t)�

Z

f(u):v0n(t)dx+ krv0n(t)k22 + kv0n(t)k

mm

=1

2(g0 � rvn)(t)�

1

2g(t) krvn(t)k22 ;

where

En(t) =1

2

8<:kv0n(t)k22 +0@1� tZ

0

g(s)ds

1A krvn(t)k22 + (g � rvn)(t)9=; ;

is the functional energy associated to the problem (2:1):

It�s clear by using (G2), that

d

dtEn(t)�

Z

f(u):v0n(t)dx+ krv0n(t)k22 + kv0n(t)k

mm � 0; 8t � 0:

Which implies, by using Young�s inequality, for all � > 0

d

dtEn(t) + krv0n(t)k

22 + kv0n(t)k

mm �

1

4�kf(u)k22 + � kv0n(t)k

22 : (2.17)

Integrating (2:17) over [0; t] ; (t < T ), we obtain

En(t) +

tZ0

krv0n(s)k22 ds+

tZ0

kv0n(s)kmm ds

� 1

4�

TZ0

kf(u)k22 ds+ �

TZ0

kv0n(s)k22 ds+

1

2

�ku1nk22 + kru0nk

22

�:

Since f 2 H1(0; T;H10 ()); we deduce

En(t) +

tZ0

krv0n(s)k22 ds+

tZ0

kv0n(s)kmm ds � CT

�ku1nk22 + kru0nk

22

�; (2.18)

for, � small enough and every n � 1, where CT > 0 is positive constant independent of n:Then, by the de�nition of En(t) and by using (2:11) ; (2:12), we get

kv0n(t)k22 +

0@1� tZ0

g(s)ds

1A krvn(t)k22 + (g � rvn)(t) � KT ; (2.19)

andtZ0

kv0n(s)kmm ds � KT ; (2.20)



and

tZ0

krv0n(s)k22 ds � KT ; (2.21)

where KT = CT�ku1nk22 + kru0nk

22

�; by (2:19), we get tn = T; 8n:

However, the insu¢ cient regularity of the nonlinear operator, jvtjm�2 vt; with the presence of theviscoelastic term and strong damping, we must prove in the next a prior estimates that, the family of

approximations vn de�ned in (2:13) is compact in the strong topology and by using compactness of

the embedding H1([0; T ] ; H1()) ,! L2([0; T ] ; L2()); we can extract a subsequence of vn denoted

also by v� such that v0� converges strongly in L

2([0; T ] ; L2()): To do this, it�s su¢ ces to prove

that v0n is bounded in L

1([0; T ] ; H10 ()) and v

00

n is bounded in L1([0; T ] ; L2()), then by using

Aubin-Lions Lemma, our conclusion holds.

The second a priori estimate:

Substituting � = wj in (2:14) and taking ��wj = �jwj, multiplying by '0nj (t) and summing up the

product result with respect to j; we get by Green�s formula.

Z

rv00n:rv0ndx+Z

�v0n:�v0ndx+

Z

�vn:�v0ndx�

Z

tZ0

g(t� s)�vn:�v0ndsdx

+nXi=1

Z

@

@xi

�jv0nj

m�2v0n

� @v0n@xi

dx (2.22)

=

Z

rf(u):rv0ndx:

As in [26], we have

nXi=1

Z

@

@xi

�jv0nj

m�2v0n

� @v0n@xi

dx

= (m� 1)nXi=1

Z

@

@xi

�jv0nj

m�22

@v0n@xi

�2dx (2.23)

=4(m� 1)

m2

nXi=1

Z

�@

@xi

�jv0nj

m�22 v0n

��2dx:



Also, the fourth term in the left hand side of (2:22) can be written as followsZ

tZ0

g(t� s)�vn:�v0ndsdx = �1

2g(t) k�vnk22 +

1

2(g0 ��vn)

�12

d

dt

8<:(g ��vn)� k�vnk22tZ0

g(s)ds

9=; : (2.24)

Therefore, (2:22) becomes

1

2

d

dt

24krv0nk22 + (g ��vn) + k�vnk220@1� tZ

0

g(s)ds

1A35+4(m� 1)

m2

nXi=1

Z

�@

@xi

�jv0nj

m�22 v0n

��2dx (2.25)

+ k�v0nk22 +

1

2g(t) k�vnk22 �

1

2(g0 ��vn)

=

Z

rf(u):rv0ndx:

Let us de�ne the energy term

Kn(t) =1

2

24krv0nk22 + (g ��vn) + k�vnk220@1� tZ

0

g(s)ds

1A35 (2.26)

Then, it�s clear that (2:25) takes the form

d

dtKn(t)�

Z

rf(u):rv0ndx+ k�v0nk22

+nXi=1

Z

�@

@xi

�jv0nj

m�22 v0n

��2dx

(2.27)

= �12(g(t) k�vnk22) +

1

2(g0 ��vn):

Using (G1) ; (G2) and integrating (2:27) over [0; t] ; we obtain

Kn(t) +nXi=1

tZ0

Z

�@

@xi

�jv0nj

m�22 v0n

��2dxds+

tZ0

k�v0nk22 ds

�TZ0

Z

rf(u):rv0ndxds+1

2(krv1nk22 + k�v0nk

22): (2.28)



Obviously, by using Young�s inequality, we get

TZ0

Z

rf(u):rv0ndxds �1

4�

TZ0

krf(u)k22 ds+ �

TZ0

krv0nk22 ds:

Inserting the above estimate into (2:28); to get

Kn(t) +

nXi=1

tZ0

Z

�@

@xi

�jv0nj

m�22 v0n

��2dxds+

tZ0

k�v0nk22 ds

� CT (krv1nk22 + k�v0nk22):

Thus,

Kn(t) +nXi=1

tZ0

Z

�@

@xi

�jv0nj

m�22 v0n

��2dxds+

tZ0

k�v0nk22 ds � CT ;

for, � small enough and every n � 1, where CT > 0 is positive constant independent of n: Therefore,this equivalent by the de�nition of Kn(t)

krv0nk22 + (g ��vn) + k�vnk

22

0@1� tZ0

g(s)ds

1A � CT ; (2.29)

andnXi=1

tZ0

Z

�@

@xi

�jv0nj

m�22 v0n

��2dxds � CT ; (2.30)

andtZ0

k�v0nk22 ds � CT : (2.31)

Then, from (2:29), (2:30) and (2:31) ; we conclude

v0n is bounded in L1([0; T ] ; H1

0 ()); (2.32)

vn is bounded in L1([0; T ] ; H2()); (2.33)

@

@xi

�jv0nj

m�22 v0n

�is bounded in L2([0; T ] ; L2()); i = 1; :::; n: (2.34)

The third a priori estimate:

It�s clear that

d

dt

24 tZ0

g(t� s)�vn(s)ds

35 = g(0)�vn +

tZ0

g0(t� s)�vn(s)ds: (2.35)



Performing an integration by parts in (2:35) we �nd that

d

dt

24 tZ0

g(t� s)�vn(s)ds

35 = g(t)�un0 +

tZ0

g(t� s)�v0n(s)ds: (2.36)

Now, returning to (2:14), di¤erentiating throughout with respect to t, and using (2:36), we obtainZ

(v000n (t)��v00n(t)��v0n(t)) �dx

+

Z

0@ tZ0

g(t� s)�v0n(s)ds+ g(t)�un0 + (m� 1)(jv0n(t)jm�2

v00n(t))

1A �dx

=

Z

(f(u))0)�dx; (2.37)

where, (f(u))0 =@f

@t. By substitution of � = v00h(t) in (2:37); yieldsZ

v000n (t)v00n(t)dx�

Z

�v00n(t)v00n(t)dx�

Z

�v0n(t)v00n(t)dx

+

Z

0@ tZ0

g(t� s)�v0n(s)ds+ g(t)�un0

1A v00n(t)dx

+(m� 1)Z

jv0n(t)jm�2

v00n(t)v00n(t)dx (2.38)

=

0Z

(f(u))0 v00n(t)dx:

The fourth term in the left hand side of (2:38) can be analyzed as follows. It�s clear that Lemma

2.1.3 implies

Z

tZ0

g(t� s)�v0n(s):v00n(t)dsdx = �

Z

tZ0

g(t� s)rv0n(s):rv00n(t)dsdx

=1

2

d

dt(g � rv0n) (t)�

1

2

d

dt

tZ0

g(s) krv0n(t)k22 ds

(2.39)

�12(g0 � rv0n) (t) +

1

2g(t) krv0n(t)k

22 :



Also,

(m� 1)hjv0n(t)jm�2

v00n(t); v00n(t)i =

4(m� 1)m2

Z

�@

@t

�jv0nj

m�22 v0n(t)

��2dx: (2.40)

Inserting (2:39) and (2:40) into (2:38), we obtain

1

2

d

dtkv00nk

22 + krv00nk

22 +

1

2

d

dtkrv0nk

22 +

1

2

d

dt

8<:(g � rv0n) (t)� krv0nk22tZ0

g(s)ds

9=;+4(m� 1)

m2

Z

�@

@t

�jv0nj

m�22 v0n(t)

��2dx+ (g(t)�un0)

Z

v00n(t)dx (2.41)

=

Z

(f(u))0 v00n(t)dx+1

2(g0 � rv0n) (t)�

1

2g(t) krv0nk

22 :

Let us denote by

n(t) =1

2

8<:kv00nk22 ++(g � rv0n) (t) + (1�tZ0

g(s)ds) krv0nk22

9=; : (2.42)

We obtain, from (2:41) ; (G2) that

d

dtn(t)�

Z

(f(u))0 :v00n(t)dx+4(m� 1)

m2

Z

�@

@t

�jv0nj

m�22 v0n(t)

��2dx

+ krv00nk22 + (g(0)�un0)

Z

v00n(t)dx

=1

2(g0 � rv0n) (t)�

1

2g(t) krv0n(t)k

22

� 0:

Integration the above estimate over [0; t], we conclude

n(t) +4(m� 1)

m2

tZ0

Z

�@

@t

�jv0nj

m�22 v0n(t)

��2dxds

+

tZ0

krv00nk22 ds+ (g(0)�un0)

tZ0

Z

v00n(s)dxds

� 1

2kv00n0k

22 +

1

2krvn1k22 +

1

4�

TZ0

(f(u))0 22ds+ �

TZ0

kv00nk22 ds:



Then, for � small enough, we deduce

n(t) +4(m� 1)

m2

tZ0

Z

�@

@t

�jv0nj

m�22 v0n

��2dxds

+

tZ0

krv00nk22 ds+ (g(0)�un0)

tZ0

Z

v00ndxds (2.43)

� CT

�kv00n0k

22 + krvn1k

22

�:

In order to estimate the term kv00n0k22, taking t = 0 in (2:14), we �nd

kv00n0k22 =

Z

�vn1:v00n0dx+

Z

�vn0:v00n0dx

�Z

jvn1jm�2 vn1:v00n0dx+Z

f(u(0)):v00n0dx:

Thanks to Cauchy-Schwartz inequality (Lemma 1.2.10), we write

kv00n0k22 � kv00n0k2

0BB@k�vn1k2 + k�vn0k2 + kf(u(0))k2 +0@Z

jv1nj2(m�1) dx

1A12

1CCA ;

which implies, by using (2:11) and (2:12) ; that

kv00n0k2 � k�vn1k2 + k�vn0k2 + kf(u(0))k2 +

0@Z

jv1nj2(m�1) dx

1A12

� C:

(2.44)

Then,

n(t) � LT ; (2.45)

4(m� 1)m2

tZ0

Z

�@

@t

�jv0nj

m�22 v0n

��2dxds � LT ; (2.46)

andtZ0

krv00nk22 ds � LT ; (2.47)

andtZ0

Z

v00ndxds � LT : (2.48)



where LT > 0:

From (2:11); (2:12) and (2:44)� (2:48) ; we deduce

v00n is bounded in L1([0; T ] ; L2()); (2.49)

vn is bounded in L1([0; T ] ; H10 ()); (2.50)

@

@t

�jv0nj

m�22 v0n

�is bounded in L2([0; T ] ; L2()); i = 1; :::; n: (2.51)

3.Pass to the limit:

By the �rst, the second and the third estimates, we obtain

vn is bounded in L1([0; T ] ; H2() \H10 ()); (2.52)


0 ()); (2.53)

v00n is bounded in L1([0; T ] ; L2()); (2.54)

v0n is bounded in Lm((0; T )� ): (2.55)

Therefore, up to a subsequence, and by using the (Theorem 1.1.5), we observe that there exists a

subsequence v� of vn and a function v that we my pass to the limit in (2:14); we obtain a weak

solution v of (2:1) with the above regularity

v� *� v in L1([0; T ] ; H1

0 () \H2 ());

v0� *� v0in L1([0; T ] ; H1

0 () \ Lm());

v00� *� v

00in L1([0; T ] ; L2()):

By using the fact that

L1([0; T ] ; L2()) ,! L2([0; T ] ; L2());

L1([0; T ] ; H10 ()) ,! L2([0; T ] ; H1

0 ()):

We get


0 ());

v00n is bounded in L2([0; T ] ; L2());

therefore,

v0n is bounded in H1([0; T ] ; H1()): (2.56)

Consequently, since the embedding

H1([0; T ] ; H1()) ,! L2([0; T ] ; L2())



is compact, then we can extract a subsequence v0� such that

v0� ! v0 in L2([0; T ] ; L2()). (2.57)

which implies

v0� ! v0 a.e on (0; T )� .

By (2:20), we have

v0n is bounded in Lm([0; T ]� ):

and by using Theorem 1.2.2

jv0� jm�2

v0� * # in Lm0([0; T ] ; Lm

0())

wherem

m� 1 = m0:

The estimates (2:34) and (2:46) imply that

jv0� jm�22 v0� * � in H1([0; T ] ; H1()

by using the fact that, the mapping u 7�! jujm�2 u is continuous, (Lemma 1.2.9) and since the weaktopology is separate, we deduce

# = jv0jm�2 v0 (2.58)

� = jv0jm�22 v0 (2.59)

Then, by using the uniqueness of limit, we deduce

jv0� jm�2

v0� * jv0jm�2 v0 in Lm0([0; T ] ; Lm

0()) (2.60)

jv0� jm�22 v0� * jv0j

m�22 v0 in H1([0; T ] ; H1() (2.61)

Now, we will pass to the limit in (2:14), by the same techniques as in [26]:

Taking � = wj, n = � and �xed j < �;Z

v00� (t):wjdx+

Z

rv� (t):rwjdx+Z

rv0� (t):rwjdx

�Z

tZ0

g(t� s)rv� (s):rwjdsdx+Z

jv0� (t)jm�2

v0� (t):wjdx (2.62)

=

Z

f(u):wjdx:



We obtain, by using the property of continuous of the operator in the distributions spaceZ

v00� (t):wjdx *�Z

v00(t):wjdx; in D0 (0; T )

Z

rv� (t):rwjdx *�Z

rv(t):rwjdx; in L1 (0; T )Z

rv0� (t):rwjdx *�Z

rv0(t):rwjdx; in L1 (0; T )

Z

tZ0

g(t� s)rv� (s):rwjdsdx *�Z

tZ0

g(t� s)rv(s):rwjdsdx; in L1 (0; T )Z

jv0� (t)jm�2

v0� (t):wjdx *�Z

jv0(t)jm�2 v0(t):wjdx; in L1 (0; T )

We deduce from (2:62), thatZ

v00(t):wjdx+

Z

rv(t):rwjdx+Z

rv0(t):rwjdx

�Z

tZ0

g(t� s)rv(s):rwjdsdx+Z

jv0(t)jm�2 v0(t):wjdx (2.63)

=

Z

f(u):wjdx:

Since, the basis wj (j = 1; :::) is dense in H10 () \H2 () ; we can generalize (2:63) ; as followsZ

v00(t):'dx+

Z

rv(t):r'dx+Z

rv0(t):r'dx

�Z

tZ0

g(t� s)rv(s):r'dsdx+Z

jv0(t)jm�2 v0(t):'dx

=

Z

f(u):'dx; 8' 2 H10 () \H2 () :

Then,

v 2 L1�[0; T ] ; H2() \H1

0 ()�;

vt 2 L1�[0; T ] ; H1

0 ()�;

vtt 2 L1�[0; T ] ; L2()

�;

vt 2 Lm ([0; T ]� ()) :

This complete the our proof of existence.



Uniqueness:

Let v1; v2 two solutions of (2:1), and let w = v1 � v2 satisfying :

w00 ��w ��w0 +tZ0

g(t� s)�wds+�jv01j

m�2v01 � jv02j

m�2v02

�= 0: (2.64)

Multiplying (2:64), by w0and integrating over , we get

1

2

d

dt

0@kw0n(t)k22 +0@1� tZ

0

g(s)ds

1A krwn(t)k22 + (g � rwn)(t)1A

+

Z

�jv01j

m�2v01 � jv02j

m�2v02

�w0dx

= �krw0n(t)k22 +

1

2(g0 � rwn)(t)�

1

2g(t) krwn(t)k22 :

Denote by

J(t) = kw0n(t)k22 +

0@1� tZ0

g(s)ds

1A krwn(t)k22 + (g � rwn)(t): (2.65)

Since the function y 7! jyjm�2 y is increasing, we haveZ

�jv01j

m�2v01 � jv02j

m�2v02

�w0dx � 0

and since1

2(g0 � rwn)(t) � 0;

we deduce �d

dtJ(t) � 0

�: (2.66)

This implies that J(t) is uniformly bounded by J(0) and is decreasing in t; since w(0) = 0; we

obtain w = 0 and v1 = v2:

Proof of Lemma 2.1.1.

As in [14]; since D() = H2(), we approximate, u0; u1 by sequences (u�0) ; (u�1) in D(), and uby a sequence (u�) in C ([0; T ] ;D()), for the problem (2:1). Lemma 2.1.2 guarantees the existenceof a sequence of unique solutions (v�) satisfying (2:6)� (2:9) : Now, to complete the proof of Lemma2.1.1, we proceed to show that the sequence (v�) is Cauchy in YT equipped with the norm

kuk2YT = kuk2H + kutk

2Lm([0;T ]�) :



where

kuk2H = max0�t�T

8<:Z

�u2t + l jruj2

�(x; t)dx

9=;Denote w = v� � v� for �; � given. Then w is a solution of the Cauchy problem:8>>>>>>>>>>>><>>>>>>>>>>>>:

wtt ��w ��wt + L(w) + k(v�t )� k(v�t )

= f(u�)� f(u�); x 2 ; t > 0

w(0; x) = u�0 � u�0; wt(0; x) = u�1 � u�1; x 2

w(t; x) = 0; x 2 �; t > 0

; (2.67)

where,

k(v�t ) = jv�t jm�2 v�t

f(u�) = ju�jp�2 u�

L(w) =

tZ0

g(t� s)�w(s; x)ds:

The energy equality reads as

1

2

d

dt

8<:kwtk22 +0@1� tZ

0

g(s)ds

1A krwk22 + (g � rw)(t)9=;

+

Z

�k(v�t )� k(v�t )

�wtdx+ krwtk22

(2.68)

=

Z

�f(u�)� f(u�)

�wtdx+

1

2(g0 � rw) (s)ds� 1

2krw(s)k22

tZ0

g(s)ds:

The term, Z

�k(v�t )� k(v�t )

�wtdx =

Z

�jv�t j

m�2 v�t ��v�t ��m�2 v�t��v�t � v�t

�dx

is nonnegative.

We need to estimate��Z

(f(u)� f(u)) (v � v) dx

�� C(kukH + kukH)p�2 ku� ukH kv � vkH ;



ful�lled for u; u; v; v 2 H10 (); where C is a constant depending on ; l; p only. Then, Hölder�s

inequality yields, for1

q+1

n+1

2= 1 (q =

2n

n� 2);��Z

�f(u�)� f(u�)

�wtdx

�� =

��Z

�ju�jp�2 u� �

��u��p�2 u��v�t � v�t

�dx

�� C

u� � u� Lq

v�t � v�t

L2

�ku�kp�2n(p�2) +

u� p�2n(p�2)

�:

(2.69)

The Sobolev embedding Lq ,! H10 () gives u� � u�

Lq()

� C ru� �ru�

L2():

Then,

ku�kp�2n(p�2) + u� p�2

n(p�2) � C(ku�kp�2L2() + u� p�2

L2()):

The necessity to estimate ku�kn(p�2) by the energy norm kukH requires a restriction on p. Namely,

we need n(p� 2) � 2n

n� 2 ; then the Sobolev embedding Lq ,! H1

0 () gives

ku�kp�2n(p�2) � kukp�2H :

Therefore, (2:69) takes the form��Z

�f(u�)� f(u�)

�wtdx

�� C

v�t � v�t

L2()

ru� �ru� L2()

�kru�kp�2L2() +

ru� p�2L2()

�(2.70)

under the fact thattZ0

(g0 � rw) (s)ds � 0;

we conclude

�tZ0

(g0 � rw) (s)ds+ (g � rw) (t) + krw(t)k22

tZ0

g(s)ds � 0:

Thus,

1

2kw(t; :)k2H �

1

2kw(0; :)k2H + C

tZ0

ru� �ru� L2()

kwt(s; :)kH ds:

The Gronwell Lemma and Young�s inequality guarantee that

kw(t; :)kH � kw(0; :)kH + CT u� � u�

C([0;T ];H)

:



Since v�(t; :)� v�(t; :) H� C

v�(0; :)� v�(0; :) H+ CT

u� � u� C([0;T ];H)

; (2.71)

then fv�g is a Cauchy sequence in C([0; t] ; H); since fu�g and fv�(0; :)g are Cauchy sequences inC([0; T ] ; H) and H, respectively.

Now, we shall prove that fv�t g is a Cauchy sequence, in Lm ([0; T ]� ), to control the normkv�t k

2Lm([0;T ]�) : By the following algebraic inequality�

� j�jm�2 � � j�jm�2�(�� ) � C j�� jm ; (2.72)

which holds for any real �; � and c, we getZ

�k(v�t )� k(v�t )

�wtdx =

Z

�v�t jv�t j

m�2 � v�t

��v�t ��m�2��v�t � v�t

�dx

� C��v�t � v�t

��2Lm([0;T ]�)

:

This estimate combined with (2:68) gives v�t � v�t

2Lm([0;t]�)

� C v�(0; :)� v�(0; :)

Lm([0;t]�)

+CR

tZ0

u� � u� Lm([0;t]�)

v�(t; :)� v�(t; :) Lm([0;t]�) ds:

So by using Gromwell Lemma, we obtain fv�t g is a Cauchy sequence, in Lm ([0; T ]� ) and hencefv�g is a Cauchy sequence in YT : Let v its limit in YT and by Lemma 2.1.2, v is a weak solution of(2:1).

Now, we are ready to show the local existence of the problem (P )

Proof of Theorem 2.1.1.

Let (u0; u1) 2 (H10 ())

2; and

R2 =�kru0k22 + ku1k

22

�:

For any T > 0; consider

MT =�u 2 YT : u(0) = u0; ut(0) = u1 and kukYT � R

:

Let� : MT !MT

u 7! v = �(u):



We will prove as in [13] that,

(i) �(MT ) �MT :

(ii) � is contraction in MT :

Beginning by the �rst assertion. By Lemma 2.1.1, for any u 2 MT we may de�ne v = �(u);

the unique solution of problem (2:1). We claim that, for a suitable T > 0, � is contractive map

satisfying

�(MT ) �MT :

Let u 2MT ; the corresponding solution v = �(u) satis�es for all t 2 [0; T ] the energy identity :

1

2

8<:kv0(t)k22 +0@1� tZ

0

g(s)ds

1A krv(t)k22 + (g � rv)(t)9=;

+

tZ0

krv0(s)k22 ds+tZ0

kv0(s)kmm)ds

(2.73)

=1

2

�kv1k22 + krv0k

22

�+

tZ0

Z

ju(s)jp�2 u(s)v0(t)dxds:

We get

1

2kv(t)k2YT �

1

2kv(0)k2YT +

tZ0

Z

ju(s)jp�2 u(s)v0(t)dxds: (2.74)

We estimate the last term in the right-hand side in (2:74) as follows: thanks to Hölder�s, Young�s

inequalities, we have Z

ju(s)jp�2 u(s)v0(t)dx � C kukpYT kvkYT ;

then,

kv(t)k2YT � kv(0)k2YT+ CRp

tZ0

kvkYT ds;

where C depending only on T; R: Recalling that u0; u1 converge, then

kv(t)kYT � kv(0)kYT + CRpT:

Choosing T su¢ ciently small, we getkvkYT � R; which shows that

�(MT ) �MT :



Now, we prove that � is contraction inMT : Taking w1 and w2 inMT , subtracting the two equations

in (2:1); for v1 = �(w1) and v2 = �(w2); and setting v = v1 � v2; we obtain for all � 2 H10 () and

a.e. t 2 [0; T ]

Z

vtt:�dx+

Z

rvr�dx+Z

rvtr�dx+Z

tZ0

g(t� s)rvr�dsdx

+

Z

�jvtjm�2 vt

��dx

=

Z

�jw1jp�2w1 � jw2jp�2w2

��dx: (2.75)

Therefore, by taking � = vt in (2:75) and using the same techniques as above, we obtain

kv(t; :)k2YT � C

tZ0

�kw1kp�2YT

+ kw2kp�2YT

�kw1 � w2kYT kv(s; :)k

2YTds: (2.76)

It�s easy to see that

kv(t; :)k2Yt = k�(w1)� �(w2)k2Yt� � kw1 � w2k2Yt ; (2.77)

for some 0 < � < 1 where � = 2CTRP�2:

Finally by the contraction mapping theorem together with (2:77); we obtain that there exists a

unique weak solution u of u = �(u) and as �(u) 2 YT we have u 2 YT : So there exists a unique

weak solution u to our problem (P ) de�ned on [0; T ] ; The main statement of Theorem 2.1.1 is

proved.

Remark 2.1.1 Let us mention that in our problem (P ) the existence of the term source ( f(u) =

jujp�2 u ) in the right hand forces us to use the contraction mapping theorem. Since we assume alittle restriction on the initial data. To this end, let us mention again that our result holds by the

well depth method, by choosing the initial data satisfying a more restrictions.


Chapter 3

Global Existence and Energy Decay

Abstract

In this chapter, we prove that the solution obtained in the second chapter (Local solution ) is

global in time: In addition, we show that the energy of solutions decays exponentially if m = 2 and

polynomial if m > 2, provided that the initial data are small enough. The existence of the source

term�jujp�2 u

�forces us to use the potential well depth method in which the concept of so-called

stable set appears. We will make use of arguments in [44] with the necessary modi�cations imposed

by the nature of our problem.

39

Chapter 3. Global Existence and Energy Decay 40

3.1 Global Existence Result

In order to state and prove our results, we introduce the functional

I(t) = I(u(t)) =

0@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)� b ku(t)kpp ; (3.1)

and

J(t) = J(u(t)) =1

2

0@1� tZ0

g(s)ds

1A kru(t)k22 + 12(g � ru)(t)� b

pku(t)kpp ; (3.2)

for u(t; x) 2 H10 () ; t � 0:

As in [19], the potential well depth, is de�ned as

d = infu2H1

0 ()�f0gsup��0

J (�u) : (3.3)

The functional energy associated to (P ) is de�ned as fallows

E(u(t); ut(t)) = E(t) =1

2kut(t)k22 + J(t): (3.4)

Now, we introduce the stable set as follows:

W =�u 2 H1

0 () : J(u) < d; I(u) > 0[ f0g : (3.5)

We will prove the invariance of the set W: That is if for some t0 > 0 if u(t0) 2 W; then u(t) 2 W;

8t � t0:

Lemma 3.1.1 d is positive constant.

Proof. We have

J(�u) =�2

2

0@0@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)1A� b

p�p ku(t)kpp : (3.6)

Using (G1); (G2) to get

J(�u) � K(�);

where K(�) =�2

2l kruk22 �

b

p�p kukpp :

Section 3.1. Global Existence Result


By di¤erentiating the second term in the last equality with respect to �; to get

d

d�K(�) = �l kruk22 � b�p�1 kukpp : (3.7)

For, �1 = 0 and �2 =

l kruk22b kukpp

! 1p�2

; then we have

d

d�K(�) = 0:

As K(�1) = 0, we have

K(�2) =1

2

l kruk22b kukpp

! 2p�2

l kruk22 �b

p

l kruk22b kukpp

! pp�2

kukpp

=1

2b�2p�2 (l)

pp�2�kukpp

� �2p�2 �kruk22� p

p�2

�1pb�2p�2 (l)

pp�2�kukpp

�� 2p�2 �kruk22� p

p�2

= (l)p

P�2 b�2p�2�1

2� 1p

�kruk

2pp�22 kuk

�2pp�2p : (3.8)

By Sobolev-Poincaré�s inequality, we deduce that K(�2) > 0: Then, we obtain

sup fJ(�u); � � 0g � sup fK(�); � � 0g

> 0: (3.9)

Then, by the de�nition of d; we conclude that d > 0:

Lemma 3.1.2 ([19]) W is a bounded neighbourhood of 0 in H10 ().

Proof. For u 2 W; and u 6= 0; we have

J(t) =1

2

0@1� tZ0

g(s)ds

1A kru(t)k22 + 12(g � ru)(t)� b

pku(t)kpp

=

�p� 22p

�240@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)35+ 1

pI(u(t))

��p� 22p

�240@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)35 : (3.10)



By using (G1) and (G2) then (3:10) becomes

J(t) ��p� 22p

�0@1� tZ0

g(s)ds

1A kru(t)k22� l

�p� 22p

�kru(t)k22

then,

kru(t)k22 � 1

l

�2p

p� 2

�J(t)

<1

l

�2p

p� 2

�d = R:

Consequently, 8u 2 W we have u 2 B where

B =�u 2 H1

0 () : kru(t)k22 < R

: (3.11)


Now, we will show that our local solution u(t; x) is global in time, for this purpose it su¢ ces to

prove that the norm of the solution is bounded, independently of t; this is equivalent to prove the

following theorem.

Theorem 3.1.1 Suppose that (G1) ; (G2) and (2:3) hold. if u0 2 W; u1 2 H10 () and

bCp�l

�2p

(p� 2) lE(0)�p�2

2

< 1; (3.12)

where C� is the best Poincaré�s constant. Then the local solution u(t; x) is global in time.

Remark 3.1.1 Let us remark, that if there exists t0 2 [0; T ) such that u(t0) 2 W and ut(t0) 2H10 () and condition (3:12) holds for t0: Then the same result of theorem 3.1.1 stays true.

Before we prove our results, we need the following Lemma, which means that, our energy is uniformly

bounded and decreasing along the trajectories.

Lemma 3.1.3 ([44]) Suppose that (G1) ; (G2), (2:3) hold, and let (u0; u1) 2 (H10 ())

2. Let u(t; x)

be the solution of (P ) ; then the modi�ed energy E(t) is non-increasing function for almost every

t 2 [0; T ) ; andd

dtE(t) = �a kut(t)kmm +

1

2(g0 � ru) (t)� 1

2g(t) kru(t)k22 � w krut(t)k22

(3.13)

� 0; 8t 2 [0; T ) :



Proof. By multiplying the di¤erential equation in (P ) by ut and integration over we obtain

d

dt

8<:12 kut(t)k22 + 120@1� tZ

0

g(s)ds

1A kru(t)k22 + 12(g � ru)(t)� b

pku(t)kpp

9=;= �a kut(t)kmm +

1

2(g0 � ru) (t)� 1

2g(t) kru(t)k22 � w krut(t)k22

� 1

2(g0 � ru) (t) � 0; 8t 2 [0; T ) :

By the de�nition of E(t), we conclude

d

dtE(t) � 0: (3.14)


The following lemma tells us that if the initial data ( or for some t0 > 0) is in the sat W , then the

solution stays there forever.

Lemma 3.1.4 ([44]) Suppose that (G1) ; (G2) ; (2:3) and (3:12) hold. If u0 2 W; u1 2 H10 (),

then the solution u(t) 2 W; 8t � 0.

Proof. Since u0 2 W , then

I(0) = kru0k22 � ku0kpp > 0;

consequently, by continuity, there exists Tm � T such that

I(u (t)) =

0@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)� b ku(t)kpp � 0; 8t 2 [0; Tm] :

This gives

J(t) =1

2

0@1� tZ0

g(s)ds

1A kru(t)k22 + 12(g � ru)(t)� b

pku(t)kpp

=

�p� 22p

�240@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)35+ 1

pI(u(t))

��p� 22p

�240@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)35 : (3.15)



By using (3:1) ; (3:15) and the fact thattR0

g(s)ds �1R0

g(s)ds; we easily see that

kru(t)k22 � 1

l

�2p

p� 2

�J(t)

� 1

l

�2p

p� 2

�E(u(t)) (3.16)

� 1

l

�2p

p� 2

�E(0); 8t 2 [0; Tm] :

We then exploit (G1) ; (3:12) ; (3:16) ; and we note that the embedding H10 () ,! Lp (), we have

ku(t)kp � C kru(t)k2 (3.17)

for 2 < p � 2n

n� 2 if n � 3, or p > 2 if n = 1; 2; and C = C (n; p; ) :

Consequently, we have

b ku(t)kpp � bCp� kru(t)kp2 ; 8t 2 [0; Tm]

� bCp� kru(t)kp�22 kru(t)k22

� bCp�lkru(t)kp�22 l kru(t)k22

� �l kru(t)k22 ;

(3.18)

which means by the de�nition of l

b ku(t)kpp � �

0@1� tZ0

g(s)ds

1A kru(t)k22<

0@1� tZ0

g(s)ds

1A kru(t)k22 ; 8t 2 [0; Tm] :

where

� =bCp�l

�2p

(p� 2) lE(0)�p�2

2

: (3.19)

Therefore,

I(t) =

0@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)� b ku(t)kpp > 0: (3.20)



for all t 2 [0; Tm] ;By taking the fact that

limt7!Tm

bCp�l

�2p

(p� 2) lE(0)�p�2

2

� � < 1: (3.21)

This shows that the solution u(t) 2 W; for all t 2 [0; Tm] : By repeating this procedure Tm extendedto T:


In order to prove theorem 3.1.1, it su¢ ces to show that the following norm

kru(t)k2 + kut(t)k2 ; (3.22)

is bounded independently of t.

To achieve this, we use (3:4) ; (3:14) and (3:15) to get

E(0) � E(t) = J(t) +1

2kut(t)k22

��p� 22p

�240@1� tZ0

g(s)ds

1A kru(t)k22 + (g � ru)(t)35

+1

2kut(t)k22 +

1

pI(t)

��p� 22p

��l kru(t)k22 + (g � ru)(t)

�+1

2kut(t)k22 +

1

pI(t)

(3.23)

��p� 22p

��l kru(t)k22 +

1

2kut(t)k22

�;

since I(t) and (g � ru)(t) are positive, hence

kru(t)k22 + kut(t)k22 � CE(0);

where C is a positive constant depending only on p and l.

This completes the proof of theorem 3.1.1.



The following lemma is very useful

Lemma 3.1.5 ([44]) Suppose that (2:3) and (3:12) hold. Then

b ku(t)kpp � (1� �)

0@1� tZ0

g(s)ds

1A kru(t)k22 (3.24)

where � = 1� �:

3.2 Decay of Solutions

We can now state the asymptotic behavior of the solution of (P ).

Theorem 3.2.1 Suppose that (G1) ; (G2) and (2:3) hold. Assume further that u0 2 W and u1 2H10 () satisfying (3:12) : Then the global solution satis�es

E(t) � E (0) exp (��t) ; 8t � 0 if m = 2; (3.25)

or

E(t) ��E(0)�r +K0rt

�s; 8t � 0 if m > 2, (3.26)

where � and K0 are constants independent of t; r =m

2� 1 and s = 2

2�m:

The following Lemma will play a decisive role in the proof of our result. The proof of this lemma

was given in Nakao [34].

Lemma 3.2.1 ( [37]) Let '(t) be a nonincreasing and nonnegative function de�ned on [0; T ] ;

T > 1; satisfying

'1+r(t) � k0 (' (t)� ' (t+ 1)) ; t 2 [0; T ] ;

for k0 > 1 and r � 0: Then we have, for each t 2 [0; T ] ;8>><>>:' (t) � ' (0) exp

��k [t� 1]+

�; r = 0

' (t) ��' (0)�r + k0r [t� 1]+

�1r ; r > 0

; (3.27)

where [t� 1]+ = max ft� 1; 0g ; and k = ln�

k0k0 � 1

�:

Section 3.2. Decay of Solutions



Multiplying the �rst equation in (P ) ; by ut and integrate over ; to obtain

d

dtE(t) + w krutk22 + a kutkmm =

1

2(g0 � ru) (t)� 1

2g(t) kru(t)k22 :

Then, integrate the last equality over [t; t+ 1] to get

E(t+ 1)� E(t) + w

t+1Zt

krutk22 ds+ a

t+1Zt

kutkmm ds

=

t+1Zt

1

2(g0 � ru) (s)ds�

t+1Zt

1

2g(s) kru(t)k22 ds:

Therefore,

E(t)� E(t+ 1) = Fm(t)� 12

t+1Zt

(g0 � ru) (s)ds+ 12

t+1Zt

g(t) kru(t)k22 ds; (3.28)

where

Fm(t) = a

t+1Zt

kutkmm ds+ w

t+1Zt

krutk22 ds: (3.29)

Using Poincaré�s inequality to �nd

t+1Zt

kutk22 ds � C ()

t+1Zt

kutk2m ds: (3.30)

Exploiting Hölder�s inequality, we obtain

t+1Zt

kutk2m ds �

0@ t+1Zt

ds

1Am�2m0@ t+1Z

t

�kutk2m

�m2

1A2m

ds

�

0@ t+1Zt

�kutk2m

�m2

1A2m

ds: (3.31)

Combining (3:29); (3:30); and (3:31); we obtain, for a constant C1; depending on

t+1Zt

kutk22 ds � C1F2(t); C1 > 0: (3.32)



By applying the mean value theorem, ( Theorem 1.3.3. in chapter1), we get for some t1 2�t; t+

1

4

�;

t2 2�t+

3

4; t+ 1

�kut(ti)k2 � 2C ()

12 F (t); i = 1; 2: (3.33)

Hence, by (G2) and sincet+1Zt

krutk22 ds � C2F (t)2; C2 > 0 (3.34)

there exist t1 2�t; t+

1

4

�; t2 2

�t+

3

4; t+ 1

�such that

krut(ti)k22 � 4C ()F (t)2; i = 1; 2: (3.35)

Next, we multiply the �rst equation in (P ) by u and integrate over � [t1; t2] to obtaint2Zt1

240@1� tZ0

g(�)d�

1A kru(t)k22 ds� b kukpp

35 ds= �

t2Zt1

Z

u:uttdxds� w

t2Zt1

Z

ru:rutdxds� a

t2Zt1

Z

u: jutjm�2 utdxds

+

t2Zt1

sZ0

g(s� �)

Z

ru(s): [ru(�)�ru(s)] dxd�ds.

Obviously,t2Zt1

I(s)ds = �t2Zt1

Z

u:uttdxds� w

t2Zt1

Z

ru:rutdxds� a

t2Zt1

Z

u: jutjm�2 utdxds

+

t2Zt1

sZ0

g(s� �)

Z

ru(s): [ru(�)�ru(s)] dxd�ds

+

t2Zt1

(g � ru) (s)ds: (3.36)

Note that by integrating by parts, to obtain��t2Zt1

Z

u:uttdxds

�� =

��24Z

utudx

35t2t1

�t2Zt1

Z

u2tdxds

��=

��Z

ut(t2)u(t2)dx�Z

ut(t1)u(t1)dx�t2Zt1

kutk22 ds

�� ;Section 3.2. Decay of Solutions


Using Hölder�s and Poincaré�s inequalities, we get��t2Zt1

Z

u:uttdxds

�� C2�

2Xi=1

krut(ti)k2 kru(ti)k2 + C2�

t2Zt1

krutk22 dt: (3.37)

By using Hölder�s inequality once again, we have��t2Zt1

Z

ru:rutdxds

�� t2Zt1

kruk2 krutk2 ds (3.38)

Furthermore, by (3:35) and (3:16), we have

krut(ti)k2 kru(ti)k2 � C3 (C())12 F (t) sup

t1�s�t2E(s)

12 ; (3.39)

where, C3 = 2�

2p

l (p� 2)

�12

:

From (3:34) we have by Hölder�s inequality

t2Zt1

kruk2 krutk2 dt �t2Zt1

E(s)12

�1

l

�2p

p� 2

��12

krutk2 ds

� 1

2C3 sup

t1�s�t2E(s)

12

t2Zt1

krutk2 ds;

which implies

t2Zt1

krutk2 dt �

0@ t2Zt1

1dt

1A120@ t2Zt1

krutk22 dt

1A12

�p3C22

F (t):

Then,t2Zt1

kruk2 krutk2 dt � C4F (t) supt1�s�t2

E(s)12 (3.40)

where C4 =C3p3C24

: Therefore (3:37) ; becomes��t2Zt1

Z

u:uttdxds

�� 2C2�C3F (t) supt1�s�t2E(s)

12 + C2�C2F (t)

2: (3.41)



We then exploit Young�s inequality to estimate

t2Zt1

Z

tZ0

g(s� �)ru(t): [ru(s)�ru(t)] d�dxdt

(3.42)

� �

t2Zt1

tZ0

g(s� �) kruk22 d�dt+1

4�

t2Zt1

(g � ru) (t)dt; 8� > 0:

Now, the third term in the right-hand side of (3:36), can be estimated as follows

t2Zt1

Z

jutjm�2 ut:udxds �t2Zt1

Z

jutjm�1 : juj dxds:

By Hölder�s inequality, we �nd

t2Zt1

Z

jutjm�1 : juj dxds �t2Zt1

26640@Z

jutjm dx

1Am�1m0@Z

jujm dx

1A1m

3775 ds

=

t2Zt1

kutkm�1m kukm ds:

By Sobolev-Poincaré�s inequality, we have

t2Zt1

kutkm�1m kukm ds � C ()

t2Zt1

kutkm�1m kruk2 ds;

for 2 < m � 2n

n� 2 if n � 3, or 2 � m <1 if n = 1; 2:



Using Hölder�s inequality, and since t1; t2 2 [t; t+ 1] and E(t) decreasing in time, we conclude fromthe last inequality, (3:16) and (3:29) ; that

t2Zt1

kutkm�1m kukm ds � C ()

�2p

l (p� 2)

�12

t2Zt1

kutkm�1m (J(u))12 ds

� C ()

�2p

l (p� 2)

�12

t2Zt1

kutkm�1m (E(u))12 ds

� C ()

�2p

l (p� 2)

�12

supt1�s�t2

(E(u))12 �

0@ t2Zt1

kutkmm ds

1Am�1m0@ t2Zt1

ds

1A1m

(3.43)

��1

a

�m�1m

C () supt1�s�t2

(E(t))12

�2p

l (p� 2)

�12

F (t)m�1

Then, taking into account (3:41)� (3:43), estimate (3:36) takes the form

t2Zt1

I(t)dt ��2C2� +

p3C24

w

�C3F (t) sup

t1�s�t2E(s)

12 + C2�C2F (t)

2

+a1m

2C3C () sup

t1�s�t2(E(t))

12 F (t)m�1

(3.44)

+�

t2Zt1

tZ0

g(t� s) kruk22 dsdt+�1

4�+ 1

� t2Zt1

(g � ru) (t)dt:

Moreover, from (3:4) and (3:10), we see that

E(t) =1

2kutk22 dt+ J(t)

=1

2kutk22 +

�p� 22p

�0@1� tZ0

g(s)ds

1A kruk22+

�p� 22p

�(g � ru) (t) + 1

pI(t): (3.45)



By integrating (3:45) over [t1; t2] ; we obtain

t2Zt1

E(t)dt =1

2

t2Zt1

kutk22 dt+�p� 22p

� t2Zt1

0@1� tZ0

g(s)ds

1A kruk22 dt+

�p� 22p

� t2Zt1

(g � ru) (t)dt+ 1p

t2Zt1

I(t)dt; (3.46)

which implies by exploiting (3:32)

t2Zt1

E(t)dt � C12(F (t))2 +

1

p

t2Zt1

I(t)dt+

�p� 22p

� t2Zt1

(g � ru) (t)dt

(3.47)

+

�p� 22p

� t2Zt1

0@1� tZ0

g(s)ds

1A kruk22 dt:By using (3:11), Lemma 3.1.5, we see that0@1� tZ

0

g(s)ds

1A kruk22 � 1

�I(t): (3.48)

Therefore, (3:47), takes the form

t2Zt1

E(t)dt � C ()

2(F (t))2 +

�p� 22p

� t2Zt1

(g � ru) (t)dt

+

�1

p+p� 22p�

� t2Zt1

I(t)dt: (3.49)

Again an integration of (3:14) over [s; t2] ; s 2 [0; t2] gives

E(s) = E(t2) + a

t2Zs

kut(t)kmm d� +1

2

t2Zs

g(�) kru(t)k22 d�

�12

t2Zs

�g0 � ru

�(t)d� + w

t2Zs

krut(t)k22 d� (3.50)

By using the fact that t2 � t1 �1

2, we have

t2Zt1

E(s)ds �t2Zt1

E(t2)ds �1

2E(t2): (3.51)



The fourth term in (3:44), can be handled as

tZ0

g(t� s) kruk22 ds = kruk22

tZ0

g(t� s)ds

(3.52)

� 2p (1� l)

l (p� 2) E(t):

Thus,

t2Zt1

tZ0

g(t� s) kruk22 dsdt � 2p (1� l)

l (p� 2)

t2Zt1

E(t)dt

� p (1� l)

l (p� 2)E(t1)

� p (1� l)

l (p� 2)E(t): (3.53)

Hence, by (3:53) ; we obtain from (3:44)

t2Zt1

I(t)dt ��2C2� +

p3C24

w

�C3F (t) sup

t1�s�t2E(s)

12 + C2�C2F (t)

2

+a1m

2C3C () sup

t1�s�t2(E(t))

12 F (t)m�1

+�p (1� l)

l (p� 2)E(t) +�1

4�+ 1

� t2Zt1

(g � ru) (t)dt: (3.54)

From (3:50) and (3:51) we have

E(t) � 2

t2Zt1

E(s)ds+ a

t+1Zt

kut(t)kmm d� +1

2

t+1Zt

g(�) kru(t)k22 d�

�12

t+1Zt

(g0 � ru) (t)d� + w

t+1Zt

krut(t)k22 d� : (3.55)



Obviously, (3:49) and (3:55) give us

E(t) � 2

0@C ()2

(F (t))2 +

�p� 22p

� t2Zt1

(g � ru) (t)dt+�1

p+p� 22p�

� t2Zt1

I(t)dt

1A+a

t2Zt1

kut(t)kmm dt+1

2

t2Zt1

g(t) kru(t)k22 dt�1

2

t2Zt1

(g0 � ru) (t)dt

+w

t2Zt1

krut(t)k22 dt:

Consequently, plugging the estimate (3:54) into the above estimate, we conclude

E(t) � C () (F (t))2 +

�p� 2p

� t2Zt1

(g � ru) (t)dt

+2

�1

p+p� 22p�

��2C2� +

p3C24

w

�C3F (t) sup

t1�s�t2E(s)

12 + C2�C2F (t)

2

�

+

�1

p+p� 22p�

�a1mC3C () sup

t1�s�t2(E(t))

12 F (t)m�1

+2

�1

p+p� 22p�

�24�p (1� l)

l (p� 2)E(t) +�1

4�+ 1

� t2Zt1

(g � ru) (t)dt)

35(3.56)

+Fm(t)� 12

t2Zt1

(g0 � ru) (t)dt+ 12

t2Zt1

g(t) kru(t)k22 dt:

We also have, by the Poincaré�s inequality

ku(s)k2 � C kru(s)k2

� C

�2p

l (p� 2)

�12

E(t)12 ; (3.57)

Choosing � small enough so that

1� 2�1

p+p� 22p�

��p (1� l)

l (p� 2) > 0; (3.58)



we deduce, from (3:56) that there exists K > 0 such that

E(t) � K

�F (t)2 + E(t)

12F (t) + E(t)

12F (t)m�1 + F (t)m

�

+1

2

t+1Zt

g(s) kru(s)k22 ds�1

2

t+1Zt

(g0 � ru) (s)ds

(3.59)

+

��p� 2p

�+ 2

�1

4�+ 1

��1

p+p� 22p�

�� t+1Zt

(g � ru) (s)ds

Using (G2) again we can write

t2Zt1

(g � ru) (t)dt � ��t2Zt1

(g0 � ru) (t)dt; � > 0:

Then, we obtain, from (3:59),

E(t) � K

�F (t)2 + E(t)

12F (t) + E(t)

12F (t)m�1 + F (t)m

�(3.60)

+1

2

t+1Zt

g(s) kru(s)k22 ds��1 +

1

2

� t+1Zt

(g0 � ru) (s)ds:

where �1 = �

��p� 2p

�+ 2

�1

p+p� 22p�

��1

4�+ 1

��:

An appropriate use of Young�s inequality in (3:60), we can �nd K1 > 0 such that

E(t) � K1

�F (t)2 + F (t)2(m�1) + F (t)m

�(3.61)

+

2412

t+1Zt


1

2

� t+1Zt

(g0 � ru) (s)ds

35 ;for K1 a positive constant.



Using (G2) again to get

E(t) � K1

�F (t)2 + F (t)2(m�1) + F (t)m

�

+

24��1 + 12� t+1Z

t


1

2

� t+1Zt

(g0 � ru) (s)ds

35� K1

�F (t)2 + F (t)2(m�1) + F (t)m

�(3.62)

+(1 + 2�1)

2412

t+1Zt


2

t+1Zt

(g0 � ru) (s)ds

35At this end we distinguish two cases:

Case 1. m = 2. In this case we use (3:28) and (3:62), we can �nd K2 > 0 such that

E(t) � K1 F (t)2

+(1 + 2�1)

2412

t+1Zt


2

t+1Zt

(g0 � ru) (s)ds

35� K2 [E(t)� E(t+ 1)] : (3.63)

Since E(t) is nonincreasing and nonnegative function, an application of Lemma 3.2.1 yields

E(t) � K2 [E(t)� E(t+ 1)] ; t � 0; (3.64)

which implies that

E(t) � E (0) exp�� [t� 1]+

�; on [0;1) ; (3.65)

where � = ln�

K2

K2 � 1

�:

Case 2. m > 2: In this case we, again use (3:28) and (3:62) to arrive at

F (t)2 =

24(E(t)� E(t+ 1))� 12

t+1Zt

g(s) kru(s)k22 ds+1

2

t+1Zt

(g0 � ru) (s)ds

352m

: (3.66)

We then use the algebraic inequality

(a+ b)m2 � 2

m2

�am2 + b

m2

�; m � 2: (3.67)



To infer from (3:62) ; and by using (3:67) ; that

[E(t)]m2 � K3

�1 + F (t)2(m�2) + F (t)m�2

�m2 F (t)m

+2m2 (1 + 2�1)

m2

2412

t+1Zt


2

t+1Zt

(g0 � ru) (s)ds

35m2

� K3

�1 + F (t)2(m�2) + F (t)m�2

�m2 � [E(t)� E(t+ 1)]

(3.68)

+2m2 (1 + 2�1)

m2

2412

t+1Zt


2

t+1Zt

(g0 � ru) (s)ds

35m2

where K3 = 2m2 K1: We use (3:28) to obtain0@1

2

t+1Zt

g(s) kruk22 ds�1

2

t+1Zt

(g0 � ru) (s)ds

1Am2

� (E(t)� E(t+ 1))m2 (3.69)

A combination of (3:68) ; (3:69) yields

[E(t)]m2 � K3

�1 + F (t)2(m�2) + F (t)m�2

�m2 � (E(t)� E(t+ 1))

+2m2 (1 + 2�1)

m2 [E(t)� E(t+ 1)]

m2�1 [E(t)� E(t+ 1)]

��K3

�1 + F (t)2(m�2) + F (t)m�2

�m2 + 2

m2 (1 + 2�1)

m2 [E(t)� E(t+ 1)]

m2�1��

[E(t)� E(t+ 1)] (3.70)

By using (3:62) ; the estimate (3:70) takes the form

[E(t)]m2 �

nK32

mh1 + E(0)(m�2) + (E(0))

m2�1i+ 2

m2 (1 + 2�1)

m2 (E(0))

m2�1o�

(E(t)� E(t+ 1))

� K0 (E(t)� E(t+ 1)) : (3.71)



Again, using Lemma 3.2.1, we conclude

E(t) ��E(0)�r +K0r [t� 1]+

�s; (3.72)

with r =m

2� 1 > 0; s = 2

2�mand K0 is some given positive constant.



Chapter 4

Exponential Growth

Abstract

Our goal in this chapter is to prove that when the initial energy is negative and p > m, then, the

solution with the Lp�norm grows as an exponential function provided that1R0

g(s)ds <p� 2p� 1 , by

using carefully the arguments of the method used in [16], with necessary modi�cation imposed by

the nature of our problem.

59

Chapter 4. Exponential Growth 60

4.1 Growth result

Our result reads as follows.

Theorem 4.1.1 Suppose that m � 2 and m < p � 1, if n = 1; 2; m < p � 2 (n� 1)n� 2 ; if n � 3:

Assume further that E(0) < 0 and1R0

g(s)ds <p� 2p� 1 holds: Then the unique local solution of problem

(P ) grows exponentially.

Proof. We set

H(t) = �E(t): (4.1)

By multiplying the �rst equations in (P ) by �ut, integrating over and using Lemma 2.1.3, weobtain

� d

dt

8<:12 kutk22 + 120@1� tZ

0

g(s)ds

1A kruk22 + 12 (g � ru) (t)� b

pkukpp

9=;= a kutkmm �

1

2(g0 � ru) (t) + 1

2g(t) kruk22 + w krutk22 : (4.2)

By the de�nition of H(t); (4:2) rewritten as

H 0(t) = a kutkmm �1

2(g0 � ru) (t) + 1

2g(t) kruk22 + w krutk22 � 0; 8t � 0: (4.3)

Consequently, E(0) < 0; we have

H(0) = �12ku1k22 �

1

2kru0k22 +

b

pku0kpp > 0: (4.4)

It�s clear that by (4:1), we have

H(0) � H(t); 8t � 0: (4.5)

Using (G2) ; to get

H(t)� b

pkukpp = �

2412kutk22 +

1

2

0@1� tZ0

g(s)ds

1A kruk22 + 12 (g � ru) (t)35

� 0; 8t � 0: (4.6)

One implies

0 < H(0) � H(t) � b

pkukpp : (4.7)

Section 4.1. Growth result


Let us de�ne the functional

L(t) = H(t) + "

Z

utudx+ "w

2kruk22 : (4.8)

for " small to be chosen later.

By taking the time derivative of (4:8) ; we obtain

L0(t) = H 0(t) + "

Z

uutt(t; x)dx+ " kutk22 + "w

Z

rutrudx

=

�w krutk22 + a kutkmm �

1

2(g0 � ru) (t) + 1

2g(t) kruk22

�

+" kutk22 + "w

Z

rutrudx+ "

Z

uttudx: (4.9)

Using the �rst equations in (P ), to obtainZ

uuttdx = b kukpp � kruk22 � !

Z

rutrudx� a

Z

jutjm�2 utudx

+

Z

rutZ0

g(t� s)ru(s; x)dsdx: (4.10)

Inserting (4:10) into (4:9) to get

L0(t) = w krutk22 + a kutkmm �1

2(g0 � ru) (t) + 1

2g(t) kruk22

+" kutk22 � " kruk22 + "

tZ0

g(t� s)

Z

ru:ru(s)dxds

+"b kukpp � "a

Z

jutjm�2 utudx: (4.11)

By using (G2) ; the last equality takes the form

L0(t) � w krutk22 + a kutkmm + " kutk22 � " kruk22 + "b kukpp(4.12)

+"

tZ0

g(t� s)

Z

ru:ru(s)dxds� "a

Z

jutjm�2 utud�:

To estimate the last term in the right-hand side of (4:12) ; we use the following Young�s inequality

XY � �r

rXr +

��q

qY q; X; Y � 0; (4.13)



for all � > 0 be chosen later;1

r+1

q= 1; with r = m and q =

m

m� 1 :So we have Z

jutjm�2 utudx �Z

jutjm�1 juj dx

� �m

mkukmm +

�m� 1m

��(

�mm�1) kutkmm ; 8t � 0: (4.14)

Therefore, the estimate (4:12) takes the form

L0(t) � w krutk22 + a kutkmm + " kutk22 � " kruk22 + "b kukpp

+"

tZ0

g(t� s)

Z

ru:ru(s)dxds

�"a�m

mkukmm � "a

�m� 1m

��

� �mm�1

�kutkmm

� w krutk22 + a kutkmm + " kutk22 � " kruk22 + "b kukpp

+" kruk22

tZ0

g(s)ds+ "

tZ0

g(t� s)

Z

ru (t) [ru (s)�ru (t)] dxds

�"a�m

mkukmm � "a

�m� 1m

��

� �mm�1

�kutkmm : (4.15)

Using Cauchy-Schwarz and Young�s inequalities to obtain

L0(t) � w krutk22 + a

�1� "

�m� 1m

��

� �mm�1

��kutkmm + " kutk22

�" kruk22 + "b kukpp + " kruk22

tZ0

g(s)ds

�"tZ0

g(t� s) kruk2 kru(s)�ru (t)k2 ds� "a�m

mkukmm

� w krutk22 + a

�1� "

�m� 1m

��

� �mm�1

��kutkmm + " kutk22 + "b kukpp

(4.16)

+"

0@12

tZ0

g(s)ds� 1

1A kruk22 � "1

2(g � ru(t)� "a

�m

mkukmm :



Using assumptions to substitute for b kukpp : Hence, (4:16) becomes


�1� "

�m� 1m

��

� �mm�1

��kutkmm + " kutk22

+"

0@pH(t) + p

2kutk22 +

p

2(g � ru) (t) + p

2

0@1� tZ0

g(s)ds

1A kruk221A

+"

0@12

tZ0

g(s)ds� 1

1A kruk22 � "1

2(g � ru(t)� "a

�m

mkukmm :

� w krutk22 + a

�1� "

�m� 1m

��

� �mm�1

��kutkmm + "

�1 +

p

2

�kutk22

(4.17)

+"a1 kruk22 + "a2(g � ru(t)� "a�m

mkukmm + "pH(t):

where a1 =�1� p

2

� 1R0

g(s)ds+

�p� 22

�> 0; a2 =

p� 12

> 0:

In order to undervalue L0(t) with terms of E(t) and since p > m, we have from the embedding

Lp () ,! Lm () ;

kukmm � C kukmp � C�kukpp

�mp; 8t � 0: (4.18)

for some positive constant C depending on only. Since 0 <m

p< 1; we use the algebraic inequality

zk � (z + 1) ��1 +

1

w

�(z + w) ; 8z � 0; 0 < k � 1; w > 0;

to �nd �kukpp

�mp � K

�kukpp +H(0)

�; 8t � 0; (4.19)

where K = 1 +1

H(0)> 0; then by (4:7) we have

kukmm � C

�1 +

b

p

�kukpp ; 8t � 0: (4.20)

Inserting (4:20) into (4:17), to get


�1� "

�m� 1m

��

� �mm�1

��kutkmm + "

�1 +

p

2

�kutk22

(4.21)

+"a1 kruk22 + "a2(g � ru(t)� "C1 kukpp + "pH(t):

where C1 = aC�m

m

�1 +

b

p

�> 0:



By using (4:1) and by the same statements as in [16], we have

2H(t) = �kutk22 � kruk22 +

tZ0

g(s)ds kruk22 � (g � ru) (t) +2b

pkukpp

(4.22)

� �kutk22 � kruk22 � (g � ru) (t) +

2b

pkukpp ; 8t � 0:

Adding and substituting the value 2a3H(t) from (4:21), and choosing � small enough such that

a3 < min fa1; a2g ; we obtain


�1� "

�m� 1m

��

� �mm�1

��kutkmm

+"�1 +

p

2� a3

�kutk22 + " (a1 � a3) kruk22

+" (a2 � a3) (g � ru(t) + "

�2b

pa3 � C1

�kukpp

+" (p� 2a3)H(t): (4.23)

Now, once � is �xed, we can choose " small enough such that

1� "

�m� 1m

��

� �mm�1

�> 0; and L(0) > 0: (4.24)

Therefore, (4:23) takes the form

L0(t) � "�nH(t) + kutk22 + kruk

22 + (g � ru(t)) + kuk

pp

o; (4.25)

for some � > 0:

Now, using (G2), Young�s and Poincaré�s inequalities in (4:8) to get

L(t) � �1�H(t) + kutk22 + kruk

22

; (4.26)

for some �1 > 0: Since, H(t) > 0; we have from (4:1)

� 12kutk22 �

1

2

0@1� tZ0

g(s)ds

1A kruk22 � 12 (g � ru) (t) + b

pkukpp > 0; 8t � 0: (4.27)

Then,

1

2

0@1� tZ0

g(s)ds

1A kruk22 <b

pkukpp

<b

pkukpp + (g � ru) (t): (4.28)



In the other hand, using (G1) ; to get

1

2(1� l) kruk22 � 1

2

0@1� tZ0

g(s)ds

1A kruk22<

b

pkukpp + (g � ru) (t): (4.29)

Consequently,

kruk22 <2b

pkukpp + 2 (g � ru) (t) + 2l kruk

22 ; b; l > 0: (4.30)

Inserting (4:30) into (4:26) ; to see that there exists a positive constant � such that

L(t) � �

�H(t) + kutk22 + kruk

22 + (g � ru) (t) +

b

pkukpp

�; 8t � 0: (4.31)

From inequalities (4:25) and (4:31) we obtain the di¤erential inequality

L0(t)

L(t)� �; for some � > 0; 8t � 0: (4.32)

Integration of (4:32) ; between 0 and t gives us

L(t) � L(0) exp (�t) ; 8t � 0; (4.33)

From (4:8) and for " small enough, we have

L(t) � H(t) � b

pkukpp : (4.34)

By (4:33) and (4:34) ; we have

kukpp � C exp (�t) ; C > 0; 8t � 0: (4.35)

Therefore, we conclude that the solution in the Lp�norm growths exponentially.

4.1. Growth result

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