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THESE
presentee a
L’UNIVERSITE DE VALENCIENNES ET DU HAINAUT-CAMBRESIS
ECOLE DOCTORALE REGIONALE SCIENCES POUR L’INGENIEURLILLE NORD-DE-FRANCE - 072
en vue d’obtenir le titre de
DOCTEUR DE L’UNIVERSITE
Specialite : MATHEMATIQUES APPLIQUEES
par
Julie VALEIN
Stabilite de quelques problemes d’evolution
Soutenue le 20 Novembre 2008 devant le jury compose de
Directeur de these : Serge NICAISE, Universite de Valenciennes et du Hainaut-CambresisRapporteurs : Kaıs AMMARI, Faculte des Sciences de Monastir
Marius TUCSNAK, Universite Henri Poincare Nancy 1Examinateurs : Felix ALI MEHMETI, Universite de Valenciennes et du Hainaut-Cambresis
Joachim VON BELOW, Universite du Littoral Cote d’OpaleOlivier GOUBET, Universite de Picardie Jules Verne
Stabilite de quelques problemes d’evolution
Dans cette these nous etudions la stabilisation de quelques equations d’evolution par retro-action (feedback). Tout d’abord, nous considerons la stabilisation de l’equation des ondes surdes reseaux 1-d par des feedbacks situes aux noeuds. Dans le premier chapitre, en supposant quele poids du feedback avec retard est plus petit que celui sans retard, nous donnons des condi-tions spectrales pour obtenir la stabilite forte, exponentielle ou polynomiale en nous ramenant al’etude d’une inegalite d’observabilite pour le probleme conservatif. Dans le second chapitre, noustransferons des inegalites d’observabilite a poids deja existantes pour un autre probleme conser-vatif en inegalites d’observabilite faibles pour le systeme dissipe sans retard. Grace a une inegalited’interpolation, nous obtenons des taux de decroissance explicites qui dependent des proprietesgeometriques et topologiques du reseau. Nous developpons ensuite, dans le chapitre 3, une theorieabstraite pour les equations d’evolution du second ordre avec retard generalisant les resultats duchapitre 1. Nous etudions le cas ou le retard depend du temps pour les equations des ondes etde la chaleur dans le chapitre 4. En emettant certaines hypotheses sur ce retard et en utilisantune fonctionnelle de Lyapunov appropriee, nous prouvons que l’energie est exponentiellementdecroissante et nous donnons explicitement son taux de decroissance. Enfin, nous montrons dansle chapitre 5, qu’une technique de filtrage permet d’obtenir une decroissance quasi-exponentiellede l’equation des ondes discretisee en espace par differences finies avec un amortissement interne.
Mots-cles : Theorie du controle, stabilisation, equations d’evolution, terme de retard, reseaux,semi-discretisation.
Stability of some evolution problems
In this PhD thesis we study the stabilization of some evolution equations by feedback laws.First we consider the stabilization of the wave equation on 1-d networks with nodal feedbacks.In chapter 1, assuming that the weight of the feedback without delay is smaller than the onewith delay, we give spectral conditions to obtain the strong, exponential or polynomial stability,by studying an observability inequality for the conservative system. In chapter 2 we transferknown observability results for another conservative system into a weighted observability estimatefor the dissipative one without delay. Thanks to an interpolation inequality, we obtain explicitdecay rates which depend on the geometric and topological properties of the network. Then wedevelop, in chapter 3, an abstract theory for second order evolution equation with delay, whichgeneralizes the results of chapter 1. We study the case where the delay depends on time forthe heat and wave equations in chapter 4. Using some assumptions about the delay and anappropriate Lyapunov functional, we prove that the energy is exponentially decreasing and wegive explicitely its decay rate. Finally, we show, in chapter 4, that a filtering technique allows toobtain a quasi-exponential decay of a finite difference space discretization of the wave equationby pointwise interior stabilization.
Key words : Control theory, stabilization, evolution equations, delay term, network, semi-discretization.
Specialite : Mathematiques Appliquees
Laboratoire de Mathematiques et leurs Applications (LAMAV), Universite de Valenciennes et duHainaut-Cambresis, Le Mont-Houy, 59313 Valenciennes Cedex 9
Remerciements
En premier lieu, je tiens a remercier chaleureusement Serge Nicaise pour sa disponibilite,son dynamisme et sa gentillesse. Travailler avec lui fut un grand plaisir qui, je l’espere,pourra se poursuivre. Il a su me guider avec un enthousiasme constant et communicatif,et m’encourager pendant ces trois annees. Il m’a temoigne sa confiance en m’incitant aparticiper a de nombreuses conferences. Ses grandes qualites scientifiques et humaines ontete indispensables a l’elaboration de cette these. Pour tout cela, je ne l’en remercieraijamais assez.
Ma gratitude va a Kaıs Ammari et a Marius Tucsnak d’avoir accepte de rapporter surmes travaux. Leur lecture attentive et leurs remarques judicieuses ont ete precieuses.
Je suis egalement reconnaissante a Felix Ali Mehmeti, Joachim von Below et OlivierGoubet pour leur participation a mon jury.
J’apprecie l’interet que tous ont porte a mon travail. C’est pour moi un grand honneurd’avoir un tel jury.
J’exprime ma profonde et sincere reconnaissance a Enrique Zuazua pour m’avoir ac-cueillie a l’Universidad Autonoma de Madrid. Ces six mois de travail en Espagne m’ontbeaucoup appris. J’ai particulierement apprecie sa grande culture mathematique et sa fa-culte a exprimer simplement des idees parfois complexes. J’ai ete sensible a l’accueil cordialet a la gentillesse des membres du departement de mathematiques de cette universite.
Un grand merci a Sylvain Ervedoza pour toutes les discussions mathematiques ou autresque nous avons eues a Madrid et qui se poursuivent aujourd’hui.
Les membres du LAMAV m’ont accueillie comme une collegue a part entiere et m’ontpermis de passer trois annees tres agreables. Ils m’ont fait confiance en me proposant desenseignements interessants et enrichissants ; leurs appuis et leurs conseils pedagogiquesm’ont aidee dans mes fonctions nouvelles d’enseignante. Je les en remercie vivement, ainsique Nabila Daifi pour son aide administrative precieuse et sa bonne humeur.
Je veux egalement faire part de mon amitie a Isabelle, ma collegue de bureau.
Je tiens a remercier les informaticiens du « couloir d’a cote » : Celia, Christophe, Dana,Dimitri, Eric, Fred, Nico, Patrick, pour tous les bons moments de detente passes ensemble.
A toutes les personnes que j’ai rencontrees lors des conferences et avec qui j’ai eu defructueuses discussions, merci.
A tous mes amis exterieurs a l’universite qui m’ont permis de m’evader de mes travauxet de decompresser : Yann, Ludo, Claire, Laure, Ludo, Anais, Caro, Sophie, et a monprofesseur de danse Sophie Cirier, un grand merci.
Enfin, je souhaite de tout coeur remercier ma famille : mes parents, ma soeur Sarah etmes grands-parents, pour leur amour et leur soutien sans faille. Je les remercie de m’avoirsupportee et encouragee pendant les moments de doute. Je n’aurai jamais pu faire cettethese sans eux.
Table des matieres
Introduction 5
1 Stabilization of the wave equation on 1-d networks with a delay term inthe nodal feedbacks 251.1 Introduction/Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.2 Well posedness of the problem . . . . . . . . . . . . . . . . . . . . . . . . . 271.3 The energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
1.3.1 Decay of the energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.3.2 Problem without damping . . . . . . . . . . . . . . . . . . . . . . . 381.3.3 Decay of the energy to 0 . . . . . . . . . . . . . . . . . . . . . . . . 381.3.4 Counterexample to the stability of the system . . . . . . . . . . . . 45
1.4 A regularity result and an a priori estimate . . . . . . . . . . . . . . . . . . 471.5 The exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
1.5.1 An observability inequality . . . . . . . . . . . . . . . . . . . . . . . 581.5.2 The stability result . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
1.6 The polynomial stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621.6.1 An observability estimate . . . . . . . . . . . . . . . . . . . . . . . 621.6.2 Polynomial decay of the energy . . . . . . . . . . . . . . . . . . . . 67
1.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 681.7.1 One string with an interior damping . . . . . . . . . . . . . . . . . 681.7.2 A star shaped network . . . . . . . . . . . . . . . . . . . . . . . . . 711.7.3 More complex networks . . . . . . . . . . . . . . . . . . . . . . . . 76
2 Weak stabilization of the wave equation on 1-d networks 852.1 Introduction and main results . . . . . . . . . . . . . . . . . . . . . . . . . 852.2 The weighted observability inequality . . . . . . . . . . . . . . . . . . . . . 91
2.2.1 Preliminaries about the Dirichlet problem . . . . . . . . . . . . . . 912.2.2 The weighted observability inequality . . . . . . . . . . . . . . . . . 922.2.3 Proof of Theorem 2.2.1 . . . . . . . . . . . . . . . . . . . . . . . . . 94
2.3 The stabilization result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 972.3.1 An interpolation inequality . . . . . . . . . . . . . . . . . . . . . . . 972.3.2 The main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
2.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
1
2.4.1 The star-shaped network with N strings . . . . . . . . . . . . . . . 1072.4.2 A non star-shaped tree . . . . . . . . . . . . . . . . . . . . . . . . . 113
3 Stabilization of second order evolution equations with unbounded feed-back with delay 1153.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1153.2 Well posedness of the problem . . . . . . . . . . . . . . . . . . . . . . . . . 1173.3 The energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
3.3.1 Decay of the energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 1213.3.2 Decay of the energy to 0 . . . . . . . . . . . . . . . . . . . . . . . . 123
3.4 The exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.4.1 A priori estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1253.4.2 The stability result . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
3.5 The polynomial stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1293.6 Checking the observability inequalities . . . . . . . . . . . . . . . . . . . . 132
3.6.1 Preliminaries about Ingham’s inequality . . . . . . . . . . . . . . . 1323.6.2 A first observability inequality . . . . . . . . . . . . . . . . . . . . . 1363.6.3 A second observability inequality . . . . . . . . . . . . . . . . . . . 139
3.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1403.7.1 A wave equation on 1-d networks with nodal feedbacks . . . . . . . 1403.7.2 One Euler-Bernoulli beam with interior damping . . . . . . . . . . 1433.7.3 Examples with distributed damping terms . . . . . . . . . . . . . . 148
4 Stability of the heat and of the wave equations with boundary time-varying delays 1574.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1574.2 Exponential stability of the delayed heat equation . . . . . . . . . . . . . . 158
4.2.1 Well posedness of the problem . . . . . . . . . . . . . . . . . . . . . 1594.2.2 The decay of the energy . . . . . . . . . . . . . . . . . . . . . . . . 1664.2.3 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . 167
4.3 Exponential stability of the delayed wave equation . . . . . . . . . . . . . . 1704.3.1 Well posedness of the problem . . . . . . . . . . . . . . . . . . . . . 1714.3.2 The decay of the energy . . . . . . . . . . . . . . . . . . . . . . . . 1774.3.3 Exponential stability . . . . . . . . . . . . . . . . . . . . . . . . . . 178
5 Quasi exponential decay of a finite difference space discretization of the1-d wave equation by pointwise interior stabilization 1835.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1835.2 The continuous and discrete problems . . . . . . . . . . . . . . . . . . . . . 1845.3 Decay of the energy to 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1885.4 Non exponential decay of the discrete energy . . . . . . . . . . . . . . . . . 1905.5 Filtering technique . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
5.5.1 Interior observability of the discrete wave equation . . . . . . . . . 195
2
5.5.2 Some estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1995.5.3 “Quasi” exponential decay . . . . . . . . . . . . . . . . . . . . . . . 205
Conclusion 211
Bibliographie 213
3
4
Introduction
La Theorie du Controle des Equations aux Derivees Partielles intervient dans differentscontextes et de plusieurs manieres.
Les problemes de controlabilite, d’observabilite et de stabilite des Equations aux DeriveesPartielles ont fait l’objet, recemment, de nombreux travaux. Dans cette these nous noussommes interesses a l’etude de la stabilisation de quelques equations d’evolution.
Le probleme de controlabilite peut se formuler simplement : considerons un systemed’evolution decrit par des equations differentielles ordinaires ou par des equations auxderivees partielles et un intervalle de temps [0, T ]. Peut-on amener les solutions d’un etatinitial (au temps t = 0) a un etat final (au temps t = T ) en agissant par un controleapproprie applique sur le bord ou dans une partie du domaine dans laquelle l’equationevolue ?
Il y a eu d’intensives recherches sur le sujet depuis les trois dernieres decennies. Nousrenvoyons par exemple aux livres de J.-L. Lions [70,71], de Lasiecka et Triggiani [67] et deFattorini [43], aux articles de Russell [100] et de Zuazua [111,113]...
Dans un cadre fonctionnel approprie, le probleme de controlabilite est equivalent a celuid’observabilite qui consiste a analyser si l’energie totale des solutions peut etre evaluee aumoyen de mesures partielles sur un sous-ensemble du domaine ou du bord. Pour obtenir desestimees d’observabilite il existe diverses methodes, comme la technique des multiplicateurs(par exemple [61,70]), l’analyse microlocale ( [15,25]), les inegalites de Carleman ( [28,44,46,69]) ou encore les criteres frequentiels ( [26,77]), les criteres spectraux ( [73,95]) ou lesinegalites d’Ingham ( [53, 57, 62]).
La stabilisation a pour but d’attenuer les vibrations par retro-action (feedback) ; elleconsiste donc a garantir la decroissance de l’energie des solutions vers 0 de facon plus oumoins rapide par un mecanisme de dissipation.
Plus precisement, le probleme de stabilisation auquel on s’interesse revient a determinerle comportement asymptotique de l’energie que l’on note E(t) (c’est la norme des solutionsdans l’espace d’etat), a etudier sa limite afin de determiner si cette limite est nulle oupas, et, si cette limite est nulle, a donner une estimation de la vitesse de decroissance del’energie vers zero.
On peut noter des differences entre les problemes de controlabilite et ceux de stabilite.D’une part, dans les premiers le temps varie dans un intervalle fini [0, T ], alors que pour lesproblemes de stabilisation le temps t tend vers l’infini. D’autre part, dans les problemes de
5
controlabilite, le controle peut entrer dans le systeme librement en boucle ouverte, tandisque pour les problemes de stabilisation, le controle est de la forme feedback ou en bouclefermee.
Malgre cela, les liens entre les problemes de stabilisation et de controlabilite sont etroitset l’on demontre certaines implications entre ces deux problemes (voir par exemple [71,84,100]). Cependant, la stabilisation ne peut pas toujours etre obtenue comme consequencede la controlabilite, et c’est pour cela que son etude est souvent faite independamment etdirectement.
Il existe plusieurs degres de stabilite que l’on peut etudier. Le premier degre consiste aanalyser simplement la decroissance de l’energie des solutions vers zero, i.e. :
E(t) → 0, lorsque t→ +∞.
C’est ce que l’on appelle la stabilisation forte.Pour le second, on s’interesse a la decroissance de l’energie la plus rapide, c’est-a-dire
lorsque celle-ci tend vers 0 de maniere exponentielle, i.e. :
E(t) ≤ Ce−δt, ∀t > 0,
ou C et δ sont des constantes positives avec C qui depend des donnees initiales.Quant au troisieme, il etudie des situations intermediaires, dans lesquelles la decroissance
des solutions n’est pas exponentielle, mais du type polynomial par exemple :
E(t) ≤ C
tα, ∀t > 0,
ou C et α sont des constantes positives avec C qui depend des donnees initiales. Dans cecas, il faut prendre des donnees initiales plus regulieres, dans le domaine de l’operateur.
Nos travaux s’orientent dans trois directions : la stabilisation sur des reseaux 1-d, lastabilisation avec un terme de retard, et l’aspect numerique.
L’etude des equations aux derivees partielles sur des structures en forme de reseau ou degraphe a connu de reelles avancees depuis le debut des annees 80 ( [4,76,83]). Recemment,de nombreux auteurs se sont interesses aux problemes de controle sur les reseaux unidimen-sionnels, nous pouvons citer par exemple les livres [38], [63], [64] et [65]. Ces etudes utilisentdes resultats dans plusieurs domaines : series de Fourier non-harmoniques, approximationsDiophantiennes, theorie des graphes, techniques de propagation des ondes. Dans [38], ilest demontre des resultats de controlabilite en fonction des proprietes d’irrationnalite desrapports des longueurs des cordes du reseau. Dans le cas des reseaux en forme d’etoile oud’arbre, la controlabilite est prouvee en utilisant la formule de D’Alembert, tandis que pourles reseaux generaux (qui peuvent contenir des circuits), les auteurs utilisent le theoremede Beurling-Malliavin. Dans les deux premiers chapitres de cette these, nous nous sommesinteresses a la stabilisation de l’equation des ondes sur un reseau de cordes en utilisantdeux methodes differentes que l’on presentera plus tard.
6
Une recherche tres active s’est amorcee recemment sur les problemes de stabilisationavec effet retard. Les phenomenes de retard (en temps) apparaissent dans de nombreusesapplications, par exemple en biologie [51], en mecanique [1] ou encore en automatique [101].Il est bien connu que le terme avec retard dans le feedback peut etre la cause d’instabilite[16–18] : un retard arbitrairement petit dans le feedback peut destabiliser le systeme commele montre [39–41,75,85,98,110]. Mais, a contrario, un terme de retard peut aussi ameliorer laperformance du systeme [1,101]. Les problemes de stabilite de systemes avec retard revetentdonc une importance non negligeable et les chapitres 1, 3 et 4 de ce travail traitent de cesquestions. Pour regler les problemes d’instabilite qui peuvent survenir dans les systemesavec retard, nous avons considere dans ces trois chapitres deux types de dissipation : unpremier operateur de feedback sans retard et un second avec retard. Pour avoir la stabilitede ces systemes, nous avons suppose que le feedback avec retard est majore par celui sansretard (dans un sens que nous preciserons plus tard) dans le but de compenser les effetsd’instabilite qui peuvent intervenir. Cette idee a ete introduite par Xu, Yung et Li [110]et Nicaise et Pignotti [85]. Dans le premier chapitre nous avons considere l’equation desondes sur un reseau 1-d avec un terme de retard dans les feedbacks. Dans le chapitre 3nous avons generalise cette approche dans le cadre des equations d’evolution abstraites dusecond ordre avec un terme de retard dans les feedbacks (non bornes). Dans le chapitre 4,nous avons etudie le cas ou le retard depend du temps (et donc n’est plus constant) pourles equations unidimensionnelles de la chaleur et des ondes.
Une autre question interessante dans le domaine du controle et de la stabilisation portesur l’etude des approximations numeriques associees au systeme considere. En pratique,le modele continu est approche par un modele discret en espace ou en temps (par lamethode des differences finies, des elements finis ou des volumes finis). Dans de nombreuxproblemes de controle, le schema numerique a tendance a mal se comporter et, en general,le controle du probleme discret ne converge pas vers le controle du probleme continu.Ce phenomene est du aux modes etrangers que le schema numerique introduit a hautesfrequences ; le schema genere des oscillations a hautes frequences qui n’existent pas dans lemodele continu. Par consequent, controler un modele numerique discretise ne garantit pasd’obtenir une bonne approximation numerique du controle du modele continu. L’inegalited’observabilite discrete n’est en fait pas uniforme par rapport au pas de discretisation,en raison de l’existence des solutions a hautes frequences dont la vitesse de groupe depropagation est de l’ordre du parametre de discretisation. Il se produit le meme phenomenedans le cas de la stabilisation avec un taux de decroissance de l’energie discretisee non-uniforme par rapport au pas de discretisation [14, 96, 102, 103]. Ce phenomene a ete misen evidence dans les annees 1990 ( [49, 50]), et ces auteurs ont propose des techniquestres efficaces afin de restaurer la convergence du controle discret, comme la regularisationde Tychonoff [49, 50], l’algorithme bi-grille [47, 79], la methode des elements finis mixtes[14,30,31,48,78], la viscosite numerique [96,102] ou le filtrage des hautes frequences [56,112].Ces methodes ont pour effet de redresser le spectre discret pour les hautes frequences. Nousavons etudie, dans le dernier chapitre de cette these, la stabilisation de l’equation des ondesdiscretisee en espace par differences finies avec un amortissement en un point interieur et
7
avons prouve une decroissance quasi-exponentielle en filtrant les hautes frequences.
Nous allons maintenant rappeler brievement les principaux outils et methodes employes.Nous utilisons principalement, dans les chapitres 1, 3 et 5, la methodologie de Ammari etTuscnak [11] pour les equations d’evolution du second ordre. Les resultats de stabilite sontbases sur l’obtention d’inegalites d’observabilite pour le probleme conservatif associe.
Considerons donc l’equation d’evolution suivante
ω(t) + Aω(t) +BB∗ω(t) = 0, t > 0ω(0) = ω0, ω(0) = ω1,
(1)
ou A : D(A) → H est un operateur positif, auto-adjoint avec inverse compact dans un
espace de Hilbert H , U est un espace de Hilbert (identifie a son dual) et B ∈ L(U, D(A12 )′).
L’energie de ω(t) a l’instant t est definie par
E(t) =1
2
(
‖ω(t)‖2H +
∥
∥
∥A
12ω(t)
∥
∥
∥
2
H
)
.
C’est un systeme dissipatif, qui verifie formellement
E ′(t) = −‖B∗ω(t)‖2U ≤ 0. (2)
Nous allons expliquer sur cet exemple la methodologie employee dans les chapitres 1,3 et 5, sans faire aucune demonstration mais en donnant les outils principaux. La stabilitede cet exemple a deja ete etudiee dans [11] et nos rappels suivent celle-ci.
Notre premiere preoccupation est l’existence et l’unicite des solutions de (1). Pour cela,nous reecrivons (1) comme un systeme du premier ordre
U ′(t) = AU(t), t > 0U(0) = U0 = (ω0, ω1),
(3)
ou A : D(A) → H et H un espace de Hilbert. L’outil principal est alors la theorie dessemi-groupes et en particulier le Theoreme de Lumer-Phillips. Rappelons tout d’abord ladefinition d’un semi-groupe fortement continu (voir [74, 93] pour plus de details) :
Definition 0.0.1. Une famille S(t) (0 ≤ t < ∞) d’operateurs lineaires bornes dans unespace de Banach H est appelee un semi-groupe fortement continu (ou C0 semi-groupe) si
i) S(t1 + t2) = S(t1)S(t2), ∀t1, t2 ≥ 0,ii) S(0) = I,iii) Pour tout x ∈ H, S(t)x est continu en t sur [0,∞).La famille S(t) est appelee C0 semi-groupe de contractions (ou semi-groupe de contrac-
tions) si de plus ‖S(t)‖ ≤ 1 pour tout t ≥ 0.
8
Pour un tel semi-groupe, on definit un operateur A de domaine D(A) qui est l’ensembledes points x tel que la limite
Ax = limh→0
S(h)x− x
h,
existe. L’operateur A est alors appele le generateur infinitesimal du semi-groupe S(t). Pourun operateur A donne, si A coıncide avec le generateur infinitesimal de S(t), on dit alorsqu’il genere un semi-groupe fortement continu S(t), t ≥ 0.
Le resultat suivant est connu sous le nom de Theoreme de Lumer-Phillips :
Theoreme 0.0.1 (Theoreme de Lumer-Phillips). Soit A un operateur lineaire de domaineD(A) dense dans un espace de Hilbert H. Si
i) A est dissipatif, i.e. Re (Ax, x) ≥ 0 pour tout x ∈ D(A)ii) pour tout λ > 0, λI − A est surjectif,
alors A genere un C0 semi-groupe de contractions sur H.
Si un operateur verifie les conditions i) et ii), on dit aussi qu’il est m-dissipatif.Dans notre exemple, nous devons donc verifier que A est m-dissipatif pour obtenir
l’existence et l’unicite des solutions de (3). De plus, si U0 ∈ H, alors U ∈ C([0, +∞), H)et si U0 ∈ D(A), alors U ∈ C([0, +∞), D(A)) ∩ C1([0, +∞),H).
Cette methode sera appliquee dans les trois premiers chapitres. Pour le chapitre 4(ou nous etudions la stabilisation de l’equation des ondes et de la chaleur avec un retarddependant du temps), l’operateur A(t) de (3) dependra du temps avec un domaine D(A(t))qui sera lui independant du temps. Nous utiliserons alors la technique de normes variablesde Kato (voir [59, 60] et le chapitre 4 pour plus de details).
Une fois l’existence et l’unicite des solutions de notre systeme (1) prouvees ainsi que ladecroissance de l’energie (qui est la norme des solutions dans H), nous etudions la stabiliteforte, c’est-a-dire nous donnons des conditions necessaires et suffisantes pour que l’energietende vers zero. Pour cela, nous utilisons dans les chapitres 1 et 5 le principe d’invariancede LaSalle et dans le chapitre 3 un resultat de Arendt et Batty [12].
Rappelons ce qu’est le principe d’invariance de LaSalle. Dans ce paragraphe, (Z, d)designe un espace metrique complet.
Definition 0.0.2. Un systeme dynamique sur Z est une famille S(t)t≥0 d’applicationssur Z telle que S(t) ∈ C(Z, Z), ∀t ≥ 0 et verifiant i)-iii) de la Definition 0.0.1.
Definition 0.0.3. Soit z ∈ Z. L’ensemble
ω(z) = y ∈ Z : ∃tn → ∞, S(tn)z → y lorsque n→ ∞
est appele ensemble ω-limite de z.
Remarque 1. Pour tout z ∈ Z et t ≥ 0, S(t)(ω(z)) ⊂ ω(z).
De plus, si⋃
t≥0
S(t)z est relativement compact dans Z, alors S(t)(ω(z)) = ω(z) 6= ∅.
9
Definition 0.0.4. Une fonction φ ∈ C(Z, R) est dite fonction de Lyapounov pour S(t)t≥0
si on aφ(S(t)z) ≤ φ(z), ∀z ∈ Z, ∀t ≥ 0.
Apres ces quelques rappels, nous pouvons enoncer le principe d’invariance de LaSalle :
Theoreme 0.0.2 (Principe d’invariance de LaSalle). Soit φ une fonction de Lyapounovpour S(t) et z ∈ Z tel que ∪t≥0S(t)z soit relativement compact dans Z. Alors :
(i) L = limt→∞
φ(S(t)z) existe
(ii) φ(y) = L pour tout y ∈ ω(z).
Dans notre exemple, nous appliquerons le principe d’invariance de LaSalle a l’ensemblerelativement compact ∪t≥0S(t)U0, ou S(t) designe le C0 semi-groupe de contractions generepar A et a la fonctionnelle de Lyapounov φ = ‖.‖H.
Il nous reste alors a examiner la stabilite exponentielle ou polynomiale de notre systemedissipe (1). En suivant [11], cette etude est basee sur l’obtention d’une inegalite d’obser-vabilite du probleme conservatif associe a (1). Celui-ci est simplement obtenu en prenantcomme operateur de feedback B = 0, i.e.
φ(t) + Aφ(t) = 0, t > 0
φ(0) = ω0, φ(0) = ω1,(4)
et pour donnees initiales les donnees initiales de ω solution de (1). Nous verifions alorsfacilement que l’energie des solutions de ce systeme est constante en temps t ; c’est pourcela que nous parlons de systeme conservatif. Les inegalites d’observabilite sont alors dedeux types
∥
∥
∥A
12ω0
∥
∥
∥
2
H+ ‖ω1‖2
H ≤ C
∫ T
0
‖(B∗φ)′(t)‖2U dt (5)
et
‖ω0‖2
D(A1−m
2 )+ ‖ω1‖2
D(A−m2 )
≤ C
∫ T
0
‖(B∗φ)′(t)‖2U dt, (6)
amenant deux types de decroissance : exponentielle ou polynomiale. Ces inegalites d’obser-vabilite sont prouvees, dans les chapitres 1, 3 (et 5 pour une version discrete) en developpantles solutions du probleme conservatif associe en series de Fourier et en utilisant des criteresspectraux associes a l’operateur A
12 et l’inegalite d’Ingham [57] :
Theoreme 0.0.3 (Inegalite d’Ingham). Soit (λn)n∈Z une suite de nombres reels verifiantla condition suivante dite condition du gap
∃γ > 0, ∀n ∈ Z, λn+1 − λn ≥ γ. (7)
Alors pour tout T > 2π/γ, il existe deux constantes positives C1 et C2 telles que la fonctionf(t) =
∑
n∈Zane
iλnt satisfait l’inegalite d’Ingham suivante :
C1
∑
n∈Z
|an|2 ≤∫ T
0
|f(t)|2 dt ≤ C2
∑
n∈Z
|an|2 .
10
Cependant il arrive dans de tres nombreux cas que la condition du gap (7) des valeurs
propres de l’operateur A12 ne soit pas verifiee, comme par exemple dans le cas de reseaux
de cordes (voir chapitre 1). Nous pouvons alors utiliser l’inegalite d’Ingham generalisee quiest valable si la suite (λn)n satisfait (7) par blocs, c’est-a-dire si nous avons
∃M > 0, ∃γ > 0, ∀n ∈ Z, λn+M − λn ≥Mγ > 0 (8)
(voir [13, 62] et les chapitres 1 et 3 pour plus de details).Une autre maniere d’obtenir les inegalites d’observabilite (5) et (6) est d’utiliser la
formule de D’Alembert (voir [38]).Une fois ces inegalites d’observabilite prouvees, pour obtenir les resultats de stabilisa-
tion, les auteurs de [11] et [6–9] decomposent la solution ω de (1) sous la forme
ω = φ+ ψ, (9)
ou φ est solution du probleme conservatif (4) et ψ est solution de
ψ(t) + Aψ(t) = −BB∗ω(t), t > 0
ψ(0) = 0, ψ(0) = 0,(10)
avec donnees initiales nulles et pour laquelle ils montrent un resultat de regularite.Avec ce resultat de regularite et avec (5) et (2), ils obtiennent l’estimee suivante :
E(0) − E(T ) ≥ CE(T ),
et doncE(T ) ≤ γE(0), avec 0 < γ < 1,
ce qui amene a la decroissance exponentielle de l’energie en appliquant cette estimee surdes intervalles de temps [(m− 1)T, mT ] (ou m ∈ N) successivement. Ceci est possible carle systeme (1) est invariant par translation en temps.
Pour la decroissance polynomiale, une technique similaire est employee en utilisant leLemme 5.2 de [10]. Une fois encore, le fait que le systeme soit invariant par translation entemps est essentiel.
Cette methode a ete utilisee au cours des chapitres 1, 2, 3 et egalement dans une versiondiscrete dans le chapitre 5.
En revanche, nous ne pouvons pas appliquer cette methode au probleme de la stabili-sation de l’equation des ondes (ou de la chaleur) avec un terme de retard dependant dutemps dans les feedbacks, puisque le systeme n’est alors plus invariant par translation entemps. Aussi, dans le chapitre 4, nous avons choisi une autre methode et l’introduction defonctionnelles de Lyapounov appropriees.
Notons que les chapitres de cette these correspondent a des articles qui ont ete publies( [89]) ou soumis ( [87, 88, 90, 106]). Nous avons donc garde la structure generale de cesarticles ; seules les references ont ete regroupees dans une bibliographie commune.
11
Nous allons maintenant presenter les differents chapitres de cette these.
Dans le premier chapitre, nous etudions la stabilisation de l’equation des ondes sur unreseau unidimensionnel de N branches ej avec un terme de retard dans les feedbacks situesaux noeuds. Plus precisement, le deplacement uj le long de la corde ej verifie l’equationdes ondes sur cette corde ej de longueur lj :
∂2uj∂t2
(x, t) − ∂2uj∂x2
(x, t) = 0, 0 < x < lj , t > 0, ∀j ∈ 1, ..., N. (11)
Nous supposons la continuite en tous les noeuds interieurs, i.e. en tous les sommets deVint :
uj(v, t) = ul(v, t) = u(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0, (12)
ou ici Ev est l’ensemble des branches qui relient le sommet v. Nous fixons une partition desnoeuds exterieurs :
Vext = D ∪N ∪ Vcext,ou nous imposons les conditions de Dirichlet aux noeuds de D
ujv(v, t) = 0 ∀v ∈ D, t > 0, (13)
ou jv est le seul element de Ev pour v ∈ Vext, les conditions de Neumann aux noeuds de N∂ujv∂njv
(v, t) = 0 ∀v ∈ N , t > 0,
ou ∂ujv/∂njv(v, .) designe la derivee normale exterieure de uj au sommet v. De plus nousconsiderons un sous-ensemble Vcint de l’ensemble des noeuds interieurs Vint. Pour tous lesnoeuds interieurs qui ne sont pas dissipes la loi de Kirchhoff est verifiee :
∑
j∈Ev
∂uj∂nj
(v, t) = 0 ∀v ∈ Vint\Vcint, t > 0. (14)
Enfin pour tous les noeuds dissipes, c’est-a-dire pour tous les sommets de
Vc = Vcint ∪ Vcext,
nous imposons deux types de dissipation, une sans retard et avec un poids α(v)1 ≥ 0, et une
avec retard τv > 0 et avec un poids α(v)2 ≥ 0 :
∑
j∈Ev
∂uj∂nj
(v, t) = −(
α(v)1
∂u
∂t(v, t) + α
(v)2
∂u
∂t(v, t− τv)
)
∀v ∈ Vc, t > 0. (15)
Notons que α(v)1 , α
(v)2 et τv dependent du noeud v dissipe. Enfin les donnees initiales sont
u(t = 0) = u(0),∂u
∂t(t = 0) = u(1), (16)
12
ainsi que∂u
∂t(v, t− τv) = f 0
v (t− τv) ∀v ∈ Vc, 0 < t < τv,
a cause du retard dans le systeme.En reecrivant ce systeme en un systeme du premier ordre (3) et en appliquant le
theoreme de Lumer-Phillips comme explique dans le debut de l’introduction, nous prou-vons que ce systeme est bien pose (c’est-a-dire qu’il admet une solution unique qui dependcontinument des donnees initiales) sous la condition
α(v)2 ≤ α
(v)1 , ∀v ∈ Vc. (17)
Nous devons cependant restreindre cette hypothese a
α(v)2 < α
(v)1 , ∀v ∈ Vc (18)
pour obtenir la decroissance stricte de l’energie
E(t) :=1
2
N∑
j=1
∫ lj
0
(
(
∂uj∂t
)2
+
(
∂uj∂x
)2)
dx+∑
v∈Vc
ξ(v)
2
(
∫ 1
0
(
∂u
∂t(v, t− τvρ)
)2
dρ
)
,
ou ξ(v) est une constante positive satisfaisant (qui existe par (18))
τvα(v)2 < ξ(v) < τv(2α
(v)1 − α
(v)2 ), ∀v ∈ Vc. (19)
En utilisant le principe d’invariance de LaSalle et sous la condition (18), nous donnons alorsune condition necessaire et suffisante pour que l’energie tende vers zero quand t→ ∞, c’est-a-dire pour avoir la stabilite forte. Cette condition est une condition spectrale associee auprobleme conservatif (4) (c’est-a-dire pour α
(v)2 = α
(v)1 = 0). Par exemple, dans le cas le
plus simple ou, si nous notons (λ2k)k et (ϕk)k les valeurs propres et vecteurs propres du
systeme conservatif associe, le gap simple (7) est verifie et les valeurs propres sont simples,cette condition necessaire et suffisante de stabilite forte s’ecrit
∀k ∈ N,∑
v∈Vc
|ϕk(v)|2 > 0.
Nous avons des conditions similaires dans le cas ou les valeurs propres sont multiples et oule gap generalise (8) est verifie.
Dans le cas ou (18) n’est pas verifie, nous avons prouve sur un exemple que le systemeest instable en exhibant une suite de retards et une suite de points ou il y a dissipationpour lesquelles les solutions du systeme dissipe ont une energie constante.
L’etude de la stabilite exponentielle et polynomiale est basee, comme explique pre-cedemment, sur l’obtention d’inegalites d’observabilite pour le probleme conservatif endeveloppant les solutions en series de Fourier et en utilisant l’inegalite d’Ingham. Nous
13
donnons une condition necessaire et suffisante, de type spectral, pour obtenir l’inegalited’observabilite correspondante a (5), c’est-a-dire
∥
∥u(0)∥
∥
2
V+∥
∥u(1)∥
∥
2
L2(R)≤ C
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt, (20)
ou φ est la solution correspondant a (4) et C, T sont des constantes strictement positives.Dans le cas ou le gap simple (7) est verifie et les valeurs propres sont simples, cette conditionnecessaire et suffisante devient :
∃α > 0, ∀k ∈ N,∑
v∈Vc
|ϕk(v)|2 ≥ α. (21)
De la meme maniere la condition spectrale plus faible
∃m ∈ N, ∃α > 0, ∀k ∈ N,∑
v∈Vc
|ϕk(v)|2 ≥α
k2m(22)
est equivalente a l’inegalite d’observabilite correspondante a (6), c’est-a-dire
∑
k≥1
1
k2m(a2kλ
2k + b2k) ≤ C
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt, (23)
ou u(0) =∑
k≥1 akϕk, u(1) =
∑
k≥1 bkϕk, φ est la solution correspondant a (4), C, T sontdes constantes strictement positives et m ∈ N∗.
Pour obtenir les resultats de stabilite exponentielle et polynomiale il suffit alors dedemontrer un resultat de regularite pour le systeme correspondant a (10) et d’utiliser,comme explique precedemment, la technique de [11]. Ce resultat de regularite pour lesysteme correspondant a (10) est assez technique et utilise notamment des constructionslocales.
Ainsi la decroissance exponentielle de l’energie provient de l’inegalite d’observabilite(20), et donc de (21) dans le cas ou le gap simple (7) est verifie et les valeurs propres sontsimples.
L’inegalite d’observabilite (23) (et donc (22) dans le cas ou le gap simple (7) est verifieet les valeurs propres sont simples) implique la decroissance polynomiale de l’energie enutilisant en plus un lemme d’interpolation.
Tous ces resultats sont illustres d’exemples concrets. Nous remarquons que le type destabilite depend des conditions aux bords et des proprietes d’irrationnalite des rapportsdes longueurs du reseau. Pour des reseaux complexes, comportant par exemple des boucles,nous utilisons un resultat de [20, 81] qui donne explicitement le spectre de l’operateur deLaplace via des proprietes algebriques du reseau dans le cas ou les branches sont de memeslongueurs.
Dans le premier chapitre, les conditions suffisantes (et necessaires) pour obtenir lastabilite sont des conditions spectrales ; elles requierent le calcul explicite du spectre de
14
l’operateur de Laplace sur le reseau, ce qui n’est pas evident pour des reseaux quelconques.De plus, notre analyse est limitee a des inegalites d’observabilite fortes (amenant a desresultats de decroissance exponentielle ou polynomiale), qui ont lieu pour une classe res-treinte de reseau. Dans le chapitre 2, nous repondons a ces limites pour des reseaux ar-bitraires dans lesquels nous pouvons demontrer des inegalites d’observabilite plus faiblesamenant a des taux de decroissance plus lents.
Dans le chapitre 2, nous etudions la stabilisation de l’equation des ondes sur des reseaux1-d en ne dissipant qu’en un seul noeud exterieur (que l’on suppose etre v1) et nousconsiderons le meme systeme que dans le chapitre 1, c’est-a-dire nous considerons (11)-
(16) mais sans retard (i.e. α(v1)2 = 0 et α
(v1)1 = α) et avec
N = ∅, Vc = v1 , Vcint = ∅.
Le but de ce chapitre est de developper une methode systematique pour obtenir des tauxde decroissance sur des reseaux quelconques et de donner un resultat general permettant detransformer un resultat d’observabilite pour un systeme conservatif en stabilisation pour lesysteme dissipe. Contrairement au chapitre precedent, nous ne prenons pas pour le systemeconservatif le systeme correspondant a (4) (i.e. le systeme (11)-(16) avec α
(v1)1 = α = 0 et
α(v1)2 = 0), mais le systeme (11)-(14) avec la condition de Dirichlet au noeud v1 :
ψ1(0, t) = 0 (24)
a la place de la condition de Neumann.L’etude de l’observabilite de ce probleme est motivee par les problemes de controle et
a deja ete realisee ( [33–38]).Le but est donc de montrer le lien entre les resultats d’observabilite existants pour la
solution ψ de (11)-(14) et (24) et la stabilisation du systeme dissipe. Plus precisement,notons (λ2
n)n≥1 la suite des valeurs propres correspondant au probleme (11)-(14) et (24)et (ϕDn )n≥1 les vecteurs propres correspondants a (λ2
n) formant une base orthonormale deL2(R). Sous des conditions sur la topologie du reseau et les longueurs des branches, Dageret Zuazua [33–38] ont prouve des inegalites d’observabilite a poids pour (11)-(14) et (24)de la forme
ED∗ (ψ, 0) :=
∑
n≥1
c2n(λ2nψ
20,n + ψ2
1,n) ≤ C
∫ T
0
∣
∣
∣
∣
∂ψ1
∂x(0, t)
∣
∣
∣
∣
2
dt, (25)
pour une constante positive C, ou ψ0,n, ψ1,n sont les coefficients de Fourier des donneesinitiales de ψ dans la base (ϕDn )n et avec des poids positifs (c2n)n≥1 dependants des proprietesdu reseau.
Cette inegalite d’observabilite a poids est prouvee de differentes manieres : par laformule de D’Alembert (ce qui evite de calculer le spectre) pour les arbres ou par ledeveloppement en series de Fourier des solutions et le theoreme de Beurling-Malliavinpour les reseaux plus complexes, comprenant des boucles par exemple.
La cle de ce chapitre reside dans le fait que l’on peut obtenir, de maniere systematique,l’inegalite d’observabilite a poids pour la solution du probleme dissipe directement par
15
(25). Ceci nous evite de refaire toute l’analyse (en particulier en theorie des nombres),assez subtile, deja effectuee sur les poids c2n dans [38].
Pour cela, nous decomposons la solution du systeme dissipe u comme la somme de w,une solution de (11)-(14) et (24) avec donnee initiale appropriee (u(0) − u
(0)1 (0)ϕ, u(1)) (ou
ϕ est une fonction reguliere donnee telle que ϕ1(0) = 1), et d’un reste. Appliquant (25) aw, nous obtenons l’estimee d’observabilite a poids pour u solution du probleme dissipe
ED∗ (w, 0) + u
(0)1 (0)2 ≤ CT
∫ T
0
(
∂u1
∂t(0, t)
)2
dt, (26)
ou ED∗ (w, 0) est defini par (25), avec des poids (c2n)n dependant du reseau. Si les poids c2n
sont non nuls pour tout n ∈ N∗, l’energie du systeme dissipatif tend vers 0 quand t → ∞.Cependant, en general, les poids tendent vers 0 quand n → ∞, la quantite est plus faibleque la norme dans l’espace d’energie et le taux de decroissance n’est pas exponentiel.
Il est important de souligner que (26) est vrai sous les memes hypotheses sur le reseauque pour (25) pour le probleme de Dirichlet (11)-(14) et (24). Nous n’avons donc pas besoind’hypothese supplementaire sur le reseau.
Pour obtenir les proprietes de decroissance a partir de (26), nous regardons cetteinegalite comme une estimee d’observabilite faible dans laquelle l’energie observee E−(0)
est egale, pour parler rapidemment, a ED∗ (w, 0) + u
(0)1 (0)2. En pratique nous prenons sou-
vent, si necessaire, des points situes sur l’enveloppe convexe de c2n au lieu des poids c2neux-memes dans la definition de ED
∗ . L’energie observee E− est plus faible que la normedans l’espace d’energie des donnees initiales qui est necessaire pour prouver la decroissanceexponentielle, et par consequent nous obtenons des taux de decroissance plus faibles (lo-garithmique par exemple). Pour obtenir des taux de decroissance explicites a partir del’inegalite d’observabilite faible nous utilisons une inegalite d’interpolation qui est une va-riante de celle de Begout et Soria [19] et une generalisation de l’inegalite d’Holder. Pourcela nous avons besoin de supposer plus de regularite des donnees initiales. Pour etre plusprecis nous considerons les donnees initiales (u(0), u(1)) ∈ Xs := [D(A), D(A0)]1−s pour0 < s < 1/2. Nous en deduisons une inegalite d’interpolation de la forme
1 ≤ Φs
(
E−(0)
CEu(0)
)
∥
∥(u(0), u(1))∥
∥
2
Xs
C ′Eu(0),
ou Φs est une fonction croissante qui depend de s et de l’energie E− (et donc des poids c2n).L’inegalite d’interpolation precedente entraıne
E−(0) ≥ CEu(0)Φ−1s
(
Eu(0)
C ′ ‖(u(0), u(1))‖2Xs
)
.
Avec la derivee de l’energie et (26), nous obtenons
Eu(0) −Eu(T ) ≥ CEu(0)Φ−1s
(
Eu(0)
C ′ ‖(u(0), u(1))‖2Xs
)
,
16
ce qui implique, par la technique de Ammari et Tucsnak [11]
∀t > 0, Eu(t) ≤ CΦs
(
1
t+ 1
)
∥
∥(u(0), u(1))∥
∥
2
Xs. (27)
Evidemment, le taux de decroissance dans (27) depend du comportement de la fonctionΦs pres de 0. Donc, dans le but de determiner le taux de decroissance explicite nous avonsbesoin d’une description precise de la fonction Φs, qui depend de s et des energies E etE− et donc, des poids (c2n)n de (26). Ces poids dependent de la topologie du reseau et desproprietes de theorie des nombres des longueurs des cordes.
Cette approche nous permet d’obtenir de maniere systematique des taux de decroissancepour l’energie des solutions regulieres du systeme comme une consequence des proprietesd’observabilite d’un systeme conservatif.
Dans les deux premiers chapitres, nous avons demontre des resultats de stabilite sur desreseaux en utilisant des inegalites d’observabilite pour des systemes conservatifs (differentsdans ces deux chapitres). Le premier chapitre traite en plus de stabilisation de systeme avecretard. En mettant en parallele ce premier chapitre et le travail de Nicaise et Pignotti [85]sur les problemes avec retard pour l’equation des ondes sur des domaines de Rn, n ≥ 1,nous avons remarque que les methodes developpees pour demontrer l’existence de solutionet trouver les taux de decroissance presentent des similarites. Cette observation nous aamenes a considerer un systeme abstrait et general qui contient une grande classe deproblemes avec des feedbacks avec retard, permettant de retrouver ces resultats et d’enobtenir d’autres, dans le meme esprit que l’article de Ammari et Tucsnak [11].
Dans le troisieme chapitre, nous considerons ainsi le systeme (1) mais avec un retarddans le feedback, c’est-a-dire
ω(t) + Aω(t) +B1B∗1 ω(t) +B2B
∗2 ω(t− τ) = 0, t > 0
ω(0) = ω0, ω(0) = ω1,B∗
2 ω(t− τ) = f 0(t− τ), 0 < t < τ,(28)
ou τ > 0 est le retard, A : D(A) → H est un operateur positif, auto-adjoint avec inversecompact dans un espace de Hilbert H , U1, U2 sont des espaces de Hilbert (identifies a leurdual) et Bi ∈ L(U, D(A1/2)′), i = 1, 2. Notons V = D(A1/2).
La premiere question naturelle a se poser est l’existence et l’unicite des solutions. Pourcela, nous reecrivons ce systeme comme un systeme du premier ordre (3) et nous appli-quons le theoreme de Lumer-Phillips (comme explique dans le debut de l’introduction), ensupposant la condition suivante
∃0 < α ≤ 1, ∀u ∈ V, ‖B∗2u‖2
U2≤ α ‖B∗
1u‖2U1. (29)
Cette condition est coherente avec (17) en prenant B∗i ϕ = (
√
α(v)i ϕ(v))v∈Vc pour ϕ ∈
D(A1/2). Comme dans le premier chapitre, nous devons restreindre notre hypothese a
∃0 < α < 1, ∀u ∈ V, ‖B∗2u‖2
U2≤ α ‖B∗
1u‖2U1, (30)
17
pour obtenir la decroissance stricte de l’energie
E(t) :=1
2
(
∥
∥
∥A
12ω∥
∥
∥
2
H+ ‖ω‖2
H + τξ
∫ 1
0
‖B∗2 ω(t− τρ)‖2
U2dρ
)
, (31)
ou ξ est une constante positive satisfaisant
1 < ξ <2
α− 1, (32)
qui existe puisque 0 < α < 1. De plus, cette hypothese semble realiste car, sans cettehypothese, il existe des cas ou des instabilites peuvent apparaitre, comme nous l’avons vudans le chapitre 1 (voir aussi [85, 110]).
Sous cette condition et en utilisant un resultat de [12] (voir aussi [105]) nous obtenonsune condition necessaire et suffisante pour la stabilite forte du systeme, qui est
Pour tout vecteur propre non nul ϕ ∈ D(A) de A, B∗1ϕ 6= 0.
Remarquons que cette derniere condition est independante du retard et par consequent sousla condition (30), notre systeme est fortement stable si et seulement si le meme systemesans retard est fortement stable.
Dans la troisieme etape, sous la condition (30) et une hypothese de borne venant de [11]entre la resolvante de A et les operateurs B1 et B2 :
si β > 0 est fixe et Cβ = λ ∈ C |ℜλ = β , la fonction
λ ∈ Cβ → H(λ) = λB∗(λ2I + A)−1B ∈ L(U) est bornee, (33)
ou B = (B1, B2) ∈ L(U, V ′) avec U = U1 × U2, nous prouvons que la decroissanceexponentielle du systeme (28) provient de l’estimee d’observabilite (5) pour le problemeconservatif (4), en utilisant la decomposition (9) et la methode de [11] comme explique dansle debut de l’introduction. Une fois de plus, cette estimee d’observabilite est independantedu terme avec retard B2B
∗2ω(t − τ) et par consequent, sous les conditions (30) et (33), la
decroissance exponentielle de (1) implique la decroissance exponentielle de (28). Malgretout nous donnons la dependance de la decroissance par rapport au retard, en particuliernous montrons que si le retard augmente le taux de decroissance diminue.
Une analyse similaire pour la decroissance polynomiale est effectuee en prenant l’estimeed’observabilite plus faible (6) et en utilisant le lemme technique 5.2 de [11]. Une fois deplus, nous montrons que si le retard augmente le taux de decroissance diminue.
Pour pouvoir appliquer plus facilement ces resultats, nous donnons la preuve de cesestimees d’observabilite (5) et (6) en ecrivant les solutions en series de Fourier, en utilisantl’inegalite d’Ingham (classique ou generalisee) et une reduction a des conditions entre lesvecteurs propres de A et l’operateur de feedback B∗
1 . La condition necessaire et suffisantepour obtenir (5), dans le cas ou le gap simple (7) est verifie et les valeurs propres sontsimples, est exactement (21) dans le cas des reseaux (chapitre 1).
Nous terminons ce chapitre en illustrant ces resultats d’exemples dans lesquels notrecadre abstrait s’applique. A notre connaissance, ces exemples, a l’exception du premier qui
18
reprend le chapitre 1, sont nouveaux, puisque en particulier notre cadre abstrait permet deconsiderer des operateurs de feedback B1 et B2 differents, a condition qu’ils verifient (30).Nous etudions par exemple l’equation des poutres d’Euler-Bernoulli avec des termes dedamping internes et localises en un point, l’equation des ondes avec des termes de dampingrepartis (1-d et multi-d) ou encore l’equation des ondes sur un reseau avec des termes dedamping repartis.
Dans le chapitre 4, nous etudions la stabilisation des equations de la chaleur et desondes avec un terme de retard qui depend du temps (et non plus constant comme au-paravant). Les systemes n’etant plus invariants par translation en temps, les techniquesutilisees jusqu’alors (c’est-a-dire se ramener a une inegalite d’observabilite) ne sont plusapplicables. Nous allons donc utiliser une methode differente en introduisant des fonction-nelles de Lyapounov appropriees. Plus precisement nous considerons l’equation des ondessuivante
utt(x, t) − auxx(x, t) = 0, 0 < x < π, t > 0,u(0, t) = 0, t > 0,
ux(π, t) = −α1ut(π, t) − α2ut(π, t− τ(t)), t > 0,u(x, 0) = u0(x), ut(x, 0) = u1(x), 0 < x < π,ut(π, t− τ(0)) = f 0(t− τ(0)), 0 < t < τ(0),
(34)
avec un parametre constant a > 0 et ou α1, α2 sont des nombres reels positifs, ou le retardτ(t) est une fonction du temps qui satisfait
∀t > 0, τ (t) ≤ d < 1, (35)
∃M > 0, ∀t > 0, 0 < τ0 ≤ τ(t) ≤M, (36)
et∀T > 0, τ ∈W 2,∞([0, T ]). (37)
Comme pour les chapitres precedents, nous nous interessons tout d’abord a l’existenceet l’unicite des solutions de ce systeme et pour cela nous reecrivons notre systeme en unsysteme du premier ordre (3) et nous supposons que α1 et α2 verifient
α22 ≤ (1 − d)α2
1. (38)
Cependant a cause du retard dependant du temps, nous n’utiliserons pas le theoreme deLumer-Phillips car l’operateur A de (3) depend du temps. Mais comme le domaine del’operateur A(t) est constant en temps, nous employons la technique de normes variablesde Kato [59, 60] en prenant pour produit scalaire de H = φ ∈ H1(0, π) : φ(0) = 0 ×L2(0, π) × L2(0, 1) un produit scalaire qui depend du temps
⟨
uωz
,
uωz
⟩
t
=
∫ π
0
(auxux + ωω)dx+ qτ(t)
∫ 1
0
z(ρ)z(ρ)dρ,
19
ou q est une constante positive choisie telle que
Ψq =1
2
(
q − 2aα1 −aµ1α2
−aα2 −q(1 − d)
)
soit negative (ce qui est garanti par (35) et (38)), et avec norme associee ‖.‖t .Comme pour les chapitres 1 et 3, nous restreignons l’hypothese (38) en
α22 < (1 − d)α2
1. (39)
pour obtenir la decroissance stricte de l’energie E(t), qui correspond a
E(t) =1
2
∥
∥
∥(u, ut, ut(π, t− τ(t)ρ))T
∥
∥
∥
2
t, (40)
ou q est une constante positive choisie telle que Ψq soit definie negative (ce qui est garantipar (35) et (39)). Nous remarquons que dans le cas ou le retard est constant en temps, i.e.τ(t) = τ pour tout t > 0 (et donc d = 0), nous retrouvons les resultats des chapitres 1 et 3et de [85] pour la meme energie. En effet, l’energie est alors strictement decroissante sousla condition (18), ce qui correspond a (39) pour d = 0.
Sous les hypotheses (35), (36) et (39), nous prouvons la stabilite exponentielle del’equation des ondes (34) en utilisant la fonctionnelle de Lyapounov suivante
E(t) = E(t) + γ
(
2
∫ π
0
xutuxdx+ E2(t)
)
, (41)
ou γ > 0 est un parametre fixe suffisamment petit, E est l’energie standard definie par (40)avec q une constante positive fixee telle que Ψq soit definie negative et ou E2 est definie par
E2(t) = q
∫ t
t−τ(t)e2δ(s−t)u2
t (π, s)ds = qτ(t)
∫ 1
0
e−2δτ(t)ρu2t (π, t− τ(t)ρ)dρ, (42)
avec δ > 0.La fonctionnelle de Lyapounov E(t) + 2γ
∫ π
0xutuxdx est standard dans les problemes
avec conditions au bord avec memoire (voir par exemple [86]). Nous avons ajoute deuxtermes a l’energie standard E(t) pour prendre en compte la dependance de τ par rapporta t. De plus nous remarquons que les energies E et E sont equivalentes.
De cette maniere, sous les conditions (35), (36) et (39), nous prouvons que ce systemeest exponentiellement stable et nous donnons egalement le taux exact de decroissance defacon explicite. Il depend du retard τ(t) (et plus particulierement de M et d), de α1, α2,de a et de δ. Nous remarquons que nous pouvons choisir δ tel que la decroissance soit aussirapide que possible pour une fonction τ fixee. Cependant notre taux de decroissance vadiminuer si le maximum M de τ augmente, ce qui est coherent avec l’etude du chapitre 3.
Nous effectuons egalement dans ce chapitre 4 la meme etude pour l’equation de lachaleur.
20
Dans le cinquieme et dernier chapitre nous nous interessons a la stabilisation de l’equationdes ondes discretisee en espace par differences finies avec un amortissement en un pointinterieur. Le principal probleme pour l’etude de la stabilisation des approximations nume-riques d’un systeme est que le taux de decroissance n’est pas uniforme par rapport au pasde discretisation. Il faut donc utiliser une methode pour redresser le spectre discret pourles hautes frequences.
Plus precisement, nous considerons l’equation des ondes sur un intervalle de longueur1 avec un amortissement en ξ ∈ (0, 1)
ytt(x, t) − yxx(x, t) = 0 0 < x < 1, t > 0,y(0, t) = 0, yx(1, t) = 0 t > 0,y(ξ−, t) = y(ξ+, t) t > 0,yx(ξ−, t) − yx(ξ+, t) = −αyt(ξ, t) t > 0,y(t = 0) = y(0), yt(t = 0) = y(1) 0 < x < 1,
(43)
ou (y(0), y(1)) ∈ V × L2(0, 1), V = y ∈ H1(0, 1) ; y(0) = 0 et α est une constantepositive. Ce systeme (43) est bien pose dans l’espace d’energie V × L2(0, 1) (voir [6] ou ledebut de l’introduction).
L’energie de la solution du systeme (43) est donnee par
E(t) =1
2
∫ 1
0
(|yt(x, t)|2 + |yx(x, t)|2)dx
et verifie la loi de dissipation suivante
dE(t)
dt= −α |yt(ξ, t)|2 . (44)
De plus, nous savons que (voir [6]) limt→∞
E(t) = 0 pour toute donnee initiale dans V ×L2(0, 1)
si et seulement si
ξ 6= 2p
2q + 1, ∀p, q ∈ N (45)
et, comme nous l’avons signale precedemment, la decroissance exponentielle de la solutionde (43) est equivalente a une estimee d’observabilite pour le systeme conservatif associe(4). Dans ce cas, l’estimee d’observabilite est verifiee si et seulement si ξ est un nombrerationnel qui admet une decomposition en fraction irreductible de la forme
ξ =p
q, ou p est impair,
et par consequent sous cette condition, le systeme (43) est exponentiellement stable dansl’espace d’energie. Nous supposons donc que ξ est fixe et verifie ces conditions.
Dans ce chapitre nous nous interessons a la discretisation en espace par differences finiesde (43). Soit N ∈ N∗, h = 1
N+1et considerons la subdivision de (0, 1) donnee par
0 = x0 < x1 = h < ... < xj−1 < xj = jh < xj+1 < ... < xN < xN+1 = 1,
21
i.e. xj = jh pour tout j = 0, ..., N + 1. Comme ξ n’est pas necessairement egal a jhpour tout j, nous fixons jN ∈ N ∩ (0, N + 1) tel que xjN → ξ quand N → ∞. La semi-discretisation en espace de (43) est la suivante : pour l’equation des ondes, nous obtenons
y′′j −yj+1 − 2yj + yj−1
h2= 0, t > 0, j = 1, ..., N, j 6= jN , (46)
les conditions au bord (Dirichlet et Neumann) deviennent
y0 = 0, yN+1 − yN = 0, t > 0, (47)
l’approximation naturelle de la condition de transmission est
yjN+1 − 2yjN + yjN−1
h= αy′jN , t > 0, (48)
et les donnees initiales du probleme discretise sont
yj(t = 0) = y(0)j , y′j(t = 0) = y
(1)j , j = 1, ..., N. (49)
Ici yj(t) est une approximation de y(xj, t), y etant la solution de (43), a condition que
les conditions initiales (y(0)j , y
(1)j ), j = 0, ..., N + 1 soient des approximations des donnees
initiales dans (43). Notons yh = (yj)j, y(0)h = (y
(0)j )j et y
(1)h = (y
(1)j )j . La stabilisation du
meme type de probleme mais avec un amortissement au bord (en x = 1) a ete effectueedans [103] (ou avec un amortissement interne localise dans [102]) en ajoutant une viscositenumerique.
Nous introduisons l’energie de ce probleme discretise par
Eh(t) =h
2
N∑
j=0, j 6=jN
∣
∣y′j(t)∣
∣
2+h
2
N∑
j=0
∣
∣
∣
∣
yj+1(t) − yj(t)
h
∣
∣
∣
∣
2
, (50)
qui est une discretisation de l’energie continue E, qui est decroissante et qui verifie
E ′h(t) = −α(y′jN (t))2.
En utilisant le principe d’invariance de LaSalle, nous montrons que l’energie discretisee Ehtend vers 0 quand t→ ∞ si et seulement si
jNh 6= (2 − h)l
2k + 1, ∀k = 0, ..., N − 1, l ∈ N.
Nous remarquons que cette condition est une version discrete de (45) pour le modelecontinu.
Ensuite, nous montrons, comme pour ce type de probleme (voir [14, 96, 102, 103] parexemple), que la decroissance exponentielle de l’energie discrete de (46)-(49) n’est pasuniforme par rapport au pas de discretisation, en prenant jN = N p
q, ou p est impair et N
22
est un multiple de q. Ceci est du a l’existence des modes etrangers que le schema numeriqueintroduit a haute frequence et qui n’apparaissent pas dans le modele continu.
Pour surmonter cet obstacle, nous filtrons les hautes frequences et en consequence nousintroduisons la classe Ch(γ) des solutions du probleme discretise generees par les vecteurspropres du probleme conservatif discretise
u′′j − uj+1−2uj+uj−1
h2 = 0 t > 0, j = 1, ..., Nu0 = 0, uN+1 − uN = 0 t > 0
uj(t = 0) = y(0)j , u′j(t = 0) = y
(1)j j = 1, ..., N,
(51)
associes aux valeurs propres telles que λh2 ≤ γ. Ainsi la classe Ch(γ) est definie par
Ch(γ) :=
∑
λk, h≤ γ
h2
akϕk, h avec ak ∈ R
.
Nous supposons dans la suite que jN =[
p(2N+1)2q
]
∈ N et donc xjN → ξ = pq. Nous pouvons
alors prouver une inegalite d’observabilite discrete uniforme (en h) pour les solutions uhde (51) dans la classe Ch(γ)
(T − 2)EjN (uh, 0) ≤ C
∫ T
0
∣
∣u′jN (t)∣
∣
2dt,
pour un certain γ, puisqu’en filtrant nous pouvons utiliser l’inegalite d’Ingham. En effet,le gap (7) des valeurs propres entrant dans le developpement de Fourier des solutions de(51) dans la classe Ch(γ) est verifie.
Grace a cette inegalite d’observabilite uniforme, nous pouvons prouver la decroissancequasi-exponentielle de l’energie Eh de (46)-(49), en effectuant la decomposition yh = uh+whet en prouvant certaines estimees par la technique des multiplicateurs. De plus, sans filtrer,la decroissance quasi-exponentielle n’est pas uniforme par rapport a h, ce qui montrel’avantage de cette technique. Nous parlons de decroissance quasi-exponentielle car nouspouvons majorer Eh(t) par Ke−ωtEh(yh, 0) (qui correspond a la decroissance exponentielle)plus un terme residuel qui tend vers 0 quand h → 0, en prenant des donnees initialessuffisamment regulieres. Par consequent cette estimee tend vers la stabilite exponentielleE(y, t) ≤ Ke−ωtE(y, 0) quand h → 0 ; c’est dans ce sens que cette estimee est quasi-optimale. De plus, le membre de droite de notre estimee de decroissance quasi-exponentiellese comporte comme une fonction exponentielle decroissante pour t entre 0 et − lnh
ω+ c et
est constante (proportionnellement a h) pour t suffisamment grand.Afin d’illustrer ces resultats, nous avons effectue des tests numeriques qui montrent
que, sans filtrer nous n’avons pas de decroissance exponentielle (nous prenons un “grand”vecteur propre) mais que filtrer (en prenant un “petit” vecteur propre) permet de retablirune decroissance exponentielle.
23
24
Chapitre 1
Stabilization of the wave equation on1-d networks with a delay term inthe nodal feedbacks
1.1 Introduction/Notations
Time delay effects arise in many practical problems, see for instance [1, 51, 101] forbiological, electrical engineering, or mechanical applications. Furthermore it is well knownthat they can induce some instabilities [39, 40, 42, 85, 110], or on the contrary improve theperformance of the system [1, 101].
Recently, control problems on 1-d networks are paying attention of many authors,see [38, 65] and the references cited there. We here investigate the effect of time delay inboundary and/or transmission stabilization of the wave equation in 1−d networks. To ourknowledge, the analysis of this effect to 1 − d networks is not yet done.
Before going on, let us recall some definitions and notation about 1 − d networks usedin the whole chapter. We refer to [2, 3, 20–23,82, 91] for more details.
Definition 1.1.1. A 1 − d network R is a connected set of Rn, n ≥ 1 defined by
R =N⋃
j=1
ej
where ej is a curve that we identify with the interval (0, lj), lj > 0, and such that for k 6= j,ej ∩ ek is either empty or a common extremity called a vertex or a node (here ej means theclosure of ej).
For a function u : R −→ R, we set uj = u|ejthe restriction of u to the edge ej .
We denote by E = ej ; 1 ≤ j ≤ N the set of edges of R and by V the set of verticesof R. For a fixed vertex v, let
Ev = j ∈ 1, ..., N ; v ∈ ej
25
be the set of edges having v as vertex. If card (Ev) = 1, v is an exterior node, while if card(Ev) ≥ 2, v is an interior node. We set Vext the set of exterior nodes and Vint the set ofinterior nodes. For v ∈ Vext, the single element of Ev is denoted by jv.
We now fix a partition of Vext :
Vext = D ∪N ∪ Vcext.
Clearly we will impose Dirichlet boundary condition at the nodes of D, Neumann boundarycondition at the nodes of N and finally a feedback boundary condition at the nodes of Vcext.We further fix a subset Vcint of Vint, where a feedback transmission condition will be imposed.For shortness, we denote by Vc the set of controlled nodes, namely
Vc = Vcint ∪ Vcext.
We also suppose that D 6= ∅ ; so that the H1 semi-norm becomes a norm.We can now formulate our initial/boundary value problem :
∂2uj
∂t2(x, t) − ∂2uj
∂x2 (x, t) = 0 0 < x < lj, t > 0,∀j ∈ 1, ..., N,
uj(v, t) = ul(v, t) = u(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂uj
∂nj(v, t) = −(α
(v)1
∂u∂t
(v, t) + α(v)2
∂u∂t
(v, t− τv)) ∀v ∈ Vc, t > 0,
∑
j∈Ev
∂uj
∂nj(v, t) = 0 ∀v ∈ Vint\Vcint, t > 0,
ujv(v, t) = 0 ∀v ∈ D, t > 0,∂ujv
∂njv(v, t) = 0 ∀v ∈ N , t > 0,
u(t = 0) = u(0), ∂u∂t
(t = 0) = u(1),∂u∂t
(v, t− τv) = f 0v (t− τv) ∀v ∈ Vc, 0 < t < τv,
(1.1)
where α(v)1 , α
(v)2 ≥ 0 are fixed nonnegative real numbers, the delay τv > 0 is also supposed
to be fixed and∂uj
∂nj(v, ·) means the outward normal (space) derivative of uj at the vertex
v.Note that uj represents the displacement of the string ej .Remark that the condition ∂u
∂t(v, t− τv) = f 0
v (t− τv) for v ∈ Vc, 0 < t < τv denotes aninitial value in the past, but is necessary due to the delay equation.
In the absence of delay, i.e. α(v)2 = 0 for all v ∈ Vc, the above problem has been
considered by some authors in some particular situations, for instance Ammari and Tucsnak[11], Ammari, Henrot and Tucsnak [6], Ammari and Jellouli [7, 8], Ammari, Jellouli andKhenissi [9] and Xu, Liu and Liu [109]. In these papers, some sufficient conditions are
given in order to guarantee some stabilities of the system. On the contrary, if α(v)1 = 0
that is if we have only the delay part in the boundary/transmission condition, system (1.1)may become unstable. See, for instance Datko, Lagnese and Polis [42] for the example ofa string. Therefore it is interesting to seek for stabilization results in general 1-d networks
26
when the parameters α(v)1 and α
(v)2 are both nonzero. In the special case of one string and
a feedback law at one extremity, this problem has been studied by Xu, Yung and Li [110],where the authors use a spectral analysis. For the wave equation in higher dimensionalspace domain, we refer to [85].
In accordance with [85, 110], assuming that
α(v)2 ≤ α
(v)1 , ∀v ∈ Vc,
we show the decay of an appropriate energy. We further give a necessary and sufficientcondition for the decay to zero of the energy. If the above condition does not hold, weconjecture that the energy does not decay. We do not investigate this problem in its fullgenerality but study it in a particular case.
Now ifα
(v)2 < α
(v)1 , ∀v ∈ Vc,
we first give a sufficient condition for the exponential decay of the energy. We secondlyfind a sufficient condition for the polynomial decay of the energy.
Our method is based on the use of observability estimates of the problem withoutdamping. Here we have chosen to obtain these observability estimates by a frequencydomain method. The use of other techniques like the D’Alembert representation formula[7,38] may avoid the use of the frequency domain method but give quite often non optimaldecay rates for the energy. Note finally that the observability estimate is independent ofthe delay term.
The chapter is organized as follows. After the recall of some definitions and notation,we show in the second section that our problem is well posed. Then in section 1.3, weprove the decay of an appropriate energy and give a necessary and sufficient conditionwhich guarantees the decay to 0 of the energy. Section 1.4 is devoted to the proof of aregularity result and an a priori estimate used for the stability results. In section 1.5 wegive a sufficient condition for the exponential stability of our system. Similarly section 1.6is concerned with a sufficient condition for the polynomial stability of our system. Finallywe end up with some illustrative examples in section 1.7.
In the whole chapter the notation a . b means that there exists a positive constant Cindependent of a and b such that a ≤ C b. The notation a ∼ b means that a . b and b . ahold simultaneously.
1.2 Well posedness of the problem
We aim to show that problem (1.1) is well-posed. For that purpose, we use semi-grouptheory and an idea from [85].
For future uses, we introduce the spatial operator associated with the system similarto (1.1) but without damping. Introduce
L2(R) = u : R → R; uj ∈ L2(0, lj), ∀j = 1, · · · , N,
27
which is a Hilbert space for the natural inner product. Its associated norm will be denotedby ‖ · ‖L2(R). Let further V be the Hilbert space
V := φ ∈N∏
j=1
H1(0, lj) : φj(v) = φk(v) ∀j, k ∈ Ev, ∀v ∈ Vint ; φjv(v) = 0 ∀v ∈ D,
equipped with the inner product
< φ, φ >V =
N∑
j=1
∫ lj
0
∂φj∂x
∂φj∂x
dx.
For shortness for u ∈ L1(R) = u : R → R; uj ∈ L1(0, lj), ∀j = 1, · · · , N, we oftenwrite
∫
Ru =
N∑
j=1
∫ lj
0
uj(x) dx.
Now we introduce the operator A from L2(R) into itself by
D(A) := u ∈ V ∩N∏
j=1
H2(0, lj) :∑
j∈Ev
∂uj∂nj
(v) = 0, ∀v ∈ Vint ;
∂ujv∂njv
(v) = 0, ∀v ∈ N ∪ Vcext,
(Au)j = −∂2uj∂x2
∀j = 1, · · · , N, ∀u ∈ D(A).
This operator is a positive selfadjoint operator since it is the Friedrichs extension of thetriple (L2(R), V, a), where the bilinear form a is defined by
a(u, v) =N∑
j=1
∫ lj
0
∂uj∂x
∂vj∂x
dx, ∀u, v ∈ V.
Let further set X = V ∩N∏
j=1
H2(0, lj), which is a Hilbert space with the inner product
(u, v)X = (u, v)L2(R) + (∆u,∆v)L2(R), ∀u, v ∈ X,
where we have set
(∆u)j =∂2uj∂x2
∀j = 1, · · · , N, u ∈ X.
Now we come back to our system (1.1) and transform it as follows. For all v ∈ Vc letus introduce the auxiliary variable zv(ρ, t) = ∂u
∂t(v, t− τvρ) for ρ ∈ (0, 1) and t > 0. In this
28
manner, we eliminate the delay term in (1.1) and problem (1.1) is equivalent to
∂2uj
∂t2(x, t) − ∂2uj
∂x2 (x, t) = 0 0 < x < lj , t > 0, ∀j ∈ 1, ..., N,τv
∂zv
∂t(ρ, t) + ∂zv
∂ρ(ρ, t) = 0 0 < ρ < 1, t > 0, ∀v ∈ Vc,
uj(v, t) = ul(v, t) = u(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂uj
∂nj(v, t) = −(α
(v)1
∂u∂t
(v, t) + α(v)2 zv(1, t)) ∀v ∈ Vc, t > 0,
∑
j∈Ev
∂uj
∂nj(v, t) = 0 ∀v ∈ Vint\Vcint, t > 0,
ujv(v, t) = 0 ∀v ∈ D, t > 0,∂ujv
∂njv(v, t) = 0 ∀v ∈ N , t > 0,
zv(0, t) = ∂u∂t
(v, t) ∀v ∈ Vc, t > 0,u(t = 0) = u(0), ∂u
∂t(t = 0) = u(1),
zv(ρ, 0) = f 0v (−τvρ) ∀v ∈ Vc, 0 < ρ < 1.
(1.2)
Note that zv satisfies a transport equation in the t, ρ variables, with an initial datum att = 0 and ρ = 0.
If we introduce z = (zv)v∈Vc and
U :=
(
u,∂u
∂t, z
)⊤,
then U satisfies
U ′ =
(
∂u
∂t,∂2u
∂t2,∂z
∂t
)⊤=
(
∂u
∂t, ∆u, −
(
1
τv
∂zv∂ρ
)
v∈Vc
)⊤.
Consequently the problem (1.2) may be rewritten as the first order evolution equation
U ′ = AU,U(0) = (u(0), u(1), (f 0(−τv.))v)⊤ = U0,
(1.3)
where the operator A is defined by
A
uwz
:=
w∆u
−( 1τv∂zv
∂ρ)v
with domain
D(A) := (u, w, z) ∈ (V ∩N∏
j=1
H2(0, lj)) × V ×H1(0, 1)Vc :
∑
j∈Ev
∂uj∂nj
(v) = − (α(v)1 w(v) + α
(v)2 zv(1)) ∀v ∈ Vc ;
∑
j∈Ev
∂uj
∂nj(v) = 0 ∀v ∈ Vint\Vcint ;
∂ujv
∂njv(v) = 0 ∀v ∈ N ; zv(0) = w(v) ∀v ∈ Vc,
29
where Vc is the number of nodes of Vc.Now introduce the Hilbert space
H := V × L2(R) × L2(0, 1)Vc ,
equipped with the usual inner product
⟨
uwz
,
uwz
⟩
=N∑
j=1
∫ lj
0
(
∂uj∂x
∂uj∂x
+ wjwj
)
dx+∑
v∈Vc
∫ 1
0
zv(ρ)zv(ρ)dρ.
Lemma 1.2.1. D(A) is dense in H.
Proof: Let (f, g, h)⊤ ∈ H be orthogonal to all elements of D(A), namely
0 =
⟨
uwz
,
fgh
⟩
=N∑
j=1
∫ lj
0
(∂uj∂x
∂fj∂x
+ wjgj)dx+∑
v∈Vc
∫ 1
0
zv(ρ)hv(ρ)dρ,
for all (u, w, z)⊤ ∈ D(A).We first take u = 0 and w = 0 and z ∈ D(0, 1)Vc . As (0, 0, z) ∈ D(A), we get
∑
v∈Vc
∫ 1
0
zv(ρ)hv(ρ)dρ = 0.
Since D(0, 1) is dense in L2(0, 1), we deduce that h = 0.
In the same manner since
N∏
j=1
D(0, lj) is dense in
N∏
j=1
L2(0, lj), by taking u = 0, z = 0
and w ∈N∏
j=1
D(0, lj) we see that g = 0.
The above orthogonality condition is then reduced to
0 =N∑
j=1
∫ lj
0
∂uj∂x
∂fj∂x
dx, ∀(u, w, z) ∈ D(A).
By restricting ourselves to w = 0 and z = 0, we obtain
N∑
j=1
∫ lj
0
∂uj∂x
∂fj∂x
dx = 0, ∀(u, 0, 0) ∈ D(A).
But we easily check that (u, 0, 0) ∈ D(A) if and only if u ∈ D(A). Since it is well knownthat D(A) is dense in V (equipped with the inner product < ., . >V ), we conclude thatf = 0.
30
Let us now suppose thatα
(v)2 ≤ α
(v)1 , ∀v ∈ Vc. (1.4)
Under this condition, we will show that the operator A generates a C0-semi-group in H .For that purpose, we choose positive real numbers ξv such that
τvα(v)2 ≤ ξ(v) ≤ τv(2α
(v)1 − α
(v)2 ), ∀v ∈ Vc. (1.5)
These constants exist due to the condition (1.4).We now introduce the following inner product on H
⟨
uwz
,
uwz
⟩
H
=N∑
j=1
∫ lj
0
(
∂uj∂x
∂uj∂x
+ wjwj
)
dx+∑
v∈Vc
ξ(v)
(∫ 1
0
zv(ρ)zv(ρ)dρ
)
.
This inner product is clearly equivalent to the usual inner product of H .
Theorem 1.2.2. For an initial datum U0 ∈ H, there exists a unique solution U ∈C([0, +∞), H) to problem (1.3). Moreover, if U0 ∈ D(A), then
U ∈ C([0, +∞), D(A)) ∩ C1([0, +∞), H).
Proof: By Lumer-Phillips’ theorem, it suffices to show that A is dissipative and maximalmonotone.
We first prove that A is dissipative. Take U = (u, w, z)⊤ ∈ D(A). Then
(AU, U) =
⟨
w∆u
−( 1τv∂zv
∂ρ)v
,
uwz
⟩
H
=N∑
j=1
∫ lj
0
(
∂wj∂x
∂uj∂x
+∂2uj∂x2
wj
)
dx+∑
v∈Vc
ξ(v)
(∫ 1
0
− 1
τv
∂zv∂ρ
(ρ)zv(ρ)dρ
)
.
By integrating by parts, we obtain
(AU, U) =N∑
j=1
∫ lj
0
(−wj∂2uj∂x2
+∂2uj∂x2
wj)dx+N∑
j=1
[wj∂uj∂x
]lj0 −
∑
v∈Vc
ξ(v)
τv
(∫ 1
0
∂zv∂ρ
(ρ)zv(ρ)dρ
)
=
N∑
j=1
[wj∂uj∂x
]lj0 −
∑
v∈Vc
ξ(v)
τv
(∫ 1
0
∂zv∂ρ
(ρ)zv(ρ)dρ
)
.
Again an integration by parts leads to∫ 1
0
∂zv∂ρ
(ρ)zv(ρ)dρ =1
2(z2v(1) − z2
v(0)).
31
Moreover by the boundary/transmission conditions satisfied by (u, w, z)⊤ ∈ D(A), wehave
N∑
j=1
[wj∂uj∂x
]lj0 =
∑
v∈V
∑
j∈Ev
wj(v)∂uj∂nj
(v)
=∑
v∈Vc
∑
j∈Ev
wj(v)∂uj
∂nj(v) +
∑
v∈Dwjv(v)
∂ujv
∂njv(v) +
∑
v∈Nwjv(v)
∂ujv
∂njv(v)
+∑
v∈Vint\Vcint
∑
j∈Ev
wj(v)∂uj
∂nj(v)
=∑
v∈Vc
(∑
j∈Ev
∂uj
∂nj(v))wj(v) +
∑
v∈Vint\Vcint
(∑
j∈Ev
∂uj
∂nj(v))wj(v)
=∑
v∈Vc
−(α(v)1 w(v) + α
(v)2 zv(1))zv(0)
= −∑
v∈Vc
(α(v)1 zv(0)2 + α
(v)2 zv(1)zv(0)).
These properties yield
(AU, U) = −∑
v∈Vc
(
α(v)1 zv(0)2 + α
(v)2 zv(1)zv(0)
)
−∑
v∈Vc
ξ(v)
2τv(z2v(1) − z2
v(0))
= −∑
v∈Vc
(
(α(v)1 − ξ(v)
2τv)zv(0)2 + ξ(v)
2τvz2v(1) + α
(v)2 zv(1)zv(0)
)
.
By Cauchy-Schwarz’s inequality we have
−α(v)2 zv(1)zv(0) ≤ α
(v)2
2z2v(1) +
α(v)2
2z2v(0)
and therefore
(AU, U) ≤ −∑
v∈Vc
(
(α(v)1 − ξ(v)
2τv− α
(v)2
2)zv(0)2 + (
ξ(v)
2τv− α
(v)2
2)z2v(1)
)
with α(v)1 − ξ(v)
2τv− α
(v)2
2≥ 0 and ξ(v)
2τv− α
(v)2
2≥ 0 because α
(v)1 and α
(v)2 satisfy condition (1.5).
This shows that (AU, U) ≤ 0 and then the dissipativeness of A.Let us now prove that A is maximal monotone, i.e. that λI −A is surjective for some
λ > 0.Let (f, g, h)⊤ ∈ H . We look for U = (u, w, z)⊤ ∈ D(A) solution of
(λI −A)
uwz
=
fgh
(1.6)
32
or equivalently
λuj − wj = fj ∀j ∈ 1, ..., N,λwj − ∂2uj
∂x2 = gj ∀j ∈ 1, ..., N,λzv + 1
τv∂zv
∂ρ= hv ∀v ∈ Vc.
(1.7)
Suppose that we have found u with the appropriate regularity. Then for all j ∈1, ..., N, we have
wj := λuj − fj ∈ H1(0, lj) (1.8)
with wjv(v) = λujv(v) − fjv(v) = 0 for v ∈ D.We can then determine z since w(v) = zv(0). Indeed, for v ∈ Vc, zv satisfies the
differential equation
λzv +1
τv
∂zv∂ρ
= hv
and the boundary condition
zv(0) = w(v) = λu(v) − f(v).
Therefore zv is explicitly given by
zv(ρ) = λu(v)e−λτvρ − f(v)e−λτvρ + τve−λτvρ
∫ ρ
0
eλτvσhv(σ)dσ.
This means that once u is found with the appropriate properties, we can find z and w.Note that in particular we have
zv(1) = λu(v)e−λτv − f(v)e−λτv + τve−λτv
∫ 1
0
eλτvσhv(σ)dσ
= λu(v)e−λτv + z0v(v)
where z0v(v) = −f(v)e−λτv + τve
−λτv∫ 1
0eλτvσhv(σ)dσ is a fixed real number depending only
on f and h.It remains to find u. By (1.7) and (1.8), uj must satisfy
λ2uj −∂2uj∂x2
= gj + λfj.
Multiplying this identity by a test function φj , integrating in space and using integrationby parts, we obtain
N∑
j=1
∫ lj
0
(λ2uj −∂2uj∂x2
)φjdx =
N∑
j=1
∫ lj
0
(λ2ujφj +∂uj∂x
∂φj∂x
)dx−N∑
j=1
[∂uj∂x
φj ]lj0
=N∑
j=1
∫ lj
0
(λ2ujφj +∂uj∂x
∂φj∂x
)dx−∑
v∈V
∑
j∈Ev
∂uj∂nj
(v)φj(v).
33
But using the fact that (u, w, z)⊤ must belong to D(A), we have
∑
v∈V
∑
j∈Ev
∂uj
∂nj(v)φj(v) =
∑
v∈Vc
∑
j∈Ev
∂uj
∂nj(v)φj(v) +
∑
v∈D
∂ujv
∂njv(v)φjv(v)
+∑
v∈N
∂ujv
∂njv(v)φjv(v) +
∑
v∈Vint\Vcint
∑
j∈Ev
∂uj
∂nj(v)φj(v)
=∑
v∈Vc
(∑
j∈Ev
∂uj
∂nj(v))φ(v)
= −∑
v∈Vc
(α(v)1 wj(v) + α
(v)2 zv(1))φ(v).
Using the above expression for zv(1) we arrive at the problem
N∑
j=1
∫ lj
0
(λ2ujφj +∂uj∂x
∂φj∂x
)dx +∑
v∈Vc
(α(v)1 + α
(v)2 e−λτv)λu(v)φ(v)
=
N∑
j=1
∫ lj
0
(gj + λfj)φjdx
+∑
v∈Vc
(α(v)1 f(v) − α
(v)2 z0
v(v))φ(v),∀φ ∈ V.
(1.9)
This problem has a unique solution u ∈ V by Lax-Milgram’s lemma, because the left-hand
side of (1.9) is coercive on V . If we consider φ ∈N∏
j=1
D(0, lj) ⊂ V , then u satisfies
λ2uj −∂2uj∂x2
= gj + λfj in D′(0, lj) ∀j = 1, · · · , N.
This directly implies that u ∈N∏
j=1
H2(0, lj) and then u ∈ V ∩N∏
j=1
H2(0, lj). Coming back to
(1.9) and by integrating by parts, we find
∑
v∈Vc
(
∑
j∈Ev
∂uj
∂nj(v) + (α
(v)1 + α
(v)2 e−λτv)λu(v) + (α
(v)2 z0
v(v) − α(v)1 f(v))
)
φ(v)
= −∑
v∈N
∂uj
∂nj(v)φj(v) −
∑
v∈Vint\Vcint
(∑
j∈Ev
∂uj
∂nj(v))φj(v), ∀φ ∈ V.
Consequently, by taking particular test functions φ, we obtain
∂ujv∂x
(v) = 0 ∀v ∈ N ,
∑
j∈Ev
∂uj∂nj
(v) = 0 ∀v ∈ Vint\Vcint
34
∑
j∈Ev
∂uj∂nj
(v) = −(α(v)1 + α
(v)2 e−λτv)λu(v) − (α
(v)2 z0
v(v) − α(v)1 f(v))
= −α(v)2 zv(1) − α
(v)1 (λu(v) + f(v))
= −(α(v)2 zv(1) + α
(v)1 w(v)) ∀v ∈ Vc.
In summary we have found (u, w, z)⊤ ∈ D(A) satisfying (1.6).
1.3 The energy
We now restrict the hypothesis (1.4) to obtain the decay of the energy. Namely wesuppose that
α(v)2 < α
(v)1 , ∀v ∈ Vc. (1.10)
Let us choose the following energy (which corresponds to the inner product on H)
E(t) :=1
2
N∑
j=1
∫ lj
0
(
(
∂uj∂t
)2
+
(
∂uj∂x
)2)
dx+∑
v∈Vc
ξ(v)
2
(∫ 1
0
(∂u
∂t(v, t− τvρ))
2dρ
)
(1.11)
where ξ(v) is a positive constant satisfying (that exists due to (1.10))
τvα(v)2 < ξ(v) < τv(2α
(v)1 − α
(v)2 ), ∀v ∈ Vc. (1.12)
1.3.1 Decay of the energy
Proposition 1.3.1. For all regular solution of problem (1.1), the energy is non increasing
and there exists two positive constants C1 and C2 depending only on ξ(v), α(v)1 ,
α(v)2 and τv such that
−C2
∑
v∈Vc
(
(
∂u
∂t(v, t)
)2
+
(
∂u
∂t(v, t− τv)
)2)
≤ E ′(t)
≤ −C1
∑
v∈Vc
(
(
∂u
∂t(v, t)
)2
+
(
∂u
∂t(v, t− τv)
)2)
. (1.13)
Proof: Deriving (1.11) and integrating by parts in space, we obtain
E ′(t) =
N∑
j=1
∫ lj
0
(∂uj∂t
∂2uj∂t2
+∂uj∂x
∂2uj∂x∂t
)dx
+∑
v∈Vc
ξ(v)(
∫ 1
0
∂u
∂t(v, t− τvρ)
∂2u
∂t2(v, t− τvρ)dρ)
35
and then
E ′(t) =N∑
j=1
∫ lj
0
(∂2uj∂t2
− ∂u2j
∂x2)∂uj∂t
dx+N∑
j=1
[∂uj∂t
∂uj∂x
]lj0
+∑
v∈Vc
ξ(v)(
∫ 1
0
∂u
∂t(v, t− τvρ)
∂2u
∂t2(v, t− τvρ)dρ)
=∑
v∈V
∑
j∈Ev
∂uj∂nj
(v)∂uj∂t
(v) +∑
v∈Vc
ξ(v)
∫ 1
0
∂u
∂t(v, t− τvρ)
∂2u
∂t2(v, t− τvρ)dρ
=∑
v∈Vc
(∑
j∈Ev
∂uj∂nj
(v))∂u
∂t(v) +
∑
v∈D
∂ujv∂njv
(v)∂ujv∂t
(v) +∑
v∈N
∂ujv∂njv
(v)∂ujv∂t
(v)
+∑
v∈Vint\Vcint
(∑
j∈Ev
∂uj∂nj
(v))∂u
∂t(v)
+∑
v∈Vc
ξ(v)
∫ 1
0
∂u
∂t(v, t− τvρ)
∂2u
∂t2(v, t− τvρ)dρ
= −∑
v∈Vc
(α(v)1 (
∂u
∂t(v, t))2 + α
(v)2
∂u
∂t(v, t− τv)
∂u
∂t(v, t))
+∑
v∈Vc
ξ(v)
∫ 1
0
∂u
∂t(v, t− τvρ)
∂2u
∂t2(v, t− τvρ)dρ.
Now for all v ∈ Vc, recalling that zv(ρ, t) = ∂u∂t
(v, t− τvρ), we see that
∫ 1
0
∂u
∂t(v, t− τvρ)
∂2u
∂t2(v, t− τvρ)dρ =
∫ 1
0
zv(ρ, t)∂zv∂t
(ρ, t)dρ
= − 1
τv
∫ 1
0
zv(ρ, t)∂zv∂ρ
(ρ, t)dρ.
By an integration by parts in ρ, we obtain
∫ 1
0
∂u
∂t(v, t− τvρ)
∂2u
∂t2(v, t− τvρ)dρ = − 1
2τv
(
(∂u
∂t(v, t− τv))
2 − (∂u
∂t(v, t))2
)
.
Therefore, we have
E ′(t) = −∑
v∈Vc
[α(v)1 (
∂u
∂t(v, t))2 + α
(v)2
∂u
∂t(v, t− τv)
∂u
∂t(v, t)
+ξ(v)
2τv((∂u
∂t(v, t− τv))
2 − (∂u
∂t(v, t))2)]
= −∑
v∈Vc
[(α(v)1 − ξ(v)
2τv)(∂u
∂t(v, t))2 + α
(v)2
∂u
∂t(v, t− τv)
∂u
∂t(v, t)
+ξ(v)
2τv(∂u
∂t(v, t− τv))
2].
36
Cauchy-Schwarz’s inequality yields
E ′(t) ≤ −∑
v∈Vc
(
(α(v)1 − ξ(v)
2τv− α
(v)2
2)(∂u
∂t(v, t))2 + (
ξ(v)
2τv− α
(v)2
2)(∂u
∂t(v, t− τv))
2
)
E ′(t) ≥ −∑
v∈Vc
(
(α(v)1 − ξ(v)
2τv+α
(v)2
2)(∂u
∂t(v, t))2 + (
ξ(v)
2τv+α
(v)2
2)(∂u
∂t(v, t− τv))
2
)
.
The first estimate leads to
E ′(t) ≤ −C1
∑
v∈Vc
(
(∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2
)
with
C1 = min
(α(v)1 − ξ(v)
2τv− α
(v)2
2), (
ξ(v)
2τv− α
(v)2
2) : v ∈ Vc
which is positive according to the assumption (1.12). The second one yields
E ′(t) ≥ −C2
∑
v∈Vc
(
(∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2
)
with
C2 = max
(α(v)1 − ξ(v)
2τv+α
(v)2
2), (
ξ(v)
2τv+α
(v)2
2) : v ∈ Vc
which is also positive due to (1.12).
We have just shown that under the assumption (1.10), the energy decays. But we wouldlike to obtain strong stability of the system, in other words, the decay to 0 of the energy.This is the goal of the remainder of this section. But before going on, let us make the nextremark that will be useful later on.
Remark 1.3.2. Integrating the expression (1.13) between 0 and T , we obtain
∑
v∈Vc
∫ T
0
(
(∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2
)
dt . E(0) − E(T ) . E(0))
and therefore∫ T
0
(
(∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2
)
dt . E(0), ∀v ∈ Vc.
This estimate implies that α(v)1
∂u∂t
(v, .)+α(v)2
∂u∂t
(v, .−τv) belongs to L2(0, T ) for all v ∈ Vc,with the estimate∥
∥
∥
∥
α(v)1
∂u
∂t(v, .) + α
(v)2
∂u
∂t(v, .− τv)
∥
∥
∥
∥
2
L2(0, T )
=
∫ T
0
(α(v)1
∂u
∂t(v, t) + α
(v)2
∂u
∂t(v, t− τv))
2dt
≤ 2maxv∈Vc
(α(v)1 )2
∫ T
0
(∂u
∂t(v, t)2 +
∂u
∂t(v, t− τv)
2)dt . E(0) < +∞.
37
1.3.2 Problem without damping
In the sequel we need to consider the problem without damping
∂2φj
∂t2(x, t) − ∂2φj
∂x2 (x, t) = 0 0 < x < lj , t > 0, ∀j ∈ 1, ..., Nφj(v, t) = φl(v, t) = φ(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂φj
∂nj(v, t) = 0 ∀v ∈ Vint, t > 0,
φjv(v, t) = 0 ∀v ∈ D, t > 0,∂φjv
∂njv(v, t) = 0 ∀v ∈ N ∪ Vcext, t > 0,
φ(t = 0) = u(0), ∂φ∂t
(t = 0) = u(1).
(1.14)
It is well known that this problem is well posed in the natural energy space (see forinstance [3]).
Lemma 1.3.3. Suppose that (u(0), u(1)) ∈ V ×N∏
j=1
L2(0, lj). Then problem (1.14) admits a
unique solution
φ ∈ C(0, T ; V ) ∩ C1(0, T ;
N∏
j=1
L2(0, lj)).
This problem is obviously conservative, its energy is constant.
1.3.3 Decay of the energy to 0
We look at the spectral problem associated with problem (1.14), in other words
−λ2φj(x) − ∂2φj
∂x2 (x) = 0 0 < x < lj , ∀j ∈ 1, ..., N,φj(v) = φl(v) = φ(v) ∀j, l ∈ Ev, v ∈ Vint,∑
j∈Ev
∂φj
∂nj(v) = 0 ∀v ∈ Vint,
φjv(v) = 0 ∀v ∈ D,∂φjv
∂njv(v) = 0 ∀v ∈ N ∪ Vcext.
This system corresponds to an eigenvalue problem of the positive selfadjoint operatorA defined above. Let us then denote by λ2
kk≥1 the set of eigenvalues counted withouttheir multiplicities, i.e. λk 6= λl, ∀k 6= l, where without any restriction, we may supposethat λk > 0. For all k ∈ N∗, let lk be the multiplicity of the eigenvalue λ2
k (remark thatlk ≤ 2N, ∀k ∈ N∗) and let ϕk, i1≤i≤lk be the orthonormal eigenvectors associated withthe eigenvalue λ2
k.
38
Definition 1.3.4. For k ≥ 1 and v ∈ Vc, we denote by Mv(λ2k) the following matrix of
size lk
Mv(λ2k) :=
ϕ2k, 1(v) ϕk, 1(v)ϕk, 2(v) ... ϕk,1(v)ϕk, lk(v)
ϕk, 1(v)ϕk,2(v) ϕ2k, 2(v) · · · ϕk,2(v)ϕk, lk(v)
......
. . ....
ϕk,1(v)ϕk, lk(v) ϕk, 2(v)ϕk, lk(v) · · · ϕ2k, lk
(v)
.
Moreover, let M(λ2k) be the matrix of size lk
M(λ2k) :=
∑
v∈Vc
Mv(λ2k).
Now we recall that the following generalized gap condition holds, namely from Propo-sition 6.2 of [38], we have
∃γ > 0, ∀k ≥ 1, λk+N+1 − λk ≥ (N + 1)γ. (1.15)
From this property we will deduce an inequality of Ingham’s type. Namely fix a positivereal number γ′ ≤ γ and denote by Ak, k = 1, · · · , N + 1 the set of natural numbers msatisfying (see for instance [13])
λm − λm−1 ≥ γ′
λn − λn−1 < γ′ for m+ 1 ≤ n ≤ m+ k − 1,λm+k − λm+k−1 ≥ γ′.
Then one easily checks that the sets Ak+j, j = 0, · · · , k − 1, k = 1, · · · , N + 1, form apartition of N∗.
Now for m ∈ Ak, we recall that the finite differences em+j(t), j = 0, · · · , k − 1, corres-ponding to the exponential functions eiλm+jt, j = 0, · · · , k − 1 are given by
em+j(t) =
m+j∑
p=m
m+j∏
q=m, q 6=p(λp − λq)
−1eiλpt.
Write for shortness, e−n(t) the same finite differences functions corresponding to −λn.Now we are ready to recall the next inequality of Ingham’s type, see for instance Theo-
rem 1.5 of [13] :
Theorem 1.3.5. If the sequence (λn)n≥1 satisfies (1.15), then for all sequence (an)n∈Z∗
(where Z∗ = Z \ 0), the function
f(t) =∑
n∈Z∗
anen(t),
satisfies the estimates∫ T
0
|f(t)|2 ∼∑
n∈Z∗
|an|2, (1.16)
for T > 2π/γ.
39
Going back to the original functions eiλnt, the above equivalence (1.16) means that, forT > 2π/γ, the function (from now on λ−n = −λn)
f(t) =∑
n∈Z∗
αneiλnt,
satisfies the estimates∫ T
0
|f(t)|2 ∼N+1∑
k=1
∑
|n|∈Ak
‖B−1n Cn‖2
2, (1.17)
where ‖ · ‖2 means the Euclidean norm of the vector, for n ∈ Ak, the vector Cn is given by
Cn = (αn, · · · , αn+k−1)⊤,
and the k × k matrix Bn allows to pass from the coefficients an to αn, namely
Cn = Bn · (an, · · · , an+k−1)⊤,
and is given by Bn = (Bn,ij)1≤i,j≤k with
Bn,ij =
n+j−1∏
q=n, q 6=n+i−1
(λn+i−1 − λq)−1 if i ≤ j, (i, j) 6= (1, 1),
1 if i = j = 1,0 if i > j.
We proceed similarly for n ≤ −1, but the indices being decreasing from n to n− k + 1.
Remark 1.3.6. Notice that if the standard gap condition
∃γ > 0, ∀k ≥ 1, λk+1 − λk ≥ γ (1.18)
holds, then A1 = Z∗ and B1 = 1 and in that case the next equivalence holds (see [57]) :
∫ T
0
|f(t)|2 ∼∑
n∈Z∗
|αn|2.
We are now ready to give a necessary and sufficient condition that guarantees the decayto 0 of the energy.
Proposition 1.3.7. For all initial data in H, we have
limt→∞
E(t) = 0 (1.19)
if and only if the operator A satisfies
λmin(M(λ2k)) > 0, ∀k ∈ N∗, (1.20)
where λmin(M) denotes the smallest eigenvalue of the matrix M.
40
Proof: ⇐ Let us show that (1.20) implies (1.19) : Let S(t) be the semi-group of contrac-tions generated by the operator A.
It suffices to show that
limt→∞
S(t)
u(0)
u(1)
(f 0(−τv.))v
= 0, ∀
u(0)
u(1)
(f 0(−τv.))v
∈ D(A).
Let us fix U0 =
u(0)
u(1)
(f 0(−τv.))v
∈ D(A). Since D(A) is compactly embedded into H , the
setorb(U0) =
⋃
t≥0
S(t)U0
is precompact in H . Indeed, for any sequence (tn)n, since U0 ∈ D(A), one has S(tn)U0 ∈D(A) and
‖S(tn)U0‖D(A) = ‖S(tn)U0‖H + ‖AS(tn)U0‖H = ‖S(tn)U0‖H + ‖S(tn)AU0‖≤ ‖U0‖H + ‖AU0‖H = cst.
Therefore the sequence S(tn)U0 is bounded in D(A) and by the compact embedding ofD(A) into H , there exists a subsequence, still denote by S(tn)U0 which converges in H. Inthis case, the ω-limit of U0 defined by
ω(U0) = U ∈ H : ∃(tn), tn → ∞, S(tn)U0 → U, t→ ∞is non empty.
On the other hand, if Φ ∈ ω(U0), then
S(t)Φ ∈ ω(U0).
Note further that one readily checks that S(t)Φ is of the form
S(t)Φ =
φ(., t)∂φ∂t
(., t)ψ
,
for some φ ∈ C([0,∞);V ) ∩ C1([0,∞);L2(R)) and ψ ∈ C([0,∞);L2(0, 1)Vc).We can now apply LaSalle’s invariance principle [32] with the relatively compact set
⋃
t≥0
S(t)U0 and the Lyapounov functional φ = ‖.‖H . Since
φ(0)
φ(1)
φ(2)
and S(t)
φ(0)
φ(1)
φ(2)
=
φ(., t)∂φ∂t
(., t)ψ
belong to ω(U0), we find that
∥
∥
∥
∥
∥
∥
φ(0)
φ(1)
φ(2)
∥
∥
∥
∥
∥
∥
H
=
∥
∥
∥
∥
∥
∥
φ(., t)∂φ∂t
(., t)ψ
∥
∥
∥
∥
∥
∥
H
= L ∀t ≥ 0.
41
Therefore φ satisfies problem (1.2) with initial conditions φ(·, 0) = φ(0) and∂φ∂t
(·, 0) = φ(1). Moreover by (1.13) we have
0 = L2 − L2 =
∥
∥
∥
∥
∥
∥
φ(., t)∂φ∂t
(., t)ψ
∥
∥
∥
∥
∥
∥
2
H
−
∥
∥
∥
∥
∥
∥
φ(0)
φ(1)
φ(2)
∥
∥
∥
∥
∥
∥
2
H
≤ −C1
∑
v∈Vc
∫ t
0
(ψv(0, s)2 + ψv(1, s)
2)ds ≤ 0.
In other words, it holds
∑
v∈Vc
∫ t
0
(ψv(0, s)2 + ψv(1, s)
2)ds = 0
which implies thatψv(0, t) = ψv(1, t) = 0 ∀t ≥ 0, ∀v ∈ Vc. (1.21)
In particular, this implies that φ is solution of problem (1.14) with initial data φ(·, 0) = φ(0)
and ∂φ∂t
(·, 0) = φ(1) because 0 = ψv(0, t) = ∂φ∂t
(v, t) and 0 = ψv(1, t) = ∂φ∂t
(v, t− τv), whichmeans that in (1.2) the damping terms disappear.
Let us now write
φ(0) =∑
k≥1
lk∑
i=1
ak, iϕk, i,
φ(1) =∑
k≥1
lk∑
i=1
bk, iϕk, i,
where (λk, iak, i)i, k, (bk, i)i, k ∈ l2(N∗). Then φ is given by
φ(·, t) =∑
k≥1
lk∑
i=1
(
ak, i cos(λkt) +bk, iλk
sin(λkt)
)
ϕk, i.
Consequently by (1.21) for v ∈ Vc and j ∈ Ev
0 =∂φj∂t
(v, t) =∑
k≥1
lk∑
i=1
(−ak, iλk sin(λkt) + bk, i cos(λkt))ϕk, i(v).
By grouping the terms corresponding to the same eigenvalue, we get
0 =∂φj∂t
(v, t) =∑
k≥1
(
lk∑
i=1
− ak, iϕk, i(v)
)
λk sin(λkt) +∑
k≥1
(
lk∑
i=1
bk, iϕk, i(v)
)
cos(λkt)
=∑
n∈Z∗
αn(v)eiλnt,
42
where
αk(v) =1
2
((
lk∑
i=1
bk, iϕk, i(v)
)
+ i
(
lk∑
i=1
ak, iϕk, i(v)
)
λk
)
, ∀k ≥ 1,
α−k(v) =1
2
((
lk∑
i=1
bk, iϕk, i(v)
)
− i
(
lk∑
i=1
ak, iϕk, i(v)
)
λk
)
, ∀k ≥ 1.
Integrating this identity between 0 and T > 0 sufficiently large and using Ingham’s inequa-lity (1.17), we obtain (with the notation introduced above)
0 =
∫ T
0
(
∂φj∂t
(v, t)
)2
dt &
N+1∑
k=1
∑
|n|∈Ak
‖B−1n Cn(v)‖2
2,
where Cn(v) is defined as Cn using αn(v) instead of αn. Summing on v ∈ Vc :
0 &∑
v∈Vc
N+1∑
k=1
∑
|n|∈Ak
‖B−1n Cn(v)‖2
2 ≥ 0.
This implies that for all k = 1, · · · , N + 1 and all |n| ∈ Ak, we have
∑
v∈Vc
‖B−1n Cn(v)‖2
2 = 0.
Since B−1n is invertible, there exits γn > 0 such that
‖B−1n Cn(v)‖2 ≥ γn‖Cn(v)‖2.
From this estimate we deduce that for all k = 1, · · · , N + 1 and all |n| ∈ Ak, we have
∑
v∈Vc
‖Cn(v)‖22 = 0.
Since λk 6= 0, we necessarily have
∀k ≥ 1,∑
v∈Vc
(
lk∑
i=1
ak, iϕk, i(v)
)2
= 0 and∑
v∈Vc
(
lk∑
i=1
bk, iϕk, i(v)
)2
= 0.
For a fixed k ≥ 1, if we set b =
bk, 1...
bk, lk
, then
∑
v∈Vc
(
lk∑
i=1
bk, iϕk, i(v)
)2
= tbM(λ2k)b.
43
As a consequence if λmin(M(λ2k)) > 0, we obtain that bk, 1 = ... = bk, lk = 0. In the same
manner we have ak, 1 = ... = ak, lk = 0.We have proved that φ(0) = 0 = φ(1).
Moreover φ(2) = 0 because ψv satisfies the transport equation
∂ψv
∂t= − 1
τv
∂ψv
∂ρ,
ψv(0, t) = ψv(1, t) = 0,ψv(ρ, 0) = φ(2),
for all v ∈ Vc.We have shown that for all
(
φ(0), φ(1), φ(2))⊤ ∈ ω(U0), we have φ(0) = 0 = φ(1) = φ(2).
Consequently limt→∞
S(t)U0 = 0 and then limt→∞
E(t) = 0.
⇒ Let us show that (1.19) implies (1.20). For that purpose we use a contradictionargument. Suppose that there exists k > 0 such that λmin(M(λ2
k)) = 0. This means that
there exists a = (ak, 1, · · · , ak, lk)⊤ 6= 0, such that
∑
v∈Vc
(
lk∑
i=1
ak, iϕk, i(v)
)2
= 0.
Let us set
u(., t) =
(
lk∑
i=1
ak, iϕk, i
)
cos(λkt).
Then u is solution of (1.1) and satisfies
E(t) = 12
N∑
j=1
∫ lj
0
(
(∂uj∂t
)2 + (∂uj∂x
)2
)
dx+∑
v∈Vc
ξ(v)
2
(∫ 1
0
(∂uj∂t
(v, t− τvρ))2dρ
)
= 12
N∑
j=1
∫ lj
0
(
(∂uj∂t
)2 + (∂uj∂x
)2
)
dx
= E(u(0))
because
∀v ∈ Vc, ∀t,∂uj∂t
(v, t) =
(
lk∑
i=1
ak, iϕk, i(v)
)
(−λk) sin(λkt) = 0.
This means that we have obtained a solution of problem (1.1) with a constant energy,which contradicts (1.19).
Remark 1.3.8. 1. Notice that the condition (1.20) is independent of the choice of theorthonormal basis of eigenvectors associated with the eigenvalue λ2
k. Indeed if ϕk, ilki=1 is
44
another orthonormal basis of eigenvectors associated with the eigenvalue λ2k, then there
exists an orthogonal matrix O ∈ Rlk×lk such that
ϕk, 1...
ϕk, lk
= O
ϕk, 1...
ϕk, lk
.
Consequently the matrix M(λ2k) built as M(λ2
k) by using ϕk, ilki=1 instead of ϕk, ilki=1 isgiven by
M(λ2k) = OM(λ2
k)O⊤,
and therefore λmin(M(λ2k)) = λmin(M(λ2
k)).2. If lk = 1, then the condition (1.20) is reduced to
∀k ∈ N∗,∑
v∈Vc
|ϕk(v)|2 > 0
because M(λ2k) is the matrix of size 1 equals to
∑
v∈Vc
|ϕk(v)|2 .
1.3.4 Counterexample to the stability of the system
In this section (and in the remainder of the chapter) we have made the hypothesis(1.10). As in [110], we may expect non-stability results if this condition fails.
In [110], the authors consider the wave equation on a string of length π and useda boundary control. They show that if (1.10) does not hold then non-stabilities appear.Since their problem enters in our framework, this is a first counterexample. As a secondcounterexample, we consider the wave equation on a string of length π but with an interiorcontrol. Namely we consider the problem
∂2u∂t2
(x, t) − ∂2u∂x2 (x, t) = 0 0 < x < π, t > 0,
∂u∂x
(ξ−, t) − ∂u∂x
(ξ+, t) = −(α1∂u∂t
(ξ, t) + α2∂u∂t
(ξ, t− τ)) t > 0,u(ξ−, t) = u(ξ+, t) t > 0,u(0, t) = 0 t > 0,∂u∂x
(π, t) = 0 t > 0,u(t = 0) = u(0), ∂u
∂t(t = 0) = u(1) 0 < x < π,
∂u∂t
(ξ, t− τ) = f 0(t− τ) 0 < t < τ.
(1.22)
Lemma 1.3.9. A complex number λ ∈ C is called an eigenvalue associated with system(1.22) if and only if λ satisfies
(α1 + α2e−λτ ) cosh(λ(ξ − π)) sinh(λξ) + cosh(λπ) = 0. (1.23)
45
Proof: Setting u(t, .) = eλtϕ, we see that u is solution of (1.22) if and only if ϕ satisfies
λ2ϕ(x) − ∂2ϕ∂x2 (x) = 0 0 < x < π,
∂ϕ∂x
(ξ−) − ∂ϕ∂x
(ξ+) = −(α1 + α2e−λτ )λϕ(ξ),
ϕ(ξ−) = ϕ(ξ+),ϕ(0) = 0,∂ϕ∂x
(π) = 0.
We then obtain
ϕ(x) =
A sinh(λx) in (0, ξ)A1 cosh(λ(x− π)) in (ξ, π)
where A, A1 are real constants. The continuity ϕ(ξ−) = ϕ(ξ+) and ∂ϕ∂x
(ξ−) − ∂ϕ∂x
(ξ+) =−(α1 + α2e
−λτ )λϕ(ξ) imply that
(
sinh(λξ) − cosh(λ(ξ − π))cosh(λξ) + (α1 + α2e
−λτ ) sinh(λξ) − sinh(λ(ξ − π))
)(
AA1
)
= 0.
Therefore a non trivial solution exists if and only if
det
(
sinh(λξ) − cosh(λ(ξ − π))cosh(λξ) + (α1 + α2e
−λτ ) sinh(λξ) − sinh(λ(ξ − π))
)
= 0,
and simple calculations lead to the characteristic equation (1.23).
The characteristic equation (1.23) is equivalent to
∆(λ) :=(α1 + α2e−λτ )e2λ(ξ−π) − (α1 + α2e
−λτ − 2)e−2λπ
− (α1 + α2e−λτ )e−2λξ + α2e
−λτ + 2 + α1 = 0.
Take an interior control ξ and a delay τ such that
ξ
π=
2m+ 1
2n+ 1, τ =
2(2k + 1)
2n+ 1π
where n, m, k ∈ Z. Now we look for λ in the form
λ = η + i2n+ 1
2, η ∈ R.
For such a λ, we have
∆(λ) = (α1 − α2e−ητ )e2η(ξ−π) + (α1 − α2e
−ητ − 2)e−2ηπ
+(α1 − α2e−ητ )e−2ηξ − α2e
−ητ + 2 + α1
=: ∆0(η).
If we suppose thatα2 ≥ α1 ≥ 0.
46
Then
∆0(0) = (α1 − α2) + (α1 − α2 − 2) + (α1 − α2) − α2 + 2 + α1 = 4(α1 − α2) ≤ 0
andlim
η−→+∞∆0(η) = α1 + 2 > 0.
By the mean value theorem, there exists η ≥ 0 such that ∆0(η) = 0. Therefore λ =η+ 2n+1
2i, n ∈ Z, is an eigenvalue of system (1.22) with Re(λ) = η ≥ 0. The system is then
unstable for the countable set of delays τ and of control points ξ in the above form.
1.4 A regularity result and an a priori estimate
We consider the following problem with non-homogeneous transmission conditions
∂2wj
∂t2(x, t) − ∂2wj
∂x2 (x, t) = 0 0 < xj < lj , t > 0, ∀j ∈ 1, ..., Nwj(v, t) = wl(v, t) = w(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂wj
∂nj(v, t) = kv(t) ∀v ∈ Vc, t > 0,
∑
j∈Ev
∂wj
∂nj(v, t) = 0 ∀v ∈ Vint\Vcint, t > 0,
wjv(v, t) = 0 ∀v ∈ D, t > 0,∂wjv
∂njv(v, t) = 0 ∀v ∈ N , t > 0,
w(t = 0) = 0, ∂w∂t
(t = 0) = 0
(1.24)
This system modelizes the vibrations of a network of strings with local forces at the nodesof Vc. The next proposition gives existence and regularity results for the solution of problem(1.24).
Proposition 1.4.1. Let T > 0 be fixed. Suppose that kv ∈ L2(0, T ) for all v ∈ Vc.
Then the problem (1.24) admits a unique solution w ∈N∏
j=1
H1((0, lj) × (0, T )). Moreover
w(v, .) ∈ H1(0, T ) for all v ∈ Vc and∑
v∈Vc
‖w(v, .)‖H1(0, T ) .∑
v∈Vc
‖kv‖L2(0, T ) ∀v ∈ Vc. (1.25)
The proof of this proposition is relatively technical and requires some preliminary re-sults.
We first consider the following problem
wtt(x, t) − wxx(x, t) = 0 in (0, 1) × (0, T ),w(1, t) = 0 on (0, T ),
wx(0, t) = k(t) on (0, T ),w(x, 0) = 0, wt(x, 0) = 0 on (0, 1).
(1.26)
47
Lemma 1.4.2. Assume that k ∈ L2(0, T ). Then problem (1.26) has a unique solutionw ∈ H1((0, 1) × (0, T )) which satisfies
‖w‖H1((0, 1)×(0, T )) . ‖k‖L2(0, T ) .
Moreover w(0, .) ∈ H1(0, T ) and satisfies
‖w(0, .)‖H1(0, T ) . ‖k‖L2(0, T ) .
Proof: We extend k by 0 on R \ [0, T ] because (1.26) is reversible in time.Let w(x, λ) where λ = γ+ iη, γ > 0, η ∈ R, be the Laplace transform of w with respect
to t. Then w satisfies
λ2w(x, λ) − ∂2w∂x2 (x, λ) = 0 in (0, 1),
w(1, λ) = 0,∂w∂x
(0, λ) = k(λ),
where ℜ λ > 0. Consequently w(x, λ) = a cosh(λ(x − 1)) + b sinh(λ(x − 1)), with twocomplex numbers a and b.Since w(1, λ) = a = 0 and ∂w
∂x(0, λ) = bλ cosh(λ) = k(λ), we deduce that
w(x, λ) =sinh(λ(x− 1))
λ cosh(λ)k(λ).
because cosh(λ) 6= 0.
• Existence of w : w(x, t) = L−1( sinh(λ(x−1))λ cosh(λ)
) ⋆ k where L−1 denotes the inverse Laplacetransform.
• Uniqueness of w : if w1, w2 are two solutions of (1.26), then w = w1 − w2 satisfiesthe wave equation in (0, 1) with homogeneous boundary and initial conditions. Thereforew1 − w2 = 0, which proves the uniqueness of the solution of (1.26).
• Regularity : Let γ > 0 be fixed and set Cγ := λ ∈ C ; ℜλ = γ. Define H(x, λ) =sinh(λ(x−1))
cosh(λ), for λ ∈ Cγ. We clearly have
‖H‖L∞((0, 1)×Cγ ) ≤ coth(γ).
Therefore
‖w(x, λ)‖L2((0, 1)×Rη) =
∥
∥
∥
∥
H(x, λ)
λk(λ)
∥
∥
∥
∥
L2((0, 1)×Rη)
≤ coth(γ)
γ
∥
∥
∥k(λ)
∥
∥
∥
L2(Rη), (1.27)
which implies that w ∈ L2((0, 1) × (0, T )) with
‖w‖L2((0, 1)×(0, T )) . ‖k‖L2(0, T ) . (1.28)
Indeed, we have
w(x, λ) =
∫ ∞
0
e−(γ+iη)tw(x, t)dt =
∫ +∞
−∞e−γte−iηtw(x, t)dt
=F(e−γ.w)(η) = F(w1)(η)
48
where F denotes the Fourier transform and where we have set w1 = e−γ.w. Therefore
‖w‖L2((0, 1)×(0, T )) ∼ ‖w1‖L2((0, 1)×(0, T )) as e−γT ≤ e−γt ≤ 1 on (0, T )
≤ ‖w1‖L2((0, 1)×R)
∼ ‖F(w1)(η)‖L2((0, 1)×Rη) by Plancherel’s formula
∼ ‖w(x, λ)‖L2((0, 1)×Rη)
while‖k‖L2(0, T ) = ‖k‖L2(R) ∼ ‖ke−γ.‖L2(R)
∼ ‖F(e−γ.k)(η)‖L2(Rη)
=∥
∥
∥k(λ)
∥
∥
∥
L2(Rη).
These two equivalences and the estimate (1.27) lead to (1.28).Since
‖λw(x, λ)‖L2((0, 1)×Rη) ≤ coth(γ)∥
∥
∥k(λ)
∥
∥
∥
L2(Rη)
we deduce that w ∈ H1(0, T ; L2(0, 1)) and
‖w‖H1(0, T ;L2(0, 1)) . ‖k‖L2(0, T ) .
Indeed∥
∥
∂w1
∂t
∥
∥
L2((0, 1)×(0, T ))≤
∥
∥
∂w1
∂t
∥
∥
L2((0, 1)×R)
∼∥
∥F(∂w1
∂t)(η)
∥
∥
L2((0, 1)×Rη)by Plancherel’s formula
∼ ‖(iη)F(w1)(η)‖L2((0, 1)×Rη)
∼ ‖(iη)w(x, λ)‖L2((0, 1)×Rη)
and‖λw(x, λ)‖2
L2((0, 1)×Rη) =∫ 1
0
∫
Rη|λ|2 w(x, λ)2dλdx
=∫ 1
0
∫
Rη(γ2 + η2)w(x, λ)2dλdx
≥ ‖(iη)w(x, λ)‖2L2((0, 1)×Rη) .
Therefore∥
∥
∂w1
∂t
∥
∥
L2((0, 1)×(0, T )). ‖λw(x, λ)‖L2((0, 1)×Rη)
.∥
∥
∥k(λ)
∥
∥
∥
L2(Rη). ‖k‖L2(0, T ) .
We finally conclude that∥
∥
∂w∂t
∥
∥
L2((0, 1)×(0, T ))=
∥
∥eγ. ∂w1
∂t+ γw
∥
∥
L2((0, 1)×(0, T ))
.∥
∥
∂w1
∂t
∥
∥
L2((0, 1)×(0, T ))+ ‖w‖L2((0, 1)×(0, T ))
. ‖k‖L2(0, T ) .
In a similar manner we have
‖λw(0, λ)‖L2(Rη) ≤ coth(γ)∥
∥
∥k(λ)
∥
∥
∥
L2(Rη)
49
which implies that w(0, .) ∈ H1(0, T ) and satisfies
‖w(0, .)‖H1(0, T ) . ‖k‖L2(0, T ) .
Finally ∂w∂x
(x, λ) = cosh(λ(x−1))cosh(λ)
k(λ). But the standard estimate |cosh z|≤ cosh(Re z), ∀z ∈ C implies that
∣
∣
∣
∣
cosh(λ(x− 1))
cosh(λ)
∣
∣
∣
∣
≤ cosh(γ(1 − x))
sinh(γ)≤ cosh(γ)
sinh(γ)= coth γ ;
therefore∥
∥
∥
∥
∂w(x, λ)
∂x
∥
∥
∥
∥
L2((0, 1)×Rη)
≤ coth(γ)∥
∥
∥k(λ)∥
∥
∥
L2(Rη)
which leads to ∂w∂x
∈ L2((0, 1) × (0, T )) with
∥
∥
∥
∥
∂w
∂x
∥
∥
∥
∥
L2((0, 1)×(0, T ))
. ‖k‖L2(0, T ) .
Indeed∥
∥
∂w∂x
∥
∥
L2((0, 1)×(0, T ))∼
∥
∥
∂w1
∂x
∥
∥
L2((0, 1)×(0, T ))≤∥
∥
∂w1
∂x
∥
∥
L2((0, 1)×R)
∼∥
∥F(∂w1
∂x)(η)
∥
∥
L2((0, 1)×Rη)by Plancherel’s formula
∼∥
∥
∂w∂x
∥
∥
L2((0, 1)×Rη).
It remains to check the initial conditions. We remark that
L−1
(
sinh(λ(x− 1))
λ cosh(λ)
)
=4
π
+∞∑
n=1
(−1)n+1
2n− 1sin
(
(2n− 1)π(x− 1)
2
)
sin
(
(2n− 1)πt
2
)
=F (x, t).
Note further that F ∈ L∞([0, T ] ; L2(0, 1)). Therefore
w(x, t) =
∫ t
0
F (x, t− s)k(s)ds.
Consequently we directly see thatw(x, 0) = 0
(the trace having a meaning because w ∈ H1((0, 1) × (0, T ))). Moreover
∂w
∂t(x, t) = F (x, 0)k(t) +
∫ t
0
∂F
∂t(x, t− s)k(s)ds =
∫ t
0
∂F
∂t(x, t− s)k(s)ds.
Consequently∥
∥
∂w∂t
(., t)∥
∥
V ′ =∥
∥
∥
∫ t
0∂F∂t
(x, t− s)k(s)ds∥
∥
∥
V ′
≤∫ t
0
∥
∥
∂F∂t
(x, t− s)∥
∥
V ′ |k(s)| ds,
50
where V ′ is the dual space of V with V = u ∈ H1(0, 1); u(1) = 0. But we may write
∂F
∂t(x, t) =
4
π
π
2
+∞∑
n=1
(−1)n+1 sin
(
(2n− 1)π(x− 1)
2
)
cos
(
(2n− 1)πt
2
)
=
+∞∑
n=1
an(t) sin
(
(2n− 1)π(x− 1)
2
)
,
where an(t) = 2(−1)n+1 cos( (2n−1)πt2
) and sin( (2n−1)π(x−1)2
)n is an orthogonal basis ofL2(0, 1). This implies that
∥
∥
∥
∥
∂F
∂t(x, t)
∥
∥
∥
∥
V ′
∼+∞∑
n=1
a2n(t)n
−2,
with a2n(t) ≤ 4. Therefore
∥
∥
∥
∥
∂F
∂t(x, t)
∥
∥
∥
∥
V ′
. 1
and consequently∥
∥
∂w∂t
(·, t)∥
∥
V ′ .∫ t
0|k(s)| ds
. (∫ t
0ds)
12 (∫ t
0|k(s)|2 ds) 1
2
.√t ‖k‖L2(0, 1) .
This shows that∂w
∂t∈ C([0, T ] ; V ′)
and moreover∂w
∂t(x, 0) = 0.
This ends the proof of the Lemma.
We now make a local construction. Namely for a fixed v ∈ Vc, we consider w(v) solutionof
∂2w(v)j
∂t2(x, t) − ∂2w
(v)j
∂x2 (x, t) = 0 0 < x < lv, t > 0, ∀j ∈ Ev,w
(v)j (v, t) = w
(v)l (v, t) ∀j, l ∈ Ev, t > 0,
∑
j∈Ev
∂w(v)j
∂nj(v, t) = kv(t) t > 0,
w(v)j (lv, t) = 0 ∀j ∈ Ev, t > 0,
w(v)(t = 0) = 0, ∂w(v)
∂t(t = 0) = 0,
(1.29)
where lv = minj∈Ev lj (without loss of generality we may identify v with the extremity 0for all edges of Ev).
The unique solution of this system is simply w(v)j (x, t) = w(x/lv, t/lv), where w is
solution of problem (1.26) with k(t) = − lvEvkv(lvt), when Ev is the cardinal of Ev. By
Lemma 1.4.2, we directly obtain the
51
Lemma 1.4.3. The system (1.29) admits a unique solution w(v)j ∈ H1((0, lv) × (0, T )),
j ∈ Ev such that∥
∥
∥w
(v)j
∥
∥
∥
H1((0, lv)×(0, T )). ‖kv‖L2(0, T ) .
Moreover w(v)j (v, .) ∈ H1(0, T ) with the estimate
∥
∥
∥w
(v)j (v, .)
∥
∥
∥
H1(0, T ). ‖kv‖L2(0, T ) .
Let us now set (assuming for the moment that w exists)
ω(x, t) := w(x, t) −∑
v∈Vc
η(v)(x)w(v)(x, t)
where w(v) is solution of problem (1.29) and η(v) is a cut-off function such that suppη(v) = Ev, η(v) is equal to 1 in a neighborhood of v and is 0 outside a larger neighborhoodof v. Then we easily see that ω is solution of the following system
∂2ωj
∂t2(x, t) − ∂2ωj
∂x2 (x, t) = hj(x, t) 0 < xj < lj, t > 0, ∀j ∈ 1, ..., N,ωj(v, t) = ωl(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂ωj
∂nj(v, t) = 0 ∀v ∈ Vc ∪ Vcint, t > 0,
ωjv(v, t) = 0 ∀v ∈ D, t > 0,∂ωjv
∂njv(v, t) = 0 ∀v ∈ N , t > 0,
ω(t = 0) = 0, ∂ω∂t
(t = 0) = 0,
(1.30)
where hj is given by
hj(x, t) :=∑
v∈Vc : j∈Ev
(
∂2η(v)j
∂x2(x)wj
(v)(x, t) + 2∂η
(v)j
∂x(x)
∂w(v)j
∂x(x, t)
)
.
We have then transformed the system with nonhomogeneous transmission conditions toa system with nonhomogeneous right-hand sides. This system can be written in the form
∂2ω
∂t2+ Aω = h,
where the operator A was defined before. Since A is a positive selfadjoint operator onL2(R), we directly obtain the
Lemma 1.4.4. The solution ω of (1.30) has the regularity
ω ∈ C(0, T ; V ) ∩ C1(0, T ; L2(R))
and is given by the so-called constant variation formula
ω(·, t) =∑
k≥1
lk∑
i=1
1
λk
(∫ t
0
sin((t− s)λk)(h(·, s), ϕk,i)L2(R)ds
)
ϕk,i.
52
Lemmas 1.4.3 and 1.4.4 guarantee the existence of a unique solution w ∈N∏
j=1
H1((0, lj)
× (0, T )) of problem (1.24) given by
w(x, t) = ω(x, t) +∑
v∈Vc
η(v)(x)w(v)(x, t),
and satisfyingN∑
j=1
‖wj‖H1((0, lj)×(0, T )) .∑
v∈Vc
‖kv‖L2(0,T ). (1.31)
It remains to show the regularity at the nodes of Vc. Fix one vertex v ∈ Vc and a cut-offfunction χ(v) such that
χ(v)j ≡ 1 on [0, lv/3], χ
(v)j ≡ 0 on [2lv/3, lj], ∀j ∈ Ev,
where we have identified v to 0. Let us now set
W = χ(v)w.
This function W is solution of the following wave equation :
∂2Wj
∂t2(x, t) − ∂2Wj
∂x2 (x, t) = hj(x, t) 0 < x < lv, t > 0, ∀j ∈ Ev,Wj(0, t) = Wl(0, t) ∀j, l ∈ Ev, t > 0,∑
j∈Ev
∂Wj
∂x(0, t) = kv(t) t > 0,
Wj(lv, t) = 0 ∀j ∈ Ev, t > 0,W (t = 0) = 0, ∂W
∂t(t = 0) = 0,
(1.32)
where hj is given by
hj(x, t) := −∑
v∈Vc : j∈Ev
(
∂2χ(v)j
∂x2(x)wj(x, t) + 2
∂χ(v)j
∂x(x)
∂wj∂x
(x, t)
)
.
According to the estimate (1.31) satisfied by w, we have
‖hj‖L2((0,lv)×(0,T )) .∑
v′∈Vc
‖kv′‖L2(0,T ).
Recalling that Ev is the cardinal of Ev, we then may write
Ev = jii=1,··· ,Ev .
Introduce
V1 =Ev∑
i=1
Wji,
Vi = Wji −Wj1, ∀i = 2, · · · , Ev.
53
We remark that Vi, i = 2, · · · , Ev is solution of the wave equation with Dirichlet boundarycondition at 0 and lv, while V1 is solution of the wave equation with Dirichlet boundarycondition at lv and Neumann boundary condition at 0, namely
∂2V1
∂t2(x, t) − ∂2V1
∂x2 (x, t) = g(x, t) 0 < x < lv, t > 0,∂V1
∂x(0, t) = kv(t) t > 0,
V1(lv, t) = 0 t > 0,V1(t = 0) = 0, ∂V1
∂t(t = 0) = 0,
(1.33)
where g =∑
j∈Evhj and then satisfies
‖g‖L2((0,lv)×(0,T )) .∑
v′∈Vc
‖kv′‖L2(0,T ). (1.34)
But Lemma 1.4.5 below shows that
‖V1(0, ·)‖H1(0,T ) . ‖g‖L2((0,lv)×(0,T )) + ‖kv‖L2(0,T ).
Since
Wj1 =1
Ev(V1 −
Ev∑
i=2
Vi),
Wji = Vi +Wj1 , ∀i = 2, · · · , Ev,
we conclude that
‖Wji(0, ·)‖H1(0,T ) . ‖g‖L2((0,lv)×(0,T )) + ‖kv‖L2(0,T ), ∀i = 1, · · · , Ev.
Using (1.34), we obtain the estimate (1.25) from Proposition 1.4.1.
Lemma 1.4.5. Let V1 ∈ H1((0, lv) × (0, T )) be a solution of (1.33) with g ∈ L2((0, lv) ×(0, T )) and kv ∈ L2(0, T ). Then
‖∂V1
∂t(0, ·)‖L2(0,T ) . ‖g‖L2((0,lv)×(0,T )) + ‖kv‖L2(0,T ).
Proof: Let us denote by V the solution of
∂2V∂t2
(x, t) − ∂2V∂x2 (x, t) = g(x, t) 0 < x < lv, t > 0,
∂V∂x
(0, t) = kv(t) t > 0,V (lv, t) = 0 t > 0,V (·, t = 0) = 0, ∂V
∂t(·, t = 0) = 0,
(1.35)
where g (resp. kv) means the extension of g (resp. kv) by zero outside (0, T ). Since V = V1
on (0, lv) × (0, T ), it suffices to show that
‖∂V∂t
(0, ·)‖L2(0,T ) . ‖g‖L2((0,lv)×(0,T )) + ‖kv‖L2(0,T ). (1.36)
54
For that purpose we use a multiplier technique. Namely we multiply the wave equationsatisfied by V by (lv − x)(2T − t)Vx(x, t) (here and below for shortness we write Vx = ∂V
∂x)
and integrate the result on (0, lv) × (0, 2T ). This yields∫ lv
0
∫ 2T
0
g(x, t)(lv − x)(2T − t)Vx(x, t) dxdt = I1 − I2, (1.37)
where
I1 =
∫ lv
0
∫ 2T
0
Vtt(lv − x)(2T − t)Vx dxdt,
I2 =
∫ lv
0
∫ 2T
0
Vxx(lv − x)(2T − t)Vx dxdt.
For the first term I1 an integration by parts in time yields (no boundary terms occurbecause Vt(t = 0) = 0)
I1 =
∫ lv
0
∫ 2T
0
Vt(lv − x)Vx(x, t) dxdt− I3, (1.38)
where
I3 =
∫ lv
0
∫ 2T
0
Vt(lv − x)(2T − t)Vxt(x, t) dxdt.
For I3, an integration by parts in space and Leibniz’s rule lead to
I3 = −∫ lv
0
∫ 2T
0
∂
∂x(Vt(lv − x))(2T − t)Vt dxdt
+
[∫ 2T
0
(lv − x))(2T − t)|Vt|2 dt]lv
0
=
∫ lv
0
∫ 2T
0
|Vt|2(2T − t) dxdt− I3
−∫ 2T
0
(2T − t)lv|Vt(0, t)|2 dt.
This shows that
I3 =1
2
∫ lv
0
∫ 2T
0
|Vt|2(2T − t) dxdt− lv2
∫ 2T
0
(2T − t)|Vt(0, t)|2 dt.
Inserting this expression in (1.38) we find that
I1 =
∫ lv
0
∫ 2T
0
Vt(lv − x)Vx(x, t) dxdt (1.39)
−1
2
∫ lv
0
∫ 2T
0
|Vt|2(2T − t) dxdt
+lv2
∫ 2T
0
(2T − t)|Vt(0, t)|2 dt.
55
Similarly for the second term I2 an integration by parts in space yields
I2 = −∫ lv
0
∫ 2T
0
Vx(2T − t)∂
∂x((lv − x)Vx) dxdt
+
[∫ 2T
0
(2T − t)(lv − x)|Vx|2 dt]lv
0
= −I2 +
∫ lv
0
∫ 2T
0
|Vx|2(2T − t) dxdt
−lv∫ 2T
0
(2T − t)|Vx(0, t)|2 dt.
This means that
I2 =1
2
∫ lv
0
∫ 2T
0
|Vx|2(2T − t) dxdt (1.40)
− lv2
∫ 2T
0
(2T − t)|Vx(0, t)|2 dt.
Inserting (1.39) and (1.40) into the identity (1.37), we have obtained that∫ lv
0
∫ 2T
0
g(x, t)(lv − x)(2T − t)Vx(x, t) dxdt =
∫ lv
0
∫ 2T
0
Vt(lv − x)Vx(x, t) dxdt
−1
2
∫ lv
0
∫ 2T
0
|Vt|2(2T − t) dxdt
+lv2
∫ 2T
0
(2T − t)|Vt(0, t)|2 dt.
−1
2
∫ lv
0
∫ 2T
0
|Vx|2(2T − t) dxdt
+lv2
∫ 2T
0
(2T − t)|Vx(0, t)|2 dt.
Reminding the Neumann boundary condition satisfied by V we get
lv2
∫ 2T
0
(2T − t)|Vt(0, t)|2 dt =
∫ lv
0
∫ 2T
0
g(x, t)(lv − x)(2T − t)Vx(x, t) dxdt
−∫ lv
0
∫ 2T
0
Vt(lv − x)Vx(x, t) dxdt
+1
2
∫ lv
0
∫ 2T
0
|Vt|2(2T − t) dxdt
+1
2
∫ lv
0
∫ 2T
0
|Vx|2(2T − t) dxdt
− lv2
∫ T
0
(2T − t)|kv(t)|2 dt.
56
By Cauchy-Schwarz’s inequality we obtain finally
∫ 2T
0
(2T − t)|Vt(0, t)|2 dt . ‖g‖2L2((0,lv)×(0,T )) + ‖V ‖2
H1((0,lv)×(0,2T )).
This leads to the conclusion due to the estimate (1.34) and the estimate (consequence ofour previous considerations, see (1.31))
‖V ‖H1((0,lv)×(0,2T )) . ‖g‖L2((0,lv)×(0,T )) + ‖kv‖L2(0,T ),
and since 2T − t ≥ T on (0, T ).
Now we prove an a priori estimate which uses the trace regularity result of Proposition1.4.1 and that will be useful to prove our stability results for problem (1.1).
Let u ∈ C(0, T ; V )∩C1(0, T ; L2(R)) be the solution of (1.1). Then it can be splittedup in the form
u = φ+ ψ
where φ is solution of problem without damping (1.14), and ψ satisfies
∂2ψj
∂t2(x, t) − ∂2ψj
∂x2 (x, t) = 0 0 < x < lj , t > 0, ∀j ∈ 1, ..., N,ψj(v, t) = ψl(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂ψj
∂nj(v, t) = −(α
(v)1
∂u∂t
(v, t) + α(v)2
∂u∂t
(v, t− τv)) ∀v ∈ Vc, t > 0,
∑
j∈Ev
∂ψj
∂nj(v, t) = 0 ∀v ∈ Vint\Vcint, t > 0,
ψjv(v, t) = 0 ∀v ∈ D, t > 0,∂ψjv
∂njv(v, t) = 0 ∀v ∈ N , t > 0,
ψ(t = 0) = 0, ∂ψ∂t
(t = 0) = 0.(1.41)
In other words ψ is solution of problem (1.24) with
kv(t) = −(
α(v)1
∂u
∂t(v, t) + α
(v)2
∂u
∂t(v, t− τv)
)
. (1.42)
For all v ∈ Vc, we may write∥
∥
∥
∥
∂φ
∂t(v, .)
∥
∥
∥
∥
L2(0, T )
≤∥
∥
∥
∥
∂u
∂t(v, .)
∥
∥
∥
∥
L2(0, T )
+
∥
∥
∥
∥
∂ψ
∂t(v, .)
∥
∥
∥
∥
L2(0, T )
.
Now by Remark 1.3.2, kv = −(α(v)1
∂u∂t
(v, ·) + α(v)2
∂u∂t
(v, · − τv)) ∈ L2(0, T ) for all v ∈ Vc.Then we can apply Proposition 1.4.1 to ψ and obtain
∑
v∈Vc
∥
∥
∂ψ∂t
(v, .)∥
∥
L2(0, T ).
∑
v∈Vc
∥
∥
∥α
(v)1
∂u∂t
(v, .) + α(v)2
∂u∂t
(v, .− τv)∥
∥
∥
L2(0, T )
.∑
v∈Vc
(∥
∥
∂u∂t
(v, .)∥
∥
L2(0, T )+∥
∥
∂u∂t
(v, .− τv)∥
∥
L2(0, T )).
57
The two above estimates yield∑
v∈Vc
∥
∥
∂φ∂t
(v, .)∥
∥
L2(0, T )≤
∑
v∈Vc
∥
∥
∂u∂t
(v, .)∥
∥
L2(0, T )+∑
v∈Vc
∥
∥
∂ψ∂t
(v, .)∥
∥
L2(0, T )
.∑
v∈Vc
(∥
∥
∂u∂t
(v, .)∥
∥
L2(0, T )+∥
∥
∂u∂t
(v, .− τv)∥
∥
L2(0, T )).
This directly leads to the next a priori bound :
Lemma 1.4.6. Suppose that (u(0), u(1), (f 0(−τv·))v∈Vc) ∈ H. Then the solutions u of (1.1)with initial data (u(0), u(1), (f 0(−τv·))v∈Vc) and φ of (1.14) with initial data (u(0), u(1))(which belongs to V × L2(R)) satisfy
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt .∑
v∈Vc
∫ T
0
(
(∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2
)
dt.
1.5 The exponential stability
Our approach is based (as for the polynomial stability) on a trace regularity result(Proposition 1.4.1 and Lemma 1.4.6) and on an observability inequality of problem withoutdamping (1.14).
1.5.1 An observability inequality
Proposition 1.5.1. Let (ϕk, i)1≤i≤lk, k≥1 be the orthonormal basis of the operator A. Let φbe the solution of (1.14) with (u(0), u(1)) ∈ V ×L2(R). Then there exists a time T > 0 anda constant C > 0 (depending on T ) such that
∥
∥u(0)∥
∥
2
V+∥
∥u(1)∥
∥
2
L2(R)≤ C
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt (1.43)
if and only if∃α > 0, ∀k = 1, · · · , N + 1, ∀n ∈ Ak, λmin(Mn) ≥ α, (1.44)
where the matrix Mn is defined by
Mn =∑
v∈Vc
Φn(v)⊤B−⊤
n B−1n Φn(v),
the matrix Φn(v) of size k × Ln, where Ln =∑k
i=1 ln+i−1, is given as follows : for alli = 1, · · · , k, we set
(Φn(v))ij =
ϕn+i−1,j−Ln,i−1(v) if Ln,i−1 < j ≤ Ln,i,
0 else,
where Ln,0 = 0 and for i ≥ 1, Ln,i =∑i
i′=1 ln+i′−1.
58
Proof: We first show that (1.44)⇒(1.43). Writting u(0) =∑
k≥1
∑lki=1 ak, iϕk, i and u(1) =
∑
k≥1
∑lki=1 bk, iϕk, i where (λkak, i)i, k, (bk, i)i, k ∈ l2(N∗), then the solution φ of problem
(1.14) is given by
φ(·, t) =∑
k≥1
lk∑
i=1
(
ak, i cos(λkt) +bk, iλk
sin(λkt)
)
ϕk, i.
Consequently for any v ∈ Vc, we get
∂φ
∂t(v, t) =
∑
k≥1
lk∑
i=1
(−ak, iλk sin(λkt) + bk, i cos(λkt))ϕk, i(v).
Putting together the terms corresponding to the same eigenvalue, we obtain
∂φ
∂t(v, t) =
∑
k≥1
(
lk∑
i=1
− ak, iϕk, i(v)
)
λk sin(λkt) +∑
k≥1
(
lk∑
i=1
bk, iϕk, i(v)
)
cos(λkt).
Using the notation introduced in Proposition 1.3.7, this is equivalent to
∂φ
∂t(v, t) =
∑
n∈Z∗
αn(v)eiλnt.
Integrating the square of this identity between 0 and T > 0 and using Ingham’s inequality(1.17) for T large enough, and summing on v ∈ Vc, we get
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt &
N+1∑
k=1
∑
|n|∈Ak
∑
v∈Vc
‖B−1n Cn(v)‖2
2.
But for all n ∈ Ak, setting
An =(
λnan, 1, · · · , λnan, ln , λn+1an+1,1, · · · , λn+1an+1, ln+1 , · · · ,λn+k−1an+k−1, 1, · · · , λn+k−1an+k−1, ln+k−1
)⊤,
Bn =(
bn, 1, · · · , bn, ln , bn+1,1, · · · , bn+1, ln+1, · · · , bn+k−1, 1, · · · , bn+k−1, ln+k−1
)⊤,
we readily check that
∑
v∈Vc
‖B−1n Cn(v)‖2
2 =1
4(B⊤
n MnBn + A⊤nMnAn).
Hence the assumption (1.44) yields (because Mn is a symmetric matrix)
∑
v∈Vc
‖B−1n Cn(v)‖2
2 & ‖Bn‖22 + ‖An‖2
2.
59
Therefore under our hypothesis, we have
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt &∑
k≥1
lk∑
i=1
(a2k, iλ
2k + b2k, i)
&∥
∥u(0)∥
∥
2
V+∥
∥u(1)∥
∥
2
L2(R)
because (ϕk,i)k,i is an orthonormal basis associated with the operator A.It remains to show that (1.43)⇒(1.44).Let k = 1, · · · , N + 1 and n ∈ Ak be fixed. Take u(0) =
∑n+k−1m=n
∑lmi=1 am, iϕm, i and
u(1) =∑n+k−1
m=n
∑lmi=1 bm, iϕm, i. Then the solution φ of problem (1.14) is given by
φ(·, t) =
n+k−1∑
m=n
lm∑
i=1
(
am, i cos(λmt) +bm, iλm
sin(λmt)
)
ϕm, i.
Then for v ∈ Vc
∂φ
∂t(v, t) =
n+k−1∑
m=n
lm∑
i=1
(−am, iλm sin(λmt) + bm, i cos(λmt))ϕm, i(v).
Applying again Ingham’s inequality, we get for T large enough
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt ∼ B⊤n MnBn + A⊤
nMnAn.
By (1.43), we obtain
B⊤n MnBn + A⊤
nMnAn ≥ Cn+k−1∑
m=n
lm∑
i=1
(a2m, iλ
2n + b2m, i),
for some C > 0. Hence we conclude that
λmin(Mn) ≥ C.
This ends the proof.
Remark 1.5.2. If the standard gap condition (1.18) holds, then the condition (1.44)reduces to
∃α > 0, ∀k ≥ 1, λmin(M(λ2k)) ≥ α. (1.45)
In particular if lk = 1, then the condition (1.45) becomes
∀k ≥ 1,∑
v∈Vc
|ϕk(v)|2 ≥ α.
60
1.5.2 The stability result
We are now ready to give a sufficient condition that guarantees the exponential stabilityof (1.1).
Theorem 1.5.3. The system (1.1) is exponentially stable in the energy space if (1.43)holds.
Proof: Let u be a solution of (1.1) with initial data (u(0), u(1), (f 0(−τv·))v∈Vc) ∈ D(A).Integrating the inequality (1.13) of Proposition 1.3.1 between 0 and T where T is
sufficiently large, we obtain
E(0) − E(T ) &∑
v∈Vc
∫ T
0
[(∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2]dt
&1
2
∑
v∈Vc
∫ T
0
[(∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2]dt+1
2
∑
v∈Vc
∫ T
0
(∂u
∂t(v, t− τv))
2dt
&∑
v∈Vc
∫ T
0
(∂φ
∂t(v, t))2dt+
∑
v∈Vc
∫ T
0
(∂u
∂t(v, t− τv))
2dt by Lemma 1.4.6
&∥
∥u(0)∥
∥
2
V+∥
∥u(1)∥
∥
2
L2(R)+∑
v∈Vc
∫ T
0
(∂u
∂t(v, t− τv))
2dt
by assumption
&∥
∥u(0)∥
∥
2
V+∥
∥u(1)∥
∥
2
L2(R)+∑
v∈Vc
τv
∫ 1
0
(∂u
∂t(v, −τvρ))2dρ.
Indeed, for T > τv, by changes of variables, we have∫ T
0
(∂u
∂t(v, t− τv))
2dt =
∫ T−τv
−τv(∂u
∂t(v, t))2dt
≥∫ 0
−τv(∂u
∂t(v, t))2dt = τv
∫ 1
0
(∂u
∂t(v, −τvρ))2dρ.
(1.46)
The previous inequality directly implies that
E(0) −E(T ) &1
2
(
‖u(0)‖2V + ‖u(1)‖2
L2(R) +∑
v∈Vc
ξ(v)
2
∫ 1
0
(∂u
∂t(v, −τvρ))2dρ
)
.
This means that for T large enough
E(0) −E(T ) ≥ CE(0)
for some C > 0. Since our system is invariant by translation and the energy is decreasing,it is well known that the above estimate implies the existence of C1 > 0 and C2 > 0 suchthat
E(t) ≤ C1E(0)e−C2t, ∀t ≥ 0. (1.47)
Hence the energy decays exponentially. By density of D(A) into H , we deduce that (1.47)holds for any initial data in H .
61
1.6 The polynomial stability
1.6.1 An observability estimate
Proposition 1.6.1. Let (ϕk, i)1≤i≤lk, k≥1 be an orthonormal basis of eigenvectors of theoperator A. Let m ∈ N∗. Let φ be the solution of (1.14) with (u(0), u(1)) ∈ V ×L2(R). Thenthere exists a time T > 0 and a constant C > 0 such that
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt ≥ C
(
∑
k≥1
1
k2m
lk∑
i=1
(a2k, iλ
2k + b2k, i)
)
(1.48)
where u(0) =∑
k≥1
lk∑
i=1
ak, iϕk, i and u(1) =∑
k≥1
lk∑
i=1
bk, iϕk, i if and only if
∃m ∈ N∗, ∃α > 0, ∀k = 1, · · · , N + 1, ∀n ∈ Ak, λmin(Mn) ≥ α
n2m. (1.49)
Proof: The proof is similar to the one of Proposition 1.5.1 and is therefore omitted.
Remark 1.6.2. As before, if the standard gap condition (1.18) holds, then the condition(1.49) reduces to
∃m ∈ N∗, ∃α > 0, ∀k ≥ 1, λmin(M(λ2k)) ≥
α
k2m, (1.50)
and if lk = 1, then condition (1.50) is nothing else than
∃α > 0, ∀k ≥ 1,∑
v∈Vc
|ϕk(v)|2 ≥α
k2m.
Let (u(0), u(1), (f 0(−τv.))) ∈ D(A). By the so-called Weyl’s formula (see for instance[20, 82]), we have
λk ∼kπ
L
where L =∑N
j=1 lj. This implies that
∑
k≥1
1k2m
lk∑
i=1
(a2k, iλ
2k + b2k, i) ∼
∑
k≥1
lk∑
i=1
(a2k, iλ
2(1−m)k + b2k, iλ
−2mk )
∼∥
∥u(0)∥
∥
2
D(A1−m
2 )+∥
∥u(1)∥
∥
2
D(A−m2 )
because, for u =∑
k≥1
∑lki=1 uk,iϕk,i, we have ‖u‖2
D(As) ∼∑
k≥1
∑lki=1 λ
4sk u
2k,i for all s ∈ R.
Therefore the observability estimate (1.48) is equivalent to
∑
v∈Vc
∫ T
0
(
∂φ
∂t(v, t)
)2
dt &∥
∥u(0)∥
∥
2
D(A1−m
2 )+∥
∥u(1)∥
∥
2
D(A−m2 ).
62
Now using Lemma 1.4.6, we obtain
∥
∥u(0)∥
∥
2
D(A1−m
2 )+∥
∥u(1)∥
∥
2
D(A−m2 )
.∑
v∈Vc
∫ T
0
(
(∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2
)
dt.
On the other hand integrating the inequality (1.13) of Proposition 1.3.1 between 0 and Twhere T is sufficiently large, we have
E(0) −E(T ) &∑
v∈Vc
∫ T
0
((∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2)dt
&1
2
∑
v∈Vc
∫ T
0
((∂u
∂t(v, t))2 + (
∂u
∂t(v, t− τv))
2)dt
+1
2
∑
v∈Vc
∫ T
0
(∂u
∂t(v, t− τv))
2dt.
Therefore
E(0) − E(T ) &∥
∥u(0)∥
∥
2
D(A1−m
2 )+∥
∥u(1)∥
∥
2
D(A−m2 )
+∑
v∈Vc
∫ T
0
(∂u
∂t(v, t− τv))
2dt
&∥
∥(u(0), u(1))∥
∥
2
X−m+∑
v∈Vc
τv
∫ 1
0
(∂u
∂t(v, −τvρ))2dρ,
where X−m = D(A1−m
2 )×D(A−m2 ) and due to (1.46). This finally shows that the observa-
bility estimate (1.48) implies that
E(T ) ≤ E(0) −K1
(
∥
∥(u(0), u(1))∥
∥
2
X−m+∑
v∈Vc
∫ 1
0
(∂u
∂t(v, −τvρ))2dρ
)
, (1.51)
for some K1 > 0.Now we recall the following interpolation result.
Lemma 1.6.3. For (u(0), u(1)) ∈ D(A) × V , we have
∥
∥u(0)∥
∥
m+1
D(A12 )
.∥
∥u(0)∥
∥
m
D(A)
∥
∥u(0)∥
∥
D(A1−m
2 ),
∥
∥u(1)∥
∥
m+1
D(A0).
∥
∥u(1)∥
∥
m
D(A12 )
∥
∥u(1)∥
∥
D(A−m2 )
.
Proof: Direct consequence of the equivalence ‖u‖2D(As) ∼
∑
k≥1
∑lki=1 |uk,i|2 λ4s
k for all s ∈R, when u =
∑
k≥1
∑lki=1 uk,iϕk,i and of Holder’s inequality.
Corollary 1.6.4. For all u ∈ X, it holds
‖u‖m+1V . ‖u‖mX ‖u‖
D(A1−m
2 ). (1.52)
63
Proof: We fix a sequence of cut-off functions ηv, v ∈ V satisfying
∑
v∈Vηv = 1 on R, ηv ≡ 1 near v,
and such that the support of ηv is included into Sv, where Sv is the star-shaped networkmade of the edges ej , j ∈ Ev.
Denote by Av the Laplace operator defined on Sv with Dirichlet boundary conditionon all nodes of Sv except at v, where we impose Neumann or transmission conditions. Letus show that if (1.52) holds on Sv for ηvu, then it holds on R. Indeed a convex inequalityyields
‖u‖m+1V .
∑
v∈V‖ηvu‖m+1
V .
Now using (1.52) on Sv for ηvu, we get
‖u‖m+1V .
∑
v∈V‖ηvu‖mXv
‖ηvu‖D(A
1−m2
v ), (1.53)
the norm Xv being defined as the norm X but on Sv. By Leibniz’s rule, we directly have
‖ηvu‖Xv. ‖u‖X .
Therefore it remains to estimate ‖ηvu‖D(A
1−m2
v ). For that purpose, we use a duality argu-
ment, namely
‖ηvu‖D(A
1−m2
v )= sup
w∈D(Am−1
2v )
∫
Svηvuw
‖w‖D(A
m−12
v )
= sup
w∈D(Am−1
2v )
∫
Svuηvw
‖w‖D(A
m−12
v )
= sup
w∈D(Am−1
2v )
∫
R uηvw
‖w‖D(A
m−12
v )
,
by extending ηvw by zero outside Sv. Using the duality in R, we obtain
‖ηvu‖D(A
1−m2
v )≤ ‖u‖
D(A1−m
2 )sup
w∈D(Am−1
2v )
‖ηvw‖D(A
m−12 )
‖w‖D(A
m−12
v )
.
But again Leibniz’s rule yields
sup
w∈D(Am−1
2v )
‖ηvw‖D(A
m−12 )
‖w‖D(A
m−12
v )
. 1,
64
which shows that‖ηvu‖
D(A1−m
2v )
. ‖u‖D(A
1−m2 )
.
This estimate in (1.53) shows that (1.52) holds.We are reduced to show that (1.52) holds on Sv for ηvu. For that purpose, without loss
of generality we may assume that v is identified with 0 for all edges of Ev and that suppηv ⊂ ∪j∈Ev [0, 2lj/3]. For shortness write U = ηvu. Now for any j ∈ Ev, we introduced thefollowing extension of Uj :
EjUj(x) =
Uj(x) if x ∈ (0, lj),∑n−1
i=0 νiUj(−2ix) if x ∈ (−2−(n−1)lj , 0),
where U is extended by zero outside its support and the real numbers νi are the uniquesolution of the system
∑n−1i=0 νi = 1
−∑n−1i=0 2iνi = 1
∑n−1i=0 22iνi = 1
∑n−1i=0 2−2kiνi = 1, ∀k = 1, · · · , n− 3,
and finally n = m+52
if m is odd and n = m+42
if m is even.With the help of these extension operators, we obtain an extension of U ∈ Xv to a func-
tion EU , which belongs toD(Av) (due to the three first properties of the νi), where Av is thepositive Laplace operator on the star shaped network Sv = ∪j∈Ev(0, lj)
⋃∪j∈Ev(−2−(n−1)lj , 0),with interior vertex v and Dirichlet boundary conditions at all other vertices. ApplyingLemma 1.6.3 to EU on the network Sv, we may write
‖EU‖m+1
D(A12v )
. ‖EU‖mD(Av) ‖EU‖D(A
1−m2
v ).
Now using the fact that the norm in D(A12v ) is equivalent to the H1 semi-norm and since
EU is equal to U on Sv, we obtain
‖U‖m+1
D(A12v )
. ‖EU‖mD(Av) ‖EU‖D(A
1−m2
v ). (1.54)
This means that it remains to estimate the right-hand side of this estimate. First byconstruction, we easily check that
‖EU‖D(Av) . ‖U‖Xv. (1.55)
As before to estimate the second factor, we use a duality argument. First we remark that
for w ∈ D(Am−1
2v ), we have
∫
Sv
EUw =∑
j∈Ev
∫ lj
0
Uj(x)wj(x) dx+∑
j∈Ev
∫ 0
−2−(n−1)lj
(Eu)−j(x)w−j(x) dx,
65
where for shortness we write w−j the restriction of w to the edge (−2−(n−1)lj, 0). By changesof variables, we obtain
∫
Sv
Euw =∑
j∈Ev
∫ lj
0
Uj(x)(Fw)j(x) dx,
where
(Fw)j(x) = wj(x) + χj(x)
n−1∑
i=0
νi2−iUj(−2−ix), ∀x ∈ (0, lj),
the cut-off function χj being fixed such that χj ≡ 1 on [0, 2lj/3] and χj ≡ 0 on [5lj/6, lj] (re-minding that Uj(x) ≡ 0 for x > 2lj/3). Now we notice that the conditions on νi guarantees
that Fw belongs to D(Am−1
2v ) and by Leibniz’s rule we have
‖Fw‖D(A
m−12
v ). ‖w‖
D(Am−1
2v )
.
By duality we conclude that
‖EU‖D(A
1−m2
v ). ‖U‖
D(A1−m
2v )
.
This estimate and (1.55) in (1.54) show the requested estimate (1.52) on Sv for ηvu.
Now let (u(0), u(1), (f 0(−τv.))) ∈ D(A) be fixed (and different from 0). By a convexity
inequality and since V × L2(R) = D(A12 ) ×D(A0), we have
∥
∥(u(0), u(1))∥
∥
m+1
V×L2(R).∥
∥u(0)∥
∥
m+1
D(A12 )
+∥
∥u(1)∥
∥
m+1
D(A0).
Using Lemma 1.6.3 and Corollary 1.6.4, we get
∥
∥(u(0), u(1))∥
∥
m+1
V×L2(R).∥
∥u(0)∥
∥
m
X
∥
∥u(0)∥
∥
D(A1−m
2 )+∥
∥u(1)∥
∥
m
D(A12 )
∥
∥u(1)∥
∥
D(A−m2 )
. (∥
∥u(0)∥
∥
m
X+∥
∥u(1)∥
∥
m
D(A12 )
)(∥
∥u(0)∥
∥
D(A1−m
2 )+∥
∥u(1)∥
∥
D(A−m2 )
)
. (∥
∥u(0)∥
∥
X+∥
∥u(1)∥
∥
D(A12 )
)m(∥
∥u(0)∥
∥
D(A1−m
2 )+∥
∥u(1)∥
∥
D(A−m2 )
)
.∥
∥(u(0), u(1))∥
∥
m
X×D(A12 )
∥
∥(u(0), u(1))∥
∥
D(A1−m
2 )×D(A−m2 )
.∥
∥(u(0), u(1))∥
∥
m
X×D(A12 )
∥
∥(u(0), u(1))∥
∥
X−m.
This inequality is equivalent to
∥
∥(u(0), u(1))∥
∥
2
X−m&
∥
∥(u(0), u(1))∥
∥
2m+2
V×L2(R)
‖(u(0), u(1))‖2m
X×D(A12 )
.
Using the trivial inequality∥
∥(u(0), u(1))∥
∥
X×D(A12 )
≤∥
∥(u(0), u(1), (f 0(−τv.)))∥
∥
X×D(A12 )×H1(0, 1)Vc
.∥
∥(u(0), u(1), (f 0(−τv.)))∥
∥
D(A),
66
we finally obtain
∥
∥(u(0), u(1))∥
∥
2
X−m&
∥
∥(u(0), u(1))∥
∥
2m+2
V×L2(R)
‖(u(0), u(1), (f 0(−τv.)))‖2mD(A)
. (1.56)
1.6.2 Polynomial decay of the energy
The proof of our stability result requires the next technical Lemma proved in Lemma5.2 of [10].
Lemma 1.6.5. Let (εk)k be a sequence of positive real numbers satisfying
εk+1 ≤ εk − Cε2+αk+1 , ∀k ≥ 0, (1.57)
where C > 0 and α > −1. Then there exists a positive constant M (depending on α andC) such that
εk ≤M
(1 + k)1
1+α
, ∀k ≥ 0.
Theorem 1.6.6. Let u be a solution of (1.1) with (u(0), u(1), (f 0(−τv·))) ∈ D(A). If (1.48)holds, then the energy decays polynomially, i.e.
E(t) ≤ C
(1 + t)1m
∥
∥(u(0), u(1), (f 0(−τv·)))∥
∥
2
D(A), ∀t ≥ 0, (1.58)
for some C > 0.
Proof: Introduce the modified energy
E(t) =1
2‖U(t)‖2
D(A) =1
2(‖U(t)‖2
H + ‖AU(t)‖2H).
As in Proposition 1.3.1, this energy is decaying.Combining the estimates (1.51) and (1.56), we obtain
E(T ) ≤ E(0) −K2
∥
∥(u(0), u(1))∥
∥
2m+2
V×L2(R)
E(0)m+∑
v∈Vc
∫ 1
0
(∂u
∂t(v, −τvρ))2dρ
,
for some K2 > 0, or equivalently
E(T ) ≤ E(0) −K2
∥
∥(u(0), u(1))∥
∥
2m+2
V×L2(R)
E(0)m+∥
∥(f 0v (−τv.))
∥
∥
2
L2(0, 1)Vc
.
Using the trivial estimate∥
∥(f 0v (−τv.))
∥
∥
2m+2
L2(0, 1)Vc≤
∥
∥(f 0v (−τv.))
∥
∥
2
L2(0, 1)Vc
∥
∥(f 0v (−τv.))
∥
∥
2m
H1(0, 1)Vc
.∥
∥(f 0v (−τv.))
∥
∥
2
L2(0, 1)VcE(0)m,
67
the above inequality becomes
E(T ) ≤ E(0) −K2
∥
∥(u(0), u(1))∥
∥
2m+2
V×L2(R)+ ‖(f 0
v (−τv.))‖2m+2L2(0, 1)Vc
E(0)m
≤ E(0) −K3E(0)m+1
E(0)m,
with K3 > 0. Since the energy of our system is decaying, we obtain
E(T ) ≤ E(0) −K3E(T )m+1
E(0)m. (1.59)
We now follow the method used in [10]. The estimate (1.59) being valid on the intervals[kT, (k + 1)T ], for any k ≥ 0, we have
E((k + 1)T ) ≤ E(kT ) −K3E((k + 1)T )m+1
E(kT )m. (1.60)
We set
εk =E(kT )
E(0).
By dividing (1.60) by E(0), we obtain
εk+1 ≤ εk −K3εm+1k+1 , (1.61)
because E(kT ) ≤ E(0). By Lemma 1.6.5 with α = m− 1 > −1 (since m > 0), there existsa constant M > 0 such that
εk ≤M
(1 + k)1m
, ∀k ≥ 0,
or equivalently
E(kT ) ≤ M
(1 + k)1m
E(0).
This estimate and again the decay of the energy lead to the estimate (1.58).
1.7 Examples
Our aim is to give some concrete examples that illustrate our stability results.
1.7.1 One string with an interior damping
We consider a homogeneous string of length π with an interior damping at ξ. Two typesof boundary conditions will be considered.
68
Mixed boundary conditions
We study the following problem (see Fig. 1.1)
∂2u∂t2
(x, t) − ∂2u∂x2 (x, t) = 0 0 < x < π, t > 0,
∂u∂x
(ξ−, t) − ∂u∂x
(ξ+, t) = −(α1∂u∂t
(ξ, t) + α2∂u∂t
(ξ, t− τ)) t > 0,u(0, t) = 0, ∂u
∂x(π, t) = 0 t > 0,
u(t = 0) = u(0), ∂u∂t
(t = 0) = u(1) 0 < x < π,∂u∂t
(ξ, t− τ) = f 0(t− τ) 0 < t < τ.
(1.62)
Here contrary to subsection 1.3.4, we suppose that α2 < α1.
0D
π ξ
N
Fig. 1.1 –
It is well known that the eigenvectors associated with problem (1.62) without damping
are ϕk(x) =√
2π
sin((k+ 12)x) of eigenvalue (k+ 1
2)2, k ≥ 0 of multiplicity 1. We then have
to look at∑
v∈Vc
|ϕk(v)|2 =2
πsin2
(
(k +1
2)ξ
)
.
For the exponential decay we will use the next result proved in Lemma 2.9 of [97].
Lemma 1.7.1. s is a rational number with an irreducible fraction
s =p
q, where p is odd (1.63)
if and only if there exists α > 0 such that∣
∣
∣sin(
(π
2+ kπ)s
)∣
∣
∣ > α, ∀k ∈ N. (1.64)
Applying Proposition 1.3.7 and Theorem 1.5.3 (and the above Lemma), we obtain
Theorem 1.7.2. 1) The energy of system (1.62) tends to 0 for all initial data in H if andonly if
ξ
π6= 2p
2q + 1, ∀p, q ∈ N.
2) The system (1.62) is exponentially stable in the energy space if ξπ
is a rational numberwith an irreducible fraction
ξ
π=p
q, where p is odd.
If we consider the system (1.62) without delay (i.e. α2 = 0), we find the results of [6]obtained by a similar method.
69
Dirichlet boundary conditions
We here consider the problem (see Fig. 1.2)
∂2u∂t2
(x, t) − ∂2u∂x2 (x, t) = 0 0 < x < π, t > 0,
∂u∂x
(ξ−, t) − ∂u∂x
(ξ+, t) = −(α1∂u∂t
(ξ, t) + α2∂u∂t
(ξ, t− τ)) t > 0,u(0, t) = 0, u(π, t) = 0 t > 0,u(t = 0) = u(0), ∂u
∂t(t = 0) = u(1) 0 < x < π,
∂u∂t
(ξ, t− τ) = f 0(t− τ) 0 < t < τ.
(1.65)
0D
π ξ
D
Fig. 1.2 –
The eigenvectors of problem without damping associated with problem (1.65) are
ϕk(x) =√
2π
sin(kx) of eigenvalue k2, k ≥ 1 of multiplicity 1. We then have to consider
∑
v∈Vc
|ϕk(v)|2 =2
πsin2(kξ).
Denote by S the set of all real numbers ρ such that ρ /∈ Q and if [0, a1, ..., an, ...] isthe expansion of ρ as a continued fraction, then the sequence (an) is bounded. It is wellknown that S is uncountable and that its Lebesgue measure is zero. Roughly speaking, theset S contains all irrational numbers which are badly approximated by rational numbers.In particular, by the Euler-Lagrange theorem, S contains all irrational quadratic numbers(i.e. the roots of a second order equation with rational coefficients). By a classical result,we have the
Lemma 1.7.3. If s ∈ S, then there exists a positive constant C such that
|sin(kπs)| ≥ C
k, ∀k ≥ 1.
By Proposition 1.3.7 and Theorem 1.6.6, we then obtain
Theorem 1.7.4. 1) The energy of system (1.65) decays to 0 for all initial data in H ifand only if
ξ
π/∈ Q.
2) If ξπ∈ S, then for all initial data in D(A), the energy of system (1.65) decays polyno-
mially like 1/t.
Without delay, we find the results of [11].
70
Remark 1.7.5. These two examples show that the boundary conditions influence thestability of the system because we do not have the same hypotheses for the decay to 0 ofthe energy ; moreover for mixed boundary conditions, we may have an exponential stability,while for Dirichlet boundary conditions, we cannot have an exponential stability but havea polynomial stability.
1.7.2 A star shaped network
Dirichlet boundary conditions at all extremities
We take Dirichlet boundary conditions at all extremities and put a damping at theinterior node (see Fig. 1.3).
D
D
D
D
Fig. 1.3 –
In appropriate coordinates, any eigenvector of the problem without damping is of theform
ϕj(x) = aj sin(λx), 1 ≤ j ≤ N,
for some constants aj . The transmission conditions at the interior node then lead to
a1 sin(λl1) = ... = aN sin(λlN), (1.66)N∑
j=1
aj cos(λlj) = 0. (1.67)
We now suppose that
∀i, j ∈ 1, ..., N, i 6= j,lilj/∈ Q. (1.68)
In that case we cannot have sin(λli) = sin(λlj) = 0 for i 6= j. Indeed if λli = pπ, p ∈ Z andλlj = qπ, q ∈ Z, then li
lj= p
q∈ Q, which contradicts our assumption.
Therefore, there exists j ∈ 1, ..., N, say j = 1 such that sin(λlj) 6= 0. But then a1 6= 0.Indeed if a1 = 0, then
a2 sin(λl2) = ... = aN sin(λlN) = 0.
As we cannot have sin(λli) = sin(λlj) = 0 for i 6= j, all aj are equal to zero except one, sayaN for example and then sin(λlN) = 0. By the last transmission condition (1.67) we wouldhave aN cos(λlN) = 0 and then aN = 0, which is impossible.
71
Under the assumption (1.68), we have
sin(λlj) 6= 0, ∀j = 1, · · · , N,
and the transmission condition (1.67) yields the characteristic equation
N∑
j=1
cot(λlj) = 0. (1.69)
From this equation, we deduce that all eigenvalues are simple. Hence
∑
v∈Vc
|ϕk(v)|2 = a21 sin2(λl1) > 0,
when ϕk is the eigenvector associated with λ and by Proposition 1.3.7 we directly concludethe
Proposition 1.7.6. If (1.68) holds, then the energy of our system tends to 0 for all initialdata in H.
Without delay we find the result of Proposition 2.1 of [7].After calculations and normalization, we find that
a21 sin2(λl1) =
2N∑
j=1
ljsin2(λlj)
.
In its full generality, it is difficult to find the behavior of a21 sin2(λl1) from the characteristic
equation (1.69). We therefore restrict ourselves to some particular cases.We first suppose that the lengths of the edges are equal to 1. Then easy calculations
allow to show that the set of eigenvalues of the operator A is made of two sequences : Firstλ2k = (π
2+ kπ)2, k ∈ N is an eigenvalue of multiplicity 1 with associated eigenvector
(ϕk)j(x) =
√
1
2sin(
(π
2+ kπ)x
)
, ∀j ∈ 1, ..., N.
Secondly, λ2k = k2π2, k ∈ N∗ is of multiplicity N − 1 with orthonormal eigenvectors given
by
ϕk, 1 = sin(kπx)
1−10...0
, ϕk, 2 =2√3
sin(kπx)
1212
−1...0
, · · · ,
72
ϕk, i =
√2i√
1 + isin(kπx)
1i1i...1i
−10...0
, · · · , ϕk,N−1 =
√
2(N − 1)√N
sin(kπx)
1N−1
1N−1...1
N−1
−1
.
where for shortness we write ϕk,1 as a vector with N components, the jth componentcorresponding to the restriction of ϕk,1 to the edge j.
If the feedback law is applied only at the interior node, as for the eigenvalue λ2k = k2π2,
Mvint(k2π2) = 0 (because sin(kπ) = 0), the eigenvalues of Mvint
(k2π2) are 0. ThereforeProposition 1.3.7 yields
Proposition 1.7.7. If the lengths of the star shaped network are one and the feedback lawis applied at the interior node, the energy does not decay to 0.
We then need to add some interior controls to stabilize the system. On each edge ejexcept one (for instance the first one), we put a control at ξj (see Fig. 1.4).
D
D
D
D
D
ξ2
ξ N
Fig. 1.4 –
Then for the eigenvalue λ2k = k2π2, we readily check that
M(k2π2) = Mvint(k2π2) +
N∑
j=2
Mξj (k2π2)
satifies
η⊤M(k2π2)η =
N∑
j=2
sin2(kπξj)(χ⊤j η)
2,
where the vectors χj = ((χj)i)Ni=2, j = 2, . . . , N are non zero vectors, which are independent
of k and of ξj , j = 2, . . . , N and satisfy
(χj)i = 0, ∀i < j.
73
Consequently if for all j ∈ 2, ..., N, ξj ∈ S, then sin2(kπξj) & 1k2 and therefore
η⊤M(k2π2)η &1
k2
N∑
j=2
(χ⊤j η)
2.
The above properties of χj imply that (∑N
j=2(χ⊤j η)
2)1/2 is a norm on RN−1 and therefore
η⊤M(k2π2)η &1
k2‖η‖2
2,
or equivalently λmin(M(k2π2)) & 1k2 .
On the other hand for λk = π2
+ kπ, we see that
∑
v∈Vc
|ϕk(v)|2 =1
2sin2
(π
2+ kπ
)
+1
2
N∑
j=2
sin2(
(π
2+ kπ)ξj
)
≥ 1
2.
We then conclude the
Proposition 1.7.8. If the lengths of the star shaped network are one and the feedback lawis applied at the interior node, and at N − 1 interior points ξj and if
ξj ∈ S, ∀j ∈ 2, ..., N,
then for any (u(0), u(1), (f 0(−τv.))) ∈ D(A), the energy of the system decays like 1t.
Now we assume that N = 2n is even and that
li = l, ∀i = 1, · · · , n, li = l′, ∀i = n + 1, · · · , 2n andl
l′6∈ Q. (1.70)
Under this assumption, (1.69) is equivalent to
cot(λl) + cot(λl′) = 0,
and then tosin(λ(l + l′)) = 0.
This means that a first set of eigenvalues is given by
λk =kπ
l + l′, k ∈ N∗.
A second set is made of the roots of sin(λl) = 0. Since ll′6∈ Q, we deduce that ai = 0,
for all i = n + 1, · · · , 2n. Consequently for all k ∈ N∗, k2π2
l2is of multiplicity n − 1, its
74
associated orthonormal eigenvectors being given by :
ϕk,1(x) =1√lsin
(
kπx
l
)
(1,−1, 0, · · · , 0, 0, · · · , 0)⊤ ,
ϕk,2(x) =2√3l
sin
(
kπx
l
)(
1
2,1
2,−1, 0, · · · , 0, 0, · · · , 0
)⊤,
...
ϕk,n−1(x) =
√
2(n− 1)√nl
sin
(
kπx
l
)(
1
n− 1,
1
n− 1, · · · , 1
n− 1,−1, 0, · · · , 0
)⊤.
By symmetry a third set of eigenvalues is made of the numbers k2π2
(l′)2of multiplicity n−1
for all k ∈ N∗.Note that for this example, the standard gap condition holds.Since the eigenvectors associated with the eigenvalues of the second and third sets are
zero at the interior node, if we impose a damping only at this interior node, the energy willnot tend to zero. Therefore some interior control points have to be added. More preciselywe impose a feedback law at some points ξi of the edge ei, for i = 2, · · · , n and fori = n + 2, · · · , 2n. By direct calculations, the matrix
M(
k2π2
l2
)
= Mvint
(
k2π2
l2
)
+
n∑
j=2
Mξj
(
k2π2
l2
)
+
2n∑
j=n+2
Mξj
(
k2π2
l2
)
satisfies
η⊤M(
k2π2
l2
)
η = l−1n∑
j=2
sin2
(
kπξjl
)
(χ⊤j η)
2,
where the vectors χj = ((χj)i)ni=2, j = 2, . . . , n satisfy the same property than before. Hence
if for all j ∈ 2, ..., n, ξjl∈ S, then sin2(
kπξjl
) & 1k2 and therefore λmin(M(k
2π2
l2)) & 1
k2 .
By symmetry if for all j ∈ n + 2, ..., 2n, ξjl′
∈ S, then the matrix M(k2π2
(l′)2) satisfies
λmin(M(k2π2
(l′)2)) & 1
k2 .
Finally for the eigenvalue λ2k = k2π2
(l+l′)2, k ∈ N∗, since we have sin(λkl
′) = (−1)k+1
sin(λkl), we deduce that
a21 sin2 (λkl1) =
2 sin2 (λkl)
n(l + l′)=
2
n(l + l′)sin2
(
kπl
l + l′
)
.
By Lemma 1.7.3 if ll+l′
∈ S, then its associated matrix M( k2π2
(l+l′)2) satisfies λmin(M
( k2π2
(l+l′)2)) & 1
k2 .Theorem 1.6.6 then leads to the
Theorem 1.7.9. Consider a star shaped network with Dirichlet boundary condition satis-fying (1.70) and feedback laws at the interior node as well as at the point ξi of the edge ei,
75
for i = 2, · · · , n and for i = n + 2, · · · , 2n. If ξil∈ S, for all i = 2, · · · , n, ξi
l′∈ S, for all
i = n + 2, · · · , 2n and ll+l′
∈ S, then for all (u(0), u(1), (f 0(−τv.))) ∈ D(A), the energy of
our system decays polynomially like 1t.
Note that the two previous examples do not enter into the setting of [7].
Mixed boundary conditions
We suppose that the star shaped network is made of 4 edges, with e1 and e4 of samelength l, and e2 and e3 of same length l′. At the extremities of e1 and e2, we imposeDirichlet condition and at the extremities of e3 and e4, Neumann condition. We control atthe interior node. In this case the eigenvalues of the problem without damping are k2π2
4(l+l′)2
of multiplicity 1.
D
D
N
N
l
l
l’
l’
Fig. 1.5 –
Similar considerations as above allow to show the following results :
Proposition 1.7.10. 1) If ll′/∈ Q, then the energy of our system tends to 0.
2) Suppose that ll′
/∈ Q and that ll+l′
∈ S and l′
l+l′∈ S, then for all (u(0), u(1),
(f 0(−τv.))) ∈ D(A), the energy decays as 1√t.
Remark 1.7.11. For instance if l =√
2 and l′ = 1−√
2, then ll′/∈ Q, l
l+l′∈ S and l′
l+l′∈ S.
1.7.3 More complex networks
In this subsection, we assume that R is a network whose edges are all of same length, i.e.lj = 1, for all j ∈ 1, ..., N. In that case, the spectrum of the Laplace operator is explicitlyknow via the algebraic properties of the network [20, 81]. More precisely, introduce theadjacency matrix C of the vertices of the network
C = (cs, t)s, t∈V\D,
where
∀s, t ∈ V \ D, cs, t =
card(Es∩Et)√card(Es)
√card(Et)
if Es ∩ Et 6= ∅0 otherwise.
We may now recall the following result proved in [81] :
76
Theorem 1.7.12. Under the above assumption, we have
Sp(A) = S1 ∪ S2, where
S1 = k2π2 with multiplicity N − #(V \ D), k ∈ N∗ of associated eigenvector ϕk, j(x) =cj sin(kπx), for some constants cj, andS2 = λ2 : cos λ ∈ Sp(C)∩] − 1, 1[. Moreover ϕ is a eigenvector of A associated with λ2
if and only if (ϕ(v))v∈V\D is an eigenvector of the matrix C of eigenvalue cosλ.
Note that this Theorem implies that the standard gap condition holds for networkswith edges which are of length one (or more generally rational numbers).
Before going on, let us make a more precise relation between the orthonormal eigen-vectors of an eigenvalue λ2 from S2 and the eigenvectors of the matrix C of eigenvaluecosλ :
Lemma 1.7.13. Let λ2 ∈ S2. There exists a positive definite symmetric matrix E(λ) ∈Rm×m, where m is the cardinal of V \D, such that for all eigenvectors ϕ, ϕ′ associated withλ2, we have
(ϕ, ϕ′)L2(R) = (ϕ′(v))⊤v∈V\DE(λ)(ϕ(v))v∈V\D. (1.71)
Moreover this matrix E(λ) is uniformly positive definite and uniformly bounded, in thesense that there exists a positive constant C (independent of λ) such that
λmin(E(λ)) ≥ C, ‖E(λ)‖2 ≤ C−1.
Proof: For all j = 1, · · · , N we may write
ϕj(x) = aj sin(λx) + bj sin(λ(1 − x)), ∀x ∈ (0, 1), (1.72)
where
aj =ϕ(v)
sin λ, bj =
ϕ(v′)
sinλ, (1.73)
when v (resp. v′) is the vertex corresponding to the extremity 1 (resp. 0) of the edge ej .We use the same relations for ϕ′ using a′j and b′j .
Now by direct calculations, we see that
(ϕ, ϕ′)L2(R) =1
2
N∑
j=1
(aj bj)(B(λ) + Br(λ))
(
a′jb′j
)
,
where the 2 × 2 matrices B(λ) and Br(λ) are given by
B(λ) =
(
1 − cosλ− cosλ 1
)
, Br(λ) =1
2λ
(
sin(2λ) 2 sinλ2 sinλ sin(2λ)
)
.
This proves the identity (1.71) due to the relation (1.73).
77
Let us now remark that B(λ) depends only on cosλ and then on cosλ0 with λ0 ∈ (0, π),when λ = λ0 +2kπ or λ = −λ0 +2(k+1)π, for some k ∈ N. The eigenvalues of B(λ) being1 ± cosλ, this matrix is uniformly positive definite, i.e. there exists a positive constant C(independent of λ) such that
(a b)B(λ)
(
ab
)
≥ C(a2 + b2), ∀a, b ∈ R.
We further remark that the matrix Br(λ) is a remainder since
‖Br(λ)‖2 .1
λ.
Therefore we introduce the matrix F(λ0) ∈ Rm×m as follows :
(ξv)⊤v∈V\DF(λ0)(ξ
′v)v∈V\D =
1
2
N∑
j=1
(aj bj)B(λ)
(
a′jb′j
)
,
∀(ξv)v∈V\D, (ξ′v)v∈V\D ∈ Rm,
with the relation
aj =ξv
sinλ, bj =
ξv′
sinλ. (1.74)
From the uniform positiveness of B(λ) and the above relations between aj , bj and ξv, wedirectly deduce that F(λ0) is uniformly positive definite.
The previous considerations clearly show that
E(λ) = F(λ0) + Fr(λ),
with
‖Fr(λ)‖2 .1
λ.
Therefore there exists Λ > 0 such that E(λ) is uniformly positive definite, for all λ > Λ.It remains to consider the case 0 < λ ≤ Λ. But in that case we see that
(ξv)⊤v∈V\DE(λ)(ξ′v)v∈V\D = (ϕ, ϕ′)L2(R), ∀(ξv)v∈V\D, (ξ
′v)v∈V\D ∈ Rm,
when ϕ is given by (1.72) when aj , bj are defined by (1.74) (and similarly for ϕ′). As theabove right-hand side is an inner product on L2(R), the left-hand side is also an innerproduct in Rm. Hence the matrix E(λ) is positive definite. The uniformness follows fromthe fact that the interval (0,Λ] contains a finite number of λ such that λ2 ∈ S2.
The uniform boundedness of E(λ) is proved in the same manner.
Corollary 1.7.14. Assume that Vc = V \ D. Then for any λ2 ∈ S2, its associated matrixM(λ2) is uniformly positive definite, i.e.
λmin(M(λ2)) & 1.
78
Proof: Assume that λ2 is of multiplicity l and denote by ϕi, i = 1, · · · , l the associatedorthonormal eigenvectors. Now we introduce the vectors
(ϕi(v))v∈V\D = E(λ)1/2(ϕi(v))v∈V\D, ∀i = 1, · · · , l.According to the relation (1.71) these vectors are orthonormal :
(ϕi(v))⊤v∈V\D(ϕj(v))v∈V\D = δij , ∀i, j = 1, · · · , l.
Now we simply remark that
(ξi)⊤M(λ2)(ξi) =
∑
i,j
ξiξj(ϕj(v))⊤v (ϕi(v))v,
and consequently
(ξi)⊤M(λ2)(ξi) =
∑
i,j
ξiξj(ϕj(v))⊤v E(λ)−1(ϕi(v))v
= (ϕ(v))⊤v E(λ)−1(ϕ(v))v,
where (ϕ(v))v =∑
i ξi(ϕi(v))v. By the uniform boundedness of E(λ), we deduce that
(ξi)⊤M(λ2)(ξi) & (ϕ(v))⊤v (ϕ(v))v
&∑
i,j
ξiξj(ϕj(v))⊤v (ϕi(v))v =
∑
i
ξ2i ,
this last identity following from the orthonormality of the vectors (ϕi(v))v.
From this Corollary we see that if we control at least at all nodes of V \ D then theassumption for the exponential stability (and obviously polynomial stability) holds for alleigenvalues of S2. In that case we only need to manage the eigenvalues of S1. Note furtherthat if S1 is not empty, then some additional interior controls are necessary since theeigenvectors associated with such eigenvalues are zero at the nodes.
Now if we want to control on a subset of V \D then the assumption for the exponentialstability does not necessary hold for the eigenvalues of S2. Let us then describe how weproceed in that case. For a fixed λ2 ∈ S2, we denote by (ϕappi (v))v∈V\D ∈ Rm, i = 1, · · · , l,the eigenvectors of C of eigenvalue cosλ such that their corresponding eigenvectors on thenetwork are approximated orthonormal, namely
(ϕappi (v))⊤v∈V\DF(λ0)(ϕappj (v))v∈V\D = δij ,
This basis can be computed for all λ since it depends only on λ0 (which form a finite set).Denote by Mapp(λ2
0), the matrix build as M(λ2) by replacing (ϕi(v))v∈Vc by (ϕappi (v))v∈Vc ,namely
Mapp(λ20) =
(ϕapp1 (v))⊤v∈Vc
...(ϕappl (v))⊤v∈Vc
· ((ϕapp1 (v))v∈Vc . . . (ϕappl (v))v∈Vc) .
Now we can state the
79
Lemma 1.7.15. Let λ2 ∈ S2 be fixed. If Mapp(λ20) is positive definite and if C has no
eigenvector associated with cosλ0 identically equal to zero at the nodes of Vc, then M(λ2)is uniformly positive definite.
Proof: As before we show that
M(λ2) = Mapp(λ20) + Mr(λ
2),
where
‖Mr(λ2)‖2 .
1
λ.
Consequently for λ large enough, the uniform positive definiteness follows from the positivedefiniteness of Mapp(λ2
0). On the contrary for small λ, it suffices to remark that M(λ) hasan eigenvalue equal to zero if and only if C has an eigenvector associated with cosλ = cos λ0
identically equal to zero at the nodes of Vc. Since we have assumed that such eigenvectorsdo not exist, M(λ2) is positive definite and the conclusion follows by finiteness.
Note that our above Lemma has only to be used for multiple eigenvalues. Indeed assumethat λ2 ∈ S2 is simple, then denote by (ϕ(v))v∈V\D its eigenvector such that
(ϕ(v))⊤v∈V\DE(λ)(ϕ(v))v∈V\D = 1.
Now consider any computed eigenvector (ψv)v∈V\D of C that depends only on cosλ andthen on λ0. This eigenvector then satisfies
(ϕ(v))v∈V\D = µ(ψv)v∈V\D,
with µ ∈ R such that µ2 = 1c(λ)
, when
c(λ) = (ψv)⊤v∈V\DE(λ)(ψv)v∈V\D.
Since c(λ) remains uniformly bounded from below and from above, it is equivalent to checkthe uniform definite positiveness of M(λ2) using (ϕ(v))v∈V\D or using (ψv)v∈V\D.
A first example
Consider the tree described by figure 1.6.By Theorem 1.7.12, the eigenvalues of A are k2π2 of multiplicity 5−4 = 1, and λ2 such
that cos(λ) ∈ Sp(C)∩] − 1, 1[.1st case : We easily check that the eigenvector associated with k2π2 is given by
ϕk(x) =
√
2
3sin(kπx)
10
(−1)k
10
.
80
D D
N
N
e1
e2
e3 e4
e5
1
2
3
4
Fig. 1.6 –
Since the eigenvector is zero at all nodes of the network, feedbacks at the nodes are notsufficient to stabilize the system. But if we take a control at ξ1 on the edge e1 for instance,we have
ϕ21(ξ1) =
2
3sin2(kπξ1) &
1
k2for ξ1 ∈ S.
2d case : The eigenvalues of the matrix
C =
0 1√3
13
01√3
0 0 013
0 0 1√3
0 0 1√3
0
are −16+ 1
6
√13, −1
6− 1
6
√13, 1
6+ 1
6
√13, 1
6− 1
6
√13 ∈]−1, 1[ (of multiplicity 1) of associated
eigenvectors
(ϕ(v))v∈V\D ≃
−0.75−10.751
,
1.3−1−1.3
1
,
1.31
1.31
,
−0.751
−0.751
,
respectively. Then we can consider a feedback at any node of V \ D, for instance at thenode number 4 (see Fig. 1.7).
D D
N
e1
e2
e3 e4
e5
1
2
3
4
ξ1
C
Fig. 1.7 –
We then have the next result.
81
Proposition 1.7.16. If we control at ξ1 ∈ S on the edge e1 and at one of the vertices ofV \ D, for all (u(0), u(1), (f 0(−τv.))) ∈ D(A), the energy decays like 1
t.
A second example
Consider the tree as described in figure 1.8.
D
N
e1
e2
e3 e
e
1
2
3
e4
e5
6
7
4
5
6
7
N
N
N
Fig. 1.8 –
By Theorem 1.7.12, the operator A has only the eigenvalues λ2 such that cos λ ∈Sp(C)∩]−1, 1[. Therefore by Corollary 1.7.14, if we control at all nodes except the Dirichletone, then we obtain an exponential decay. Without delay we find the result of [109].
Note that it suffices to control at the nodes 5, 6 and 7. Indeed the eigenvalues of thematrix
C =
0 13
13
0 0 0 013
0 0 1√3
1√3
0 013
0 0 0 0 1√3
1√3
0 1√3
0 0 0 0 0
0 1√3
0 0 0 0 0
0 0 1√3
0 0 0 0
0 0 1√3
0 0 0 0
are approximatively 0.9, −0.9, 0.8, −0.8, 0, 0, 0 ∈ [−1, 1] of eigenvector
(ϕ(v))v∈V\D ≃
1.151.61.61111
,
1.15−1.6−1.6
1111
,
0−1.41.4−1−111
,
01.4−1.4−1−111
,
−1.7001010
,
82
−1.7001001
,
000−1100
respectively. Therefore if we control at the nodes 5, 6 and 7 the assumption for the exponen-tial stability holds for the simple eigenvalues corresponding to 0.9, −0.9, 0.8,−0.8. Now for cosλ = 0, i.e. λ = π
2+ kπ, we see that C has no eigenvectors which are zero
at the nodes 5, 6 and 7. Furthermore we easily check that F(π2
4) is the diagonal matrix
with entries (3/2, 3/2, 3/2, 1/2, 1/2, 1/2, 1/2) and then
ϕapp1 =√
2
000−1100
, ϕapp2 =√
2
−100
1/21/210
, ϕapp3 =√
2
−2/900
1/91/9−7/9
1
.
By direct calculations we obtain
Mapp(π2
4) = 2
1 12
19
12
54
−1318
19
−1318
−13181
.
Since this matrix is positive definite, by Lemma 1.7.15 we deduce the uniform positivedefiniteness of M((π
2+ kπ)2).
In other words, we have proved the
Proposition 1.7.17. If the feedback law is at the vertices 5, 6 and 7, then the energydecays exponentially in the energy space.
Note that if we impose Dirichlet boundary conditions at the nodes 4 and 6, then thesystem is no more exponentially stable.
A network with a circuit
Similar considerations than before allow to prove the following result :
Proposition 1.7.18. Consider the network described by figure 1.9. If the feedback law isat ξ1 ∈ S on the edge e2, at ξ2 ∈ S on the edge e3, at ξ3 ∈ S on the edge e4 and at thevertex 6 or 7, then for all (u(0), u(1), (f 0(−τv.))) ∈ D(A), the energy decays like 1
t.
83
D
D
D
N
N
e1
e2
e3
e4
e5
e6
e8
e10
e9
e7
2
3
4
5
6
7
1
ξ
ξ
ξ1
2
3
C
Fig. 1.9 –
84
Chapitre 2
Weak stabilization of the waveequation on 1-d networks
2.1 Introduction and main results
In this chapter, we consider a planar network of elastic strings that undergoes smallperpendicular vibrations. Recently, the control, observation and stabilization problems ofthese networks have been the object of intensive research (see [38, 65] and the referencestherein).
Here we are interested in the problem of stabilization of the network by means of adamping term located on one single exterior node. The aim of this chapter is to develop asystematic method to address this issue and to give a general result allowing to transforman observability result for the corresponding conservative system into a stabilization onefor the damped one.
Before going on, let us recall some definitions and notation about 1 − d networks usedin the chapter. We refer to [2, 20, 82] for more details.
A 1 − d network R is a connected set of Rn, n ≥ 1, defined by
R =N⋃
j=1
ej
where ej is a curve that we identify with the interval (0, lj), lj > 0, and such that for k 6= j,ej ∩ ek is either empty or a common end called a vertex or a node (here ej stands for theclosure of ej).
For a function u : R −→ R, we set uj = u|ejthe restriction of u to the edge ej .
We denote by E = ej ; 1 ≤ j ≤ N the set of edges of R and by V the set of verticesof R. For a fixed vertex v, let
Ev = j ∈ 1, ..., N ; v ∈ ej
be the set of edges having v as vertex. If card (Ev) = 1, v is an exterior node, while if card
85
(Ev) ≥ 2, v is an interior one. We denote by Vext the set of exterior nodes and by Vint theset of interior ones. For v ∈ Vext, the single element of Ev is denoted by jv.
We now fix a partition of Vext : Vext = D∪v1. In this way, we distinguish the conser-vative exterior nodes, D, in which we impose Dirichlet homogeneous boundary condition,and the one in which the damping term is effective, v1. To simplify the notation, we willassume that v1 is located at the end 0 of the edge e1.
Let uj = uj(t, x) : R × [0, lj] → R be the function describing the transversal displace-ment in time t of the string ej of length lj. Let us denote by L the sum of the lengths ofall edges of the network, the total length of the network.
We assume that the displacements uj satisfy the following system
∂2uj
∂t2(x, t) − ∂2uj
∂x2 (x, t) = 0 0 < x < lj , t > 0, ∀j ∈ 1, ..., N,uj(v, t) = ul(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂uj
∂nj(v, t) = 0 ∀v ∈ Vint, t > 0,
ujv(v, t) = 0 ∀v ∈ D, t > 0,∂u1
∂x(0, t) = ∂u1
∂t(0, t) ∀t > 0,
u(t = 0) = u(0), ∂u∂t
(t = 0) = u(1),
(2.1)
where ∂uj/∂nj(v, .) stands for the outward normal (space) derivative of uj at the vertexv. We denote by u the vector u = (uj)j=1,...,N.
The above system has been considered by several authors in some particular situations.We refer, for instance, to [11], [6], [7,8], [9] and [109], where explicit decay rates are obtainedfor networks with some special structures. We also refer to chapter 1 or [89] where theproblem is considered in the presence of delay terms in the feedback law.
The object of this chapter is not to give an additional result in a particular case, butrather to develop a systematic method allowing to address the issue in a general context. Wedo this transfering known observability results for the corresponding conservative systeminto stabilization results for the dissipative one. This provides a new proof for the existingresults mentioned above and allows getting new ones.
As mentioned above, in this chapter, we consider the case where the dissipation islocated on an external node of the network, but the method can be adapted to treat thecase where the damping term is located in several nodes, both exterior and interior ones.
In order to study system (2.1) we need a proper functional setting. We define the Hilbertspaces
L2(R) = u : R → R; uj ∈ L2(0, lj), ∀j = 1, · · · , N,and
V := φ ∈N∏
j=1
H1(0, lj) : φj(v) = φk(v) ∀j, k ∈ Ev, ∀v ∈ Vint ; φjv(v) = 0 ∀v ∈ D,
86
equipped with the natural inner products
< φ, φ >L2(R)=N∑
j=1
∫ lj
0
φjφjdx and < φ, φ >V =N∑
j=1
∫ lj
0
∂φj∂x
∂φj∂x
dx
respectively.It is well known that system (2.1) may be rewritten as the first order evolution equation
U ′ = AU,U(0) = (u(0), u(1)) = U0,
(2.2)
in the Hilbert spaceH := V × L2(R),
equipped with the usual inner product
⟨(
uw
)
,
(
uw
)⟩
:=
N∑
j=1
∫ lj
0
(
∂uj∂x
∂uj∂x
+ wjwj
)
dx.
Here U is the vector U = (u, ∂u∂t
)T , the operator A is defined by
A(
uw
)
:=
(
w∆u
)
,
∆ being the Laplace operator : ∆u =(
∂2uj
∂x2
)
j=1,...,N, and where the domain D(A) of the
operator A is defined by
D(A) := (u, w) ∈ (V ∩N∏
j=1
H2(0, lj)) × V :∂u1
∂x(0) = w1(0) ;
∑
j∈Ev
∂uj∂nj
(v) = 0, ∀v ∈ Vint.
Therefore, we know that for an initial datum U0 ∈ H , there exists a unique solutionU ∈ C([0, +∞), H) to problem (2.2). Moreover, if U0 ∈ D(A), then
U ∈ C([0, +∞), D(A)) ∩ C1([0, +∞), H).
We define the natural energy of u by
Eu(t) :=1
2
N∑
j=1
∫ lj
0
(
(
∂uj∂t
)2
+
(
∂uj∂x
)2)
dx. (2.3)
This energy satisfies
E ′u(t) = −
(
∂u1
∂t(0, t)
)2
≤ 0, (2.4)
and therefore it is decreasing.
87
Remark 2.1.1. Integrating the expression (2.4) between 0 and T , we obtain
∫ T
0
(
∂u1
∂t(0, t)
)2
dt = Eu(0) − Eu(T ) ≤ Eu(0).
Consequently, this estimate implies that ∂u1
∂t(0, .) belongs to L2(0, T ) for finite energy solu-
tions. This is a ”hidden” regularity property in the sense that it is not a direct consequenceof the regularity of finite energy solutions.
In [6–9, 11, 89] and in chapter 1, the method to obtain the stabilization of (2.1) inparticular cases is based on the use of observability estimates for the solutions φ of theconservative problem without damping, with Neumann boundary condition at the nodex = 0 :
∂2φj
∂t2(x, t) − ∂2φj
∂x2 (x, t) = 0 0 < x < lj , t > 0, ∀j ∈ 1, ..., N,φj(v, t) = φl(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂φj
∂nj(v, t) = 0 ∀v ∈ Vint, t > 0,
φjv(v, t) = 0 ∀v ∈ D, t > 0,∂φ1
∂x(0, t) = 0 ∀t > 0,
φ(t = 0) = u(0), ∂φ∂t
(t = 0) = u(1).
(2.5)
It is well known that this problem is well posed in the natural energy space. If we supposethat (u(0), u(1)) ∈ H , then problem (2.5) admits a unique solution
φ ∈ C(0, T ; V ) ∩ C1(0, T ; L2(R)).
This system is obviously conservative and its energy is constant : Eφ(t) = Eφ(0) for allt > 0. Let us then denote by (λ2
n)n≥1 the sequence of eigenvalues of (2.5) and let (ϕn)n≥1
be the corresponding eigenvectors forming an orthonormal basis of L2(R). In particularcases, in [6–9, 11, 89], under appropriate conditions, observability inequalities of the form
∑
n≥1
c2n(λ2na
2n + b2n) ≤ C
∫ T
0
∣
∣
∣
∣
∂φ1
∂t(0, t)
∣
∣
∣
∣
2
dt, (2.6)
are proved for system (2.5), where an, bn are the Fourier coefficients of the initial dataof φ in the basis (ϕn)n, and with weights c2n > 0 depending on the network. To obtainthe stabilization result from the observability inequality (2.6), the authors decompose thesolution u as the sum of φ, solution of (2.5) with the same initial data than u, and arest with vanishing initial data. Then they show regularity results of trace type for thisrest (see section 1.4). However, in these articles and in chapter 1, the analysis is limitedto strong observability inequalities (leading to exponential or polynomial decay results)which hold only for a restricted class of networks. In the present chapter we are able todeal with arbitrary networks in which weaker observability inequalities may hold leadingto arbitrarily slow decay rates.
88
Our analysis is based on the existing observability results for the following system withDirichlet boundary condition at all exterior nodes
∂2ψj
∂t2(x, t) − ∂2ψj
∂x2 (x, t) = 0 0 < x < lj , t > 0, ∀j ∈ 1, ..., N,ψj(v, t) = ψl(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂ψj
∂nj(v, t) = 0 ∀v ∈ Vint, t > 0,
ψjv(v, t) = 0 ∀v ∈ D, t > 0,ψ1(0, t) = 0 t > 0,
ψ(t = 0) = ψ(0), ∂ψ∂t
(t = 0) = ψ(1).
(2.7)
The study of the observability of this system is motivated by control problems (see [33–38]).The difference between systems (2.5) and (2.7) is the boundary condition at the end 0
of e1 : Dirichlet boundary condition for (2.7) and Neumann boundary condition for (2.5).If we suppose that (ψ(0), ψ(1)) ∈ H = V × L2(R) where
V = ψ ∈N∏
j=1
H1(0, lj) : ψj(v) = ψk(v) ∀j, k ∈ Ev, ∀v ∈ Vint ; ψjv(v) = 0, ∀v ∈ D ; ψ1(0) = 0,
then problem (2.7) admits a unique solution
ψ ∈ C(0, T ; V ) ∩ C1(0, T ; L2(R)).
This system is obviously also conservative.In this chapter, we show the link between the existing observability results for (2.7)
and the stabilization of (2.1). To be more precise, let us denote by (λ2n)n≥1 the sequence
of eigenvalues corresponding to problem (2.7) and let (ϕDn )n≥1 be the corresponding eigen-vectors forming an orthonormal basis of L2(R). Under some conditions on the topology ofthe network and the lengths of the strings entering in it, Dager and Zuazua [33–38] provedweighted observability inequalities for (2.7) of the following form
ED∗ (ψ, 0) :=
∑
n≥1
c2n(λ2nψ
20,n + ψ2
1,n) ≤ C
∫ T
0
∣
∣
∣
∣
∂ψ1
∂x(0, t)
∣
∣
∣
∣
2
dt, (2.8)
for a positive constant C, where ψ0,n, ψ1,n are the Fourier coefficients of the initial data ofψ in the basis (ϕDn )n and with positive weights (c2n)n≥1 depending on the properties of thenetwork.
To obtain this weighted observability estimate (2.8), several methods have been deve-loped in those articles. The first one, that uses the D’Alembert representation formula,applies to tree-shaped networks. Its main advantage is that it does not require computingthe spectrum of the network. This method has been later adapted in [7] and [9] to analyzethe stabilization of stars and trees. A second method uses the Fourier expansion of solu-tions and the Beurling-Malliavin Theorem (see for instance [37, 38, 89]). This applies for
89
general networks and avoids the difficulties related to applying the D’Alembert formula tomore complex networks that may contain circuits.
The key observation of this chapter is that the weighted observability estimates for usolution of (2.1) can be obtained directly from (2.8). This method is of systematic appli-cation and avoids remaking all the fine analysis already carried out in [33, 38], that uses,in particular, tools from Number Theory, to analyze the properties of the weights c2nn≥1
in terms of those of the network under consideration.Let us now explain how (2.8) can be applied directly in our context. For that we
decompose u as the sum of w, a solution of (2.7) with appropriate initial data (u(0) −u
(0)1 (0)ϕ, u(1)) (where ϕ is a given smooth function such that ϕ1(0) = 1), and a rest.
Applying (2.8) to the solution w of (2.7), we obtain the following weighted observabilityestimate for u solution of (2.1)
ED∗ (w, 0) + u
(0)1 (0)2 ≤ CT
∫ T
0
(
∂u1
∂t(0, t)
)2
dt, (2.9)
where ED∗ (w, 0) is defined by (2.8), with weights (c2n)n depending on the network. If the
weights c2n are non-trivial for all n ∈ N∗, the energy of the dissipative system tends to 0as t→ ∞. However, in general, the weights tend to 0 as n→ ∞, the observed quantity isweaker than the V × L2(R)-norm and the decay rate is not exponential.
It is important to underline that (2.9) holds under the same assumptions on the networkneeded for (2.8) to hold for the Dirichlet problem (2.7). Thus, no further analysis is requiredto establish the existence of the weights c2nn≥1 in terms of the properties of the network.
To derive decay properties out of (2.9), we view this inequality as a weak observabilityestimate in which the observed energy E−(0) is equal, roughly speaking, to ED
∗ (w, 0) +
u(0)1 (0)2. In practice we often take, if necessary, the lower convex envelop of c2n instead of
the weights c2n in the definition of ED∗ (see Section 3.1). The observed energy E− is weaker
than the V × L2(R)-norm of the initial data that would be required to prove exponentialdecay and, consequently, we obtain weaker decay rates. To obtain explicit decay rates outof this weak observability inequality we use an interpolation inequality which is a variant ofthe one from Begout and Soria [19] and which is a generalization of Holder’s inequality. Forthis to be done we need to assume more regularity of the initial data. To be more precisewe shall consider initial data (u(0), u(1)) ∈ Xs := [D(A), D(A0)]1−s for 0 < s < 1/2. Inthis way we deduce an interpolation inequality of the form
1 ≤ Φs
(
E−(0)
CEu(0)
)
∥
∥(u(0), u(1))∥
∥
2
Xs
C ′Eu(0),
where Φs is an increasing function which depends on s and on the energy E− under consi-deration.
The previous interpolation inequality implies
E−(0) ≥ CEu(0)Φ−1s
(
Eu(0)
C ′ ‖(u(0), u(1))‖2Xs
)
.
90
With (2.4) and (2.9), we obtain
Eu(0) −Eu(T ) ≥ CEu(0)Φ−1s
(
Eu(0)
C ′ ‖(u(0), u(1))‖2Xs
)
,
which implies, by the semigroup property (see Ammari and Tucsnak [11])
∀t > 0, Eu(t) ≤ CΦs
(
1
t+ 1
)
∥
∥(u(0), u(1))∥
∥
2
Xs. (2.10)
Obviously, the decay rate in (2.10) depends on the behaviour of the function Φs near 0.Thus, in order to determine the explicit decay rate we need to have a sharp descriptionof the function Φs, which depends on s and on the energies E and E− and thus, on theweights (c2n)n of (2.9) in an essential way. These weights depend on the topology of thenetwork and the number theoretical properties of the lengths of the strings entering in it.
This approach allows getting in a systematic way decay rates for the energy of smoothsolutions of the damped system as a consequence of the observability properties of theundamped one.
The analysis in this chapter is limited to networks of strings with damping in one singleend but the same methods, combined with the tools developped in [38], can be applied inother situations as, for instance :
a) networks of strings with damping in several end points ;b) networks with damping on end points and internal nodes ;c) networks of beams.The chapter is organized as follows. In the second section, we show how to pass from
the observability inequality for the conservative problem (2.7) with Dirichlet boundaryconditions at all exterior nodes to the weighted observability inequality (2.9) for (2.1). Insection 2.3 we give an interpolation inequality which is a variant of the one by Begout andSoria [19] and we apply it to obtain the explicit decay estimate of the energy. Finally weend up discussing some illustrative examples in section 2.4.
2.2 The weighted observability inequality
In this section, we prove that we can obtain a weighted observability estimate for usolution of (2.1) directly from (2.8). First of all, we recall some results about system (2.7)with Dirichlet boundary conditions at all exterior nodes.
2.2.1 Preliminaries about the Dirichlet problem
Recall that if we suppose that (ψ(0), ψ(1)) ∈ H , then problem (2.7) admits a uniquesolution
ψ ∈ C(0, T ; V ) ∩ C1(0, T ; L2(R)).
91
Denote by (λ2n)n≥1 the sequence of eigenvalues corresponding to problem (2.7) and let
(ϕDn )n≥1 be the corresponding eigenvectors forming an orthonormal basis of L2(R). Weassume now that (ψ(0), ψ(1)) ∈ H and we set
ψ(0) =∑
n≥1
ψ0,nϕDn and ψ(1) =
∑
n≥1
ψ1,nϕDn ,
where (λnψ0,n)n, (ψ1,n)n ∈ l2(N∗).In [38], a weighted observability inequality is derived, motivated by control problems.
More precisely, it is shown that, under some conditions on the topology of the networkand the lengths of the strings entering in it, for all T > T0 (where T0 > 0 is large enough),there exists a sequence of positive weights c2nn≥1 and a positive constant C such that forall solution ψ of (2.7) it holds
∑
n≥1
c2n(λ2nψ
20,n + ψ2
1,n) ≤ C
∫ T
0
∣
∣
∣
∣
∂ψ1
∂x(0, t)
∣
∣
∣
∣
2
dt. (2.11)
Notice that, for this to be true, it is essential that the time T0 > 0 to be large enough. Moreprecisely, in [38] inequality (2.11) is proved in the case of general networks for T > 2L,where L is the total length of the network. In the case of tree-like networks, (2.11) isobtained by the D’Alembert representation formula and the weights c2n are strictly positivefor every n ∈ N∗ if the tree is non-degenerate. In the case of stars, this comes down toimpose irrationality conditions on the ratio of each pair of lengths. In the case of trees,the condition of non-degeneracy of the network is a natural extension of the irrationalityone for stars and is also generically true (see [33, 36, 38] for more details). In the case ofgeneral networks, (2.11) is established by means of the Beurling-Malliavin Theorem underthe assumption that all eigenfunctions have a non-trivial Neumann trace at the observationnode. This condition is generically satisfied.
We set
ED∗ (ψ, 0) :=
1
2
∑
n≥1
c2n(λ2nψ
20,n + ψ2
1,n) (2.12)
the weighted energy for ψ at time 0. Note that (ED∗ )
12 defines a norm on H because the
weights c2n, according to the results in [38], are assumed to be positive. In the sequel, weassume that the network under consideration is such that the solutions ψ of (2.7) satisfy(2.11).
2.2.2 The weighted observability inequality
In this section, we prove that we can obtain a weighted observability estimate for usolution of (2.1) directly from (2.11). Thus, we assume that the network is such that (2.11)holds and (u(0), u(1)) ∈ H .
92
We introduce w solution of (2.7) with initial data (w(0), w(1)) = (u(0) − u(0)1 (0)ϕ, u(1)),
where ϕ is a given smooth function satisfying
ϕ1(0) = 1ϕjv(v) = 0 ∀v ∈ Dϕj(v) = ϕl(v) ∀j, l ∈ Ev, v ∈ Vint∑
j∈Ev
∂ϕ
∂nj(v) = 0 ∀v ∈ Vint.
(2.13)
In this manner, the initial data (w(0), w(1)) belong to H, i.e. w(0) satisfies the Dirichletboundary condition at the end 0 of e1. Therefore, by hypothesis, w satisfies (2.11).
We also consider ǫ, the solution of the following non-homogeneous Dirichlet problem :
∂2ǫj∂t2
(x, t) − ∂2ǫj∂x2 (x, t) = 0 ∀x ∈ (0, lj), t > 0, ∀j ∈ 1, ..., N,
ǫj(v, t) = ǫl(v, t) ∀j, l ∈ Ev, v ∈ Vint, t > 0,∑
j∈Ev
∂ǫj∂nj
(v, t) = 0 ∀v ∈ Vint, t > 0,
ǫjv(v, t) = 0 ∀v ∈ D, t > 0,ǫ1(0, t) = u1(0, t) t > 0,
ǫ(t = 0) = u(0)1 (0)ϕ, ∂ǫ
∂t(t = 0) = 0.
(2.14)
Note that ǫ satisfies a non-homogeneous Dirichlet boundary condition at x = 0. Actuallyit coincides with the value of the solution u1 of (2.1) at that point. By Remark 2.1.1,we notice that ∂u1/∂t(0, .) ∈ L2(0, T ), so that the non-homogeneous Dirichlet boundarycondition belongs to H1(0, T ).
In this way we have the decomposition
u = w + ǫ. (2.15)
In this section, we prove the following theorem
Theorem 2.2.1. Assume that the network is such that the weighted observability inequality(2.11) is satisfied for the conservative system (2.7) with Dirichlet boundary conditions atall exterior nodes. We split up u, solution of (2.1), as (2.15) where w is solution of (2.7)
with initial data (u(0) − u(0)1 (0)ϕ, u(1)) and ǫ is solution of (2.14). We define E∗(u, 0) by
E∗(u, 0) := ED∗ (w, 0) + u
(0)1 (0)2, (2.16)
where ED∗ (w, 0) is defined by (2.12). Then for all T > T0, there exists CT > 0 such that
all solution u of (2.1) satisfies the weighted observability inequality
E∗(u, 0) ≤ CT
∫ T
0
(
∂u1
∂t(0, t)
)2
dt, (2.17)
provided (u(0), u(1)) ∈ H.
93
Remark 2.2.2. Note that, according to Theorem 2.2.1, we transform the observabilityinequality for the Dirichlet problem (2.7) into a similar one for the dissipative one (2.1).Thus, in particular, estimate (2.17) holds under the same assumptions on the network thatare needed for the Dirichlet problem and that are discussed in [38]. Some examples will bediscussed in Section 4.
Note also that (E∗(u, 0))12 defines a norm in the space of initial data (u(0), u(1)) ∈ H.
Indeed, when E∗(u, 0) vanishes, u(0)1 (0) = 0. Thus (u(0), u(1)) ∈ H and then E∗(u, 0) =
ED∗ (u, 0), and, by assumption,
(
ED∗ (u, 0)
)12 defines a norm in H.
2.2.3 Proof of Theorem 2.2.1
Let (u(0), u(1)) ∈ H . We decompose u as in (2.15) where w and ǫ solve (2.7) and(2.14) respectively. Therefore, w is solution of (2.7) with initial data (w(0), w(1)) = (u(0) −u
(0)1 (0)ϕ, u(1)), and thus, by hypothesis, it satisfies (2.11).
First, we have the following lemma
Lemma 2.2.3. For all T > 0 there exists CT > 0 such that the solutions u of (2.1) and ǫof (2.14) satisfy the following estimate
∫ T
0
(
∂ǫ1∂x
(0, t)
)2
dt ≤ CT
(
∫ T
0
(
∂u1
∂t(0, t)
)2
dt+(
u(0)1 (0)
)2)
. (2.18)
Remark 2.2.4. Note that (2.18) holds for all networks since, in fact, it holds locally nearthe boundary of a single string. In this context we apply it along the string containing theextreme x = 0, getting (2.18).
Proof: First of all, we easily show that the energy Eǫ of the solution ǫ (defined as in (2.3))satisfies
d
dtEǫ(t) = −∂ǫ1
∂x(0, t)
∂u1
∂t(0, t),
and then for all t > 0
Eǫ(t) = Eǫ(0) −∫ t
0
∂ǫ1∂x
(0, s)∂u1
∂t(0, s)ds.
Therefore, by using Cauchy’s inequality, we obtain
Eǫ(t) ≤ Eǫ(0) +1
4
∫ t
0
(
∂ǫ1∂x
(0, s)
)2
ds+
∫ t
0
(
∂u1
∂t(0, s)
)2
ds. (2.19)
Secondly, for all t ∈ (0, T ], multiplying the wave equation satisfied by ǫ1 by (l1 − x)∂ǫ1∂x
and by integrating on (0, l1) × (0, t), we have
∫ t
0
∫ l1
0
(l1 − x)∂ǫ1∂x
∂2ǫ1∂t2
dxds−∫ t
0
∫ l1
0
(l1 − x)∂ǫ1∂x
∂2ǫ1∂x2
dxds = 0.
94
By integration by parts, we have
∫ t
0
∫ l1
0
(l1 − x)∂ǫ1∂x
∂2ǫ1∂t2
dxds = −∫ t
0
∫ l1
0
(l1 − x)∂2ǫ1∂xt
∂ǫ1∂tdxds
+
∫ l1
0
(l1 − x)∂ǫ1∂x
(s, x)∂ǫ1∂t
(s, x)dx
∣
∣
∣
∣
t
0
= −1
2
∫ t
0
∫ l1
0
(
∂ǫ1∂t
)2
dxds+l12
∫ t
0
(
∂u1
∂t(0, s)
)2
ds
+
∫ l1
0
(l1 − x)∂ǫ1∂x
(t, x)∂ǫ1∂t
(t, x)dx,
because ∂ǫ∂t
(t = 0) = 0, and
∫ t
0
∫ l1
0
(l1 − x)∂ǫ1∂x
∂2ǫ1∂x2
dxds =1
2
∫ t
0
∫ l1
0
(l1 − x)∂
∂x
(
(
∂ǫ1∂x
)2)
dxds
=1
2
∫ t
0
∫ l1
0
(
∂ǫ1∂x
)2
dxds− l12
∫ t
0
(
∂ǫ1∂x
(0, s)
)2
ds.
Therefore, we obtain
−1
2
∫ t
0
∫ l1
0
(
∂ǫ1∂t
)2
dxds+l12
∫ t
0
(
∂u1
∂t(0, s)
)2
ds+
∫ l1
0
(l1 − x)∂ǫ1∂x
(t, x)∂ǫ1∂t
(t, x)dx
−1
2
∫ t
0
∫ l1
0
(
∂ǫ1∂x
)2
dxds+l12
∫ t
0
(
∂ǫ1∂x
(0, s)
)2
ds = 0,
and thus∫ t
0
(
∂ǫ1∂x
(0, s)
)2
ds ≤ 2
l1
∫ t
0
Eǫ(s)ds+ 2Eǫ(t). (2.20)
By grouping (2.19) and (2.20), we find
Eǫ(t) ≤ Eǫ(0) +1
2l1
∫ t
0
Eǫ(s)ds+1
2Eǫ(t) +
∫ t
0
(
∂u1
∂t(0, s)
)2
ds.
That is to say
Eǫ(t) ≤ 2Eǫ(0) +1
l1
∫ t
0
Eǫ(s)ds+ 2
∫ t
0
(
∂u1
∂t(0, s)
)2
ds.
By Gronwall’s lemma, we obtain that, for all t ∈ (0, T ],
Eǫ(t) ≤ CT
(
Eǫ(0) +
∫ T
0
(
∂u1
∂t(0, s)
)2
ds
)
,
95
where CT depends on T . By inserting this expression into (2.20), we find
∫ T
0
(
∂ǫ1∂x
(0, t)
)2
dt ≤ CT
(
Eǫ(0) +
∫ T
0
(
∂u1
∂t(0, t)
)2
dt
)
.
But
Eǫ(0) =(
u(0)1 (0)
)2(
1
2
N∑
j=1
∫ lj
0
(
∂ϕj∂x
(x)
)2
dx
)
= C(
u(0)1 (0)
)2
,
where C is a constant, since ϕ is given as in (2.13), independently of the solution underconsideration. This proves (2.18).
Let us now return to the proof of Theorem 2.2.1.Since w satisfies (2.11), we have
ED∗ (w, 0) = ED
∗ (w, t) ≤ C
∫ T
0
(
∂w1
∂x(0, t)
)2
dt, (2.21)
where ED∗ (w, 0) is defined as in (2.12) for w solution of (2.7) with initial data (u(0) −
u(0)1 (0)ϕ, u(1)).
Moreover, we have
∫ T
0
(
∂w1
∂x(0, t)
)2
dt ≤ C
∫ T
0
(
(
∂u1
∂x(0, t)
)2
+
(
∂ǫ1∂x
(0, t)
)2)
dt.
Therefore, with (2.18) and (2.21), we obtain
ED∗ (w, 0) = ED
∗ (w, t) ≤ CT
∫ T
0
(
(
∂u1
∂t(0, t)
)2
+
(
∂u1
∂x(0, t)
)2)
dt+ CT
(
u(0)1 (0)
)2
.
(2.22)We recall that E∗(u, 0) is given by (2.16) where ED
∗ (w, 0) is defined by (2.12). Then,(2.22) becomes
E∗(u, 0) ≤ CT
∫ T
0
(
(
∂u1
∂t(0, t)
)2
+
(
∂u1
∂x(0, t)
)2)
dt+ CT
(
u(0)1 (0)
)2
. (2.23)
In fact, we can remove the last term in the right hand side of (2.23).
Lemma 2.2.5. For T large enough, the solutions u of (2.1) satisfy
E∗(u, 0) ≤ CT
∫ T
0
(
(
∂u1
∂t(0, t)
)2
+
(
∂u1
∂x(0, t)
)2)
dt,
for a positive constant CT depending on T .
96
Proof: In view of (2.23) it is sufficient to show that there exists a positive constant C > 0such that
∣
∣
∣u
(0)1 (0)
∣
∣
∣
2
≤ C
∫ T
0
(
(
∂u1
∂t(0, t)
)2
+
(
∂u1
∂x(0, t)
)2)
dt.
We argue by contradiction. If this is not the case, there exists a sequence of solutions unsuch that
∣
∣
∣u
(0)n, 1(0)
∣
∣
∣= 1, ∀n ∈ N (2.24)
and∫ T
0
(
(
∂un, 1∂t
(0, t)
)2
+
(
∂un, 1∂x
(0, t)
)2)
dx→ 0, as n→ +∞. (2.25)
In view of this and (2.23), we deduce that E∗(un, 0) are uniformly bounded. Passing weaklyto the limit in the Hilbert space H, defined as the closure of V with respect to the norm
(E∗(un, 0))12 , we obtain a limit solution u such that
∣
∣
∣u
(0)1 (0)
∣
∣
∣= 1 (2.26)
and∫ T
0
(
(
∂u1
∂t(0, t)
)2
+
(
∂u1
∂x(0, t)
)2)
dx = 0. (2.27)
The fact that (2.26) holds is a consequence of the compactness of the trace operator from Rto R (it is in fact an operator of rank one). On the other hand, (2.27) holds as a consequenceof the weak lower semicontinuity. But, by unique continuation, it is easy to see, in viewof (2.27), that for all T > 2L, the whole limit solution u vanishes. Indeed, we obtain by(2.27)
∂u1
∂t(0, t) =
∂u1
∂x(0, t) = 0,
and therefore ∂u/∂t solves system (2.7) with initial data (u(1), ∂2u(0)/∂x2). Then, we canapply (2.11) to obtain
u(1) =∂2u(0)
∂x2= 0.
Consequently, ∂u/∂t = 0 on R × (0, T ), and thus u is independent of t. Thus u = 0 onR×(0, T ) since it is harmonic on the network and fulfills the Dirichlet boundary condition.This contradicts (2.26).
This lemma proves Theorem 2.2.1, because u is solution of (2.1).
2.3 The stabilization result
2.3.1 An interpolation inequality
In this subsection, we give an interpolation result similar to the one of [19].
97
Let m ∈ [0, 1), 0 < s < 1/2 and assume that
ω : (m, ∞) → (0, ω(m)) is a convex and decreasing function with ω(∞) = 0, (2.28)
Φs : (0, ω(m)) → (0, ∞) is a concave and increasing function with Φs(0) = 0, (2.29)
∀t ∈ [1, ∞), 1 ≤ Φs(ω(t))t2s, (2.30)
The function t 7→ 1
tΦ−1s (t) is nondecreasing on (0, 1). (2.31)
Before stating the needed interpolation inequality, we recall the inverse Jensen’s in-equality, which is the inverse version of the classical Jensen’s inequality (see Lemma 2.4from [19] and Rudin [99]).
Lemma 2.3.1 (Inverse Jensen’s inequality). Let (Ω, Υ, ν) be a measure space such thatν (Ω) = 1 and let −∞ ≤ a < b ≤ ∞. Assume that
1) ϕ : (a, b) → R is a concave function,2) g ∈ L1(Ω, Υ, ν) is such that for almost every x ∈ Ω, g(x) ∈ (a, b).Then ϕ(g)+ ∈ L1(Ω, Υ, ν) and
∫
Ω
ϕ(g)dν ≤ ϕ
(∫
Ω
gdν
)
.
Under the conditions (2.29)-(2.30), we have the following result which is a generalizedHolder’s inequality, a variant of Theorem 2.1 given in [19] :
Theorem 2.3.2. Let (ω, Φs) be as above satisfying (2.29)-(2.30). Then for any f =(fn)n∈N∗ ∈ l1(N∗), f 6= 0, we have
1 ≤ Φs
∑
n≥1
|fn|ω(n)
∑
n≥1
|fn|
∑
n≥1
|fn|n2s
∑
n≥1
|fn|, (2.32)
as soon as (fnω(n))n ∈ l1(N∗) and (fnn2s)n ∈ l1(N∗).
Proof: The proof is similar to that of Theorem 2.1 of [19]. We give it for the sake ofcompleteness. By (2.30) and Cauchy-Schwarz’s inequality, we have
1 =
∑
n≥1
|fn|∑
n≥1
|fn|
2
≤
∑
n≥1
|fn|∑
n≥1
|fn|Φ
12s (ω(n))ns
2
≤
∑
n≥1
|fn|∑
n≥1
|fn|Φs(ω(n))
∑
n≥1
|fn|∑
n≥1
|fn|n2s
.
98
Now, we apply Lemma 2.3.1 with ϕ = Φs a concave function, g = ω and the discretemeasure ν =
∑
n≥1|fn|
P
n≥1|fn|δn, by noticing that
∑
n≥1
|fn|∑
n≥1
|fn|
= 1.
We then obtain (2.32).
We now give some examples of pairs (ω, Φs) satisfying (2.28)-(2.31) :
Example 2.3.3. 1. If
ω(t) =c
tp,
for some p ≥ 1, we can take Φs of the form
Φs(t) =
(
t
c
)2sp
.
We can easily prove that (ω, Φs) satisfy (2.28)-(2.29) with m = 0 and (2.30)-(2.31).2. If
ω(t) = Ce−At
where A > 2(2s+ 1) and C > 0, we can take Φs of the form
Φs(t) =
(
A
ln(
Ct
)
)2s
.
We can easily prove that Φs is an increasing function on (0, ω(0)), a concave functionon (0, ω(1/2)) (because A > 2(2s + 1)), that t 7→ 1
tΦ−1s (t) is nondecreasing on (0, 1) and
that the pair (ω, Φs) satisfies (2.30) on [1, ∞). Thus (ω, Φs) satisfy (2.28)-(2.31) withm = 1/2.
In the application of the interpolation inequality of Theorem 2.3.2 to our stabilizationproblem the weight ω is determined by the weights (c2n)n in (2.12). However, notice that, ingeneral, the weights c2n may degenerate fast and, consequently, we have to work in a moregeneral context.
Moreover, in principle, (c2n)n does not necessarily satisfy the convexity or the monotonyproperty in (2.28). To ensure that this assumption is satisfied, we introduce the notion ofthe lower convex envelop of a sequence (un)n satisfying lim infn→∞ un = 0. Roughly, it isthe ”nearest” convex and decreasing function satisfying 0 < ω(n) ≤ un for all n ∈ N. Theexistence of this function is guaranteed by the following lemma :
Lemma 2.3.4. ( [19]) Let −∞ < a < b ≤ ∞ and let ǫ : [a, b) → (0, ∞) be a continuousfunction such that lim inftրb ǫ(t) = 0. Then there exists a convex function ϕ ∈ C1
b ([a, b); R)such that 0 < ϕ ≤ ǫ and ϕ′ < 0 on [a, b).
99
The proof of this lemma can be found in [19].Now, we show how to construct the weight ω satisfying (2.28) from (un)n∈N ⊂ (0, ∞)
such that lim infn→∞ un = 0. Let ǫ ∈ C([0, ∞); R) be such that 0 < ǫ(n) ≤ un, for anyn ∈ N. Let ϕ ∈ C([0, ∞); R) be a decreasing and convex function such that for any t ≥ 0,0 < ϕ(t) ≤ ǫ(t) (which exists by the previous lemma) and consider C ⊂ [1, ∞) × [0, ∞)the closure of the convex envelop of the set (n, un); n ∈ N. Finally, fix arbitrarily t ≥ 1.Then the set Ct = C ∩ (t × R) is nonempty, closed and the previous lemma ensures thatfor any st ∈ R such that (t, st) ∈ Ct,
0 < ϕ(t) ≤ st.
So by compactness, we may define the function ω as
∀t ≥ 1, ω(t) = min st; (t, st) ∈ Ct .Finally we extend ω as a decreasing, continuous and convex function on [0, 1]. Therefore,ω satisfies (2.28) with m = 0. This function ω is called the lower convex envelop of thesequence (un)n.
In the sequel, assuming that the weights in (2.11) are such that lim infn→∞
cn = 0 and cn 6= 0
for all n ∈ N∗, we choose the function ω as follows
ω is the lower convex envelop of the sequence(
1, (c2n)n≥1
)
. (2.33)
2.3.2 The main results
In this subsection, we assume that the network is such that the weighted observabilityinequality (2.11) holds, for every solution of problem (2.7). Therefore, by Theorem 2.2.1,u satisfies (2.17). Let us make some remarks :
Remark 2.3.5. 1. If for all n ∈ N∗, we have c2n > 0, then the energy Eu tends to 0 ast→ ∞.
2. If there exists a positive constant c such that for all n ∈ N∗, c2n ≥ c > 0, thenthe energy of u solution of (2.1) is exponentially decreasing. Unfortunately this does nothold in general except, for example, for the simplest network which consists simply of asingle string and for trees with all but one damped ends (see section 1.7.3), but never fornon-trivial networks with one single damped end.
The first point can be proved by the La Salle’s invariance principle and the second oneby a classical energy method combining the observability inequality (2.17) and the energydissipation law (2.4) (see for instance chapter 1 or [89]).
Before stating the main result of this chapter, let us give a technical lemma which willbe used in the sequel. For that, we need to define Xs the interpolation space between D(A)and D(A0) :
Xs :=[
D(A), D(A0)]
1−s ,
where 0 < s < 1/2. Note that if s = 0 then X0 = D(A0) = V × L2(R) and if s = 1 thenX1 = D(A). More precisely we can identify the space Xs :
100
Lemma 2.3.6. For 0 < s < 1/2,
Xs =
(
V ∩∏
j
H1+s(0, lj)
)
×∏
j
Hs(0, lj).
Proof: First D(A) is a dense subset of V ×L2(R) and(
V ∩∏Nj=1H
2(0, lj))
×V is a dense
subset of V ×L2(R) with D(A) ⊂(
V ∩∏Nj=1H
2(0, lj))
× V . Then [D(A), V × L2(R)]1−sis a subset of
[(
V ∩N∏
j=1
H2(0, lj)
)
× V, V × L2(R)
]
1−swith the same norms.
But[(
V ∩N∏
j=1
H2(0, lj)
)
× V, V × L2(R)
]
1−s
=
(
V ∩N∏
j=1
H1+s(0, lj)
)
×N∏
j=1
Hs(0, lj),
and thus, if D(A) is dense in(
V ∩∏Nj=1H
1+s(0, lj))
×∏Nj=1H
s(0, lj), then, by a classical
result of [104],
Xs =
[(
V ∩N∏
j=1
H2(0, lj)
)
× V, V × L2(R)
]
1−s
=
(
V ∩N∏
j=1
H1+s(0, lj)
)
×N∏
j=1
Hs(0, lj).
To verify that D(A) is dense in(
V ∩∏Nj=1H
1+s(0, lj))
×∏Nj=1H
s(0, lj), let
(f, g)T ∈(
V ∩∏Nj=1H
1+s(0, lj))
×∏Nj=1H
s(0, lj) be orthogonal to all elements of D(A),
namely
0 =
⟨(
uw
)
,
(
fg
)⟩
=N∑
j=1
∫ lj
0
(∂uj∂x
∂fj∂x
+ wjgj)dx,
for all (u, w)⊤ ∈ D(A). We first take u = 0 and w ∈ ∏Nj=1 D(0, lj), where D(0, lj) is the
space of C∞ functions with compact support in (0, lj). As (0, w) ∈ D(A), we get
N∑
j=1
∫ lj
0
wjgjdx = 0.
Since∏N
j=1 D(0, lj) is dense in∏N
j=1Hs(0, lj) (because 0 < s < 1/2), we deduce that
g = 0.The above orthogonality condition is then reduced to
0 =N∑
j=1
∫ lj
0
∂uj∂x
∂fj∂x
dx, ∀(u, w) ∈ D(A).
101
By restricting ourselves to w = 0 we obtain
N∑
j=1
∫ lj
0
∂uj∂x
∂fj∂x
dx = 0, ∀(u, 0) ∈ D(A).
But we easily check that (u, 0) ∈ D(A) if and only if u ∈ D(∆), where
D(∆) := u ∈ V ∩N∏
j=1
H2(0, lj) :∑
j∈Ev
∂uj∂nj
(v) = 0, ∀v ∈ Vint ;∂u1
∂x(0) = 0.
We can verify that D(∆) is dense in V ∩ ∏Nj=1H
1+s(0, lj). Indeed, let ξ ∈ V ∩∏N
j=1H1+s(0, lj). For j ∈ 1, ..., N, we take
ξj = ξj − ξj(0)ηj − ξj(lj)ηj,
where ηj, ηj are C∞ functions satisfying
ηj(0) = 1 near 0, ηj(lj) = 0 near lj and ηj(0) = 0 near 0, ηj(lj) = 1 near lj.
Then ξj ∈ H1+s0 (0, lj). As D(0, lj) is dense into H1+s
0 (0, lj), there exits a sequence ξn,j ∈D(0, lj) such that ξn,j → ξj in H1+s
0 (0, lj) when n→ ∞. Setting
ξn,j := ξn,j + ξj(0)ηj + ξj(lj)ηj ,
we see that ξn = (ξn,j)j=1,...,N ∈ D(∆) and ξn → ξ in∏N
j=1H1+s(0, lj).
As D(∆) is dense in V ∩∏Nj=1H
1+s(0, lj), we conclude that f = 0, which proves that
D(A) is dense in(
V ∩∏Nj=1H
1+s(0, lj))
×∏Nj=1H
s(0, lj) and finishes the proof.
Then we have the following lemma :
Lemma 2.3.7. Assume that (u(0), u(1)) belongs to Xs, where 0 < s < 1/2, and (w(0), w(1)) =
(u(0)−u(0)1 (0)ϕ, u(1)) where ϕ is a given smooth function satisfying (2.13). Then there exists
a positive constant C such that
∥
∥
(
w(0), w(1))∥
∥
2
D(AsD)
+∣
∣
∣u
(0)1 (0)
∣
∣
∣
2
≤ C∥
∥
(
u(0), u(1))∥
∥
2
Xs,
where D(AsD) is the domain of the operator As with Dirichlet boundary conditions at all
exterior nodes.
Proof: First, we know that there exists C > 0 such that
∥
∥
(
w(0), w(1))∥
∥
2
D(AsD)
≤ C∥
∥
(
w(0), w(1))∥
∥
2
(V ∩QNj=1H
1+s(0, lj))×QN
j=1Hs(0, lj)
by interpolation (see [104] for example).
102
This estimate leads to the existence of a positive constant C such that
∥
∥
(
w(0), w(1))∥
∥
2
D(AsD)
≤ C∥
∥
(
w(0), w(1))∥
∥
2
(V ∩QNj=1H
1+s(0, lj))×QN
j=1Hs(0, lj)
, (2.34)
because w(0) ∈ V ∩ V .The Sobolev’s injection theorem and the fact that (w(0), w(1)) = (u(0) − u
(0)1 (0)ϕ, u(1))
imply that there exists C > 0 such that
∥
∥w(0)∥
∥
2
V ∩QNj=1H
1+s(0, lj)+∣
∣
∣u
(0)1 (0)
∣
∣
∣
2
≤ C∥
∥u(0)∥
∥
2QN
j=1H1+s(0, lj)
. (2.35)
By definition of w(1), we have also
∥
∥w(1)∥
∥
2QN
j=1Hs(0, lj)
=∥
∥u(1)∥
∥
2QN
j=1Hs(0, lj)
. (2.36)
Moreover, by the continuous injection of
[
D(A), D(A0)]
1−s = Xs
in[
N∏
j=1
H2(0, lj) ×N∏
j=1
H1(0, lj),
N∏
j=1
H1(0, lj) × L2(R)
]
1−s
=
N∏
j=1
H1+s(0, lj) ×N∏
j=1
Hs(0, lj),
there exists C > 0 such that
∥
∥
(
u(0), u(1))∥
∥
2QN
j=1H1+s(0, lj)×
QNj=1H
s(0, lj)≤ C
∥
∥
(
u(0), u(1))∥
∥
2
Xs. (2.37)
The estimates (2.34)-(2.37) prove this lemma.
The main results of the chapter are the following
Theorem 2.3.8. Assume that the weighted observability inequality (2.11) holds for everysolution of (2.7) with lim infn→∞ cn = 0 and cn 6= 0 for all n ∈ N∗. Let ω be defined by(2.33). Assume that the initial data (u(0), u(1)) belong to Xs, characterized by Lemma 2.3.6,where 0 < s < 1/2. Let Φs be a function such that the pair (ω, Φs) satisfies (2.28)-(2.31).Then there exists a constant C > 0 such that the corresponding solution u of (2.1) verifies
∀t ≥ 0, Eu(t) ≤ CΦs
(
1
t+ 1
)
∥
∥(u(0), u(1))∥
∥
2
Xs. (2.38)
Remark 2.3.9. We see that the decay rate of the energy directly depends on the behaviourof the interpolation function Φs near 0 and thus of ω and of the weights c2n as n→ ∞.
103
Proof: We split up u as in (2.15) and we decompose w(0) = u(0) − u(0)1 (0)ϕ and w(1) = u(1)
asw(0) = u(0) − u
(0)1 (0)ϕ =
∑
n≥1
w0,nϕDn and w(1) = u(1) =
∑
n≥1
w1,nϕDn ,
where (λnw0,n)n, (w1,n)n ∈ l2(N∗). We write
E−(0) =∑
n≥1
unω(n), (2.39)
where (un)n∈N∗ ∈ l1(N∗; R) is defined by
∀n ≥ 2, un =1
2
(
λ2n−1w
20,n−1 + w2
1,n−1
)
andu1 = u
(0)1 (0)2.
Observe that, by the construction (2.33) of ω, we have
E−(u, 0) ≤ E∗(u, 0).
Then, by (2.4) and (2.17) of Theorem 2.2.1, we have
Eu(0) −Eu(T ) =
∫ T
0
(
∂u1
∂t(0, t)
)2
dt ≥ CE−(0). (2.40)
Assume further that (u(0), u(1)) ∈ Xs. We define
E+(0) =∑
n≥1
unn2s. (2.41)
It follows from Theorem 2.3.2 (applied to the function f = u and the weight ω) that
1 ≤ Φs
E−(0)∑
n≥1
un
E+(0)∑
n≥1
un. (2.42)
By the so-called Weyl’s formula (see for instance [2, 20, 82]), we have
λk ∼kπ
L, (2.43)
and thus, there exist c1, c2 > 0 such that, for n large enough, we have
c1(n + 1)π
L≤ λn ≤ c2
(n+ 1)π
L.
104
Therefore, we have
E+(0) = 12
(
Lπ
)2s∑
n≥1
(
λ2nw
20,n + w2
1,n
)
(
(n + 1)π
L
)2s
+∣
∣
∣u
(0)1 (0)
∣
∣
∣
2
≤ C
(
∑
n≥1
(
λ2nw
20,n + w2
1,n
)
λ2sn +
∣
∣
∣u
(0)1 (0)
∣
∣
∣
2)
≤ C
(
∥
∥
(
w(0), w(1))∥
∥
2
D(AsD)
+∣
∣
∣u
(0)1 (0)
∣
∣
∣
2)
.
Consequently, by Lemma 2.3.7, we obtain that there exists C > 0 such that
E+(0) ≤ C∥
∥
(
u(0), u(1))∥
∥
2
Xs.
Moreover,
∑
n≥1
un =1
2
∑
n≥1
(
λ2nw
20,n + w2
1,n
)
+ u(0)1 (0)2 =
∥
∥w(0)∥
∥
2
V+ u
(0)1 (0)2 +
∥
∥u(1)∥
∥
2
L2(R).
Furthermore, there exists C ′ > 0 such that
∥
∥u(0)∥
∥
2
V=∥
∥
∥w(0) + u
(0)1 (0)ϕ
∥
∥
∥
2
V≤ C ′
(
∥
∥w(0)∥
∥
2
V+∣
∣
∣u
(0)1 (0)
∣
∣
∣
2)
= C ′(
∥
∥w(0)∥
∥
2
V+∣
∣
∣u
(0)1 (0)
∣
∣
∣
2)
because ϕ is given and w(0) ∈ V . Therefore, there exists C ′ > 0 such that
∑
n≥1
un ≥ C ′ ∥∥
(
u(0), u(1))∥
∥
2
V×L2(R).
Consequently, (2.42) becomes, by the increasing character of Φs
1 ≤ Φs
(
E−(0)
C ′ ‖(u(0), u(1))‖2V×L2(R)
)
C∥
∥(u(0), u(1))∥
∥
2
Xs
C ′ ‖(u(0), u(1))‖2V×L2(R)
,
which yields
E−(0) ≥ C ′ ∥∥(u(0), u(1))
∥
∥
2
V×L2(R)Φ−1s
∥
∥(u(0), u(1))∥
∥
2
V×L2(R)
C ‖(u(0), u(1))‖2Xs
, (2.44)
with C, C ′ > 0. From (2.40) and (2.44), it follows that there exist C, C ′ > 0,
Eu(T ) ≤ Eu(0) − C ′ ∥∥(u(0), u(1))
∥
∥
2
V×L2(R)Φ−1s
∥
∥(u(0), u(1))∥
∥
2
V×L2(R)
C ‖(u(0), u(1))‖2Xs
, (2.45)
105
for any (u(0), u(1)) ∈ Xs.We follow now the proof of Ammari and Tucsnak [11]. We rewrite (2.45) as follows :
‖(u(T ), ut(T ))‖2V×L2(R) ≤
∥
∥(u(0), u(1))∥
∥
2
V×L2(R)
−C ′ ∥∥(u(0), u(1))
∥
∥
2
V×L2(R)Φ−1s
(
‖(u(0), u(1))‖2
V ×L2(R)
C‖(u(0), u(1))‖2
Xs
)
.
This estimate remains valid in successive time-intervals [lT, (l + 1)T ]. Notice that thereexists C > 0 such that
∀t ≥ 0, ‖(u(t), ut(t))‖Xs≤ C
∥
∥
(
u(0), u(1))∥
∥
Xs, (2.46)
for 0 < s < 12
by interpolation, because it is true for s = 0 and s = 1 (see Theorem 5.1of [72]). By (2.46) and the fact that the energy is decreasing by (2.4) and that Φ−1
s isincreasing, we obtain that
‖(u((l + 1)T ), ut((l + 1)T ))‖2V×L2(R) ≤ ‖(u(lT ), ut(lT ))‖2
V×L2(R)
−C ′ ‖(u(lT ), ut(lT ))‖2V×L2(R) Φ−1
s
(
‖(u((l+1)T ), ut((l+1)T ))‖2V ×L2(R)
C‖(u(0), u(1))‖2
Xs
)
,(2.47)
for every l ∈ N ∪ 0 .Our expression (2.47) coincides with (4.16) in Ammari and Tucsnak [11] (with G = Φ−1
s
and θ = 1/2). The rest of the proof follows as in [11], where (2.31) is used, by noticing that
‖(u((l + 1)T ), ut((l + 1)T ))‖2V×L2(R)
C ‖(u(0), u(1))‖2Xs
< 1,
taking a higher constant C if necessary. Then (2.38) follows.
In addition, using (2.38) and making a particular and explicit choice of the concavefunction Φs, we obtain a more explicit dependence of the weights on the decay rate of theenergy :
Theorem 2.3.10. Assume that the weighted observability inequality (2.11) holds for everysolution of (2.7) with lim inf
n→∞cn = 0 and cn 6= 0 for all n ∈ N∗. Let ω be defined by (2.33).
We set
∀t > 0, ϕ(t) =ω(t)
t2.
Then there exists a constant C > 0 such that for any initial data (u(0), u(1)) ∈ Xs (0 < s <1/2), the corresponding solution u of (2.1) verifies
∀t ≥ 0, Eu(t) ≤C
(
ϕ−1(
1t+1
))2s
∥
∥(u(0), u(1))∥
∥
2
Xs. (2.48)
106
Proof: We set
Φs(t) =1
(ϕ−1(t))2s
where 0 < s < 1/2. Then the pair (ω, Φs) verifies (2.28)-(2.29) on (0, ∞). Indeed, as Φs
is the composition of the functions t 7→ 1/ϕ−1(t) and t 7→ t2s which are increasing andconcave functions (by Lemma 2.6 of [19]), Φs is an increasing and concave function.
Moreover, we easily check that
1
ϕ−1(ω(t))≥ 1
t
on [1, ∞). As a consequence, (2.30) holds on [1, ∞).Finally, we have
1
tΦ−1s (t) = t
1−ss ω
(
1
t12s
)
,
which is an increasing function on (0, ∞) because 0 < s < 1/2, and therefore (ω, Φs)satisfy (2.28)-(2.31). We apply now (2.38) of Theorem 2.3.8 with
Φs(t) =
(
1
ϕ−1(t)
)2s
,
to obtain the result.
2.4 Examples
In [38] the authors proved observability inequalities of type (2.11) on which our analysisis based. In the case of star-shaped networks, the weights in (2.11) depend on the irrationa-lity properties of the ratios of the lengths of the strings. In the case of tree-shaped networks,the observability inequality is proved under a condition (that is fulfilled generically withinthe class of tree-shaped networks) that generalizes the condition on the irrationality of theratios of the lengths of the strings arising in the case of stars. Finally, in the case of generalnetworks, the observability inequality holds under the condition that all eigenfunctions ofthe network are observable. This last result is a generalization of the previous one for starsand trees.
To illustrate the wide range of applications of the main result of this chapter, in thissection, we apply our previous results to some examples of particular networks : a star-shaped network and a particular tree. We obtain the weights (cn)n directly by [38] anddeduce an explicit decay rate for the corresponding dissipative system.
2.4.1 The star-shaped network with N strings
The star-shaped network with N strings is formed by N strings connected at one pointv, which constitutes a particular tree. Recall that the damping term is located on the vertex
107
v
v1
e1
v2e2
vNeN
Fig. 2.1 – A star-shaped network with N strings
v1, the origin of the controlled edge e1 of length l1. The remaining N − 1 exterior nodesare denoted by vi, i = 2, ..., N , the string that contains vi by ei and its length by li.
In [38], the authors proved the observability inequality (2.11) for the conservative system(2.7) with Dirichlet boundary conditions at all exterior nodes with the following weights
ck = maxi=2,...,N
∏
j 6=i
∣
∣
∣sin(λklj)∣
∣
∣ , ∀n ≥ 1.
First, if the ratio of any two of the lengths of the uncontrolled strings is an irrationalnumber, i.e. li/lj /∈ Q for all i 6= j, then for all k ∈ N, we have ck > 0. In this case, theenergy of u solution of the dissipative system (2.1) tends to 0 as t→ ∞. This can be easilyproved with LaSalle’s invariance principle, using the energy as Lyapunov functional, butit does not yield any explicit decay rate.
Thus assume that li/lj /∈ Q for all i 6= j. Denote by S the set of all real numbersρ such that ρ /∈ Q and so that its expansion as a continued fraction [0, a1, ..., an, ...] issuch that (an) is bounded. It is well known that S is uncountable and that its Lebesguemeasure is zero. Roughly speaking, the set S contains all irrational numbers which arebadly approximated by rational numbers. In particular, by the Euler-Lagrange theorem,S contains all irrational quadratic numbers (i.e. the roots of second order equations withrational coefficients).
We use also a well-known result asserting that, for all ǫ > 0 there exists a set Bǫ ⊂ R,such that the Lebesgue measure of R\Bǫ is equal to zero, and a constant Cǫ > 0 for which,if ξ ∈ Bǫ, then o
ξmo
≥ Cǫm1+ǫ
,
wheregη
gis the distance from η to the set Z :
oη
o= min
η−x∈Z
|x| .
108
In particular, S is contained in the sets Bǫ for every ǫ > 0 (see [38] for more details).Then, by Corollary A.10 of [38], we have
Lemma 2.4.1. 1. If for all values i, j = 2, ..., N, i 6= j, the ratios li/lj belong to S, thenthere exists a constant c > 0 such that
ck ≥c
λN−2k
, ∀k ∈ N∗.
2. If for all values i, j = 2, ..., N, i 6= j, the ratios li/lj belong to Bǫ, then there existsa constant cǫ > 0 such that
ck ≥cǫ
λN−2+ǫk
, ∀k ∈ N∗.
With more restrictive assumptions on the lengths of the uncontrolled strings, we haveanother bound for ck. Let us recall a definition of [38].
Definition 2.4.2. We say that the real numbers l1, ..., lN verify the conditions (S) if• l1, ..., lN are linearly independent over the field Q of rational numbers ;• the ratios li/lj are algebraic numbers for i, j = 1, ..., N .
In this case, we have the following result (see Corollary A.10 of [38]) :
Lemma 2.4.3. If the numbers l2, ..., lN verify the conditions (S), then for every ǫ > 0,there exists a constant cǫ > 0 such that
ck ≥cǫ
λ1+ǫk
, ∀k ∈ N∗.
Consequently, we have
Proposition 2.4.4. 1. Assume that for all values i, j = 2, ..., N, i 6= j, the ratios li/ljbelong to S. Then there exists a constant C > 0 such that for any initial data (u(0), u(1)) ∈Xs (0 < s < 1/2), the corresponding solution u of (2.1) verifies
Eu(t) ≤C
(t+ 1)s
N−2
∥
∥(u(0), u(1))∥
∥
2
Xs.
2. Assume that for all values i, j = 2, ..., N, i 6= j, the ratios li/lj belong to Bǫ forǫ > 0. Then there exists a constant Cǫ > 0 such that for any initial data (u(0), u(1)) ∈ Xs
(0 < s < 1/2), the corresponding solution u of (2.1) verifies
Eu(t) ≤C
(t+ 1)s
N−2+ǫ
∥
∥(u(0), u(1))∥
∥
2
Xs.
3. Assume that the numbers l2, ..., lN verify the conditions (S). Let ǫ > 0. Then thereexists a constant Cǫ > 0 such that for any initial data (u(0), u(1)) ∈ Xs (0 < s < 1/2), thecorresponding solution u of (2.1) verifies
Eu(t) ≤Cǫ
(t+ 1)s
1+ǫ
∥
∥(u(0), u(1))∥
∥
2
Xs.
109
Proof: We set α = N − 2 and cα = c in the first case, α = N − 2 + ǫ and cα = cǫ in thesecond one and α = 1 + ǫ and cα = cǫ in the third one. For α ≥ 1, we take
ω(t) = d2α
(
1
t
)2α
,
where dα verifies d2α ≤ 1 and c2αλ
−2αn−1 ≥ d2
αn−2α for all n ∈ N∗, which is possible because of
Weyl’s formula (2.43).We are in the situation 1 of Example 2.3.3 with p = 2α ≥ 1. Therefore, we take
Φs(t) =
(
t
d2α
)sα
Then (ω, Φs) satisfy (2.28)-(2.31) and we apply Theorem 2.3.8 to finish the proof.
Finally, if the lengths l2, ..., lN verify li/lj /∈ Q for all i 6= j, but the conditions 1, 2 and3 of Proposition 2.4.4 are not verified, it can be proved that cn verify
cn ≥ ψ(n) > 0, ∀n ∈ N∗, (2.49)
where ψ is a positive convex and decreasing function which can be smaller than cλ−αn , withα > 0. Indeed, we know by Appendix A of [38], that if we set
a(λ) =N∑
i=2
∏
j 6=i|sin(λlj)| ,
we have
a(λ) ≥ C mini=2,...,N
∏
j 6=i
o liljmi(λ)
o,
where
mi(λ) = E
(
liπλ
)
,
E(η) being the closest integer number to η : |η −E(η)| =gη
g. Therefore, it is sufficient
to bound by belowgliljm
gfor all m ∈ N and i 6= j to get a lower bound of cn.
Moreover, we know that Liouville’s numbers ξ are such that for all n ∈ N, there existsq ∈ N, q > 1 such that
0 <oξq
o<
1
qn+1.
That is why, when li/lj is a Liouville’s number (see [29]), the function ψ in (2.49) is smallerthan any negative power of λn (for example of the order of e−n), and thus the decay rate ofthe energy of u is slower than polynomial. Indeed, for instance, it is possible to constructreal numbers ξ of the form
ξ =∑
k∈N
10−ak ,
110
where (ak)k is an increasing sequence of natural numbers, which are approximated byrational ones faster than any given positive increasing function ρ :
o10apξ
oρ(10ap) < ǫ, ∀p ∈ N
for ǫ > 0 (see [38], p.66 for more details).Let us construct an irrational number x which is approximated by rational ones at an
exponential speed. The aim is to find an irrational xa such that
lim infq→∞
oxaq
o 1q
= e−a, where a is a positive constant. (2.50)
This construction uses the theory of continued fractions (see [94]). We recall here someresults about the continued fractions which are used in the sequel and we refer to [29,54,66]for more details. Let [a0, a1, ..., an, ...] be the expansion as a continued fraction of x ∈ R\Q.We set xn = [a0, a1, ..., an] = pn
qnwhere pn, qn are integers. The integers pn, qn are relatively
prime and satisfy the following relations :
p−1 = 1, p0 = a0, pn+1 = an+1pn + pn−1 (2.51)
andq−1 = 0, q0 = 1, qn+1 = an+1qn + qn−1. (2.52)
We have the following theorem which is a result of best approximation (see [29,54,66,94]) :
Theorem 2.4.5. Let x /∈ Q. Then, with the notation introduced before, we have(i) qn ≥ 2(n−1)/2
(ii)gqnx
g= |qnx− pn| and 1
2qn+1≤ g
qnxg ≤ 1
qn+1,
(iii) If 0 < q < qn+1, thengqx
g ≥ gqnx
g.
First, we show
lim infoxq
o 1q
= lim inf a− 1
qn
n+1 . (2.53)
Set l(x) = lim infgxq
g 1q and λ(x) = lim inf a
− 1qn
n+1 . By the relation (2.52), we have
an+1qn ≤ qn+1 ≤ 2an+1qn.
Then, by (ii) of Theorem 2.4.5, we obtain
lngqnx
g
qn≤ ln(1/qn+1)
qn≤ ln(1/an+1) + ln(1/qn)
qn,
and thus
lim infln
gqx
g
q≤ lim inf
ln(1/an+1)
qn,
111
which leads to l(x) ≤ λ(x). For an arbitrary q ≥ 1, let n be an integer such that qn ≤ q <qn+1. As ln
gqx
g ≤ 0 and by (ii) of Theorem 2.4.5, we have
lngqx
g
q≥ ln
gqx
g
qn≥ ln(1/2qn+1)
qn≥ ln(1/an+1) + ln(1/4qn)
qn,
which leads to
lim infln
gqx
g
q≥ lim inf
ln(1/an+1)
qn,
and to l(x) ≥ λ(x). Therefore, (2.53) holds.Thus finding xa ∈ R\Q such that (2.50) is equivalent to finding xa ∈ R\Q such that
lim inf a− 1
qn
n+1 = e−a.
We construct a such xa by induction. Set a0 = 0, which determines p0 and q0. Then, ifa0, ..., an, p0, ..., pn, q0, ..., qn are found, we choose an+1 by
an+1 = E(eaqn).
Then, pn+1 and qn+1 are imposed by the relations (2.51) and (2.52).Therefore, for all a > 0, we have found an irrational number xa such that there exists
a positive constant δ such that,
oqxa
o≥ δe−aq, ∀q ∈ N. (2.54)
Consequently, if li/lj are reals of the form xai,jfor ai,j > 0, which verify (2.54), then there
exists a positive constant C such that for all k ∈ N,
ck ≥ Ce−b(N−2)λklπ ,
where b = maxi6=j(ai,j) > 0 and l = maxj
(lj). By the Weyl’s formula, there exists c2 > 0
such that λk ≤ c2kπ/L, and thus, we obtain
c2k ≥ Ce−2c2b(N−2)k = Ce−Ak,
where A = 2c2b(N − 2).Consequently, we have
Proposition 2.4.6. Assume that for all values i, j = 2, ..., N, i 6= j, the ratios li/lj arereals of the form xai,j
for ai,j > 0 which verify (2.54). Then there exist constants C, C ′ > 0
such that for any initial data (u(0), u(1)) ∈ Xs (0 < s < 1/2), the corresponding solution uof (2.1) verifies
∀t ≥ 0, Eu(t) ≤C ′
(ln(C(1 + t)))2s
∥
∥(u(0), u(1))∥
∥
2
Xs. (2.55)
112
Proof: We apply Theorem 2.3.8 with
ω(t) = Ce−At,
for C > 0 and A > 2(2s+ 1), and
Φs(t) =
(
A
ln(
Ct
)
)2s
.
We are in the situation 2 of Example 2.3.3 and we simply apply (2.38) to obtain (2.55).
2.4.2 A non star-shaped tree
Now let us consider a tree, which is not star-shaped, having the simple structure infigure 2.2.
v1
e1
e 2
e3
e4
e5
Fig. 2.2 – A non star-shaped tree
Recall that the damping term is on the vertex v1, the origin of the damped edge e1 oflength l1. We will assume, in addition, that l4 = l2.
Recall that in [38], the authors proved the observability inequality (2.11) for the conser-vative system (2.7) with Dirichlet boundary conditions at all exterior nodes with the weightsck given by
ck = max∣
∣
∣d5(λk)
∣
∣
∣,∣
∣
∣d4(λk)
∣
∣
∣,∣
∣
∣d2(λk)
∣
∣
∣
,
whered5(λ) = − sin(λl2) sin(λl4), d4(λ) = − sin(λl2) sin(λl5),
d2(λ) = −(cos(λl3) sin(λl5) sin(λl4) + sin(λl3) cos(λl5) sin(λl4) + sin(λl3) sin(λl5) cos(λl4)).
First, if ck = 0, then∣
∣
∣d5(λk)
∣
∣
∣=∣
∣
∣d4(λk)
∣
∣
∣=∣
∣
∣d2(λk)
∣
∣
∣= 0. As l4 = l2, we obtain
sin(λl4) = 0 and sin(λl3) sin(λl5) = 0. If sin(λl3) = 0 (resp. sin(λl5) = 0), then, necessarily
113
l3/l2 (resp. l5/l2) is a rational number. Consequently, ck 6= 0 if l3/l2 and l5/l2 are irrationalnumbers and then, in this case, the energy of u solution of (2.1) decays to 0.
Secondly, applying Appendix A of [38], we know that there exists a positive constant csuch that for every k ∈ N∗,
ck ≥c
λαk
if one of the following three conditions holds• the ratios l5/l2, l3/l2 and l3/l5 belong to S and α > 4or• the ratios l5/l2, l3/l2 and l3/l5 belong to some Bǫ and α > 4 + ǫor• the numbers l3, l5, l2 satisfy the conditions (S) and α > 2 + ǫ.Consequently, by using (2.38) of Theorem 2.3.8 and the first statement of Example 2.3.3,
there exists a constant C > 0 such that for any initial data (u(0), u(1)) ∈ Xs (0 < s < 1/2),the corresponding solution u verifies
Eu(t) ≤Cǫ
(t+ 1)sα
∥
∥(u(0), u(1))∥
∥
2
Xs.
Finally, if these previous conditions do not hold, we can apply Theorem 2.3.8 to obtaina decay rate of the energy of u with an admissible pair (ω, Φs), or Theorem 2.3.10 to obtaina more explicit decay rate of the energy.
As we have seen in these examples, our method to obtain decay rates of the dissipativesystem (2.1) is of systematic application :
First, we find the weights (c2n) of the observability inequality (2.8) for the conservativeproblem (2.7) with Dirichlet boundary conditions at all exterior nodes.
Then we take, for the weight ω, the lower convex envelop of the sequence (1, (c2n)n).Finally we choose Φs such that (ω, Φs) satisfy (2.28)-(2.31) to obtain the decay rate of
the energy, given by
Φs
(
1
1 + t
)
.
Note that the study of the observability inequality (2.8) for (2.7) and the weights (c2n)have been already done in some works (see [38]). Consequently we can use directly theseresults and it is not necessary to show another observability inequality. In addition, noticethat the weighted observability inequality (2.17) holds under the same assumptions on thenetwork that are needed for the Dirichlet problem (2.7). This method allows to treat animportant class of networks.
Moreover we can extend this general principle to networks of strings with damping inseveral exterior vertices, or with damping in interior nodes.
114
Chapitre 3
Stabilization of second orderevolution equations with unboundedfeedback with delay
3.1 Introduction
Time-delay often appears in many biological, electrical engineering systems and mecha-nical applications [1, 51, 101], and in many cases, in particular for distributed parametersystems, even arbitrarily small delays in the feedback may destabilize the system, seee.g. [39–41,52, 75, 85, 89, 98, 110]. The stability issue of systems with delay is, therefore, oftheoretical and practical importance.
We further remark that some techniques developed recently in [85] and in chapter 1 inorder to obtain some existence results and decay rates have some similarities. We thereforepropose to consider an abstract setting as large as possible in order to contain a quitelarge class of problems with time delay feedbacks. In a second step we prove existence andstability results in this setting under realistic assumptions. Finally in order to show theusefulness of our approach, we give some examples where our abstract framework can beapplied. For a similar approach, we refer to the paper in preparation [5]. Without delaysuch an approach was developed in [11].
Before going on, let us present our abstract framework. Let H be a real Hilbert spacewith norm and inner product denoted respectively by ‖.‖H and (., .)H . Let A : D(A) → H
be a self-adjoint positive operator with a compact inverse in H. Let V := D(A12 ) be the
domain of A12 . Denote by D(A
12 )′ the dual space of D(A
12 ) obtained by means of the inner
product in H.Further, for i = 1, 2, let Ui be a real Hilbert space (which will be identified to its
dual space) with norm and inner product denoted respectively by ‖.‖Uiand (., .)Ui
and let
Bi ∈ L(Ui, D(A12 )′).
115
We consider the system described by
ω(t) + Aω(t) +B1u1(t) +B2u2(t− τ) = 0, t > 0ω(0) = ω0, ω(0) = ω1,
u2(t− τ) = f 0(t− τ), 0 < t < τ,(3.1)
where t ∈ [0, ∞) represents the time, τ is a positive constant which represents the delay,ω : [0, ∞) → H is the state of the system and u1 ∈ L2([0, ∞), U1), u2 ∈ L2([−τ, ∞), U2)are the input functions. Most of the linear equations modelling the vibrations of elasticstructures with distributed control with delay can be written in the form (3.1), where ωstands for the displacement field.
In many problems, coming in particular from elasticity, the input ui are given in thefeedback form ui(t) = B∗
i ω(t), which corresponds to colocated actuators and sensors. Weobtain in this way the closed loop system
ω(t) + Aω(t) +B1B∗1 ω(t) +B2B
∗2 ω(t− τ) = 0, t > 0
ω(0) = ω0, ω(0) = ω1,B∗
2 ω(t− τ) = f 0(t− τ), 0 < t < τ.(3.2)
The first natural question is the well posedness of this problem. In section 3.2 we willgive a sufficient condition that guarantees that this problem (3.2) is well-posed, where weclosely follow the approach developed in [85] for the wave equation. Secondly, we may askif this system is dissipative. We show in section 3.3 that the condition
∃0 < α < 1, ∀u ∈ V, ‖B∗2u‖2
U2≤ α ‖B∗
1u‖2U1
(3.3)
guarantees the strict decay of the energy ; under this condition, using a result from [12](see also [105]) we pertain a necessary and sufficient condition for the decay to zero of theenergy. Note that this last condition is independent of the delay and therefore under thecondition (3.3), our system is strongly stable if and only if the same system without delayis strongly stable. Note further that if (3.3) is not satisfied, there exist cases where someinstabilities may appear (see [85, 110] and chapter 1 for the wave equation). Hence thisassumption seems realistic.
In a third step, again under the condition (3.3) and a certain boundedness assumptionfrom [11] between the resolvent operator of A and of the operators B1 and B2, see condition(3.20), we prove that the exponential decay of the system (3.2) follows from a certainobservability estimate. Again this observability estimate is independent of the delay termB2B
∗2ω(t− τ) and therefore, under the conditions (3.3) and (3.20), the exponential decay
of the system (3.2) follows from the exponential decay of the same system without delay.Nevertheless we give the dependence of the decay with respect to the delay, in particularwe show that if the delay increases the decay decreases. This is the content of section 3.4.A similar analysis for the polynomial decay is performed in section 3.5 by weakening theobservability estimate. Again we show that if the delay increases the decay decreases. Inview of some applications, section 3.6 is devoted to the proof of these two observability
116
estimates by using a frequency domain method and a reduction to some conditions betweenthe eigenvectors of A and the feedback operator B∗
1 .Finally we finish this chapter by considering in section 3.7 different examples where our
abstract framework can be applied. To our knowledge, all the examples, with the exceptionof the first one, are new.
3.2 Well posedness of the problem
We aim to show that problem (3.2) is well-posed. For that purpose, we use semi-grouptheory and an idea from [85] (see also chapter 1 or [89]). Let us introduce the auxiliaryvariable z(ρ, t) = B∗
2 ω(t − τρ) for ρ ∈ (0, 1) and t > 0. Note that z verifies the transportequation for 0 < ρ < 1 and t > 0
τ ∂z∂t
+ ∂z∂ρ
= 0
z(0, t) = B∗2 ω(t)
z(ρ, 0) = B∗2 ω(−τρ) = f 0(−τρ).
(3.4)
Therefore, the problem (3.2) is equivalent to
ω(t) + Aω(t) +B1B∗1 ω(t) +B2z(1, t) = 0, t > 0
τ ∂z∂t
+ ∂z∂ρ
= 0, t > 0, 0 < ρ < 1
ω(0) = ω0, ω(0) = ω1, z(ρ, 0) = f 0(−τρ), 0 < ρ < 1z(0, t) = B∗
2 ω(t), t > 0.
(3.5)
If we introduceU := (ω, ω, z)T ,
then U satisfies
U ′ = (ω, ω, z)T =
(
ω, −Aω(t) − B1B∗1 ω(t) −B2z(1, t), −
1
τ
∂z
∂ρ
)T
.
Consequently the problem (3.2) may be rewritten as the first order evolution equation
U ′ = AUU(0) = (ω0, ω1, f
0(−τ.)), (3.6)
where the operator A is defined by
A
ωuz
=
u−Aω − B1B
∗1u−B2z(1)
− 1τ∂z∂ρ
,
with domain
D(A) := (ω, u, z) ∈ V × V ×H1((0, 1), U2); z(0) = B∗2u, Aω +B1B
∗1u+B2z(1) ∈ H.
117
Now, we introduce the Hilbert space
H = V ×H × L2((0, 1), U2)
equipped with the usual inner product⟨
ωuz
,
ωuz
⟩
=(
A12ω, A
12 ω)
H+ (u, u)H +
∫ 1
0
(z(ρ), z(ρ))U2dρ. (3.7)
Let us suppose now that
∃0 < α ≤ 1, ∀u ∈ V, ‖B∗2u‖2
U2≤ α ‖B∗
1u‖2U1. (3.8)
Under this condition, we will show that the operator A generates a C0-semigroup in H.For that purpose, we choose a positive real number ξ such that
1 ≤ ξ ≤ 2
α− 1. (3.9)
This constant exists because 0 < α ≤ 1.We now introduce the following inner product on H⟨
ωuz
,
ωuz
⟩
H
=(
A12ω, A
12 ω)
H+ (u, u)H + τξ
∫ 1
0
(z(ρ), z(ρ))U2dρ.
This new inner product is clearly equivalent to the usual inner product on H (3.7).
Theorem 3.2.1. Under the assumption (3.8), for an initial datum U0 ∈ H, there exists aunique solution U ∈ C([0, +∞), H) to system (3.6). Moreover, if U0 ∈ D(A), then
U ∈ C([0, +∞), D(A)) ∩ C1([0, +∞),H).
Proof: By Lumer-Phillips’ theorem, it suffices to show that A is m-dissipative (see Defi-nition 3.3.1 and Theorems 1.4.3 and 1.4.6 of [93]).
We first prove that A is dissipative. Take U = (ω, u, z)⊤ ∈ D(A). Then
〈AU, U〉H =
⟨
u−Aω −B1B
∗1u− B2z(1)
− 1τ∂z∂ρ
,
ωuz
⟩
H
=(
A12u, A
12ω)
H− (Aω + B1B
∗1u+B2z(1), u)H − ξ
∫ 1
0
(
∂z
∂ρ(ρ), z(ρ)
)
U2
dρ.
Since Aω +B1B∗1u+B2z(1) ∈ H, we obtain
〈AU, U〉H =(
A12u, A
12ω)
H− 〈Aω, u〉V ′, V − 〈B1B
∗1u, u〉V ′, V − 〈B2z(1), u〉V ′, V
−ξ∫ 1
0
(
∂z
∂ρ(ρ), z(ρ)
)
U2
dρ
= 〈Aω, u〉V ′, V − 〈Aω, u〉V ′, V − ‖B∗1u‖2
U1− (z(1), B∗
2u)U2
−ξ∫ 1
0
(
∂z
∂ρ(ρ), z(ρ)
)
U2
dρ,
118
by duality. By intregrating by parts, we obtain
∫ 1
0
(
∂z
∂ρ(ρ), z(ρ)
)
U2
dρ = −∫ 1
0
(
z(ρ),∂z
∂ρ(ρ)
)
U2
dρ+ (‖z(1)‖2U2
− ‖z(0)‖2U2
),
and thus∫ 1
0
(
∂z
∂ρ(ρ), z(ρ)
)
U2
dρ =1
2(‖z(1)‖2
U2− ‖B∗
2u‖2U2
).
Therefore, by Cauchy-Schwarz’s inequality, we find
〈AU, U〉H = −‖B∗1u‖2
U1− (z(1), B∗
2u)U2− ξ
2‖z(1)‖2
U2+ξ
2‖B∗
2u‖2U2
≤ −‖B∗1u‖2
U1+ (
1
2− ξ
2) ‖z(1)‖2
U2+ (
1
2+ξ
2) ‖B∗
2u‖2U2.
By (3.8), we obtain
〈AU, U〉H ≤ (α
2+ξα
2− 1) ‖B∗
1u‖2U1
+ (1
2− ξ
2) ‖z(1)‖2
U2
with α2
+ ξα2− 1 ≤ 0 and 1
2− ξ
2≤ 0 because ξ satisfies condition (3.9). This shows that
〈AU, U〉H ≤ 0 and then the dissipativeness of A.Let us now prove that λI −A is surjective for some λ > 0.Let (f, g, h)T ∈ H. We look for U = (ω, u, z)T ∈ D(A) solution of
(λI −A)
ωuz
=
fgh
or equivalently
λω − u = fλu+ Aω +B1B
∗1u+B2z(1) = g
λz + 1τ∂z∂ρ
= h.(3.10)
Suppose that we have found ω with the appropriate regularity. Then, we have
u = −f + λω ∈ V.
We can then determine z, indeed z satisfies the differential equation
λz +1
τ
∂z
∂ρ= h
and the boundary condition z(0) = B∗2u = −B∗
2f + λB∗2ω. Therefore z is explicitely given
by
z(ρ) = λB∗2ωe
−λτρ −B∗2fe
−λτρ + τe−λτρ∫ ρ
0
eλτσh(σ)dσ.
119
This means that once ω is found with the appropriate properties, we can find z and u. Inparticular, we have
z(1) = λB∗2ωe
−λτ −B∗2fe
−λτ + τe−λτ∫ 1
0
eλτσh(σ)dσ = λB∗2ωe
−λτ + z0, (3.11)
where z0 = −B∗2fe
−λτ + τe−λτ∫ 1
0eλτσh(σ)dσ is a fixed element of U2 depending only on f
and h.
It remains to find ω. By (3.10), ω must satisfy
λ2ω + Aω + λB1B∗1ω + B2z(1) = g +B1B
∗1f + λf,
and thus by (3.11),
λ2ω + Aω + λB1B∗1ω + λe−λτB2B
∗2ω = g +B1B
∗1f + λf − B2z
0 =: q,
where q ∈ V ′. We take then the duality brackets 〈., .〉V ′, V with φ ∈ V :
⟨
λ2ω + Aω + λB1B∗1ω + λe−λτB2B
∗2ω, φ
⟩
V ′, V= 〈q, φ〉V ′, V .
Moreover :⟨
λ2ω + Aω + λB1B∗1ω + λe−λτB2B
∗2ω, φ
⟩
V ′, V
= λ2 〈ω, φ〉V ′, V + 〈Aω, φ〉V ′, V + λ(〈B1B∗1ω, φ〉V ′, V + e−λτ 〈B2B
∗2ω, φ〉V ′, V )
= λ2 (ω, φ)H +(
A12ω, A
12φ)
H+ λ((B∗
1ω, B∗1φ)U1
+ e−λτ (B∗2ω, B
∗2φ)U2
)
because ω ∈ V ⊂ H . Consequently, we arrive at the problem
λ2 (ω, φ)H +(
A12ω, A
12φ)
H+λ((B∗
1ω, B∗1φ)U1
+ e−λτ (B∗2ω, B
∗2φ)U2
) = 〈q, φ〉V ′, V . (3.12)
The left hand side of (3.12) is continuous and coercive on V. Indeed, we have∣
∣
∣λ2 (ω, φ)H +
(
A12ω, A
12φ)
H+ λ((B∗
1ω, B∗1φ)U1
+ e−λτ (B∗2ω, B
∗2φ)U2
)∣
∣
∣
≤ λ2 ‖ω‖H ‖φ‖H +∥
∥
∥A
12ω∥
∥
∥
H
∥
∥
∥A
12φ∥
∥
∥
H+ λ(‖B∗
1ω‖U1‖B∗
1φ‖U1+ e−λτ ‖B∗
2ω‖U2‖B∗
2φ‖U2)
≤ Cλ2 ‖ω‖V ‖φ‖H +∥
∥
∥A
12
∥
∥
∥
2
‖ω‖V ‖φ‖V+λ(‖B∗
1‖2L(V,U1) ‖ω‖V ‖φ‖V + e−λτ ‖B∗
2‖2L(V,U2) ‖ω‖V ‖φ‖V )
≤ C ‖ω‖V ‖φ‖V ,and for φ = ω ∈ V
λ2 ‖ω‖2H +
(
A12ω, A
12ω)
H+ λ(‖B∗
1ω‖2U1
+ e−λτ ‖B∗2ω‖2
U2)
≥∥
∥
∥A
12ω∥
∥
∥
2
H≥ C ‖ω‖2
V .
Therefore, this problem (3.12) has a unique solution ω ∈ V by Lax-Milgram’s lemma.Moreover Aω+B1B
∗1u+B2z(1) = g+λf−λ2ω ∈ H. In summary, we have found (ω, u, z)T ∈
D(A) satisfying (3.10).
Remark 3.2.2. We deduce from Theorem 3.2.1 that D(A) is dense in H (see [93]).
120
3.3 The energy
We now restrict the hypothesis (3.8) to obtain the decay of the energy. Namely, wesuppose that (3.3) holds, namely
∃0 < α < 1, ∀u ∈ V, ‖B∗2u‖2
U2≤ α ‖B∗
1u‖2U1.
Let us choose the following energy (which corresponds to the inner product on H)
E(t) :=1
2
(
∥
∥
∥A
12ω∥
∥
∥
2
H+ ‖ω‖2
H + τξ
∫ 1
0
‖B∗2 ω(t− τρ)‖2
U2dρ
)
, (3.13)
where ξ is a positive constant satisfying
1 < ξ <2
α− 1, (3.14)
that exists because 0 < α < 1.
3.3.1 Decay of the energy
Proposition 3.3.1. If (3.3) holds, then for all (ω0, ω1, f0(−τ.))T ∈ D(A), the energy of
the corresponding regular solution of (3.2) (i.e. (ω, ω, B2ω(t−τρ))T ∈ C([0, +∞), D(A))∩C1([0, +∞),H)) is non-increasing and there exist two positive constants C1 and C2 depen-ding only on α and ξ such that
−C2
(
‖B∗1 ω(t)‖2
U1+ ‖B∗
2 ω(t− τ)‖2U2
)
≤ E ′(t) ≤ −C1
(
‖B∗1ω(t)‖2
U1+ ‖B∗
2ω(t− τ)‖2U2
)
.(3.15)
Proof: Deriving (3.13), we obtain
E ′(t) =(
A12ω, A
12 ω)
H+ (ω, ω)H + τξ
∫ 1
0
(B∗2ω(t− τρ), B∗
2ω(t− τρ))U2dρ
= 〈Aω, ω〉V ′,V − (ω, Aω +B1B∗1ω +B2B
∗2 ω(t− τ))H
+ξτ
∫ 1
0
(B∗2 ω(t− τρ), B∗
2ω(t− τρ))U2dρ
= 〈Aω, ω〉V ′,V − 〈ω, Aω +B1B∗1 ω +B2B
∗2 ω(t− τ)〉V, V ′
+ξτ
∫ 1
0
(B∗2 ω(t− τρ), B∗
2ω(t− τρ))U2dρ,
because Aω +B1B∗1 ω +B2B
∗2ω(t− τ) ∈ H. Then
E ′(t) = 〈Aω, ω〉V ′,V − 〈ω, Aω〉V, V ′ − 〈ω, B1B∗1 ω〉V, V ′ − 〈ω, B2B
∗2 ω(t− τ)〉V, V ′
+ξτ
∫ 1
0
(B∗2 ω(t− τρ), B∗
2 ω(t− τρ))U2dρ
= −‖B∗1 ω‖2
U1− (B∗
2 ω, B∗2 ω(t− τ))U2
+ξτ
∫ 1
0
(B∗2 ω(t− τρ), B∗
2 ω(t− τρ))U2dρ.
121
Moreover, recalling that z(ρ, t) = B∗2 ω(t− τρ), we see that
∫ 1
0
(B∗2ω(t− τρ), B∗
2ω(t− τρ))U2dρ =
∫ 1
0
(
z(ρ, t),∂z
∂t(ρ, t)
)
U2
dρ
= −1
τ
∫ 1
0
(
z(ρ, t),∂z
∂ρ(ρ, t)
)
U2
dρ,
because ∂z∂ρ
(ρ, t) = −τ ∂z∂t
(ρ, t). Then, we have
∫ 1
0
(B∗2ω(t− τρ), B∗
2ω(t− τρ))U2dρ = − 1
2τ
∫ 1
0
∂
∂ρ‖z(ρ, t)‖2
U2dρ
= − 1
2τ(‖z(1, t)‖2
U2− ‖z(0, t)‖2
U2)
= − 1
2τ(‖B∗
2 ω(t− τ)‖2U2
− ‖B∗2ω(t)‖2
U2).
Consequently,
E ′(t) = −‖B∗1ω‖2
U1− (B∗
2 ω, B∗2ω(t− τ))U2
− ξ
2‖B∗
2 ω(t− τ)‖2U2
+ξ
2‖B∗
2 ω(t)‖2U2.
Cauchy-Schwarz’s inequality yields
E ′(t) ≤ −‖B∗1ω‖2
U1+ (
1
2+ξ
2) ‖B∗
2 ω(t)‖2U2
+ (1
2− ξ
2) ‖B∗
2ω(t− τ)‖2U2
and
E ′(t) ≥ −‖B∗1 ω‖2
U1+ (−1
2+ξ
2) ‖B∗
2ω(t)‖2U2
− (1
2+ξ
2) ‖B∗
2ω(t− τ)‖2U2.
Therefore, by (3.3), these estimates leads to
E ′(t) ≤ −C1
(
‖B∗1ω(t)‖2
U1+ ‖B∗
2 ω(t− τ)‖2U2
)
with
C1 = min
(1 − ξα
2− α
2), (
ξ
2− 1
2)
which is positive according to the assumption (3.14), and to
E ′(t) ≥ −C2
(
‖B∗1 ω‖2
U1+ ‖B∗
2ω(t− τ)‖2U2
)
with
C2 = max
1,ξ + 1
2
which is also positive.
122
Remark 3.3.2. Integrating the expression (3.15) between 0 and T, we obtain
∫ T
0
(
‖B∗1ω(t)‖2
U1+ ‖B∗
2ω(t− τ)‖2U2
)
dt ≤ 1
C1
(E(0) − E(T )) ≤ 1
C1
E(0).
This estimate implies that B∗1 ω(.) ∈ L2((0, T ), U1) and B∗
2 ω(.− τ) ∈ L2((0, T ), U2).
Remark 3.3.3. If (3.3) is not satisfied, there exist cases where instabilities may appear,see chapter 1 and [85,110] for the wave equation. Hence this condition appears to be quiterealistic.
3.3.2 Decay of the energy to 0
We give a necessary and sufficient condition that guarantees the decay to 0 of theenergy.
Proposition 3.3.4. Assume that (3.3) holds. Then, for all initial data in H, we have
limt→∞
E(t) = 0 (3.16)
if and only if for any (non zero) eigenvector ϕ ∈ D(A) of A, we have
B∗1ϕ 6= 0. (3.17)
Remark 3.3.5. Notice that this necessary and sufficient condition is the same than in thecase without delay (see [105]) and therefore, the system (3.2) with delay is strongly stable(i.e. the energy tends to zero) if and only if the system without delay (i.e. for B2 = 0) isstrongly stable.
Proof: ⇐ Let us show that (3.17) implies (3.16). For that purpose we closely follow [105].First, we show that A has no eigenvalue on the imaginary axis. If it is not the case, let
iω be an eigenvalue of A where ω ∈ R. Let ϕ be an eigenvector associated with iω. Thenϕ is of the form
ϕ =
ziωz
iωe−iωτρB∗2z
,
with−ω2z + Az + iωB1B
∗1z + iωe−iωτB2B
∗2z = 0. (3.18)
It is an immediate consequence of the identity (iωI −A)ϕ = 0.First we notice that ω 6= 0 since for ω = 0, the above identity reduces to Az = 0 with
z ∈ D(A). Since by hypothesis A is invertible, we get z = 0 and therefore 0 is not aneigenvalue of A.
We now take the duality bracket 〈., .〉V ′, V between (3.18) and z ∈ V :
0 = 〈−ω2z + Az + iωB1B∗1z + iωe−iωτB2B
∗2z, z〉V ′, V
= 〈(−ω2I + A)z, z〉V ′, V + iω ‖B∗1z‖2
U1+ iωe−iωτ ‖B∗
2z‖2U2.
123
We look at the imaginary part of this expression to obtain
ω(
‖B∗1z‖2
U1+ cos(ωτ) ‖B∗
2z‖2U2
)
= 0,
which implies, because ω 6= 0,
‖B∗1z‖2
U1+ cos(ωτ) ‖B∗
2z‖2U2
= 0.
We deduce that
0 = ‖B∗1z‖2
U1+ cos(ωτ) ‖B∗
2z‖2U2
≥ ‖B∗1z‖2
U1− ‖B∗
2z‖2U2
≥ (1 − α) ‖B∗1z‖2
U1≥ 0,
by (3.3) with α < 1. Consequently
‖B∗1z‖U1
= 0
which impliesB∗
1z = 0. (3.19)
Moreover, by (3.18), (3.19) and (3.3), we have
Az = ω2z.
Therefore, z is an eigenvector of A of associated eigenvalue ω2 such that
B∗1z = 0,
which contradicts (3.17). Thus, A has no eigenvalue on the imaginary axis.Now, we can apply the main theorem of Arendt and Batty [12] : As σ(A) ∩ iR is
countable (because σ(A) is countable and what we have done previously) and as A has noeigenvalue on the imaginary axis, we obtain (3.16).
⇒ Let us show that (3.16) implies (3.17). For that purpose we use a contradictionargument. Suppose that there exists an eigenvector ϕ of A of associated eigenvalue λ2 suchthat
B∗1ϕ = 0.
Let us setω(., t) = ϕ cos(λt).
Then ω is solution of (3.2) and satisfies
E(t) = E(0)
because‖B∗
1 ω(t)‖2U1
= λ2 sin2(λt) ‖B∗1ϕ‖2
U1= 0
and‖B∗
2ω(t− τ)‖2U2
= λ2 sin2(λ(t− τ)) ‖B∗2ϕ‖2
U2
≤ αλ2 sin2(λ(t− τ)) ‖B∗1ϕ‖2
U1= 0,
by (3.3). This means that we have obtained a solution of problem (3.2) with a constantenergy, which contradicts (3.16).
124
3.4 The exponential stability
3.4.1 A priori estimate
In order to obtain the characterization of decay properties of the damped problem viaobservability inequalities for the conservative problem we will use the following assumptionfrom [11] :If β > 0 is fixed and Cβ = λ ∈ C |ℜλ = β , the function
λ ∈ Cβ → H(λ) = λB∗(λ2I + A)−1B ∈ L(U) is bounded, (3.20)
where B ∈ L(U, V ′) with U a Hilbert space.
We consider the evolution problem
y(t) + Ay(t) = Bv(t)y(0) = y(0) = 0,
(3.21)
and the following conservative system
φ(t) + Aφ(t) = 0
φ(0) = ω0, φ(0) = ω1.(3.22)
Let us recall the two following results proved in [11] :
Lemma 3.4.1. Suppose that v ∈ L2(0, T ;U) and that the solutions φ of (3.22) are suchthat B∗φ(.) ∈ H1(0, T ;U) and there exists a constant C > 0 such that
‖(B∗φ)′(.)‖L2(0,T ;U) ≤ C ‖(ω0, ω1)‖V×H , ∀(ω0, ω1) ∈ V ×H.
Then the problem (3.21) admits a unique solution having the regularity
y ∈ C(0, T ;V ) ∩ C1(0, T ;H).
Proposition 3.4.2. Suppose that v ∈ L2(0, T ;U) and that the problem (3.21) admits aunique solution having the regularity
y ∈ C(0, T ;V ) ∩ C1(0, T ;H).
Then hypothesis (3.20) holds if and only if B∗y(.) ∈ H1(0, T ;U) and there exists a constantC > 0 independent of T such that
‖(B∗y)′(.)‖L2(0,T ;U) ≤ CeβT ‖v‖L2(0,T ;U) .
125
Let ω ∈ C(0, T ;V ) ∩ C1(0, T ;H) be the solution of (3.2) with (ω0, ω1, f0(−τ.))T ∈
D(A). Then it can be split up in the form
ω = φ+ ψ,
where φ is solution of the problem without damping (3.22), and ψ satisfies
ψ(t) + Aψ(t) = −B1B∗1 ω(t) −B2B
∗2ω(t− τ)
ψ(0) = 0, ψ(0) = 0.(3.23)
We now set B = (B1B2) ∈ L(U, V ′) where U = U1 × U2. It is easy to verify that B∗ =(
B∗1
B∗2
)
∈ L(V, U). Therefore ψ is solution of
ψ(t) + Aψ(t) = Bv(t)
ψ(0) = 0, ψ(0) = 0,(3.24)
where v(t) =
(
−B∗1 ω(t)
−B∗2 ω(t− τ)
)
. In other words, ψ is solution of problem (3.21) with
B = (B1B2)
and by Remark 3.3.2
v =
(
−B∗1 ω(·)
−B∗2 ω(· − τ)
)
∈ L2((0, T ), U).
Then ψ = ω−φ ∈ C(0, T ;V )∩C1(0, T ;H). Suppose that the hypothesis (3.20) is satisfiedfor B = (B1B2) and U = U1 × U2. By applying Proposition 3.4.2, we obtain
∫ T
0
‖(B∗ψ)′‖2U dt ≤ Ce2βT
∫ T
0
‖v(t)‖2U dt,
which is equivalent to∫ T
0
(‖(B∗1ψ)′‖2
U1+ ‖(B∗
2ψ)′‖2U2
)dt ≤ Ce2βT∫ T
0
(‖B∗1 ω(t)‖2
U1+ ‖B∗
2 ω(t− τ)‖2U2
)dt.
In particular, we have∫ T
0
‖(B∗1ψ)′‖2
U1dt ≤ Ce2βT
∫ T
0
(‖B∗1 ω(t)‖2
U1+ ‖B∗
2 ω(t− τ)‖2U2
)dt.
Therefore, since ω = φ+ ψ, we have∫ T
0
‖(B∗1φ)′(t)‖2
U1dt ≤ 2
(∫ T
0
‖(B∗1ω)′(t)‖2
U1dt+
∫ T
0
‖(B∗1ψ)′(t)‖2
U1dt
)
≤ Ce2βT∫ T
0
(
‖(B∗1ω)(t)‖2
U1+ ‖(B∗
2ω)(t− τ)‖2U2
)
dt.
Thus, we have proved the following result :
126
Lemma 3.4.3. Suppose that the assumption (3.20) is satisfied for B = (B1B2), U =U1 × U2. Then the solutions ω of (3.2) and φ of (3.22) satisfy
∫ T
0
‖(B∗1φ)′(t)‖2
U1dt ≤ Ce2βT
∫ T
0
(‖B∗1 ω(t)‖2
U1+ ‖B∗
2 ω(t− τ)‖2U2
)dt,
with C > 0 independent of T .
3.4.2 The stability result
Theorem 3.4.4. Assume that the hypotheses (3.3) and (3.20) are verified for B = (B1B2),U = U1 × U2. If there exist a time T > 0 and a constant C > 0 such that the observabilityestimate
∥
∥
∥A
12ω0
∥
∥
∥
2
H+ ‖ω1‖2
H ≤ C
∫ T
0
‖(B∗1φ)′(t)‖2
U1dt (3.25)
holds, where φ is solution of (3.22), then the system (3.2) is exponentially stable in theenergy space : there exist C > 0 independent of τ and ν > 0 such that, for all initial datain H,
E(t) ≤ CE(0)e−νt, ∀t > 0. (3.26)
Proof: Let ω be a solution of (3.2) with initial data (ω0, ω1, f0(−τ ·)) ∈ D(A).
Without loss of generality, we can always assume that (3.25) holds with T > τ and Cindependent of τ .
Integrating the inequality (3.15) of Proposition 3.3.1 between 0 and T , we obtain
E(0) −E(T ) ≥ C
∫ T
0
(
‖B∗1 ω(t)‖2
U1+ ‖B∗
2 ω(t− τ)‖2U2
)
dt
≥ C
2
∫ T
0
(
‖B∗1 ω(t)‖2
U1+ ‖B∗
2 ω(t− τ)‖2U2
)
dt+C
2
∫ T
0
‖B∗2ω(t− τ)‖2
U2dt
≥ Ce−2βT
(∫ T
0
‖(B∗1φ)′(t)‖2
U1dt+
∫ T
0
‖B∗2 ω(t− τ)‖2
U2dt
)
by Lemma 3.4.3.
By assumption (3.25), we obtain
E(0) −E(T ) ≥ Ce−2βT
(
∥
∥
∥A
12ω0
∥
∥
∥
2
H+ ‖ω1‖2
H +
∫ T
0
‖B∗2 ω(t− τ)‖2
U2dt
)
,
with C independent of τ . Since T > τ , by change of variables, we have
∫ T
0
‖B∗2 ω(t− τ)‖2
U2dt =
∫ T−τ
−τ‖B∗
2ω(t)‖2U2dt
≥∫ 0
−τ‖B∗
2 ω(t)‖2U2dt = τ
∫ 1
0
‖B∗2 ω(−τρ)‖2
U2dρ.
127
The two previous inequalities and (3.14) directly imply that
E(0) − E(T ) ≥ C ′e−2βT
(
∥
∥
∥A
12ω0
∥
∥
∥
2
H+ ‖ω1‖2
H + ξτ
∫ 1
0
‖B∗2ω(−τρ)‖2
U2dρ
)
,
with C ′ = C/(2/α− 1). This means that for T > τ , we have
E(0) −E(T ) ≥ C ′e−2βTE(0).
This estimate is equivalent to
E(0) −E(T ) ≥ C ′e−2βTE(T ),
because the energy is decreasing, which leads to
E(T ) ≤ γE(0),
where γ = 11+C′e−2βT < 1. Applying this argument on [(m − 1)T, mT ], for m = 1, 2, ...
(which is valid because the system is invariant by translation in time), we will get
E(mT ) ≤ γE((m− 1)T ) ≤ ... ≤ γmE(0).
Therefore, we haveE(mT ) ≤ e−νmTE(0), m = 1, 2, ...
with ν = 1T
ln 1γ
= 1T
ln(1 + C ′e−2βT ) > 0 which depends on T and thus on τ (because
T > τ). For an arbitrary positive t, there exists m ∈ N∗ such that (m−1)T < t ≤ mT andby the non-increasing property of the energy, we conclude
E(t) ≤ E((m− 1)T ) ≤ e−ν(m−1)TE(0) ≤ 1
γe−νtE(0).
Hence the energy decays exponentially with the decay rate ν = 1T
ln 1γ
= 1T
ln(1+C ′e−2βT ) <1τ
ln(1 +C ′e−2βτ ), because T > τ . Notice that the constant C of (3.26) can be chosen suchthat C ≥ 1 + C ′ (which is independent of τ) because 1
γ= 1 + C ′e−2βT ≤ 1 + C ′.
By density of D(A) into H, we deduce that (3.26) holds for any initial data in H.
Remark 3.4.5. In the previous theorem, we have seen that the decay rate is ν = 1T
ln 1γ
=1T
ln(1 + C ′e−2βT ) < 1τ
ln(1 + C ′e−2βτ ) because T > τ and where C ′ is independent of τ .Therefore, when τ becomes larger, the decay is slower.
Remark 3.4.6. Notice that the sufficient condition (3.25) for the exponential decay ofthe energy is the same than the case without delay (see [11]). Therefore, if the hypothesis(3.20) holds and if the dissipative system without delay (i.e. with B2 = 0) is exponentiallystable, then the system (3.2) is exponentially stable.
128
3.5 The polynomial stability
In some cases, the decay of the energy is not exponential, but can be polynomial. Ouraim here is to give a sufficient condition that yields the explicit decay rate.
The proof of this stability result requires the next technical Lemma proved in [10,Lemma 5.2].
Lemma 3.5.1. Let (εk)k be a sequence of positive real numbers satisfying
εk+1 ≤ εk − Cε2+µk+1, ∀k ≥ 0, (3.27)
where C > 0 and µ > −1 are constants. Then there exists a positive constant M (dependingon µ and C) such that
εk ≤M
(1 + k)1
1+µ
, ∀k ≥ 0,
with M >(
4(1+µ)C
)1
1+µ
.
Moreover we recall the following interpolation result.
Lemma 3.5.2. For (ω0, ω1) ∈ D(A) × V , we have
‖ω0‖m+1
D(A12 )
≤ C ‖ω0‖mD(A) ‖ω0‖D(A
1−m2 )
,
‖ω1‖m+1H ≤ C ‖ω1‖m
D(A12 )‖ω1‖D(A−m
2 ),
where C > 0.
Proof: If we denote by λknn the eigenvalues of A12 counted without multiplicity, ln
the multiplicity of the eigenvalue λkn and ϕkn+j0≤j≤ln−1 the orthonormal eigenvectorsassociated with the eigenvalue λkn, this lemma is a direct consequence of the equivalence‖u‖2
D(As) ∼∑
n≥1
∑ln−1j=0 |ukn+j |2 λ4s
knfor all s ∈ R, when u =
∑
n≥1
∑ln−1j=0 ukn+jϕkn+j and
of Holder’s inequality with p = 1 + 1m
and q = m+ 1.
Since for (ω0, ω1, f0(−τ.)) ∈ D(A), ω0 is not necessarily in D(A), we can not apply
Lemma 3.5.2 to ω0, and therefore we need to make the following hypothesis : there existsC > 0 such that for all (ω0, ω1, z) ∈ D(A), we have
‖ω0‖m+1V ≤ C ‖(ω0, ω1, z)‖mD(A) ‖ω0‖
D(A1−m
2 ). (3.28)
Theorem 3.5.3. Let ω be a solution of (3.2) with (ω0, ω1, f0(−τ ·)) ∈ D(A). Assume that
the hypotheses (3.3), (3.20) and (3.28) are verified for B = (B1B2), U = U1 ×U2. If thereexist a positive real number m, a time T > 0 and a constant C > 0 such that
∫ T
0
‖(B∗1φ)′(t)‖2
U1dt ≥ C
(
‖ω0‖2
D(A1−m
2 )+ ‖ω1‖2
D(A−m2 )
)
(3.29)
129
holds where φ is solution of (3.22), then the energy decays polynomially, i.e. there existsC > 0 depending on m and τ such that, for all initial data in D(A),
E(t) ≤ C
(1 + t)1m
∥
∥(ω0, ω1, f0(−τ ·))
∥
∥
2
D(A), ∀t > 0. (3.30)
Proof: Since the hypothesis (3.20) is satisfied for B = (B1B2), U = U1 × U2, by usingLemma 3.4.3, we obtain
∫ T
0
(
‖B∗1 ω(t)‖2
U1+ ‖B∗
2 ω(t− τ)‖2U2
)
dt ≥ Ce−2βT(
‖ω0‖2
D(A1−m
2 )+ ‖ω1‖2
D(A−m2 )
)
.
On the other hand, integrating the inequality (3.15) of Proposition 3.3.1 between 0 and T ∗
for T ∗ large enough : T ∗ ≥ max(T, τ), we have
E(0) − E(T ∗) ≥ C
∫ T ∗
0
(‖B∗1ω(t)‖2
U1+ ‖B∗
2ω(t− τ)‖2U2
)dt
≥ C
2
∫ T ∗
0
(‖B∗1 ω(t)‖2
U1+ ‖B∗
2ω(t− τ)‖2U2
)dt+C
2
∫ T ∗
0
‖B∗2ω(t− τ)‖2
U2dt
≥ Ce−2βT ∗
(
‖ω0‖2
D(A1−m
2 )+ ‖ω1‖2
D(A−m2 )
+ τ
∫ 1
0
‖B∗2 ω(−τρ)‖2
U2dρ
)
,
by change of variable (because T ∗ > τ). Therefore
E(T ∗) ≤ E(0) −K1e−2βT ∗
(
‖ω0‖2
D(A1−m
2 )+ ‖ω1‖2
D(A−m2 )
+ τξ
∫ 1
0
‖B∗2ω(−τρ)‖2
U2dρ
)
,
(3.31)for some K1 > 0 independent of T ∗ and τ .
Therefore, by (3.28), the previous interpolation result of Lemma 3.5.2 and a convexityinequality, we have :
‖(ω0, ω1)‖m+1V×H ≤ C
(
‖ω0‖m+1V + ‖ω1‖m+1
H
)
≤ C(
∥
∥(ω0, ω1, f0(−τ.))
∥
∥
m
D(A)‖ω0‖
D(A1−m
2 )+ ‖ω1‖m
D(A12 )‖ω1‖D(A−m
2 )
)
≤ C∥
∥(ω0, ω1, f0(−τ.))
∥
∥
m
D(A)
(
‖ω0‖D(A
1−m2 )
+ ‖ω1‖D(A−m2 )
)
.
Denoting by X−m = D(A1−m
2 ) ×D(A−m2 ), we have shown that
‖(ω0, ω1)‖2X−m
≥ C‖(ω0, ω1)‖2m+2
V×H‖(ω0, ω1, f 0(−τ.))‖2m
D(A)
. (3.32)
Now introduce the modified energy
E(t) =1
2‖U(t)‖2
D(A) =1
2(‖U(t)‖2
H + ‖AU(t)‖2H).
130
As in Proposition 3.3.1, this energy E is decaying.Combining the estimates (3.31) and (3.32), we obtain
E(T ∗) ≤ E(0) −K2e−2βT ∗
(
‖(ω0, ω1)‖2m+2V×H
E(0)m+ ξτ
∫ 1
0
‖B∗2 ω(−τρ)‖2
U2dρ
)
,
for some K2 > 0 independent of T ∗ and τ , or equivalently
E(T ∗) ≤ E(0) −K2e−2βT ∗
(
‖(ω0, ω1)‖2m+2V×H
E(0)m+ ξτ
∥
∥f 0(−τ.)∥
∥
2
L2((0, 1), U2)
)
. (3.33)
Using the trivial estimate
(ξτ)m+1 ‖f 0(−τ.)‖2m+2L2((0, 1), U2)
= ξτ ‖f 0(−τ.)‖2L2((0, 1), U2)
(ξτ)m ‖f 0(−τ.)‖2mL2((0, 1), U2)
≤ τξ ‖f 0(−τ.)‖2L2((0, 1), U2)
E(0)m
the above inequality (3.33) becomes
E(T ∗) ≤ E(0) −K2e−2βT ∗
(
‖(ω0, ω1)‖2m+2V×H + (ξτ)m+1 ‖f 0(−τ.)‖2m+2
L2((0, 1), U2)
E(0)m
)
≤ E(0) −K ′e−2βT ∗E(0)m+1
E(0)m,
with K ′ > 0 independent of T ∗ and τ . Since the energy of our system is decaying, weobtain
E(T ∗) ≤ E(0) −K ′e−2βT ∗E(T ∗)m+1
E(0)m. (3.34)
We now follow the method used in [10]. The estimate (3.34) being valid on the intervals[kT ∗, (k + 1)T ∗], for any k ≥ 0, we have
E((k + 1)T ∗) ≤ E(kT ∗) −K ′e−2βT ∗E((k + 1)T ∗)m+1
E(kT ∗)m. (3.35)
Setting
εk =E(kT ∗)
E(0),
and dividing (3.35) by E(0), we obtain
εk+1 ≤ εk −K ′e−2βT ∗
εm+1k+1 , (3.36)
because E(kT ∗) ≤ E(0). By Lemma 3.5.1 with µ = m−1 > −1 (since m > 0), there exists
a constant M ′ > 0 (depending on m and K ′e−2βT ∗
, and verifying M ′ >(
4e2βT∗
mK ′
)1m
) such
that
εk ≤M ′
(1 + k)1m
, ∀k ≥ 0,
131
or equivalently
E(kT ∗) ≤ M ′
(1 + k)1m
E(0).
This estimate and again the decay of the energy lead to the estimate (3.30), where C =
M ′(1 + T ∗)1m .
Remark 3.5.4. Since the proof of the above theorem reveals that C = M ′(1 +T ∗)1m with
M ′ >(
4e2βT∗
mK ′
) 1m
and T ∗ > τ , the constant C depends on τ and when τ becomes larger,
the decay becomes slower.
3.6 Checking the observability inequalities
In this section, we show how to obtain the observability inequalities used in Theorems3.4.4 and 3.5.3. Our method is based on the generalized gap condition. Before givingspectral conditions to obtain exponential or polynomial decay, we recall some results aboutIngham’s inequality.
3.6.1 Preliminaries about Ingham’s inequality
Let λkk≥1 be the set of eigenvalues of A12 counted with their multiplicities (i.e. we
repeat the eigenvalues according to their multiplicities). We further rewrite the sequenceof eigenvalues λkk≥1 as follows :
λk1 < λk2 < ... < λki< ...
where k1 = 1, k2 is the lowest index of the second distinct eigenvalue, k3 is the lowestindex of the third distinct eigenvalue, etc. For all i ∈ N∗, let li be the multiplicity of theeigenvalue λki
, i.e.
λki−1 < λki= λki+1 = ... = λki+li−1 < λki+li = λki+1
.
We have k1 = 1, k2 = l1, k3 = l1 + l2, etc. Let ϕki+j0≤j≤li−1 be the orthonormal eigen-vectors associated with the eigenvalue λki
. We assume that the following generalized gapcondition holds :
∃M ∈ N∗, ∃γ0 > 0, ∀k ≥ 1, λk+M − λk ≥ Mγ0. (3.37)
Fix a positive real number γ′0 ≤ γ0 and denote by Ak, k = 1, ..., M the set of naturalnumbers km satisfying (see for instance [13])
λkm − λkm−1 ≥ γ′0λkn − λkn−1 < γ′0 for m+ 1 ≤ n ≤ m+ k − 1,λkm+k
− λkm+k−1≥ γ′0.
132
Then one easily checks that the sets Ak + j, j = 0, ..., k − 1, k = 1, ...,M form a partitionof N∗. Notice that some sets Ak may be empty because, for the generalized gap condition,the choice of M takes into account multiple eigenvalues.
Now for km ∈ Ak, we recall that the finite differences em+j(t), j = 0, · · · , k − 1, corres-
ponding to the exponential functions eiλkm+jt, j = 0, · · · , k − 1 are given by
em+j(t) =
m+j∑
p=m
m+j∏
q=m
q 6=p
(λkp − λkq)−1eiλkp t.
Write for shortness, e−n(t) the same finite differences functions corresponding to −λkn .Now we are ready to recall the next inequality of Ingham’s type, see for instance Theo-
rem 1.5 of [13] :
Theorem 3.6.1. If the sequence (λn)n≥1 satisfies (3.37), then for all sequence (an)n∈Z∗
(where Z∗ = Z \ 0), the function
f(t) =∑
n∈Z∗
anen(t),
satisfies the estimates∫ T
0
|f(t)|2dt ∼∑
n∈Z∗
|an|2, (3.38)
for T > 2πγ0
.
Going back to the original functions eiλkn t, the above equivalence (3.38) means that,for T > 2π
γ0, the function (from now on λ−kn = −λkn)
f(t) =∑
n∈Z∗
αneiλkn t,
satisfies the estimates∫ T
0
|f(t)|2dt ∼M∑
k=1
∑
|kn|∈Ak
‖B−1knCkn‖2
2, (3.39)
where ‖ · ‖2 means the Euclidean norm of the vector, for kn ∈ Ak the vector Ckn is givenby
Ckn = (αn, · · · , αn+k−1)T ,
and the k × k matrix Bkn allows to pass from the coefficients akn to αkn, namely
Ckn = Bkn · (an, · · · , an+k−1)T ,
133
and is given by Bkn = (Bkn, ij)1≤i, j≤k the matrix of size k × k such that
Bkn, ij =
n+j−1∏
q = nq 6= n+ i− 1
(λkn+i−1− λkq)
−1 if i ≤ j, (i, j) 6= (1, 1),
1 if (i, j) = (1, 1),0 if i > j.
More explicitly, we have
Bkn =
1 1λkn−λkn+1
1(λkn−λkn+1
)(λkn−λkn+2)
· · · 1(λkn−λkn+1
)···(λkn−λkn+k−1)
0 1λkn+1
−λkn
1(λkn+1
−λkn )(λkn+1−λkn+2
)· · · 1
(λkn+1−λkn )···(λkn+1
−λkn+k−1)
0 0 1(λkn+2
−λkn )(λkn+2−λkn+1
)· · · 1
(λkn+2−λkn )···(λkn+2
−λkn+k−1)
......
. . ....
0 0 0 · · · 1(λkn+k−1
−λkn )···(λkn+k−1−λkn+k−2
)
.
We proceed similarly for n ≤ −1, the indices being decreasing from n to n− k + 1.
Remark 3.6.2. If the standard gap condition
∃γ0 > 0, ∀n ≥ 1, λkn+1 − λkn ≥ γ0 (3.40)
holds, then A1 = Z∗ and B1 = 1 and in that case f(t) =∑
n∈Z∗ αneiλkn t satisfy Ingham’s
inequality (see [57]) :∫ T
0
|f(t)|2dt ∼∑
n∈Z∗
|αn|2, (3.41)
for T > 2πγ0
.
Now, let U be a separable Hilbert space (in the sequel, U will be U1). For a vector
c =
c1...cm
in Um, we set ‖.‖U, 2 the norm in Um defined by
‖c‖2U, 2 =
m∑
l=1
‖cl‖2U .
Then we obtain the inequality of Ingham’s type in U :
Proposition 3.6.3. If we have the standard gap condition (3.40), then for all sequence(an)n in U , the function
u(t) =∑
n∈Z∗
aneiλkn t
134
satisfies the estimates∫ T
0
‖u(t)‖2U dt ∼
∑
n∈Z∗
‖an‖2U ,
for T > 2πγ0
.
Proof: Since U is a separable Hilbert space, there exists a Hilbert basis (ψk)k≥1 of U .Therefore, an ∈ U can be written as
an =+∞∑
k=1
aknψk.
We truncate an as follows : forK ∈ N∗, let a(K)n =
K∑
k=1
aknψk and set uK(t) =
K∑
k=1
(∑
n∈Z∗
akneiλkn t)ψk.
Since (ψk)k≥1 is a Hilbert basis, we have by Fubini’s theorem
‖uK(t)‖2U =
K∑
k=1
∣
∣
∣
∣
∣
∑
n∈Z∗
akneiλkn t
∣
∣
∣
∣
∣
2
.
Thus, by applying Ingham’s inequality (3.41), we have
∫ T
0
‖uK(t)‖2U dt =
K∑
k=1
∫ T
0
∣
∣
∣
∣
∣
∑
n∈Z∗
akneiλknt
∣
∣
∣
∣
∣
2
dt
∼K∑
k=1
∑
n∈Z∗
(akn)2
∼∑
n∈Z∗
K∑
k=1
(akn)2.
Therefore∫ T
0
‖uK(t)‖2U dt ∼
∑
n∈Z∗
∥
∥a(K)n
∥
∥
2
U.
Since uK → u and a(K)n → an when K → +∞, we obtain the result.
In the same way, we obtain an Ingham’s type inequality in a Hilbert space U in thecase of the generalized gap condition (3.37).
Corollary 3.6.4. If the sequence (λn)n≥1 satisfies (3.37), then for all sequence (αn)n∈Z∗
in U , the function
f(t) =∑
n∈Z∗
αneiλkn t,
135
satisfies the estimates
∫ T
0
|f(t)|2dt ∼M∑
k=1
∑
|kn|∈Ak
‖B−1knCkn‖2
U,2, (3.42)
for T > 2πγ0
, where
Ckn = (αn, · · · , αn+k−1)T ∈ Uk.
3.6.2 A first observability inequality
Proposition 3.6.5. Assume that the generalized gap condition (3.37) holds and that U1 isseparable. Let φ be the solution of (3.22) with (ω0, ω1) ∈ V ×H. Then there exists a timeT > 0 and a constant C > 0 (depending on T ) such that (3.25) holds if and only if
∃γ > 0, ∀k = 1, · · · ,M, ∀kn ∈ Ak, ∀ξ ∈ RLn,∥
∥B−1kn
Φknξ∥
∥
U1, 2≥ γ ‖ξ‖2 , (3.43)
where the matrix Φkn with coefficients in U1 and size k × Ln, where Ln =∑k
i=1 ln+i−1 − 1,is given as follow : for all i = 1, · · · , k, we set
(Φkn)ij =
B∗1ϕkn+i−1+j−Ln,i−1
if Ln,i−1 < j ≤ Ln,i,0 else,
whereLn,0 = 0,
Ln,i =i∑
i′=1
ln+i′−1 − 1, ∀i ≥ 1.(3.44)
Proof: We first show that (3.43)⇒(3.25). Writting
ω0 =∑
i≥1
li−1∑
j=0
aki+jϕki+j
and
ω1 =∑
i≥1
li−1∑
j=0
bki+jϕki+j
where (λkiaki+j)i, j, (bki+j)i, j ∈ l2(N∗), then the solution φ of problem without damping
(3.22) is given by
φ(·, t) =∑
i≥1
li−1∑
j=0
(
aki+j cos(λkit) +
bki+j
λki
sin(λkit)
)
ϕki+j.
136
Consequently
(B∗1φ)′(t) =
∑
i≥1
li−1∑
j=0
(−aki+jλkisin(λki
t) + bki+j cos(λkit))B∗
1ϕki+j .
By grouping the terms corresponding to the same eigenvalue, we get
(B∗1φ)′(t) =
∑
i≥1
(
li−1∑
j=0
− aki+jB∗1ϕki+j
)
λkisin(λki
t) +∑
i≥1
lj−1∑
j=0
bki+jB∗1ϕki+j
cos(λkit)
=∑
n∈Z∗
αneiλkn t,
where
αn =1
2
((
ln−1∑
j=0
bkn+jB∗1ϕkn+j
)
+ i
(
ln−1∑
j=0
akn+jB∗1ϕkn+j
)
λkn
)
, ∀n ≥ 1,
α−n =1
2
((
ln−1∑
j=0
bkn+jB∗1ϕkn+j
)
− i
(
ln−1∑
j=0
akn+jB∗1ϕkn+j
)
λkn
)
, ∀n ≥ 1.
Integrating the square of the norm of this identity between 0 and T > 0 and usingIngham’s inequality (3.42) in U1, for T large enough, we get
∫ T
0
‖(B∗1φ)′(t)‖2
U1dt ≥ C
M∑
k=1
∑
|kn|∈Ak
∥
∥B−1knCkn
∥
∥
2
U1, 2,
where Ckn = (αn, ..., αn+k−1)T is a vector of Uk
1 .But for all kn ∈ Ak, setting
Akn =(
λknakn , · · · , λknakn+ln−1, λkn+1akn+1 , · · · , λkn+1akn+1+ln+1−1, · · · ,λkn+k−1
akn+k−1, · · · , λkn+k−1
akn+k−1+ln+k−1−1
)T,
Bkn =(
bkn , · · · , bkn+ln−1, bkn+1, · · · , bkn+1+ln+1−1, · · · , bkn+k−1, · · · ,
bkn+k−1+ln+k−1−1
)T,
we readily check that
∫ T
0
‖(B∗1φ)′(t)‖2
U1dt ≥ C
M∑
k=1
∑
|kn|∈Ak
(
∥
∥
∥B−1kn
ΦknAkn
∥
∥
∥
2
U1,2+∥
∥
∥B−1kn
ΦknBkn
∥
∥
∥
2
U1,2
)
.
137
Hence the assumption (3.43) yields
∫ T
0
‖(B∗1φ)′(t)‖2
U1dt ≥ C
M∑
k=1
∑
|kn|∈Ak
(
∥
∥
∥Akn
∥
∥
∥
2
2+∥
∥
∥Bkn
∥
∥
∥
2
2
)
= C
(
∥
∥
∥A
12ω0
∥
∥
∥
2
H+ ‖ω1‖2
H
)
because (ϕkn+i)n,i is an orthonormal basis associated with the operator A12 .
It remains to show that (3.25)⇒(3.43).Let k = 1, · · · ,M and kn ∈ Ak be fixed. Take ω0 =
∑n+k−1i=n
∑li−1j=0 aki+jϕki+j and
ω1 =∑n+k−1
i=n
∑li−1j=0 bki+jϕki+j . Then the solution φ of problem (3.22) is given by
φ(·, t) =n+k−1∑
i=n
li−1∑
j=0
(
aki+j cos(λkit) +
bki+j
λki
sin(λkit)
)
ϕki+j,
and then
(B∗1φ)′(t) =
n+k−1∑
i=n
li−1∑
j=0
(−aki+jλkisin(λki
t) + bki+j cos(λkit))B∗
1ϕki+j.
Applying again Ingham’s inequality, we get for T large enough and Akn , Bkn define above∫ T
0
‖(B∗1φ)′(t)‖2
U1dt ∼
∥
∥
∥B−1kn
ΦknAkn
∥
∥
∥
2
U1,2+∥
∥
∥B−1kn
ΦknBkn
∥
∥
∥
2
U1,2.
By (3.25), we obtain
∥
∥
∥B−1kn
ΦknAkn
∥
∥
∥
2
U1,2+∥
∥
∥B−1kn
ΦknBkn
∥
∥
∥
2
U1,2≥ C
∥
∥
∥A
12ω0
∥
∥
∥
2
H+ ‖ω1‖2
H
= Cn+k−1∑
i=n
li−1∑
j=0
(a2ki+j
λ2ki
+ b2ki+j), (3.45)
for some C > 0. Hence we conclude that∥
∥B−1kn
Φknξ∥
∥
U1, 2≥ γ ‖ξ‖2 .
This ends the proof.
Remark 3.6.6. If the standard gap condition (3.40) holds, then A1 = N∗ and B1 = 1. Inthis case, the assumption (3.43) becomes
∃γ > 0, ∀kn ≥ 1, ∀ξ ∈ Rln, ‖Φknξ‖U1≥ γ ‖ξ‖2 .
Moreover, if the standard gap condition (3.40) holds and if the eigenvalues are simple, theassumption (3.43) becomes
∃γ > 0, ∀k ≥ 1, ‖B∗1ϕk‖U1
≥ γ.
138
Remark 3.6.7. The above Proposition 3.6.5 yields a time T > 0 and a constant C > 0depending on T such that (3.25) holds but the time T and the constant C do not dependon the delay τ . Hence if the minimal time T is not strictly greater than τ , by choosingT ′ > maxT, τ, we still have
∥
∥
∥A
12ω0
∥
∥
∥
2
H+ ‖ω1‖2
H ≤ C
∫ T ′
0
‖(B∗1φ)′(t)‖2
U1dt,
with the same constant C as before and that does not depend on τ . This means that underthe generalized gap condition (3.37), the condition (3.43) is equivalent to the sufficientcondition of Theorem 3.4.4.
3.6.3 A second observability inequality
Proposition 3.6.8. Assume that the generalized gap condition (3.37) holds and that U1
is separable. Let φ be the solution of (3.22) with (ω0, ω1) ∈ V × H. Then for a fixed realnumber m > 0, there exist a time T > 0 and a constant C > 0 such that (3.29) holds ifand only if
∃γ > 0, ∀k = 1, · · · ,M, ∀kn ∈ Ak, ∀ξ ∈ RLn,∥
∥B−1kn
Φknξ∥
∥
U1, 2≥ γ
λmkn
‖ξ‖2 . (3.46)
Proof: The proof is similar to the one of Proposition 3.6.5 because
∑
n≥1
1
λ2mkn
ln−1∑
j=0
(a2kn+jλ
2kn
+ b2kn+j) =∑
n≥1
ln−1∑
j=0
(a2kn+jλ
2(1−m)kn
+ b2kn+jλ−2mkn
)
∼ ‖ω0‖2
D(A1−m
2 )+ ‖ω1‖2
D(A−m2 ).
The details are therefore omitted.
Remark 3.6.9. If the standard gap condition (3.40) holds, the assumption (3.46) becomes
∃γ > 0, ∀kn ≥ 1, ∀ξ ∈ Rln , ‖Φknξ‖U1≥ γ
λmkn
‖ξ‖2 .
Moreover, if the standard gap condition (3.40) holds and if the eigenvalues are simple, theassumption (3.46) becomes
∃γ > 0, ∀k ≥ 1, ‖B∗1ϕk‖U1
≥ γ
λmkn
.
A remark similar to Remark 3.6.7 can be made for polynomial stability.
139
3.7 Examples
We end up this chapter by considering different examples for which our abstract fra-mework can be applied. To our knowledge, all the examples, with the exception of the firstone, are new.
3.7.1 A wave equation on 1-d networks with nodal feedbacks
In this section we show that the result obtained in chapter 1 (see also [89]) enter inthe framework of this chapter. Obviously we will use the same notation for a network thatin chapter 1 that we briefly recall. We denote by E = ej ; 1 ≤ j ≤ N the set of edgesej of length lj > 0 of a given network R and V the set of vertices of R. For a functionu : R → R, we set uj = u|ej
the restriction of u to ej . For a fixed vertex v, we set
Ev = j ∈ 1, ..., N ; v ∈ ej.
If card(Ev) ≥ 2, v is an interior node. Let Vint be the set of interior nodes. If card(Ev) = 1,v is an exterior node. Let Vext be the set of exterior nodes. For v ∈ Vext, we set Ev = jv.
We now fix a partition of Vext :
Vext = D ∪N ∪ Vcext, where D 6= ∅.We actually will impose Dirichlet boundary condition at the nodes of D, Neumann boun-dary condition at the nodes of N and finally a feedback boundary condition at the nodesof Vcext. We further fix a subset Vcint of Vint where a feedback transmission condition will beimposed. By shortness, we denote by Vc the set of controlled nodes, namely
Vc = Vcint ∪ Vcext.
We here consider the following initial and boundary problem :
∂2uj
∂t2(x, t) − ∂2uj
∂x2 (x, t) = 0 0 < x < lj , t > 0, ∀j ∈ 1, ..., Nuj(v, t) = ul(v, t) = u(v, t) t > 0, ∀j, l ∈ Ev, v ∈ Vint∑
j∈Ev
∂uj
∂nj(v, t) = −α(v)
1∂u∂t
(v, t) − α(v)2
∂u∂t
(v, t− τ) t > 0, ∀v ∈ Vc∑
j∈Ev
∂uj
∂nj(v, t) = 0 t > 0, ∀v ∈ Vint\Vcint
ujv(v, t) = 0 t > 0, ∀v ∈ D∂ujv
∂njv(v, t) = 0 t > 0, ∀v ∈ N
u(t = 0) = u(0), ∂u∂t
(t = 0) = u(1)
∂u∂t
(v, t− τ) = f 0v (t− τ) ∀v ∈ Vc, 0 < t < τ,
(3.47)
where α(v)i ≥ 0 are fixed non-negative real numbers and the delay τ is positive.
140
To rewrite this problem in the form (3.2), we introduce
H = L2(R) = u : R → R; uj ∈ L2(0, lj), ∀j = 1, ..., N
and the operator
A :
D(A) → H
(ϕj)j 7→ (− d2
dx2ϕj)j(3.48)
where
D(A) = ϕ ∈ V ∩N∏
j=1
H2(0, lj) ;∑
j∈Ev
∂ϕj∂nj
(v) = 0, ∀v ∈ Vint;
∂ϕjv∂njv
(v) = 0, ∀v ∈ N ∪ Vcext, (3.49)
and
V := ϕ ∈N∏
j=1
H1(0, lj) : ϕj(v) = ϕk(v) ∀j, k ∈ Ev, ∀v ∈ Vint ; ϕjv(v) = 0, ∀v ∈ D.
The operator A is self-adjoint and positive with a compact inverse in H. Moreover
D(A12 ) = V.
We now define U = U1 = U2 = RVc , where Vc is the cardinal of Vc, with norm ‖.‖U = ‖.‖2
and the operators Bi for i = 1, 2 as
Bi :
U → D(A12 )′
(kv)v∈Vc 7→∑
v∈Vc
√
α(v)i kvδv
. (3.50)
It is easy to verify that B∗i (ϕ) = (
√
α(v)i ϕ(v))Tv∈Vc
for ϕ ∈ D(A12 ) and thus BiB
∗i (ϕ) =
∑
v∈Vcα
(v)i ϕ(v)δv for ϕ ∈ D(A
12 ). Hence the system (3.47) can be rewritten in the form
(3.2).We notice that (3.8) is here reduced to
∃0 < α ≤ 1, ∀ϕ ∈ V,∑
v∈Vc
(α(v)2 ϕ(v))2 ≤ α
∑
v∈Vc
(α(v)1 ϕ(v))2,
and therefore, the system (3.47) is well posed for α(v)2 ≤ α
(v)1 for all v ∈ Vc by Theorem
3.2.1, and the energy is decreasing for α(v)2 < α
(v)1 for all v ∈ Vc by Proposition 3.3.1.
By Proposition 6.2 of [38], the generalized gap condition (3.37) holds with M = N +1.
141
We know, by chapter 1, that the hypothesis (3.20) is satisfied. Moreover, the hypothesis(3.28) is verified because
‖ω0‖m+1V ≤ C ‖ω0‖mX ‖ω0‖
D(A1−m
2 )
≤ C ‖(ω0, ω1, z)‖mD(A) ‖ω0‖D(A
1−m2 )
,
where X = V ∩ (
N∏
j=1
H2(0, lj)), by using Corollary 1.6.4.
Now we define Ψkn(v) the matrix of size k × Ln by : for all i = 1, ..., k, we set
(Ψkn(v))ij =
ϕkn+i−1+j−Ln, i−1(v) if Ln, i−1 < j < Ln, i,
0 else,
where Ln, 0 = 1 and Ln, i =i∑
i′=1
(ln+i′−1 − 1) for i ≥ 1. Then, for all ξ ∈ RLn , we have
∥
∥B−1kn
Φknξ∥
∥
2
U1, 2=
∑
v∈Vc
αl1∥
∥B−1kn
Ψkn(v)ξ∥
∥
2
2
=∑
v∈Vc
αl1ξTΨkn(v)TB−T
knB−1kn
Ψkn(v)ξ
= ξT
(
∑
v∈Vc
αl1Ψkn(v)TB−TknB−1kn
Ψkn(v)
)
ξ.
Therefore by setting
Mkn =∑
v∈Vc
αl1Ψkn(v)TB−TknB−1kn
Ψkn(v),
we see that the assumption (3.43) becomes
∃γ > 0, ∀k ∈ 1, ...,M, ∀kn ∈ Ak, λmin(Mkn) ≥ γ,
and the assumption (3.46) becomes
∃m ∈ R∗+, ∃γ > 0, ∀k ∈ 1, ...,M, ∀kn ∈ Ak, λmin(Mkn) ≥ γ
λ2mkn
which corresponds respectively to the conditions (1.44) and (1.50) from chapter 1, becauseλk ∼ kπ
L, where L =
∑Nj=1 lj .
Note that if the standard gap condition (3.40) holds and if all eigenvalues are simple(i.e. lk = 1), then the condition (3.43) becomes
∃γ > 0, ∀k ≥ 1,∑
v∈Vc
α(v)1 |ϕk(v)|2 ≥ γ, (3.51)
142
while the conditions (3.46) becomes
∃m ∈ R∗+, ∃γ > 0, ∀k ≥ 1,
∑
v∈Vc
α(v)1 |ϕk(v)|2 ≥
γ
λ2mk
. (3.52)
Consequently, we find again all the results from chapter 1 (see for instance the examplestreated in section 1.7), here we can even precise the dependence of the decay with respectto the delay τ .
3.7.2 One Euler-Bernoulli beam with interior damping
We consider an Euler-Bernoulli beam of length 1 with interior damping and a delayterm at ξ. Two types of boundary conditions will be considered. Without delay, these twoproblems were analyzed in [10,11], where some decay rates similar to the ones proved belowwere obtained.
Mixed boundary conditions
We consider the following initial and boundary problem :
∂2ω∂t2
(x, t) + ∂4ω∂x4 (x, t) + α1
∂ω∂t
(ξ, t)δξ + α2∂ω∂t
(ξ, t− τ)δξ = 0 0 < x < 1, t > 0
ω(0, t) = ∂ω∂x
(1, t) = ∂2ω∂x2 (0, t) = ∂3ω
∂x3 (1, t) = 0 t > 0ω(x, 0) = ω0(x),
∂ω∂t
(x, 0) = ω1(x) 0 < x < 1∂ω∂t
(ξ, t− τ) = f 0(t− τ) 0 < t < τ,
(3.53)
where ξ ∈ (0, 1), α1, α2 > 0 and τ > 0. To enter into the framework of section 1, werewrite this problem in the form (3.2). For that purpose, we introduce H = L2(0, 1) andthe operator
A : D(A) → H : ϕ 7→ d4
dx4ϕ (3.54)
where D(A) = ϕ ∈ H4(0, 1) ; ϕ(0) = ∂ϕ∂x
(1) = ∂2ϕ∂x2 (0) = ∂3ϕ
∂x3 (1) = 0. The operator A isself-adjoint and positive with a compact inverse in H. We now define U = U1 = U2 = R
and the operators B1 and B2 as
Bi : U → D(A12 )′ : k 7→ √
αi k δξ, i = 1, 2. (3.55)
It is easy to verify that B∗i (ϕ) =
√αiϕ(ξ) for ϕ ∈ D(A
12 ) and thus BiB
∗i (ϕ) = αiϕ(ξ)δξ
for ϕ ∈ D(A12 ) and i = 1, 2. Then the system (3.53) can be rewritten in the form (3.2).
We notice that (3.8) is equivalent to
∃0 < α ≤ 1, α2 ≤ αα1,
and consequently, this system is well posed for α2 ≤ α1 by Theorem 3.2.1, and the energyis decreasing for α2 < α1 by Proposition 3.3.1.
Let us now state the next well known results about the spectral properties of A.
143
Proposition 3.7.1. The eigenvalues of the operator A defined in (3.54) are simple and aregiven by λ2
k = (2k+12π)4 of associated eigenvector ϕk(x) =
√2 sin(2k+1
2πx), for all k ∈ N.
Consequently the standard gap condition (3.40) holds, i.e. there exists a constant γ0 > 0such that
λk+1 − λk ≥ γ0 > 0, ∀k ≥ 0,
and moreover for all k ≥ 0, ‖B∗1ϕk‖U1
=√
2α1
∣
∣sin((kπ + π2)ξ)∣
∣ .
The hypothesis (3.20) was verified in [10]. Moreover, we have by Lemma 2.9 of [97] :
Lemma 3.7.2. ξ is a rational number with an irreductible fraction
ξ =p
q, where p is odd
if and only if there exists a constant γ > 0 such that
∀k ≥ 1,∣
∣
∣sin(
(kπ +π
2)ξ)∣
∣
∣> γ.
Therefore, by applying Proposition 3.3.4 and Theorem 3.4.4, we obtain the followingresults :
Proposition 3.7.3. Assume that α2 < α1. Then(i) The energy of system (3.53) decays to 0 if and only if
ξ 6= 2m
2k + 1, m, k ∈ N.
(ii) The energy of system (3.53) decays exponentially if ξ is a rational number with anirreductible fraction
ξ =p
q, where p is odd.
Remark 3.7.4. As mentioned before, in the case α2 = 0 we recover the results from [10].
Other boundary conditions
We here consider the following initial and boundary problem :
∂2ω∂t2
(x, t) + ∂4ω∂x4 (x, t) + α1
∂ω∂t
(ξ, t)δξ + α2∂ω∂t
(ξ, t− τ)δξ = 0 0 < x < 1, t > 0
ω(0, t) = ω(1, t) = ∂2ω∂x2 (0, t) = ∂2ω
∂x2 (1, t) = 0 t > 0ω(x, 0) = ω0(x), ∂ω
∂t(x, 0) = ω1(x) 0 < x < 1
∂ω∂t
(ξ, t− τ) = f 0(t− τ) 0 < t < τ,
(3.56)
where ξ ∈ (0, 1), α1, α2 > 0 and τ > 0. This system (3.56) is not exponentially stable ifα2 = 0 as shown in [11]. Hence we only consider the polynomial decay of system (3.56).
144
As before we rewrite this problem in the form (3.2) by introducing H = L2(0, 1) and theoperator
A : D(A) → H : ϕ 7→ d4
dx4ϕ (3.57)
with D(A) = ϕ ∈ H4(0, 1) ∩ V ; ∂2ϕ∂x2 (0) = ∂2ϕ
∂x2 (1) = 0, V = H2(0, 1) ∩ H10 (0, 1).
The operator A is self-adjoint and positive with a compact inverse in H. We then defineU = U1 = U2 = R and the operators B1, B2 by (3.55).
Then the system (3.56) can be rewritten in the form (3.2) and consequently, this systemis well posed for α2 ≤ α1 by Theorem 3.2.1, and the energy is decreasing for α2 < α1 byProposition 3.3.1.
The spectral properties of A are well known and can be summarized as follows :
Proposition 3.7.5. The eigenvalues of the operator A defined in (3.57) are simple andgiven by λ2
k = k4π4 of eigenvector ϕk(x) =√
2 sin(kπx), for all k ∈ N∗. Therefore thereexists a constant γ0 > 0 such that the standard gap condition (3.40) is verified and moreover
‖B∗1ϕk‖U1
=√
2α1 |sin(kπξ)| .
The hypothesis (3.20) was verified in [11]. Let us prove that the condition (3.28) issatisfied.
Lemma 3.7.6. Let m ∈ R∗+. Then there exists C > 0 such that for all ω0 ∈ X =
u ∈ V : u∣
∣
(0, ξ) ∈ H4(0, ξ), u∣
∣
(ξ, 1) ∈ H4(ξ, 1), ∂2u∂x2 (0) = ∂2u
∂x2 (1) = 0
, we have
‖ω0‖m+1V ≤ C ‖ω0‖mX ‖ω0‖
D(A1−m
2 ),
where the natural norm in X is given by ‖u‖2X = ‖u‖2
H4(0,ξ) + ‖u‖2H4(ξ,1).
Proof: Let us fix a cut-off function η ∈ D(0, 1) such that η = 1 in a neighbourhood of ξ,η = 0 on
[
23
+ ξ3, 1]
and η = 0 on[
0, ξ3
]
. Since (1−η)ω0 ∈ D(A), by Lemma 3.5.2, we have
‖(1 − η)ω0‖m+1V ≤ C ‖(1 − η)ω0‖mX ‖(1 − η)ω0‖
D(A1−m
2 ).
Since‖(1 − η)ω0‖
D(A1−m
2 )= sup
ϕ∈D(Am−1
2 )
((1−η)ω0, ϕ)‖ϕ‖
D(Am−1
2 )
= supϕ∈D(A
m−12 )
(ω0, (1−η)ϕ)‖ϕ‖
D(Am−1
2 )
≤ C ‖ω0‖D(A
1−m2 )
,
for some C > 0 (depending on η) we get
‖(1 − η)ω0‖m+1V ≤ C ‖ω0‖mX ‖ω0‖
D(A1−m
2 ). (3.58)
145
In a second step, we set
ω1(x) = η(ξ − x)ω0(ξ − x) if 0 < x < l1 := ξω2(x) = η(x+ ξ)ω0(x+ ξ) if 0 < x < l2 := 1 − ξ.
For any j = 1, 2, we introduce the following extension of ωj :
(Eω)j(x) = ωj(x) if x ∈ (0, lj),
(Eω)−j(x) =∑n−1
i=0 νiωj(−2ix) if x ∈ (−2−(n−1)lj, 0),
where ωj is extended by zero outside its support and the real numbers νi are the uniquesolution of the system
∑n−1i=0 νi = 1
−∑n−1i=0 2iνi = 1
∑n−1i=0 22iνi = 1
−∑n−1i=0 23iνi = 1
∑n−1i=0 2−2kiνi = 1, ∀k = 1, · · · , n− 4,
and finally n ∈ N∗ is choosen large enough such that n ≥ m+ 3.We obtain an extension of ω to a function Eω, which belongs to D(A) (due to the
four first properties of the νi), where A is the positive operator d4
dx4 on the star shaped
network S = ∪j=1, 2(0, lj)⋃∪j=1, 2(−2−(n−1)lj, 0), with interior vertex ξ (identified to 0)
and Dirichlet boundary conditions at all other vertices.Therefore, we can apply the interpolation lemma 3.5.2 to A and then write
‖Eω‖m+1
D(A12 )
≤ C ‖Eω‖mD(A) ‖Eω‖D(A1−m
2 ).
But we easily check that‖ηω0‖m+1
D(A12 )
≤ ‖Eω‖m+1
D(A12 ).
Consequently, we have
‖ηω0‖m+1
D(A12 )
≤ C ‖Eω‖mD(A) ‖Eω‖D(A1−m
2 ).
Moreover, since E is an extension operator, we have
‖Eω‖D(A) ≤ K ‖ω0‖X ,
and thus‖ηω0‖m+1
D(A12 )
≤ C ‖ω0‖mX ‖Eω‖D(A
1−m2 )
.
To estimate the last factor, we use a duality argument. We write
‖Eω‖D(A
1−m2 )
= supϕ∈D(A
m−12 )
|(Eω, ϕ)|‖ϕ‖
D(Am−1
2 )
.
146
For ϕ ∈ D(Am−1
2 ), we have
∫
SEωϕ =
∑
j=1, 2
∫ lj
0
ωj(x)ϕj(x) dx+∑
j=1, 2
∫ 0
−2−(n−1)lj
(Eω)−j(x)ϕ−j(x) dx.
By changes of variables, we obtain
∫
SEωϕ =
∑
j=1, 2
∫ lj
0
ωj(x)(Fϕ)j(x) dx,
where
(Fϕ)j(x) = ϕj(x) + χj(x)n−1∑
i=0
νi2−iϕj(−2−ix), ∀x ∈ (0, lj),
the cut-off function χj being fixed such that χj ≡ 1 on [0, 2lj/3] and χj ≡ 0 on [5lj/6, lj] (re-minding that ωj(x) ≡ 0 for x > 2lj/3). Now we notice that the conditions on νi guarantees
that Fϕ belongs to D(Am−1
2 ) and by Leibniz’s rule we have
‖Fϕ‖D(A
m−12 )
≤ C ‖ϕ‖D(A
m−12 )
.
Therefore∫
SEωϕ ≤ C ‖ω‖
D(A1−m
2 )‖ϕ‖
D(Am−1
2 ).
By duality, we conclude that
‖Eω‖D(A
1−m2 )
≤ C ‖ω‖D(A
1−m2 )
.
Consequently, with the previous inequalities, we obtain
‖ηω0‖m+1
D(A12 )
≤ C ‖ω‖mX ‖ω‖D(A
1−m2 )
. (3.59)
The conclusion follows from (3.58) and (3.59).
This lemma leads to (3.28). Indeed for (ω0, ω1, z) ∈ D(A), we have
‖ω0‖m+1V ≤ C ‖ω0‖mX ‖ω0‖
D(A1−m
2 )
≤ C ‖(ω0, ω1, z)‖mD(A) ‖ω0‖D(A
1−m2 )
.
Now, we denote by S the set of all real numbers ρ such that ρ /∈ Q and if [0, a1, ..., an, ...]is the expansion of ρ as a continued fraction, then the sequence (an) is bounded. It is wellknown that S is uncountable and that its Lebesgue measure is zero. Roughly speaking, theset S contains all irrational numbers which are badly approximated by rational numbers.In particular, by the Euler-Lagrange theorem, S contains all irrational quadratic numbers(i.e. the roots of a second order equation with rational coefficients). By a classical result,we have :
147
Lemma 3.7.7. If s ∈ S, then there exists a positive constant γ such that
|sin(kπs)| ≥ γ
k, ∀k ≥ 1.
Therefore, by applying Proposition 3.3.4 and Theorem 3.5.3 with m = 12, we obtain the
next results :
Proposition 3.7.8. Assume that α2 < α1. Then(i) The energy of system (3.56) decays to 0 if and only if ξ is irrational.(ii) The energy of system (3.56) decays polynomially like 1
(1+t)2if ξ belongs to S.
Remark 3.7.9. Again in the case α2 = 0 we recover the results from [10,11].
3.7.3 Examples with distributed damping terms
A non homogeneous string with distributed damping terms (1-d)
We consider the following initial and boundary problem :
∂2ω∂t2
(x, t) − ∂2ω∂x2 (x, t) + α1
∂ω∂t
(x, t)χ|I1 + α2∂ω∂t
(x, t− τ)χ|I2 = 0 in (0, 1) × (0, ∞)ω(0, t) = ω(1, t) = 0 t > 0
ω(x, 0) = ω0(x),∂ω∂t
(x, 0) = ω1(x) in (0, 1)∂ω∂t
(x, t− τ) = f 0(x, t− τ) in I2 × (0, τ),(3.60)
where here and below χ|I denotes the characteristic function of the set I. In the remainderof this subsection we assume that α1, α2 > 0, τ > 0 and
I2 ⊂ I1 ⊂ [0, 1].
Later we will need that
∃δ ∈ [0, 1] and ǫ > 0 : [δ, δ + ǫ] ⊂ I1. (3.61)
We rewrite this problem in the form (3.2). For that purpose, we introduce H = L2(0, 1)and the operator
A : D(A) → H : ϕ 7→ − d2
dx2ϕ (3.62)
where D(A) = H10 (0, 1) ∩ H2(0, 1) and V = D(A
12 ) = H1
0 (0, 1). The operator A is self-adjoint and positive with a compact inverse in H. We then define Ui = L2(Ii) and theoperators Bi as
Bi : Ui → H ⊂ V ′ : k 7→ √αi k χ|Ii (3.63)
where k is the extension of k by zero outside Ii (which defines an element of L2(0, 1)).
148
It is easy to verify that B∗i (ϕ) =
√αiϕ|Ii for ϕ ∈ V and thus BiB
∗i (ϕ) = αiϕ|Iiχ|Ii =
αiϕχ|Ii for ϕ ∈ V and i = 1, 2. Then the system (3.60) can be rewritten in the form (3.2).Moreover
‖B∗i ϕ‖2
Ui= αi
∫
Ii
|ϕ|2 dx.
Therefore, we notice that (3.8) is equivalent to
∃0 < α ≤ 1, α2
∫
I2
|ϕ|2 dx ≤ αα1
∫
I1
|ϕ|2 dx,
and consequently, this system is well posed for α2 ≤ α1 by Theorem 3.2.1, and the energyis decreasing for α2 < α1 by Proposition 3.3.1.
Proposition 3.7.10. The eigenvalues of the operator A defined in (3.62) are simple andgiven by λ2
k = (kπ)2 of associated eigenvector ϕk(x) =√
2 sin(kπx), for all k ∈ N∗. Hencethere exists a constant γ0 > 0 such that the standard gap condition (3.40) holds. Moreoverif (3.61) holds, then there exists γ > 0 such that ∀k ≥ 1, ‖B∗
1ϕk‖U1≥ γ.
Proof: It is well known that the eigenvectors of the operator A are ϕk(x) =√
2 sin(kπx)of eigenvalue (kπ)2, k ≥ 1 of multiplicity 1. Hence, the standard gap condition (3.40) isverified.
From the definition of B∗1 we have
‖B∗1ϕk‖2
U1= α1
∫
I1|ϕk(x)|2 dx
≥ 2α1
∫ δ+ǫ
δ|sin(kπx)|2 dx = α1
[
x− sin(2kπx)2kπ
]δ+ǫ
δ
≥ α1
(
ǫ− 1kπ
)
≥ α1ǫ2,
for k ≥ E( 2ǫπ
) + 1 =: kǫ, where E(x) is the entire part of x. This leads to the conclusionbecause for k ∈ 1, ..., kǫ, we have
∫ δ+ǫ
δ
|sin(kπx)|2 dx > 0.
Lemma 3.7.11. The operators A and B = (B1B2) ∈ L(U, V ′) where U = U1 ×U2 satisfyassumption (3.20).
Proof: Let i ∈ 1, 2 and ϕ ∈ L2(Ii). It can be easily checked that v = (λ2 + A)−1Biϕsatisfies
λ2v − d2vdx2 =
√αiϕχ|Ii
v(0) = v(1) = 0.(3.64)
Since v ∈ L2(0, 1), v can be written as
v =∞∑
k=1
ckϕk.
149
By replacing v in (3.64), ck must satisfy
ck =
√αi
λ2 + λ2k
(ϕ, ϕk) ,
and therefore
v =∞∑
k=1
√αi
λ2 + λ2k
(ϕ, ϕk)ϕk.
Moreover
‖λv‖2L2(0, 1) =
∞∑
k=1
∣
∣
∣
∣
λ
λ2 + λ2k
∣
∣
∣
∣
2
αi |(ϕ, ϕk)|2 .
Now, we set z = λλ2+λ2
k
and, if λ = β + iy, with y ∈ R, then
|z|2 =β2 + y2
(β2 − y2 + λ2k)
2 + 4β2y2≤ C(β),
where C(β) is a positive constant depending only on β. Indeed, if y2 ≤ β2+λ2k
2, then
|z|2 ≤ β2 + y2
(
β2+λ2k
2
)2 ≤ β2 +β2+λ2
k
2(
β2+λ2k
2
)2
which is bounded uniformly in k, and if y2 ≥ β2+λ2k
2, then
|z|2 ≤ β2 + y2
4β2y2,
which is a decreasing function with respect to y and thus
|z|2 ≤ β2 +β2+λ2
k
2
4β2(
β2+λ2k
2
) ,
which is again uniformly bounded in k. Therefore, we have
‖λv‖2L2(0, 1) ≤ C(β)αi
∞∑
k=1
|(ϕ, ϕk)|2 ≤ C(β)αi ‖ϕ‖2L2(Ii)
,
which leads to∥
∥λB∗j v∥
∥
2
L2(Ij)≤ αj ‖λv‖2
L2(0, 1) ≤ C(β)αiαj ‖ϕ‖2L2(Ii)
, ∀j ∈ 1, 2 .
Consequently, the operator λ → λB∗j (λI + A)−1Bi is bounded on Cβ and the lemma is
proved.
Therefore, by applying Theorem 3.4.4, we obtain :
Proposition 3.7.12. If α2 < α1 and (3.61) holds, then the energy of system (3.60) decaysexponentially.
150
The wave equation with distributed damping terms
Let Ω be an open bounded domain of Rn, n ≥ 1, with a boundary Γ of class C2. Weassume that Γ is divided into two parts ΓD and ΓN , i.e. Γ = ΓD ∪ ΓN , with ΓD ∩ ΓN = ∅and ΓD 6= ∅. Let
O2 ⊂ O1 ⊂ Ω
such that O1 is an open neighborhood of ΓN (i.e. ΓN ⊂ ∂O1). Moreover, we assume thatx0 ∈ Rn is such that
(x− x0) · ν(x) ≤ 0, ∀x ∈ ΓD. (3.65)
We consider the following initial and boundary problem :
∂2ω∂t2
(x, t) − ∆ω(x, t) + α1∂ω∂t
(x, t)χ|O1+ α2
∂ω∂t
(x, t− τ)χ|O2= 0 in Ω × (0, ∞),
ω(x, t) = 0 on ΓD × (0, ∞),∂ω∂ν
(x, t) = 0 on ΓN × (0, ∞),ω(x, 0) = ω0(x),
∂ω∂t
(x, 0) = ω1(x) in Ω,∂ω∂t
(x, t− τ) = f 0(x, t− τ) in O2 × (0, τ),(3.66)
where ∂ω∂ν
is the normal derivative of ω and α1, α2 > 0, τ > 0. In order to reformulate thisproblem in the form (3.2), we introduce H = L2(Ω) and the operator
A : D(A) → H : ϕ 7→ −∆ϕ (3.67)
where D(A) = ϕ ∈ V ∩ H2(Ω) : ∂ϕ∂ν
(x) = 0 on ΓN and V = D(A12 ) = ϕ ∈ H1(Ω) :
ϕ = 0 on ΓD. The operator A is self-adjoint and positive with a compact inverse in H.We then define Ui = L2(Oi) and the operators Bi as
Bi : Ui → H ⊂ V ′ : k 7→ √αi k χ|Oi
(3.68)
where k is the extension of k by zero outside Oi (which defines an element of L2(Ω)).It is easy to verify that B∗
i (ϕ) =√αiϕ|Oi
for ϕ ∈ V and thus BiB∗i (ϕ) = αiϕ|Oi
χ|Oifor
ϕ ∈ V and i = 1, 2. Then the system (3.66) can be rewritten in the form (3.2). Moreover
‖B∗i ϕ‖2
Ui= αi
∫
Oi
|ϕ|2 dx,
and therefore (3.8) is equivalent to
∃0 < α ≤ 1, α2
∫
O2
|ϕ|2 dx ≤ αα1
∫
O1
|ϕ|2 dx.
Consequently, this system is well posed for α2 ≤ α1 by Theorem 3.2.1, and the energy isdecreasing for α2 < α1 by Proposition 3.3.1.
To obtain the exponential decay of problem (3.66), we simply check the observabi-lity inequality (3.25) and the hypothesis (3.20), which are the aim of the two followingpropositions.
151
Proposition 3.7.13. There exists a time T0 such that for all times T > T0 there existsa positive constant C (depending on T ) for which the observability inequality (3.25) holdsfor any regular solution of problem (3.22).
Proof: This proposition is proved in [85] and [68].
Proposition 3.7.14. The operators A and B = (B1B2) ∈ L(U, V ′) where U = U1 × U2
satisfy the assumption (3.20).
Proof: Let i ∈ 1, 2 and ϕ ∈ L2(Oi). It can be easily checked that v = (λ2 + A)−1Biϕsatisfies
λ2v − ∆v =√αiϕχ|Oi
in Ω,v = 0 on ΓD,∂v∂ν
= 0 on ΓN .(3.69)
Since v ∈ L2(Ω), it can be written as
v =
∞∑
k=1
ckϕk.
By replacing v in (3.69), ck must satisfy
ck =
√αi
λ2 + λ2k
(ϕ, ϕk) ,
and therefore
v =
∞∑
k=1
√αi
λ2 + λ2k
(ϕ, ϕk)ϕk.
Moreover, by Parseval’s identity, we have
‖λv‖2L2(Ω) =
∞∑
k=1
∣
∣
∣
∣
λ
λ2 + λ2k
∣
∣
∣
∣
2
αi |(ϕ, ϕk)|2 .
Now, for λ = β + iy, with y ∈ R, we have checked in the previous subsection that∣
∣
∣
∣
λ
λ2 + λ2k
∣
∣
∣
∣
2
≤ C(β),
where C(β) is a positive constant depending only on β. Therefore, we have
‖λv‖2L2(Ω) ≤ C(β)αi
∞∑
k=1
|(ϕ, ϕk)|2 ≤ C(β)αi ‖ϕ‖2L2(Oi)
,
which leads to∥
∥λB∗j v∥
∥
2
L2(Oj)≤ αj ‖λv‖2
L2(Ω) ≤ C(β)αiαj ‖ϕ‖2L2(Oi)
, ∀j ∈ 1, 2 .
Consequently, the operator λ → λB∗j (λI + A)−1Bi is bounded on Cβ and the lemma is
proved.
Therefore, by applying Theorem 3.4.4, we obtain :
152
Proposition 3.7.15. If α2 < α1, then the energy of system (3.66) decays exponentially.
Remark 3.7.16. This result is a generalization of [85] because in [85] the authors supposedthat O2 = O1.
A beam with distributed damping terms
We consider the following initial and boundary problem :
∂2ω∂t2
(x, t) + ∂4ω∂x4 (x, t) + α1
∂ω∂t
(x, t)χ|I1 + α2∂ω∂t
(x, t− τ)χ|I2 = 0 in (0, 1) × (0, ∞),
ω(0, t) = ω(1, t) = ∂2ω∂x2 (0, t) = ∂2ω
∂x2 (1, t) = 0 t > 0,ω(x, 0) = ω0(x),
∂ω∂t
(x, 0) = ω1(x) in Ω,∂ω∂t
(x, t− τ) = f 0(x, t− τ) in I2 × (0, τ),(3.70)
where α1, α2 > 0, τ > 0 and where
I2 ⊂ I1 ⊂ [0, 1].
We rewrite this problem in the form (3.2) by introducing H = L2(0, 1), the operator A by(3.57) and the operators B1 and B2 by (3.63). Hence this system is well posed for α2 ≤ α1
by Theorem 3.2.1, and the energy is decreasing for α2 < α1 by Proposition 3.3.1.By the results of the previous subsections, we know that the standard gap condition
holds and if (3.61) holds that the eigenvector ϕk(x) =√
2 sin(kπx) of A associated withλ2k = k4π4 satisfies
‖B∗1ϕk‖U1
≥ γ,
for some γ > 0.
Lemma 3.7.17. The operators A and B = (B1B2) ∈ L(U, V ′) where U = U1 ×U2 satisfyassumption (3.20).
Proof: The proof is the same than in the previous subsection.
Therefore, by applying Proposition 3.3.4 and Theorem 3.4.4, we obtain :
Proposition 3.7.18. If α2 < α1 and (3.61) holds, then the energy of system (3.70) decaysexponentially.
A wave equation on 1-d networks with internal damping terms
In this last subsection we consider again a wave equation on a given network R butwe suppose that the feedbacks are located in the edges. Namely with the notation fromsection 3.7.1 we consider the problem :
153
∂2uj
∂t2(x, t) − ∂2uj
∂x2 (x, t) + α(j)1
∂uj
∂t(x, t)χ|I(j)1
+α(j)2
∂uj
∂t(x, t− τ)χ|I(j)2
= 0 0 < x < lj , t > 0, ∀j ∈ 1, · · · , N,uj(v, t) = ul(v, t) = u(v, t) t > 0, ∀j, l ∈ Ev, v ∈ Vint,∑
j∈Ev
∂uj
∂nj(v, t) = 0 t > 0, ∀v ∈ V\D,
ujv(v, t) = 0 t > 0, ∀v ∈ D,u(t = 0) = u(0), ∂u
∂t(t = 0) = u(1),
∂uj
∂t(x, t− τ) = f 0
j (x, t− τ) in I(j)2 × (0, τ), ∀j ∈ 1, · · · , N,
(3.71)
where α(j)i are fixed non-negative real numbers, the delay τ is positive and the intervals
I(j)i satisfy
I(j)2 ⊂ I
(j)1 ⊂ ej .
As before we can rewrite this problem in the form (3.2), by introducing H = L2(R),
the operator A defined by (3.48), Ui = L2(∪Nj=1I(j)i ) and the operators Bi for i = 1, 2 as
Bi :
U → D(A12 )′
k 7→N∑
j=1
√
α(j)i kχ|I(j)i
(3.72)
where k means the extension of k by zero outside ∪Nj=1I(j)i . It is easy to verify that
B∗i (ϕ) =
N∑
j=1
√
α(j)i ϕ|I(j)i
χ|I(j)i
for ϕ ∈ D(A12 ). Hence the system (3.71) can be rewritten in the form (3.2).
As before it is easy to see that the system (3.47) is well posed for α(j)2 ≤ α
(j)1 for
all j = 1, · · · , N by Theorem 3.2.1, and the energy is decreasing for α(j)2 < α
(j)1 for all
j = 1, · · · , N by Proposition 3.3.1.As mentioned before by Proposition 6.2 from [38], the generalized gap condition (3.37)
holds with M = N + 1, but in this general setting, the conditions (3.43) or (3.46) seemdifficult to check. Hence, for the sake of simplicity, if we suppose here that the standard gapcondition (3.40) holds and that all eigenvalues are simple (i.e. lk = 1), then the condition(3.43) becomes
∃γ > 0, ∀k ≥ 1,
N∑
j=1
α(j)1
∫
I(j)1
ϕk(x)2 dx ≥ γ, (3.73)
while the conditions (3.46) becomes
∃m ∈ R∗+, ∃γ > 0, ∀k ≥ 1,
N∑
j=1
α(j)1
∫
I(j)1
ϕk(x)2 dx ≥ γ
λ2mk
. (3.74)
154
For instance using an argument like in Proposition 3.7.10, we easily see that (3.73)
(resp. (3.74)) holds if (3.51) (resp. (3.52)) holds and if ∪Nj=1I(j)1 contains a neighborhood of
the set of control points Vc.In the same manner, we have :
Proposition 3.7.19. If there exists ε > 0 and for all j ∈ 1, · · · , N, there exists δj ∈(0, lj) such that
[δj, δj + ε] ⊂ I(j)1 ∀j ∈ 1, · · · , N,
then (3.73) holds.
Proof: For any k ∈ N∗, let ϕk be the eigenvector of A associated with the eigenvalue λ2k.
Then its restriction ϕk,j to the edge ej can be written in the form
ϕk,j(x) = ck,j cos(λkx) + dk,j sin(λkx) ∀x ∈ (0, lj),
for some real numbers ck,j and dk,j. Hence the normalization of ϕk yields
1 =N∑
j=1
∫ lj
0
|ϕk,j(x)|2 dx ≤ 2(maxjlj)
N∑
j=1
(c2k,j + d2k,j). (3.75)
On the other hand by the above expression of ϕk,j and direct calculations, we see that
∫
I(j)1
ϕk(x)2 dx ≥
∫ δj+ε
δj
ϕk(x)2 dx
≥ (c2k,j + d2k,j)(
ε
2− 1
2λk) − |ck,jdk,j|
λk
≥ (c2k,j + d2k,j)(
ε
2− 1
λk).
Therefore for k large enough such that
ε
2− 1
λk≥ ε
4,
which is equivalent to k ≥ kε, for some kε ∈ N∗, we deduce that
∫
I(j)1
ϕk(x)2 dx ≥ ε
4(c2k,j + d2
k,j).
By summing this estimate on j and using the normalization estimate (3.75), we obtain(3.73) for k ≥ kε. The proof is complete since for k ≤ kε,
N∑
j=1
α(j)1
∫
I(j)1
ϕk(x)2 dx > 0.
155
The analysis of the condition (3.73) in some particular cases reveals that the conditionof the above Proposition is far from being optimal but in its full generality we cannot easilyobtain a weaker condition.
As in the previous subsection one easily shows (see the proof of Lemma 3.7.11) thatthe operators A and B = (B1B2) ∈ L(U, V ′) where U = U1 × U2 satisfy the assumption(3.20).
In conclusion applying either Theorem 3.4.4 or Theorem 3.5.3, we obtain :
Proposition 3.7.20. Assume that α(j)2 < α
(j)1 for all j = 1, · · · , N and that the standard
gap condition (3.40) holds and that all eigenvalues are simple (i.e. lk = 1). Then(i) The energy of system (3.71) decays exponentially if (3.73) holds.
(ii) The energy of system (3.71) decays like t−1m if (3.74) holds.
156
Chapitre 4
Stability of the heat and of the waveequations with boundarytime-varying delays
4.1 Introduction
Time-delay often appears in many biological, electrical engineering systems and me-chanical applications, and in many cases, delay is a source of instability [52]. In the caseof distributed parameter systems, even arbitrarily small delays in the feedback may des-tabilize the system (see e.g. [39, 75, 85, 98]). The stability issue of systems with delay is,therefore, of theoretical and practical importance.
There are only a few works on Lyapunov-based technique for Partial Differential Equa-tions (PDEs) with delay. Most of these works analyze the case of constant delays. Thus,stability conditions and exponential bounds were derived for some scalar heat and waveequations with constant delays and with Dirichlet boundary conditions without delayin [107, 108]. Stability and instability conditions for the wave equations with constantdelay can be found in [85] and in chapter 1. The stability of linear parabolic systems withconstant coefficients and internal constant delays has been studied in [55] in the frequencydomain.
Recently the stability of PDEs with time-varying delays was analyzed in [27, 45, 92]via Lyapunov method. In the case of linear systems in the Hilbert space, the conditionsof [27,45,92] assume that the operator acting on the delayed state is bounded, which meansthat this condition can not be applied to boundary delays. These conditions were appliedto PDEs without delays in the boundary conditions (to 2D Navier-Stokes and to a scalarheat equations in [27], to a scalar heat and to a scalar wave equations in [45, 92]).
In the present chapter we analyze exponential stability of the heat and wave equationswith time-varying boundary delay. Our main novel contribution is an extension of previousresults from [85,89] and chapter 1 to time-varying delays. This extension is not straightfor-ward due to the loss of translation-invariance. In the constant delay case the exponential
157
stability was proved in [85] and in chapter 1 by using the observability inequality which cannot be applicable in the time-varying case (since the system is not invariant by translation).Hence we introduce new Lyapunov functionals with exponential terms and an additionalterm for the wave equation, which take into account the dependence of the delay with res-pect to time. For the treatment of other problems with Lyapunov technique see [45,86,92].Note further that to the best of our knowledge the heat equation with boundary delay hasnot been treated in the literature. Contrary to [85, 89] (and chapter 1), the existence re-sults do not follow from standard semi-group theory because the spatial operator dependson time due to the time-varying delay. Therefore we use the variable norm technique ofKato [59,60]. Finally for each problem we give explicit sufficient conditions that guaranteethe exponential decay and for the first time we characterize the optimal decay rate thatcan be explicitly computed once the data are given.
The chapter is mainly decomposed in two parts treating the heat equation (section4.2) and the wave equation (section 4.3). In the first subsection, we set the problem underconsideration and prove existence results by using semigroup theory. In the second subsec-tion we find sufficient conditions for the strict decay of the energy and finally in the lastsubsection we show that these conditions yield an exponential decay.
4.2 Exponential stability of the delayed heat equation
First, we consider the system described by
ut(x, t) − auxx(x, t) = 0, 0 < x < π, t > 0,u(0, t) = 0, t > 0,
ux(π, t) = −µ0u(π, t) − µ1u(π, t− τ(t)), t > 0,u(x, 0) = u0(x), 0 < x < π,
u(π, t− τ(0)) = f 0(t− τ(0)), 0 < t < τ(0),
(4.1)
with the constant parameter a > 0 and where µ0, µ1 ≥ 0 are fixed nonnegative realnumbers, the time-varying delay τ(t) satisfies
τ (t) < 1, ∀t > 0, (4.2)
and∃M > 0 : 0 < τ0 ≤ τ(t) ≤M, ∀t > 0. (4.3)
Moreover, we assume thatτ ∈W 2,∞([0, T ]), ∀T > 0. (4.4)
The boundary-value problem (4.1) describes the propagation of heat in a homogeneousone-dimensional rod with a fixed temperature at the left end. Here a stands for the heatconduction coefficient, u(x, t) is the value of the temperature field of the plant at timemoment t and location x along the rod. In the sequel, the state dependence on time t andspatial variable x is suppressed whenever possible.
158
4.2.1 Well posedness of the problem
We aim to show that problem (4.1) is well-posed. For that purpose, we use semi-grouptheory and adapt the ideas from [85].
We introduce the Hilbert space
V = φ ∈ H1(0, π) : φ(0) = 0.
We transform our system (4.1) as follows. Let us introduce the auxiliary variable z(ρ, t) =u(π, t − τ(t)ρ) for ρ ∈ (0, 1) and t > 0. Note that z verifies the transport equation for0 < ρ < 1 and t > 0
τ(t)zt(ρ, t) + (1 − τ (t)ρ)zρ(ρ, t) = 0,z(0, t) = u(π, t),
z(ρ, 0) = f 0(−τ(0)ρ).(4.5)
Therefore, the problem (4.1) is equivalent to
ut(x, t) − auxx(x, t) = 0, 0 < x < π, t > 0,τ(t)zt(ρ, t) + (1 − τ(t)ρ)zρ(ρ, t) = 0, 0 < ρ < 1, t > 0,
u(0, t) = 0, ux(π, t) = −µ0u(π, t) − µ1z(1, t), t > 0,z(0, t) = u(π, t), t > 0,u(x, 0) = u0(x), 0 < x < π,
z(ρ, 0) = f 0(−τ(0)ρ), 0 < ρ < 1.
(4.6)
If we introduceU := (u, z)⊤,
then U satisfies
Ut = (ut, zt)⊤ =
(
auxx,τ(t)ρ− 1
τ(t)zρ
)⊤.
Consequently the problem (4.1) may be rewritten as the first order evolution equation
Ut = A(t)UU(0) = (u0, f 0(−τ(0)))⊤ = U0,
(4.7)
where the time dependent operator A(t) is defined by
A(t)
(
uz
)
=
(
auxxτ(t)ρ−1τ(t)
zρ
)
,
with domainD(A(t)) := (u, z) ∈ (V ∩H2(0, π)) ×H1(0, 1) :
z(0) = u(π), ux(π) = −µ0u(π) − µ1z(1).Notice that the domain of the operator A(t) is independent of the time t, i.e.
D(A(t)) = D(A(0)), ∀t > 0. (4.8)
159
Now, we introduce the Hilbert space
H = L2(0, π) × L2(0, 1)
equipped with the usual inner product⟨(
uz
)
,
(
uz
)⟩
=
∫ π
0
uudx+
∫ 1
0
z(ρ)z(ρ)dρ.
A general theory for equations of type (4.7) has been developed using semigroup theory[59,60,93]. The simplest way to prove existence and uniqueness results is to show that thetriplet A, H, Y , with A = A(t) : t ∈ [0, T ], for some fixed T > 0 and Y = D(A(0)),forms a CD-system (or constant domain system, see [59,60]). More precisely, the followingtheorem gives the existence and uniqueness results and is proved in Theorem 1.9 of [59](see also Theorem 2.13 of [60] or [3])
Theorem 4.2.1. Assume that(i) Y = D(A(0)) is a dense subset of H,(ii) (4.8) holds,(iii) for all t ∈ [0, T ], A(t) generates a strongly continuous semigroup on H and the
family A = A(t) : t ∈ [0, T ] is stable with stability constants C and m independent of t(i.e. the semigroup (St(s))s≥0 generated by A(t) satisfies ‖St(s)u‖H ≤ Cems‖u‖H , for allu ∈ H and s ≥ 0),
(iv) ∂tA belongs to L∞∗ ([0, T ], B(Y, H)), the space of equivalent classes of essentially
bounded, strongly measure functions from [0, T ] into the set B(Y, H) of bounded operatorsfrom Y into H.
Then, problem (4.7) has a unique solution U ∈ C([0, T ], Y ) ∩ C1([0, T ], H) for anyinitial datum in Y .
Lemma 4.2.2. D(A(0)) is dense in H.
Proof: Let (f, h)⊤ ∈ H be orthogonal to all elements of D(A(0)), namely
0 =
⟨(
uz
)
,
(
fh
)⟩
=
∫ π
0
ufdx+
∫ 1
0
z(ρ)h(ρ)dρ,
for all (u, z)⊤ ∈ D(A(0)).We first take u = 0 and z ∈ D(0, 1). Since (0, z) ∈ D(A(0)), we get
∫ 1
0
z(ρ)h(ρ)dρ = 0.
Since D(0, 1) is dense in L2(0, 1), we deduce that h = 0.In the same manner, by taking z = 0 and u ∈ D(0, π) we see that f = 0.
Let us suppose now that the speed of the delay satisfies
τ (t) ≤ d < 1, ∀t > 0 (4.9)
160
and that µ0, µ1 satisfyµ2
1 ≤ (1 − d)µ20. (4.10)
Under these conditions, we will show that the operator A(t) generates a C0-semigroup inH and using the variable norm technique of Kato from [59], that problem (4.6) (and then(4.1)) has a unique solution.
For that purpose, we introduce the following time-dependent inner product on H
⟨(
uz
)
,
(
uz
)⟩
t
=
∫ π
0
uudx+ qτ(t)
∫ 1
0
z(ρ)z(ρ)dρ,
where q is a positive constant chosen later on, with associated norm denoted by ‖.‖t .Theorem 4.2.3. For an initial datum U0 ∈ H, there exists a unique solution U ∈C([0, +∞), H) to problem (4.7). Moreover, if U0 ∈ D(A(0)), then
U ∈ C([0, +∞), D(A(0))) ∩ C1([0, +∞), H).
Proof: We first prove that
‖φ‖t‖φ‖s
≤ ec
2τ0|t−s|
, ∀t, s ∈ [0, T ] (4.11)
where φ = (u, z)⊤ and c is a positive constant. For all s, t ∈ [0, T ], we have
‖φ‖2t − ‖φ‖2
s ec
τ0|t−s|
=(
1 − ec
τ0|t−s|)
∫ π
0
u2dx+ q(
τ(t) − τ(s)ec
τ0|t−s|)
∫ 1
0
z(ρ)2dρ.
We notice that 1 − ec
τ0|t−s| ≤ 0. Moreover τ(t) − τ(s)e
cτ0
|t−s| ≤ 0 for some c > 0. Indeed,
τ(t) = τ(s) + τ (a)(t− s), where a ∈ (s, t),
and thus,τ(t)
τ(s)≤ 1 +
|τ (a)|τ(s)
|t− s| .
By (4.4), τ is bounded and therefore,
τ(t)
τ(s)≤ 1 +
c
τ0|t− s| ≤ e
cτ0
|t−s|,
by (4.3), which proves (4.11).Now we calculate 〈A(t)U, U〉t for a t fixed. Take U = (u, z)⊤ ∈ D(A(t)). Then
〈A(t)U, U〉t =
⟨(
auxxτ(t)ρ−1τ(t)
zρ
)
,
(
uz
)
⟩
t
= a
∫ π
0
uxxudx− q
∫ 1
0
zρ(ρ)z(ρ)(1 − τ(t)ρ)dρ.
161
By integrating by parts in space in the first term of this right hand side, we have
〈A(t)U, U〉t = −a∫ π
0
u2xdx+ a[uux]
π0 − q
∫ 1
0
zρ(ρ)z(ρ)(1 − τ(t)ρ)dρ
= −a∫ π
0
u2xdx− aµ0u(π, t)
2 − aµ1u(π, t)u(π, t− τ(t))
−q∫ 1
0
zρ(ρ)z(ρ)(1 − τ(t)ρ)dρ.
Moreover, we have by integrating by parts in ρ :
∫ 1
0
zρ(ρ)z(ρ)(1 − τ (t)ρ)dρ =
∫ 1
0
1
2
∂
∂ρ(z(ρ)2)(1 − τ (t)ρ)dρ
=τ (t)
2
∫ 1
0
z(ρ)2dρ+1
2u2(π, t− τ(t))(1 − τ (t)) − 1
2u2(π, t).
Therefore
〈A(t)U, U〉t = −a∫ π
0
u2xdx−
qτ (t)
2
∫ 1
0
z(ρ)2dρ− aµ0u(π, t)2
−aµ1u(π, t)u(π, t− τ(t)) − q
2u(π, t− τ(t))2(1 − τ(t)) +
q
2u(π, t)2
≤ −a∫ π
0
u2xdx+ (
q
2− aµ0)u
2(π, t) − aµ1u(π, t)u(π, t− τ(t))
−q2u(π, t− τ(t))2(1 − d) +
q |τ(t)|2τ(t)
τ(t)
∫ 1
0
z(ρ)2dρ.
We can see that this inequality can be written
〈A(t)U, U〉t ≤ −a∫ π
0
u2xdx+(u(π, t), u(π, t−τ(t)))Ψq(u(π, t), u(π, t−τ(t)))⊤+κ(t) 〈U, U〉t ,
where
κ(t) =(τ(t)2 + 1)
12
2τ(t)(4.12)
and where Ψq is the 2 × 2 matrix defined by
Ψq =1
2
(
q − 2aµ0 −aµ1
−aµ1 −q(1 − d)
)
. (4.13)
Since −q(1 − d) < 0, we notice that the matrix Ψq is negative (in the sense thatXΨqX
⊤ ≤ 0, for all X = (x1, x2) ∈ R2) if and only if
q2 − 2aµ0q +a2µ2
1
1 − d≤ 0. (4.14)
162
The discriminant of this second order polynomial (in q) is
∆ = 4a2
(
µ20 −
µ21
1 − d
)
,
which is non negative if and only if (4.10) holds. Therefore, the matrix Ψq is negativefor some q > 0 if and only if (4.10) is satisfied. Hence, we choose q satisfying (4.14) orequivalently such that
aµ0 − a
√
µ20 −
µ21
1 − d≤ q ≤ aµ0 + a
√
µ20 −
µ21
1 − d.
Such a choice of q yields〈A(t)U, U〉t − κ(t) 〈U, U〉t ≤ 0, (4.15)
which proves the dissipativeness of A(t) = A(t) − κ(t)I for the inner product 〈·, ·〉t.Moreover κ′(t) = τ(t)τ (t)
2τ(t)(τ (t)+1)12− τ(t)(τ (t)2+1)
12
2τ(t)2is bounded on [0, T ] for all T > 0 (by (4.4))
and we haved
dtA(t)U =
(
0τ(t)τ(t)ρ−τ (t)(τ (t)ρ−1)
τ(t)2zρ
)
with τ(t)τ(t)ρ−τ (t)(τ (t)ρ−1)τ(t)2
bounded on [0, T ] by (4.4). Thus
d
dtA(t) ∈ L∞
∗ ([0, T ], B(D(A(0)), H)), (4.16)
the space of equivalence classes of essentially bounded, strongly measurable functions from[0, T ] into B(D(A(0)), H).
Let us prove that A(t) is maximal, i.e., that λI −A(t) is surjective for some λ > 0 andt > 0.
Let (f, h)T ∈ H. We look for U = (u, z)T ∈ D(A(t)) solution of
(λI −A(t))
(
uz
)
=
(
fh
)
or equivalently
λu− auxx = f
λz + 1−τ(t)ρτ(t)
zρ = h.(4.17)
Suppose that we have found u with the appropriate regularity. We can then determinez, indeed z satisfies the differential equation
λz +1 − τ (t)ρ
τ(t)zρ = h
163
and the boundary condition z(0) = u(π). Therefore z is explicitly given by
z(ρ) = u(π)e−λτ(t)ρ + τ(t)e−λτ(t)ρ∫ ρ
0
eλτ(t)σh(σ)dσ,
if τ (t) = 0, and
z(ρ) = u(π)eλτ(t)τ(t)
ln(1−τ (t)ρ) + eλτ(t)τ(t)
ln(1−τ(t)ρ)∫ ρ
0
h(σ)τ(t)
1 − τ(t)σe−λ
τ(t)τ(t)
ln(1−τ(t)σ)dσ,
otherwise. This means that once u is found with the appropriate properties, we can find z.In particular, we have if τ(t) = 0,
z(1) = u(π)e−λτ(t) + τ(t)e−λτ(t)∫ 1
0
eλτ(t)σh(σ)dσ = u(π)e−λτ(t) + z0,
where z0 = τ(t)e−λτ(t)∫ 1
0eλτ(t)σh(σ)dσ is a fixed real number depending only on h, and if
τ(t) 6= 0
z(1) = u(π)eλτ(t)τ(t)
ln(1−τ (t)) + eλτ(t)τ(t)
ln(1−τ (t)) ∫ 1
0h(σ)τ(t)1−τ(t)σ e
−λ τ(t)τ(t)
ln(1−τ(t)σ)dσ
= u(π)eλτ(t)τ(t)
ln(1−τ (t)) + z0,
where z0 = eλτ(t)τ(t)
ln(1−τ(t)) ∫ 1
0h(σ)τ(t)1−τ(t)σ e
−λ τ(t)τ(t)
ln(1−τ (t)σ)dσ depends only on h.
It remains to find u. By (4.17), u must satisfy
λu− auxx = f.
Multiplying this identity by a test function φ, integrating in space and using integrationby parts, we obtain
∫ π
0
(λu− auxx)φdx =
∫ π
0
(λuφ+ auxφx)dx− aux(π)φ(π) + aux(0)φ(0).
But using the fact that (u, z)⊤ must belong to D(A(t)), we have
∫ π
0
(λu− auxx)φdx =
∫ π
0
(λuφ+ auxφx)dx+ aµ0u(π)φ(π) + aµ1z(1)φ(π).
Therefore∫ π
0
(λuφ+ auxφx)dx+ aµ0u(π)φ(π) + aµ1z(1)φ(π) =
∫ π
0
fφdx.
Using the above expression for z(1), we arrive at the problem
∫ π
0
(λuφ+ auxφx)dx+ a(µ0 +µ1e−λτ(t))u(π)φ(π) =
∫ π
0
fφdx− aµ1z0φ(π), ∀φ ∈ V (4.18)
164
if τ (t) = 0, or otherwise∫ π
0
(λuφ+ auxφx)dx+ a(µ0 + µ1eλ
τ(t)τ(t)
ln(1−τ(t)))u(π)φ(π) =
∫ π
0
fφdx− aµ1z0φ(π), ∀φ ∈ V.
(4.19)These problems have a unique solution u ∈ V by Lax-Milgram’s lemma, because the left-hand side of (4.18) or (4.19) is coercive on V .
If we consider φ ∈ D(0, π) ⊂ V , then u satisfies
λu− auxx = f in D′(0, π).
This directly implies that u ∈ H2(0, π) and then u ∈ V ∩H2(0, π). Coming back to (4.18)and by integrating by parts, we find, for τ (t) = 0,
a(
ux(π) + (µ0 + µ1e−λτ(t))u(π)
)
φ(π) = −aµ1z0φ(π),
and thenux(π) = −(µ0 + µ1e
−λτ(t))u(π) − µ1z0
= −µ0u(π) − µ1(e−λτ(t)u(π) + z0)
= −µ0u(π) − µ1z(1).
We find the same result if τ (t) 6= 0.In summary we have found (u, z)⊤ ∈ D(A(t)) satisfying (4.17) and thus λI −A(t) is
surjective for some λ > 0 and t > 0. Since κ(t) > 0, we directly deduce that
λI − A(t) = (λ+ κ(t))I −A(t) is surjective (4.20)
for some λ > 0 and t > 0.Then, (4.11), (4.15) and (4.20) imply that the family A = A(t) : t ∈ [0, T ] is a stable
family of generators in H with stability constants independent of t, by Proposition 1.1from [59]. Therefore, the assumptions (i)-(iv) of Theorem 4.2.1 are verified by (4.8), (4.11),(4.15), (4.16), (4.20) and Lemma 4.2.2, and thus, the problem
Ut = A(t)U
U(0) = U0.
has a unique solution U ∈ C([0, +∞), H) and U ∈ C([0, +∞), D(A(0)))∩C1([0, +∞), H)if U0 ∈ D(A(0)). Setting
U(t) = eβ(t)U(t)
with β(t) =∫ t
0κ(s)ds, we remark that it is a solution of (4.7) because
Ut(t) = κ(t)eβ(t)U(t) + eβ(t)Ut(t)
= κ(t)eβ(t)U(t) + eβ(t)A(t)U(t)
= eβ(t)(κ(t)U(t) + A(t)U(t))
= eβ(t)A(t)U(t) = A(t)eβ(t)U(t)= A(t)U(t),
which concludes the proof.
165
4.2.2 The decay of the energy
We here suppose thatµ2
1 < (1 − d)µ20. (4.21)
Let us choose the following energy
E(t) =1
2
∫ π
0
u2(x, t)dx+qτ(t)
2
∫ 1
0
u2(π, t− τ(t)ρ)dρ, (4.22)
where q is a positive constant chosen later.
Proposition 4.2.4. Let (4.9) and (4.21) be satisfied. Then for all regular solution ofproblem (4.1), the energy is decreasing and satisfies
E ′(t) ≤ −a∫ π
0
u2x(x, t)dx+ (u(π, t), u(π, t− τ(t))) Ψq (u(π, t), u(π, t− τ(t)))⊤ < 0,
(4.23)where Ψq is the matrix defined in (4.13).
Proof: Differentiating (4.22) and by (4.1), we obtain
E ′(t) =
∫ π
0
uutdx+qτ(t)
2
∫ 1
0
u2(π, t− τ(t)ρ)dρ
+qτ(t)
∫ 1
0
u(π, t− τ(t)ρ)ut(π, t− τ(t)ρ)(1 − τ(t)ρ)dρ
= a
∫ π
0
uuxxdx+qτ(t)
2
∫ 1
0
u2(π, t− τ(t)ρ)dρ
+qτ(t)
∫ 1
0
u(π, t− τ(t)ρ)ut(π, t− τ(t)ρ)(1 − τ(t)ρ)dρ.
By integrating by parts in space, we find
a
∫ π
0
uuxxdx = −a∫ π
0
u2xdx+ a[u(x, t)ux(x, t)]
π0
= −a∫ π
0
u2xdx− aµ0u
2(π, t) − aµ1u(π, t)u(π, t− τ(t)).
Setting z(ρ, t) = u(π, t − τ(t)ρ), we see that zρ(ρ, t) = −τ(t)ut(π, t − τ(t)ρ), and byintegrating by parts in ρ, we get
∫ 1
0u(π, t− τ(t)ρ)ut(π, t− τ(t)ρ)(1 − τ (t)ρ)dρ = − 1
τ(t)
∫ 1
0z(ρ, t)zρ(ρ, t)(1 − τ(t)ρ)dρ
= − 12τ(t)
∫ 1
0∂ρ(z(ρ, t)
2)(1 − τ(t)ρ)dρ
= 12τ(t)
∫ 1
0z(ρ, t)2(−τ (t))dρ
− 12τ(t)
[z2(ρ, t)(1 − τ (t)ρ)]10= − τ(t)
2τ(t)
∫ 1
0u2(π, t− τ(t)ρ)dρ
−1−τ(t)2τ(t)
u2(π, t− τ(t)) + 12τ(t)
u2(π, t).
166
Therefore, we obtain
E ′(t) = −a∫ π
0
u2xdx−(aµ0−
q
2)u2(π, t)−aµ1u(π, t)u(π, t−τ(t))−
q
2(1−τ (t))u2(π, t−τ(t)).
We can see that this inequality can be written as
E ′(t) ≤ −a∫ π
0
u2xdx+ (u(π, t), u(π, t− τ(t)))Ψq(u(π, t), u(π, t− τ(t)))⊤.
Since −q(1 − d) < 0, Ψq is negative definite if and only if
q2 − 2aµ0q +a2µ2
1
1 − d< 0. (4.24)
The discriminant of this second order polynomial is ∆ = 4a2(
µ20 − µ2
1
1−d
)
, which is positive
if and only if (4.21) holds. Therefore, the matrix Ψq is negative definite for some q > 0 ifand only if (4.21) is satisfied, and in that case, we choose q such that
aµ0 − a
√
µ20 −
µ21
1 − d< q < aµ0 + a
√
µ20 −
µ21
1 − d, (4.25)
which concludes the proof.
4.2.3 Exponential stability
In this section, we prove the exponential stability of the heat equation (4.1) by usingthe following Lyapunov functional
E(t) = E(t) + γE2(t), (4.26)
where γ > 0 is a parameter that will be fixed small enough later on, E is the standardenergy defined by (4.22) and E2 is defined by
E2(t) = q
∫ t
t−τ(t)e2δ(s−t)u2(π, s)ds = qτ(t)
∫ 1
0
e−2δτ(t)ρu2(π, t− τ(t)ρ)dρ, (4.27)
where δ > 0 is a fixed positive real number.
Remark 4.2.5. Let us notice that the energies E and E are equivalent, since
E(t) ≤ E(t) ≤ (2γ + 1)E(t).
The result about the decay of the energy E is the following one :
167
Theorem 4.2.6. Let (4.3), (4.9) and (4.21) be satisfied. Then the energy E decays expo-nentially, more precisely there exist two positive constants α and C such that
E(t) ≤ Ce−αtE(0), ∀t > 0.
Proof: First, we differentiate E2 to have
d
dtE2(t) =
τ(t)
τ(t)E2(t) + qτ(t)
∫ 1
0
(−2δτ (t)ρ)e−2δτ(t)ρu2(π, t− τ(t)ρ)dρ+ J,
where
J = 2qτ(t)
∫ 1
0
e−2δτ(t)ρu(π, t− τ(t)ρ)ut(π, t− τ(t)ρ)(1 − τ (t)ρ)dρ.
Moreover, by noticing one more time that z(ρ, t) = u(π, t− τ(t)ρ) and by integrating byparts in ρ, we have
J = −q∫ 1
0
e−2δτ(t)ρ ∂
∂ρ(z(ρ, t)2)(1 − τ (t)ρ)dρ
= q
∫ 1
0
z2(ρ, t)(−2δτ(t)(1 − τ (t)ρ) − τ (t))e−2δτ(t)ρdρ
−qe−2δτ(t)z2(1, t)(1 − τ (t)) + qz2(0, t)
= q
∫ 1
0
(−2δτ(t)(1 − τ(t)ρ) − τ(t))u2(π, t− τ(t)ρ)e−2δτ(t)ρdρ
−qe−2δτ(t)u2(π, t− τ(t))(1 − τ (t)) + qu2(π, t).
Therefore, we have
d
dtE2(t) =
τ(t)
τ(t)E2(t) + q
∫ 1
0
(−2δτ(t) − τ(t))u2(π, t− τ(t)ρ)e−2δτ(t)ρdρ
−qe−2δτ(t)u2(π, t− τ(t))(1 − τ(t)) + qu2(π, t)
= −2δE2(t) − qe−2δτ(t)u2(π, t− τ(t))(1 − τ(t)) + qu2(π, t).
Since τ(t) < 1 (see (4.2)), we obtain
d
dtE2(t) ≤ −2δE2(t) + qu2(π, t). (4.28)
Consequently, gathering (4.23), (4.26) and (4.28), we obtain
d
dtE(t) ≤ −(aµ0 −
q
2− qγ)u2(π, t) − q
2(1 − d)u2(π, t− τ(t))
−aµ1u(π, t)u(π, t− τ(t)) − a
∫ π
0
u2x(x, t)dx− 2γδE2(t),
or equivalently
d
dtE(t) ≤ (u(π, t), u(π, t− τ(t)))Ψq(u(π, t), u(π, t− τ(t)))⊤ − a
∫ π
0
u2x(x, t)dx− 2γδE2(t),
(4.29)
168
where Ψq is the 2 × 2 matrix defined by
Ψq =1
2
(
q(1 + 2γ) − 2aµ0 −aµ1
−aµ1 −q(1 − d)
)
= Ψq + γ
(
q 00 0
)
.
Now fix q > 0 such that Ψq is negative definite (consequence of the assumption (4.21)).By a perturbation argument, we deduce that for γ > 0 small enough, Ψq is negative. Moreprecisely, we take γ = −λ
q, when λ is the greatest negative eigenvalue of Ψq, or equivalently
XΨqX⊤ ≤ λ|X|2, for all X ∈ R2. We can easily check that
λ =1
4
(
−2aµ0 + dq +√
4a2(µ20 + µ2
1) + 4a(d− 2)µ0q + (d− 2)2q2
)
< 0. (4.30)
Therefore, for γ = −λq
, we find
d
dtE(t) ≤ −2δγE2(t) − a
∫ π
0
u2x(x, t)dx. (4.31)
Since u(0, t) = 0 for all t > 0, by the min-max principle, we have
∫ π
0
u2(x, t)dx ≤ 4
∫ π
0
u2x(x, t)dx,
because the first eigenvalue of the Laplace operator with Dirichlet boundary condition at0 and Neumann boundary condition at π is −1
4. Therefore
−a∫ π
0
u2x(x, t)dx ≤ −a
4
∫ π
0
u2(x, t)dx.
This estimate in (4.31) and by the definition (4.27) of E2, we obtain
d
dtE(t) ≤ −a
4
∫ π
0
u2(x, t)dx− 2qδγτ(t)
∫ 1
0
e−2δτ(t)ρu2(π, t− τ(t)ρ)dρ.
Since τ(t) ≤ M (see (4.3)) and in view of the definition (4.22) of E(t), there exists aconstant γ′ > 0 (depending on γ and δ, namely γ′ ≤ min(a
2, 4δγe−2δM )) such that
d
dtE(t) ≤ −γ′E(t).
By applying Remark 4.2.5, we obtain
d
dtE(t) ≤ − γ′
2γ + 1E(t).
This implies thatE(t) ≤ E(0)e−αt,
169
with
α =γ′
2γ + 1≤ 1
q − 2λmin
(aq
2, −4λδe−2δM
)
.
Remark 4.2.5 leads to
E(t) ≤ E(t) ≤ E(0)e−αt ≤ (2γ + 1)E(0)e−αt.
Remark 4.2.7. In the proof of Theorem 4.2.6, we notice that we have explicitely calculatedthe decay rate of the energy, given by
α =1
q − 2λmin
(aq
2, −4λδe−2δM
)
,
where λ is given by (4.30), q by (4.25) and δ is a positive real number. Therefore, we canchoose δ so that the decay of the energy is as quick as possible. For that purpose, we noticethat the function δ → −4λδe−2δM admits a maximum at δ = 1
2Mand that this maximum
is −2λMe
. Thus the larger decay rate of the energy is given by
αmax =1
q − 2λmin
(
aq
2,−2λ
Me
)
.
Obviously, this quantity can be calculated if the data µ0, µ1 and τ are given.
4.3 Exponential stability of the delayed wave equa-
tion
We now consider the system described by
utt(x, t) − auxx(x, t) = 0, 0 < x < π, t > 0,u(0, t) = 0, t > 0,
ux(π, t) = −µ0ut(π, t) − µ1ut(π, t− τ(t)), t > 0,u(x, 0) = u0(x), ut(x, 0) = u1(x), 0 < x < π,ut(π, t− τ(0)) = f 0(t− τ(0)), 0 < t < τ(0),
(4.32)
whith the constant parameter a > 0 and where µ0, µ1 ≥ 0 are fixed nonnegative realnumbers, the time-varying delay τ(t) still satisfies (4.2), (4.3) and (4.4).
The boundary-value problem (4.32) describes the oscillations of a homogeneous stringfixed at 0 and with a feedback law at π.
170
4.3.1 Well posedness of the problem
We aim to show that problem (4.32) is well-posed. For that purpose, we use the sameideas than before.
We transform our system (4.32) as follows. Let us introduce the auxiliary variablez(ρ, t) = ut(π, t − τ(t)ρ) for ρ ∈ (0, 1) and t > 0. Note that z verifies the transportequation for 0 < ρ < 1 and t > 0 (compare with (4.5))
τ(t)zt(ρ, t) + (1 − τ (t)ρ)zρ(ρ, t) = 0,z(0, t) = ut(π, t),
z(ρ, 0) = f 0(−τ(0)ρ).(4.33)
Therefore, the problem (4.32) is equivalent to
utt(x, t) − auxx(x, t) = 0, 0 < x < π, t > 0,τ(t)zt(ρ, t) + (1 − τ(t)ρ)zρ(ρ, t) = 0, 0 < ρ < 1, t > 0,
u(0, t) = 0, ux(π, t) = −µ0ut(π, t) − µ1z(1, t), t > 0,z(0, t) = ut(π, t), t > 0,
u(x, 0) = u0(x), ut(x, 0) = u1(x), 0 < x < π,z(ρ, 0) = f 0(−τ(0)ρ), 0 < ρ < 1.
(4.34)
If we introduceU := (u, ut, z)
⊤,
then U satisfies
Ut = (ut, utt, zt)⊤ =
(
ut, auxx,τ(t)ρ− 1
τ(t)zρ
)⊤.
Consequently the problem (4.32) may be rewritten as the first order evolution equation
Ut = A(t)UU(0) = (u0, u1, f 0(−τ(0).))⊤ = U0,
(4.35)
where the time dependent operator A(t) is defined by
A(t)
uωz
=
ωauxx
τ(t)ρ−1τ(t)
zρ
,
with domain
D(A(t)) := (u, ω, z) ∈ (V ∩H2(0, π)) × V ×H1(0, 1) :z(0) = ω(π), ux(π) = −µ0ω(π) − µ1z(1),
where we recall thatV = φ ∈ H1(0, π) : φ(0) = 0.
171
Again, we notice that the domain of the operator A(t) is independent of the time t, i.e.
D(A(t)) = D(A(0)), ∀t > 0. (4.36)
Now, we introduce the Hilbert space
H = V × L2(0, π) × L2(0, 1)
equipped with the usual inner product
⟨
uωz
,
uωz
⟩
=
∫ π
0
(auxux + ωω)dx+
∫ 1
0
z(ρ)z(ρ)dρ.
Lemma 4.3.1. D(A(0)) is dense in H.
Proof: The proof is the same as the one of Lemma 1.2.1, we give it for the sake ofcompleteness. Let (f, g, h)⊤ ∈ H be orthogonal to all elements of D(A(0)), namely
0 =
⟨
uωz
,
fgh
⟩
=
∫ π
0
(auxfx + ωg)dx+
∫ 1
0
z(ρ)h(ρ)dρ,
for all (u, ω, z)⊤ ∈ D(A(0)).We first take u = 0 and w = 0 and z ∈ D(0, 1). As (0, 0, z) ∈ D(A(0)), we get
∫ 1
0
z(ρ)h(ρ)dρ = 0.
Since D(0, 1) is dense in L2(0, 1), we deduce that h = 0.In the same manner, by taking u = 0, z = 0 and ω ∈ D(0, π) we see that g = 0.The above orthogonality condition is then reduced to
0 = a
∫ π
0
uxfxdx, ∀(u, ω, z) ∈ D(A(0)).
By restricting ourselves to ω = 0 and z = 0, we obtain
∫ 1
0
uxfxdx = 0, ∀(u, 0, 0) ∈ D(A(0)).
But we easily check that (u, 0, 0) ∈ D(A(0)) if and only if u ∈ D(∆) = v ∈ H2(0, π) :v(0) = 0, v′(1) = 0, the domain of the Laplace operator with mixed boundary conditions.Since it is well known that D(∆) is dense in V (equipped with the inner product < ., . >V ),we conclude that f = 0.
172
As before we suppose that the speed of the delay satisfies (4.9) and (4.10). Under theseconditions, we will show that the operator A(t) generates a C0-semigroup in H and theunique solvability of problem (4.35).
For that purpose, we introduce the following time-dependent inner product on H
⟨
uωz
,
uωz
⟩
t
=
∫ π
0
(auxux + ωω)dx+ qτ(t)
∫ 1
0
z(ρ)z(ρ)dρ,
where q is a positive constant chosen such that Ψq is negative (guaranteed by the assump-tions (4.9) and (4.10)), with associated norm denoted by ‖.‖t .
Theorem 4.3.2. For an initial datum U0 ∈ H, there exists a unique solution U ∈C([0, +∞), H) to problem (4.35). Moreover, if U0 ∈ D(A(0)), then
U ∈ C([0, +∞), D(A(0))) ∩ C1([0, +∞), H).
Proof: We first notice that
‖φ‖t‖φ‖s
≤ ec
2τ0|t−s|
, ∀t, s ∈ [0, T ] (4.37)
where φ = (u, ω, z)⊤ and c is a positive constant. Indeed, for all s, t ∈ [0, T ], we have
‖φ‖2t − ‖φ‖2
s ec
τ0|t−s|
=(
1 − ec
τ0|t−s|)
∫ π
0
(au2x + ω2)dx+ q
(
τ(t) − τ(s)ec
τ0|t−s|)
∫ 1
0
z(ρ)2dρ,
and we conclude as in the proof of Theorem 4.2.3.Now we calculate 〈A(t)U,U〉t for a t > 0 fixed. For an arbitrary U = (u, ω, z)⊤ ∈
D(A(t)), we have
〈A(t)U, U〉t =
⟨
ωauxx
τ(t)ρ−1τ(t)
zρ
,
uωz
⟩
t
=
∫ π
0
(aωxux + auxxω)dx− q
∫ 1
0
zρ(ρ)z(ρ)(1 − τ (t)ρ)dρ.
By integrating by parts in space, we have
〈A(t)U, U〉t = a[ωux]π0 − q
∫ 1
0
zρ(ρ)z(ρ)(1 − τ(t)ρ)dρ
= −aµ0z(0)2 − aµ1z(0)z(1) − q
∫ 1
0
zρ(ρ)z(ρ)(1 − τ (t)ρ)dρ.
173
Moreover, we have by integrating by parts in ρ :
∫ 1
0
zρ(ρ)z(ρ)(1 − τ (t)ρ)dρ =
∫ 1
0
1
2
∂
∂ρ(z(ρ)2)(1 − τ (t)ρ)dρ
=τ (t)
2
∫ 1
0
z(ρ)2dρ+1
2z(1)2(1 − τ(t)) − 1
2z2(0).
These two identities yield
〈A(t)U, U〉t = −aµ0z(0)2 − aµ1z(0)z(1) − q
2z(1)2(1 − τ (t)) +
q
2z2(0) − qτ(t)
2
∫ 1
0
z(ρ)2dρ.
We can see, that this identity implies that
〈A(t)U, U〉t ≤ (z(0), z(1))Ψq(z(0), z(1))⊤ + κ(t) 〈U, U〉t ,
where Ψq is the matrix defined by (4.13) and κ(t) is given by (4.12). Since we have chosenq such that the matrix Ψq is negative, we have
〈A(t)U, U〉t − κ(t) 〈U, U〉t ≤ 0, (4.38)
which proves the dissipativeness of A(t) = A(t) − κ(t)I for the inner product 〈·, ·〉t.As in the proof of Theorem 4.2.3, we see that (4.4) implies that
d
dtA(t) ∈ L∞
∗ ([0, T ], B(D(A(0)), H)). (4.39)
Let us finally prove that A(t) is maximal, i.e., that λI − A(t) is surjective for someλ > 0 and t > 0.
Let (f, g, h)T ∈ H. We look for U = (u, ω, z)T ∈ D(A(t)) solution of
(λI −A(t))
uωz
=
fgh
or equivalently
λu− ω = fλω − auxx = g
λz + 1−τ(t)ρτ(t)
zρ = h.(4.40)
Suppose that we have found u with the appropriate regularity. Then, we have
ω = −f + λu ∈ V.
We can then determine z, indeed z satisfies the differential equation
λz +1 − τ (t)ρ
τ(t)zρ = h
174
and the boundary condition z(0) = ω(π) = −f(π) + λu(π). Therefore z is explicitly givenby
z(ρ) = λu(π)e−λτ(t)ρ − f(π)e−λτ(t)ρ + τ(t)e−λτ(t)ρ∫ ρ
0
eλτ(t)σh(σ)dσ,
if τ (t) = 0, and
z(ρ) = λu(π)eλ τ(t)
τ(t)ln(1−τ (t)ρ) − f(π)e
λ τ(t)τ(t)
ln(1−τ (t)ρ)
+eλτ(t)τ(t)
ln(1−τ(t)ρ)∫ ρ
0
h(σ)τ(t)
1 − τ(t)σe−λ
τ(t)τ(t)
ln(1−τ(t)σ)dσ,
otherwise. This means that once u is found with the appropriate properties, we can find zand ω. In particular, we have if τ (t) = 0,
z(1) = λu(π)e−λτ(t) − f(π)e−λτ(t) + τ(t)e−λτ(t)∫ 1
0
eλτ(t)σh(σ)dσ = λu(π)e−λτ(t) + z0,
where z0 = −f(π)e−λτ(t) + τ(t)e−λτ(t)∫ 1
0eλτ(t)σh(σ)dσ is a fixed real number depending
only on f and h, and if τ (t) 6= 0
z(1) = λu(π)eλ τ(t)
τ(t)ln(1−τ(t)) − f(π)e
λ τ(t)τ(t)
ln(1−τ(t))
+eλτ(t)τ(t)
ln(1−τ (t))∫ 1
0
h(σ)τ(t)
1 − τ(t)σe−λ
τ(t)τ(t)
ln(1−τ (t)σ)dσ
= λu(π)eλτ(t)τ(t)
ln(1−τ(t)) + z0,
where
z0 = −f(π)eλτ(t)τ(t)
ln(1−τ(t)) + eλτ(t)τ(t)
ln(1−τ(t))∫ 1
0
h(σ)τ(t)
1 − τ (t)σe−λ
τ(t)τ(t)
ln(1−τ (t)σ)dσ
depends only on f and h.It remains to find u. By (4.40), u must satisfy
λ2u− auxx = g + λf.
Multiplying this identity by a test function φ, integrating in space and using integrationby parts, we obtain
∫ π
0
(λ2u− auxx)φdx =
∫ π
0
(λ2uφ+ auxφx)dx− aux(π)φ(π) + aux(0)φ(0).
But using the fact that (u, ω, z)⊤ must belong to D(A(t)), we have
∫ π
0
(λ2u− auxx)φdx =
∫ π
0
(λ2uφ+ auxφx)dx+ aµ0ω(π)φ(π) + aµ1z(1)φ(π).
175
Therefore∫ π
0
(λ2uφ+ auxφx)dx+ aµ0ω(π)φ(π) + aµ1z(1)φ(π) =
∫ π
0
(g + λf)φdx.
Using the above expression for z(1) and ω = λu− f , we arrive at the problem
∫ π
0
(λ2uφ+auxφx)dx+a(µ0+µ1e−λτ(t))λu(π)φ(π) =
∫ π
0
(g+λf)φdx+a(µ0f(π)−µ1z0)φ(π), ∀φ ∈ V
(4.41)if τ (t) = 0, or otherwise
∫ π
0
(λ2uφ+ auxφx)dx+ a(µ0 + µ1eλ
τ(t)τ(t)
ln(1−τ (t)ρ))λu(π)φ(π)
=
∫ π
0
(g + λf)φdx+ a(µ0f(π) − µ1z0)φ(π),∀φ ∈ V.
(4.42)
These problems have a unique solution u ∈ V by Lax-Milgram’s lemma, because the left-hand side of (4.41) or (4.42) is coercive on V .
If we consider φ ∈ D(0, π) ⊂ V , then u satisfies
λ2u− auxx = g + λf in D′(0, π).
This directly implies that u ∈ H2(0, π) and then u ∈ V ∩H2(0, π). Coming back to (4.41)and by integrating by parts, we find, for τ (t) = 0,
a(
ux(π) + (µ0 + µ1e−λτ(t))λu(π)
)
φ(π) = a(µ0f(π) − µ1z0)φ(π),
and thenux(π) = −(µ0 + µ1e
−λτ(t))λu(π) − (µ1z0 − µ0f(π))
= −µ0(λu(π) − f(π)) − µ1(e−λτ(t)λu(π) + z0)
= −µ0ω(π) − µ1z(1).
We find the same result if τ (t) 6= 0.In summary we have found (u, ω, z)⊤ ∈ D(A(t)) satisfying (4.40) and thus λI −A(t)
is surjective for some λ > 0 and t > 0. Again as κ(t) > 0, this proves that
λI − A(t) = (λ+ κ(t))I −A(t) is surjective (4.43)
for some λ > 0 and t > 0.Then, (4.37), (4.38) and (4.43) imply that the family A = A(t) : t ∈ [0, T ] is a
stable family of generators in H with stability constants independent of t, by Proposition1.1 from [59]. Therefore, the assumptions (i)-(iv) of Theorem 4.2.1 are verified by (4.36),(4.37), (4.38), (4.39), (4.43) and Lemma 4.3.1, and thus, the problem
Ut = A(t)U
U(0) = U0.
176
has a unique solution U ∈ C([0, +∞), H) and U ∈ C([0, +∞), D(A(0)))∩C1([0, +∞), H)if U0 ∈ D(A(0)). As before, the requested solution of (4.35) is given by
U(t) = eβ(t)U(t)
with β(t) =∫ t
0κ(s)ds.
4.3.2 The decay of the energy
As for the heat equation, we restrict the hypothesis (4.10) to obtain the decay of theenergy. Namely we suppose that (4.21) holds.
Let us choose the following energy (which corresponds to the time-dependent innerproduct in H)
E(t) =1
2
∫ π
0
(
u2t (x, t) + au2
x(x, t))
dx+qτ(t)
2
∫ 1
0
u2t (π, t− τ(t)ρ)dρ, (4.44)
where q is a positive constant chosen such that Ψq is negative definite (possible if (4.9) and(4.21) hold).
Proposition 4.3.3. Let the assumptions (4.9) and (4.21) be satisfied. Then for all regularsolution of problem (4.32), the energy is decreasing and verifies
E ′(t) ≤ (ut(π, t), ut(π, t− τ(t))) Ψq (ut(π, t), ut(π, t− τ(t)))⊤ < 0, (4.45)
where Ψq is the matrix defined in (4.13).
Remark 4.3.4. In the case where the delay is constant in time, i.e. τ(t) = τ for all t > 0and thus d = 0, we recover the results from [85, 89] and chapter 1. Indeed in [85, 89], theenergy is decreasing under the assumption µ1 < µ0.
Proof: Differentiating (4.44) and integrating by parts in space, we obtain
E ′(t) =
∫ π
0
(ututt + auxuxt)dx+qτ (t)
2
∫ 1
0
u2t (π, t− τ(t)ρ)dρ
+qτ(t)
∫ 1
0
ut(π, t− τ(t)ρ)utt(π, t− τ(t)ρ)(1 − τ(t)ρ)dρ
=
∫ π
0
ut(utt − auxx)dx+ a[uxut]π0 +
qτ(t)
2
∫ 1
0
u2t (π, t− τ(t)ρ)dρ
+qτ(t)
∫ 1
0
ut(π, t− τ(t)ρ)utt(π, t− τ(t)ρ)(1 − τ(t)ρ)dρ
= aux(π, t)ut(π, t) +qτ(t)
2
∫ 1
0
u2t (π, t− τ(t)ρ)dρ
+qτ(t)
∫ 1
0
ut(π, t− τ(t)ρ)utt(π, t− τ(t)ρ)(1 − τ(t)ρ)dρ.
177
Recalling that z(ρ, t) = ut(π, t− τ(t)ρ), we see that
∫ 1
0
ut(π, t− τ(t)ρ)utt(π, t− τ(t)ρ)(1 − τ(t)ρ)dρ = − τ (t)
2τ(t)
∫ 1
0
u2t (π, t− τ(t)ρ)dρ
−1 − τ (t)
2τ(t)u2t (π, t− τ(t))
+1
2τ(t)u2t (π, t).
Therefore, we obtain
E ′(t) = aux(π, t)ut(π, t) +qτ(t)
2
∫ 1
0
u2t (π, t− τ(t)ρ)dρ
−qτ (t)2
∫ 1
0
u2t (π, t− τ(t)ρ)dρ− q
1 − τ(t)
2u2t (π, t− τ(t)) +
q
2u2t (π, t),
which implies
E ′(t) = −aµ0u2t (π, t) − aµ1ut(π, t− τ(t))ut(π, t) −
q
2(1 − τ (t))u2
t (π, t− τ(t)) +q
2u2t (π, t).
By the condition (4.9) we can see that this identity yields
E ′(t) ≤ (ut(π, t), ut(π, t− τ(t))) Ψq (ut(π, t), ut(π, t− τ(t)))⊤ .
This concludes the proof as Ψq is negative definite.
4.3.3 Exponential stability
In this section, under the assumptions (4.9) and (4.21), we prove the exponential sta-bility of the wave equation (4.32) by using the following Lyapunov functional
E(t) = E(t) + γ
(
2
∫ π
0
xutuxdx+ E2(t)
)
, (4.46)
where γ > 0 is a parameter that will be fixed small enough later on, E is the standardenergy defined by (4.44) with q a positive constant fixed such that Ψq is negative definiteand E2 is defined by
E2(t) = q
∫ t
t−τ(t)e2δ(s−t)u2
t (π, s)ds = qτ(t)
∫ 1
0
e−2δτ(t)ρu2t (π, t− τ(t)ρ)dρ, (4.47)
where δ > 0 is a fixed positive real number.The Lyapunov functional E(t) + 2γ
∫ π
0xutuxdx is standard in problems with boundary
conditions with memory (see for instance [86]). We have added the two terms to thestandard energy E(t) in order to take into account the dependence of τ with respect to t.
First we notice that the energies E and E are equivalent.
178
Lemma 4.3.5. For γ small enough, there exist two positive constants C1(γ) and C2(γ)such that
C1(γ)E(t) ≤ E(t) ≤ C2(γ)E(t).
Proof: We have
2γ
∫ π
0
xuxutdx ≤ γπ
∫ π
0
(u2x + u2
t )dx ≤ γπc
∫ π
0
(au2x + u2
t )dx,
where c = max(1, 1a) and
γqτ(t)
∫ 1
0
e−2δτ(t)ρu2t (π, t− τ(t)ρ)dρ ≤ γqτ(t)
∫ 1
0
u2t (π, t− τ(t)ρ)dρ.
Since c ≥ 1, these estimates yield
E(t) ≤ (1 + 2γcπ)E(t).
Moreover, by definition we have
γE2(t) ≥ 0
and by Cauchy-Schwarz’s inequality
2γ
∫ π
0
xuxutdx ≥ −γπ∫ π
0
(u2x + u2
t )dx.
Then
E(t) ≥ E(t) − cγπ
∫ π
0
(au2x + u2
t )dx,
and therefore, for γ small enough (γ < 12cπ
), we obtain
E(t) ≥ (1 − 2cπγ)E(t).
We are ready to state our result about the decay of the energy E :
Theorem 4.3.6. Let (4.3), (4.9) and (4.21) be satisfied. Then the energy E decays expo-nentially, more precisely there exist two positive constants α and C such that
E(t) ≤ Ce−αtE(0), ∀t > 0.
Proof: First we remark that
d
dt
(
2
∫ π
0
xutuxdx
)
= 2
∫ π
0
xuttuxdx+ 2
∫ π
0
xutuxtdx
= 2a
∫ π
0
xuxxuxdx+ 2
∫ π
0
xutuxtdx
= a
∫ π
0
x∂x(u2x)dx+ 2
∫ π
0
xutuxtdx
= −a∫ π
0
u2xdx+ aπu2
x(π, t) + 2
∫ π
0
xutuxtdx.
179
But by integrating by parts in space and by (4.32), we have∫ π
0
xutuxtdx = −∫ π
0
xuxtutdx−∫ π
0
u2tdx+ πu2
t (π, t),
that is to say
2
∫ π
0
xutuxtdx = −∫ π
0
u2tdx+ πu2
t (π, t).
Thusd
dt
(
2
∫ π
0
xutuxdx
)
= −∫ π
0
(u2t + au2
x)dx+ πu2t (π, t) + aπu2
x(π, t).
By the boundary conditions in (4.32) and Cauchy-Schwarz’s inequality, we finally find
d
dt
(
2
∫ π
0
xutuxdx
)
≤ −∫ π
0
(u2t + au2
x)dx+ π(1 + 2aµ20)u
2t (π, t) + 2aπµ2
1u2t (π, t− τ(t)).
(4.48)Then, we differentiate E2 to have
d
dtE2(t) =
τ (t)
τ(t)E2(t) + qτ(t)
∫ 1
0
(−2δτ (t)ρ)e−2δτ(t)ρu2t (π, t− τ(t)ρ)dρ+ Jw,
where
Jw = 2qτ(t)
∫ 1
0
e−2δτ(t)ρut(π, t− τ(t)ρ)utt(π, t− τ(t)ρ)(1 − τ(t)ρ)dρ.
As in the proof of Theorem 4.2.6, by integration by parts, we have
Jw = q
∫ 1
0
(−2δτ(t)(1 − τ(t)ρ) − τ (t))ut(π, t− τ(t)ρ)e−2δτ(t)ρdρ
−qe−2δτ(t)u2t (π, t− τ(t))(1 − τ(t)) + qu2
t (π, t).
These identities yield
d
dtE2(t) =
τ (t)
τ(t)E2(t) + q
∫ 1
0
(−2δτ(t) − τ (t))u2t (π, t− τ(t)ρ)e−2δτ(t)ρdρ
−qe−2δτ(t)u2t (π, t− τ(t))(1 − τ (t)) + qu2
t (π, t)
=τ (t)
τ(t)E2(t) +
−2δτ(t) − τ(t)
τ(t)E2(t) − qe−2δτ(t)u2
t (π, t− τ(t))(1 − τ(t)) + qu2t (π, t).
Since τ(t) < 1 (see (4.2)), we obtain
d
dtE2(t) ≤ −2δE2(t) + qu2
t (π, t). (4.49)
Consequently, gathering (4.46), (4.48) and (4.49), we obtain
d
dtE(t) ≤ −γ
∫ π
0
(u2t + au2
x)dx− 2γδE2(t)
+(ut(π, t), ut(π, t− τ(t))) Φq (ut(π, t), ut(π, t− τ(t)))⊤ ,
180
where Φq is the matrix defined by
Φq = 12
(
q(1 + 2γ) − 2aµ0 + 2γπ(1 + 2aµ20) −aµ1
−aµ1 4aγπµ21 − q(1 − d)
)
= Ψq + γ
(
q + π(1 + 2aµ20) 0
0 2aπµ21
)
.
Noticing that max (q + π(1 + 2aµ20), 2aπµ2
1) = q+π(1+2aµ20) by (4.21), for γ sufficiently
small, i.e. γ ≤ −λq+π(1+2aµ2
0), where λ is the greatest negative eigenvalue of Ψq given by (4.30),
Φq is negative and therefore
d
dtE(t) ≤ −γ
∫ π
0
(u2t + au2
x)dx− 2δγE2(t).
By the definition (4.47) of E2, this estimate becomes
d
dtE(t) ≤ −γ
∫ π
0
(u2t + au2
x)dx− 2δγqτ(t)e−2δτ(t)
∫ 1
0
u2t (π, t− τ(t)ρ)dρ.
Since τ(t) ≤ M (see (4.3)), in view of the definition of E, there exists a constant γ′ > 0(depending on γ and δ : γ′ ≤ 2γmin
(
1, 2δe−2δM)
) such that
d
dtE(t) ≤ −γ′E(t).
By applying Lemma 4.3.5, we arrive at
d
dtE(t) ≤ −αE(t),
where α is explicitely given by α = γ′
1+2γcπ. Therefore
E(t) ≤ E(0)e−αt,
and Lemma 4.3.5 allows to conclude the proof :
E(t) ≤ 1
1 − 2cπγE(t) ≤ 1
1 − 2cπγE(0)e−αt ≤ 1 + 2γcπ
1 − 2cπγE(0)e−αt.
Remark 4.3.7. In the case where the delay is constant in time and a = 1, we recoversome results from [85,89] and chapter 1. Moreover, in [85,89], the energy is defined by
E(t) =1
2
∫ π
0
(u2t (x, t) + u2
x(x, t))dx+ξ
2
∫ 1
0
u2t (π, t− τρ)dρ,
where ξ is a positive constant satisfying
τµ1 ≤ ξ ≤ τ(2µ0 − µ1),
under the condition (4.21), which corresponds to the definition (4.44) of E with q = ξτ.
181
Remark 4.3.8. In the proof of Theorem 4.3.6, we notice that we can explicitely calculatethe decay rate α of the energy, given by
α =2γ
1 + 2γcπmin
(
1, 2δe−2δM)
,
with
γ <1
2cπand γ ≤ −λ
q + π(1 + 2aµ20)
(by Lemma 4.3.5 and Theorem 4.3.6) and c = max(
1, 1a
)
where λ is given by (4.30), q by(4.25) and δ is a positive real number. Therefore, we can choose δ such that the decay ofthe energy is as quick as possible. By Remark 4.2.7, we get that the larger decay rate of theenergy is given by
αmax =2γ
1 + 2γcπmin
(
1,1
Me
)
.
182
Chapitre 5
Quasi exponential decay of a finitedifference space discretization of the1-d wave equation by pointwiseinterior stabilization
5.1 Introduction
Recently boundary (or internal) controllability and stability results for the wave equa-tion have been obtained using different methods, like the multipliers method, the frequencydomain method, the microlocal analysis, the differential geometry or a combination ofthem [61, 70, 114, 115]. Usually these results are based on an observability estimate. Onthe other hand, similar results at a discrete level, namely when the continuous operatoris replaced by a discrete one (obtained by a finite difference, a finite element or a finitevolume scheme, for instance) are less developed. In their pioneer works, R. Glowinski etal. in [47–50] considered the boundary controllability of the numerical approximation ofthe wave equation using the so-called HUM method. They observed a bad behavior of thecontrol of the numerical scheme, due to the spurious modes that the numerical schemeintroduces at high frequencies. The main idea is that, in general, any discrete schemegenerates spurious high frequency oscillations that do not exist at the continuous level.Therefore the discretization and the control processes do not commute ; in other words,the control of the discrete model do not converge to the control of the continuous model.Some remedies are then necessary to restore the convergence of the discrete control to thecontinuous one. In these papers different remedies were proposed, like a Tychonoff regu-larization [49, 50], a bi-grid algorithm [47] or mixed finite element method [48], providinggood numerical results. For the wave equation, rigorous convergence analyses of these curesfor some schemes, like finite differences (in space and in time), finite elements and mixedfinite elements, are made in [30,31,56,78,80,102,112,114,115]. For internal stability resultsthis phenomenon was underlined numerically in [14], where numerical simulations suggest
183
that the exponential decay of the discrete energy might be not uniformly exponential withrespect to the mesh and step sizes. As remedy the authors propose without any proof touse a mixed method. More recently the stabilization of a locally damped wave equationin [102] and for a boundary damping in [103] were rigourously analyzed by adding a nu-merical viscosity term. But to our knowledge such results are not yet proved for pointwiseinternal damping.
Consequently the main goal of our chapter can be summarized as follows :i) As for the continuous model, we furnish a necessary and sufficient condition that gua-rantees the decay to zero of the discrete energy.ii) We show that the exponential decay of the discrete energy of the discretization of thewave equation with a pointwise internal damping by finite difference is not uniform.iii) We show that a filtering technique allows to restore a quasi-optimal uniform exponen-tial decay.iv) We prove that without filtering we cannot expect quasi-exponential decay.
The schedule of the chapter is as follows : Section 5.2 is devoted to the introduction ofthe one-dimensional wave equation with an internal pointwise damping and its discretiza-tion. Then in section 5.3 we give a necessary and sufficient condition that guarantees thedecay to zero of the discrete energy. The non uniform exponential decay of this energy isproved in section 5.4. Finally the filtering method is described in section 5.5 and the quasi-optimal uniform exponential decay is established. Some numerical results are presentedthat show the advantage of the filtering technique.
5.2 The continuous and discrete problems
We consider the wave equation on an interval of length 1 with an interior damping atξ, in other words we study the solution y of the following problem
ytt(x, t) − yxx(x, t) = 0 0 < x < 1, t > 0,y(0, t) = 0, yx(1, t) = 0 t > 0,y(ξ−, t) = y(ξ+, t) t > 0,yx(ξ−, t) − yx(ξ+, t) = −αyt(ξ, t) t > 0,y(t = 0) = y(0), yt(t = 0) = y(1) 0 < x < 1,
(5.1)
where (y(0), y(1)) ∈ V × L2(0, 1), V = y ∈ H1(0, 1) ; y(0) = 0 and α is a fixed positiveconstant. It is easy to check [6] that this system (5.1) is well-posed in the energy spaceV ×L2(0, 1).More precisely, for any (y(0), y(1)) ∈ V ×L2(0, 1) there exists a unique solutiony ∈ C((0, T ), V ) ∩ C1((0, T ), L2(0, 1)) of (5.1).
The energy of the solution of system (5.1) is given by
E(t) =1
2
∫ 1
0
(|yt(x, t)|2 + |yx(x, t)|2)dx
and obeys the following dissipation law
dE(t)
dt= −α |yt(ξ, t)|2 . (5.2)
184
This implies that the energy is decreasing. Moreover, we know that (see [6] and chapter 1)limt→∞
E(t) = 0 for any initial data in V × L2(0, 1) if and only if
ξ 6= 2p
2q + 1, ∀p, q ∈ N.
Moreover, as it was proved in [6] and in chapter 1, the exponential decay property of thesolution of (5.1) is equivalent to an observability estimate for the corresponding conservativesystem
utt(x, t) − uxx(x, t) = 0 0 < x < 1, t > 0,u(0, t) = 0, ux(1, t) = 0 t > 0,u(t = 0) = y(0), ut(t = 0) = y(1) 0 < x < 1,
(5.3)
where (y(0), y(1)) ∈ V × L2(0, 1). In this case, the observability estimate holds if and onlyif ξ is a rational number with an irreducible fraction
ξ =p
q, where p is odd,
and therefore under this condition, the system (5.1) is exponentially stable in the energyspace. In the remainder of this chapter, ξ is fixed and is supposed to satisfy these twoconditions.
We now introduce the finite difference scheme used later on. For this purpose, let N bea nonnegative integer. Set h = 1
N+1and consider the subdivision of (0, 1) given by
0 = x0 < x1 = h < ... < xj−1 < xj = jh < xj+1 < ... < xN < xN+1 = 1,
i.e. xj = jh for all j = 0, ..., N + 1. As ξ is not necessarily equal to jh for some j, we fixjN ∈ N ∩ (0, N + 1) such that xjN → ξ when N → ∞.
The finite-difference space semi-discretization of system (5.1) that we consider is thefollowing
y′′j − yj+1−2yj+yj−1
h2 = 0 t > 0, j = 1, ..., N, j 6= jN ,y0 = 0, yN+1 − yN = 0 t > 0,yjN +1−2yjN
+yjN−1
h= αy′jN t > 0,
yj(t = 0) = y(0)j , y′j(t = 0) = y
(1)j j = 1, ..., N.
(5.4)
Obviously, yj(t) is an approximation of y(xj, t), y being the solution of (5.1), provided
the initial data (y(0)j , y
(1)j ), j = 0, ..., N + 1 are approximations of the initial data in (5.1).
Note further that the conditionyjN +1−2yjN
+yjN−1
h= αy′jN is a natural approximation of the
condition yx(ξ−, t) − yx(ξ+, t) = −αyt(ξ, t). Alternatively (see [24]) we could choose theapproximate condition
hy′′jN − yjN+1 − 2yjN + yjN−1
h= −αy′jN ,
185
but this choice did not allow us to prove any positive results.We now set yh = (yj)j , y
(0)h = (y
(0)j )j and y
(1)h = (y
(1)j )j .
The system (5.4) may be equivalently written
v′h = Ahvhvh(0) = vh0
, (5.5)
where vh =
(
yhy′h
)
, vh0 =
(
y(0)h
y(1)h
)
and Ah is a 2N ×2N matrix. This system is therefore
uniquely solvable (as a differential system). Let us now show that it is dissipative : introduceits energy given by
Eh(t) =h
2
N∑
j=0, j 6=jN
∣
∣y′j(t)∣
∣
2+h
2
N∑
j=0
∣
∣
∣
∣
yj+1(t) − yj(t)
h
∣
∣
∣
∣
2
, (5.6)
which is a discretization of the continuous energy E.
Lemma 5.2.1. The energy of system (5.4) is a nonincreasing function of the time t andits derivative is given by
E ′h(t) = −α(y′jN (t))2. (5.7)
Proof: Deriving (5.6) and using (5.4), we obtain
E ′h(t) = h
N∑
j=1, j 6=jNy′j(t)
(
yj+1(t) − 2yj(t) + yj−1(t)
h2
)
+h
N∑
j=0
(
yj+1(t) − yj(t)
h
)(
y′j+1(t) − y′j(t)
h
)
=1
h
N∑
j=1, j 6=jNy′j(yj+1 − yj) −
1
h
N∑
j=1, j 6=jN(yj − yj−1)y
′j +
1
h
N∑
j=0
(yj+1 − yj)(y′j+1 − y′j)
=1
h
N∑
j=0, j 6=jNy′j(yj+1 − yj) −
1
h
N∑
j=0, j 6=jN−1
(yj+1 − yj)y′j+1 +
1
h
N∑
j=0
(yj+1 − yj)(y′j+1 − y′j)
=1
hy′jN (yjN − yjN−1) −
1
hy′jN (yjN+1 − yjN )
= −y′jN (t)
(
yjN+1(t) − 2yjN (t) + yjN−1(t)
h
)
= −α(y′jN (t))2 ≤ 0 by (5.4).
This proves (5.7) and, therefore, the energy is nonincreasing.
Observe that (5.7) is the semi-discrete analogue of the dissipation law (5.2).
186
We also need to consider the finite-difference space semi-discretization of problem wi-thout damping (5.3)
u′′j − uj+1−2uj+uj−1
h2 = 0 t > 0, j = 1, ..., Nu0 = 0, uN+1 − uN = 0 t > 0
uj(t = 0) = y(0)j , u′j(t = 0) = y
(1)j j = 1, ..., N.
(5.8)
We observe that the system (5.8) can be rewritten in the following simplified form
U ′′h + AhUh = 0, t > 0
where Uh stands for the column vector (u1, ..., uN)T , Ah denotes the matrix
Ah =1
h2
2 −1 0 · · · · · · 0−1 2 −1 0 · · · 0
0. . .
. . .. . . . .
... . . . . .0 . . . 2 −10 · · · · · · · · · −1 1
entering in the finite difference discretization of the Laplacian with Dirichlet boundarycondition at the left endpoint, and Neumann one at the right endpoint. The eigenvectorsof the matrix Ah satisfy the eigenvalue system
−ϕj+1−2ϕj+ϕj−1
h2 = λϕj j = 1, ..., N,ϕ0 = 0, ϕN+1 − ϕN = 0.
(5.9)
It was shown in [58] (see also [103]) that
ϕk, hj = sin(
(2k+1)πjh2−h
)
, j = 0, 1, ..., N,
λk, h = 4h2 sin2
(
(2k+1)πh2(2−h)
)
, k = 0, 1, ..., N − 1.
(5.10)
We have the following property of the eigenvectors of (5.9) :
Lemma 5.2.2. For any eigenvector ϕ with eigenvalue λ of (5.9), the following identityholds
N∑
j=0
∣
∣
∣
∣
ϕj+1 − ϕjh
∣
∣
∣
∣
2
= λ
N∑
j=1
|ϕj|2 .
187
Proof: This result is well known (see [56]), we give its proof for the sake of completeness.Multiplying the first identity of (5.9) by ϕj and by adding on j = 1, ..., N , we obtain
−λN∑
j=1
|ϕj|2 =
N∑
j=1
(ϕj+1−2ϕj+ϕj−1
h2 )ϕj
=
N∑
j=1
(ϕj+1−ϕj
h2 )ϕj −N∑
j=1
(ϕj−ϕj−1
h2 )ϕj
=
N∑
j=1
(ϕj+1−ϕj
h2 )ϕj −N−1∑
j=0
(ϕj+1−ϕj
h2 )ϕj+1.
and we conclude by the boundary condition ϕN+1 − ϕN = 0.
The solution of (5.8) admits a Fourier development in the basis of eigenvectors of system(5.9). More precisely, every solution uh = (uj)j of (5.8) can be written as
uh(t) =
N−1∑
k=0
(
ak cos(√
λk, ht) +bk
√
λk, hsin(
√
λk, ht)
)
ϕk, h
for suitable coefficients ak, bk ∈ R, k = 0, ..., N − 1, that can be computed explicitely interms of the initial data.
The energy of the problem without damping (5.8) is given by
EjN (uh, t) =h
2
N∑
j=0
(
∣
∣u′j(t)∣
∣
2+
∣
∣
∣
∣
uj+1(t) − uj(t)
h
∣
∣
∣
∣
2)
. (5.11)
Obviously, this energy EjN (uh, .) is constant (proved like Lemma 5.2.1).
5.3 Decay of the energy to 0
Proposition 5.3.1. We havelimt→∞
Eh(t) = 0
if and only if
jNh 6= (2 − h)l
2k + 1, ∀k = 0, ..., N − 1, l ∈ N.
Proof: We use the same kind of proof that in the continuous case (see [6] or chapter 1).⇐ Let S(t) = eAht be the exponential of the matrix Ah. It suffices to show that
limt→∞
S(t)
(
y(0)h
y(1)h
)
= 0.
188
For U =
(
yjy′j
)
j
, we set
‖U‖2 =h
2
N∑
j=0,j 6=jN|y′j|2 +
h
2
N∑
j=0
∣
∣
∣
∣
yj+1 − yjh
∣
∣
∣
∣
2
.
Let us fix U0 =
(
y(0)h
y(1)h
)
∈ R2N . The set orb(U0) =⋃
t≥0
S(t)U0 is precompact in R2N
because, for any sequence (tn)n
‖S(tn)U0‖ + ‖AhS(tn)U0‖ = ‖S(tn)U0‖ + ‖S(tn)AhU0‖≤ ‖U0‖ + ‖AhU0‖ = cst.
Therefore, there exists a subsequence, still denote by S(tn)U0, which converges in R2N . Inthis case, the ω-limit of U0 defined by
ω(U0) = U ∈ R2N ; ∃tn, tn → ∞, S(tn)U0 → U, t→ ∞
is non empty. On the other hand, if Φ =
(
φ0j
φ1j
)
j
∈ ω(U0), then S(t)Φ =
(
φj(t)φ′j(t)
)
∈
ω(U0). We can now apply LaSalle’s invariance principle [32] with the relatively compact
set⋃
t≥0
S(t)U0 and the Lyapounov functional ‖.‖ . As
(
φ0j
φ1j
)
and S(t)
(
φ0j
φ1j
)
=
(
φj(t)φ′j(t)
)
belong to ω(U0), we find that∥
∥
∥
∥
(
φ0j
φ1j
)∥
∥
∥
∥
=
∥
∥
∥
∥
(
φj(t)φ′j(t)
)∥
∥
∥
∥
= L, ∀t ≥ 0.
Therefore φ(t) satisfies the problem (5.4) with initial condition φ0j and φ1
j . Moreover, itholds
0 = L2 − L2 =
∥
∥
∥
∥
(
φj(T )φ′j(T )
)∥
∥
∥
∥
2
−∥
∥
∥
∥
(
φ0j
φ1j
)∥
∥
∥
∥
2
= −α∫ T
0
∣
∣φ′jN
∣
∣
2dt,
which implies thatφ′jN
(t) = 0, ∀t > 0. (5.12)
In particular, this implies that φ is solution of problem (5.8) with initial data φ0j and φ1
j
because the damping term disappears and φ′′jN
= 0.Let us now write
φ0 =
N−1∑
k=0
akϕk, h, φ1 =
N−1∑
k=0
bkϕk, h.
Then φ is given by
φ(t) =N−1∑
k=0
(
ak cos(√
λk, ht) +bk
√
λk, hsin(
√
λk, ht)
)
ϕk, h.
189
Consequently by (5.12)
0 = φ′jN
(t) =
N−1∑
k=0
(−ak√
λk, hϕk, hjN
) sin(√
λk, ht) +
N−1∑
k=0
(bkϕk, hjN
) cos(√
λk, ht).
As the size h of the discretization is fixed, we can use Ingham’s inequality (see [57]) becausethe gap condition is verified. We find that
0 =
∫ T
0
∣
∣φ′jN
∣
∣
2dt ≥ C(h)
N−1∑
k=0
[(ak√
λk, hϕk, hjN
)2 + (bkϕk, hjN
)2] ≥ 0,
for T large enough and for a constant C(h) > 0 which depends on the size h of thediscretization. Consequently
akϕk, hjN
= 0 and bkϕk, hjN
= 0, ∀k = 0, ..., N − 1.
But for all k = 0, ..., N − 1,
ϕk, hjN6= 0 ⇔ sin
(
(2k + 1)πjNh
2 − h
)
6= 0 ⇔ jNh 6= (2 − h)l
2k + 1, ∀l ∈ N.
Therefore, if jNh 6= (2−h)l2k+1
, for all l ∈ N, k = 0, ..., N − 1, we will have ak = bk = 0, for all
k = 0, ..., N − 1 and therefore φ0 = φ1 = 0. We have shown that all (φ0, φ1)T ∈ ω(U0) isequal to zero and consequently lim
t→∞E(t) = 0.
⇒ We use a contradiction argument. Suppose that there exists k ∈ 0, ..., N−1 such
that jNh = (2−h)l2k+1
, so that ϕk, hjN = 0. Then
yh(t) = ϕk, h cos(λk, ht)
is solution of (5.4) with a constant energy.
Remark 5.3.2. Our previous result is a discrete version of the continuous one [6] thatsays that the energy of the solution of system (5.1) tends to zero if and only if
ξ 6= 2p
2q + 1, ∀p, q ∈ N.
5.4 Non exponential decay of the discrete energy
In this section, our goal is to prove the non exponential decay of solutions of (5.4).First, we need the two following lemmas.
190
Lemma 5.4.1. If there exist positive constants M and ω such that for all y(0)h and y
(1)h in
RN ,Eh(t) ≤ Me−ωtEh(0), ∀t ≥ 0 (5.13)
holds, then there exist positive constants T and C such that for all y(0)h and y
(1)h in RN , one
has
EjN (uh, 0) ≤ C
∫ T
0
∣
∣u′jN∣
∣
2dt+ Ch2
∫ T
0
|(Ahuh)jN |2 dt, (5.14)
where uh is solution of (5.8) and EjN (uh, .) is defined by (5.11).
Proof: Let us suppose that (5.13) holds. It follows from the dissipation law (5.7) that forall T > 0,
Eh(0) − Eh(T ) = α
∫ T
0
∣
∣y′jN∣
∣
2dt. (5.15)
By (5.13), for T large enough (any T ≥ ln( 4M3
)
ωsuffices), one has Eh(T ) ≤ 3
4Eh(0), and so
α
∫ T
0
∣
∣y′jN∣
∣
2dt ≥ Eh(0) − 3
4Eh(0) =
1
4Eh(0) =
1
4Eh(uh, 0). (5.16)
Now, we split up yh solution of (5.4) as follows
yh = uh + wh, (5.17)
where uh = (uj)j is solution of problem without damping (5.8) and the difference wh =(wj)j solves
w′′j − wj+1−2wj+wj−1
h2 = 0 t > 0, j = 1, ..., N, j 6= jNw0 = 0, wN+1 − wN = 0 t > 0wjN +1−2wjN
+wjN−1
h= αy′jN − h(Ahuh)jN t > 0
wj(t = 0) = 0, w′j(t = 0) = 0 j = 1, ..., N.
(5.18)
Denote by
Eh(wh, t) =h
2
N∑
j=0, j 6=jN
∣
∣w′j(t)∣
∣
2+h
2
N∑
j=0
∣
∣
∣
∣
wj+1(t) − wj(t)
h
∣
∣
∣
∣
2
. (5.19)
We easily check, like in the proof of Lemma 5.2.1, that
E ′h(wh, t) = −w′
jN(αy′jN − h(Ahuh)jN ). (5.20)
Therefore
Eh(wh, t) =
∫ t
0
E ′h(wh, t)dt = −α
∫ t
0
w′jNy′jNdt+ h
∫ t
0
w′jN
(Ahuh)jNdt
= −α∫ t
0
w′jNu′jNdt− α
∫ t
0
(w′jN
)2dt+ h
∫ t
0
w′jN
(Ahuh)jNdt,
191
by (5.17), which implies that
Eh(wh, t) + α
∫ t
0
(w′jN
)2dt = −α∫ t
0
w′jNu′jNdt+ h
∫ t
0
w′jN
(Ahuh)jNdt.
By Cauchy-Schwarz’s inequality, we deduce that
α
∫ t
0
(w′jN
)2dt ≤ −α∫ t
0
w′jNu′jNdt+ h
∫ t
0
w′jN
(Ahuh)jNdt
≤(∫ t
0
(w′jN
)2dt
)12
(
α
(∫ t
0
(u′jN )2dt
)12
+ h
(∫ t
0
|(Ahuh)jN |2 dt)
12
)
,
and therefore
α
(∫ t
0
(w′jN
)2dt
)12
≤ α
(∫ t
0
(u′jN )2dt
)12
+ h
(∫ t
0
|(Ahuh)jN |2 dt)
12
.
By a convex inequality, we obtain that
α2
∫ t
0
(w′jN
)2dt ≤ 2α2
∫ t
0
(u′jN )2dt+ 2h2
∫ t
0
|(Ahuh)jN |2 dt. (5.21)
By the estimate (5.16) and by (5.17), we find that
Eh(uh, 0) ≤ 4α
∫ T
0
∣
∣y′jN∣
∣
2dt ≤ 8α
(∫ T
0
∣
∣u′jN∣
∣
2dt+
∫ T
0
∣
∣w′jN
∣
∣
2dt
)
and so, by the estimate (5.21), we obtain
Eh(uh, 0) ≤ 8α
∫ T
0
∣
∣u′jN∣
∣
2dt+ 16α
∫ T
0
∣
∣u′jN∣
∣
2dt+
16
αh2
∫ T
0
|(Ahuh)jN |2 dt
≤ C
∫ T
0
∣
∣u′jN∣
∣
2dt+ Ch2
∫ T
0
|(Ahuh)jN |2 dt.(5.22)
The difference between (5.22) and (5.14) relies on the left-hand side, i.e. the energyEh(uh, 0) instead of EjN (uh, 0). To solve this problem, we notice that
EjN (uh, t) = Eh(uh, t) +h
2(u′jN (t))2, ∀t > 0. (5.23)
Deriving the previous identity (5.23), we have
0 = E ′jN
(uh, t) = E ′h(uh, t) + hu′jN (t)u′′jN (t), ∀t > 0,
and therefore
Eh(uh, t) = Eh(uh, 0) − h
∫ t
0
u′jN (s)u′′jN (s)ds, ∀t > 0. (5.24)
192
Moreover, since EjN (uh, .) is constant, for all ǫ > 0, we have
EjN (uh, 0) =1
ǫ
∫ ǫ
0
EjN (uh, t)dt
=1
ǫ
∫ ǫ
0
(
Eh(uh, t) +h
2(u′jN (t))2
)
dt by (5.23)
=1
ǫ
∫ ǫ
0
(Eh(uh, 0) +h
2(u′jN (t))2)dt− 1
ǫ
∫ ǫ
0
h
(∫ t
0
u′jN (s)u′′jN (s)ds
)
dt by (5.24)
≤ Eh(uh, 0) +h
2ǫ
∫ ǫ
0
(u′jN (t))2dt+1
2
∫ ǫ
0
(u′jN (s))2ds+h2
2
∫ ǫ
0
(u′′jN (s))2ds
≤ Eh(uh, 0) +h
2ǫ
∫ ǫ
0
(u′jN (t))2dt+1
2
∫ ǫ
0
(u′jN (t))2dt+h2
2
∫ ǫ
0
|(Ahuh)jN |2 dt
by (5.8). Taking ǫ = h, we have
EjN (uh, 0) ≤ Eh(uh, 0) +
∫ h
0
(u′jN (t))2dt+h2
2
∫ h
0
|(Ahuh)jN |2 dt. (5.25)
Therefore with (5.22) and for T > h, we obtain (5.14).
Lemma 5.4.2. Assume that N is a multiple of q and jN = N pq, where p is odd (so that
xjN → pq). Then for all T > 0 there exist a positive constant C(T ) and initial data such
that the solution uh of (5.8) with these initial data satisfies
EjN (uh, 0) ≥ C(T )
h2
(∫ T
0
∣
∣u′jN (t)∣
∣
2dt+ h2
∫ T
0
|(Ahuh)jN |2 dt)
.
Proof: Choosey
(0)j = ϕk, hj , y
(1)j = 0,
where k ∈ 0, ..., N − 1 will be chosen later. It is easy to check that
uh = cos(√
λk, ht)ϕk, h
solves (5.8). On the other hand, we can verify (see [103] and Lemma 5.2.2) that
EjN (uh, 0) =2 − h
8λk, h.
Moreover∫ T
0
∣
∣u′jN (t)∣
∣
2dt ≤
∫ T
0
λk, h
∣
∣
∣ϕk, hjN
∣
∣
∣
2
dt = λk, hT∣
∣
∣ϕk, hjN
∣
∣
∣
2
. (5.26)
Since we assume that N is a multiple of q, that is to say N = mq, where m ∈ N, thenjN = N p
q= mp. First, we notice that
xjN = jNh =Np
q(N + 1)=
mp
N + 1→ p
q.
193
Secondly, we have
ϕk, hjN= sin
(
(2k + 1)πjNh
2 − h
)
= sin
(
(2k + 1)πmp
2N + 1
)
because h = 1N+1
and so 2 − h = 2N+1N+1
. Now, we fix k in the form
k = N + β,
with β < 0 to determine. Therefore, by simple calculations, we get
(2k + 1)jNh
2 − h=Np
q+
2βNp
(2N + 1)q.
Thus, by choosing 2β = −2q, that is to say β = −q, then k = N − q ∈ 0, ..., N − 1 forN large enough. Again simple calculations yield
(2k + 1)jNh
2 − h= mp− p+
p
2N + 1.
As a consequence
ϕN−q, hjN
= sin
(
(mp− p)π +pπ
2N + 1
)
= ± sin
(
pπ
2N + 1
)
.
Because 2N + 1 = 2−hh, we obtain
ϕN−q, hjN
∼ ±(πp)h
2, (5.27)
when h→ 0. Thus, for N large enough, by (5.26) and (5.10), there exists C > 0 such that∫ T
0
∣
∣u′jN (t)∣
∣
2dt ≤ C
(
λN−q, hT (πp)2h2
4
)
.
Similarly we have
h2
∫ T
0
|(Ahuh)jN |2 dt ≤ h2λ2N−q, h
∣
∣
∣ϕN−q, hjN
∣
∣
∣
2
T ≤ C(h2λN−q, hT ) for a C > 0,
due to (5.27) and because h2λN−q, h ∼ 1. Indeed, (2(N−q)+1)πh2(2−h)
= π2− qπ
2N+1and thus, by
(5.10),
λN−q, h =4
h2cos2
(
qπ
2N + 1
)
∼ 4
h2. (5.28)
Therefore, there exists C(T ) > 0 such that for h small enough,∫ T
0
∣
∣u′jN (t)∣
∣
2dt+ h2
∫ T
0
|(Ahuh)jN |2 dt ≤ C(T )λN−q, hh2
≤ C(T )h2EjN (uh, 0).
The main result of this section is the following theorem
194
Theorem 5.4.3. Assume that N is a multiple of q and jN = N pq
(hence xjN → pq). Then
the decay of Eh to zero is not uniformly exponential with respect to h. More precisely theredo not exist positive constants M and ω which are independent of h such that for all h > 0and y
(0)h and y
(1)h in RN , (5.13) holds.
Proof: Lemma 5.4.2 shows that the estimate (5.14) is not uniform with respect to h. ByLemma 5.4.1, (5.13) cannot hold with M and ω independent of h.
In conclusion, the decay of Eh to zero is not uniformly exponential with respect to hdue to the existence of high frequency spurious solutions of the semi-discrete model. Toovercome this obstacle, we rule out the high frequency spurious modes.
5.5 Filtering technique
5.5.1 Interior observability of the discrete wave equation
We recall that every solution of (5.8) can be developped in Fourier series as follow
uh(t) =
N−1∑
k=0
(
ak cos(√
λk, ht) +bk
√
λk, hsin(
√
λk, ht)
)
ϕk, h
where ak, bk ∈ R, k = 0, . . . , N − 1, are the Fourier coefficients of the initial data, i.e.
y(0)h =
N−1∑
k=0
akϕk, h, y
(1)h =
N−1∑
k=0
bkϕk, h.
In order to obtain a positive counterpart to Theorem 5.4.3, we have to introduce suitablesubclasses of solutions of (5.8) generated by eigenvectors of (5.9) associated with eigenvaluessuch that λh2 ≤ γ. More precisely, introduce
Ch(γ) :=
∑
λk, h≤ γ
h2
akϕk, h with ak ∈ R
,
the space spanned by the eigenvectors ϕk, h for which λk, h ≤ γh2 .
Moreover, in order to use Ingham’s inequality, we need an estimate between the squa-reroots of consecutive eigenvalues entering in the Fourier development of the solutions of(5.8) in the class Ch(γ). This is stated in the next lemma proved in [56].
Lemma 5.5.1. Assume thatγ = 4 sin2
(πǫ
2
)
(5.29)
for some 0 ≤ ǫ < 1. Then√
λk, h −√
λk−1, h ≥ π cos(πǫ
2
)
for all eigenvalues in the range λk, hh2 ≤ γ.
195
Remark 5.5.2. Note that every 0 ≤ γ < 4 can be written in the form (5.29) for some0 ≤ ǫ < 1.
Proof: This result is well known (see [56]), we give its proof for the sake of completeness.Let us compute the gap
√
λk, h −√
λk−1, h =2
h
(
sin
(
(2k + 1)πh
2(2 − h)
)
− sin
(
(2k − 1)πh
2(2 − h)
))
=2
h
(
(2k + 1)πh
2(2 − h)− (2k − 1)πh
2(2 − h)
)
cos(ξ)
=2π
2 − hcos ξ,
for ξ ∈ [ (2k−1)πh2(2−h)
, (2k+1)πh2(2−h)
]. Moreover, the eigenvalue λk, h = 4h2 sin2( (2k+1)πh
2(2−h)) satisfies λh2 ≤
γ = 4 sin2(πǫ2) if and only if sin2( (2k+1)πh
2(2−h)) ≤ sin2(πǫ
2), that is to say if and only if
(2k + 1)πh
2(2 − h)≤ πǫ
2
because 0 < πǫ2< π
2and 0 < (2k+1)πh
2(2−h)= (2k+1)π
2(2N+1)≤ π
2(k ≤ N) and because the function
sinus is increasing on [0, π2]. Since 0 ≤ (2k−1)πh
2(2−h)≤ ξ ≤ (2k+1)πh
2(2−h)≤ π
2ǫ < π
2and because the
function cosinus is decreasing on [0, π2], we obtain
√
λk, h −√
λk−1, h =2π
2 − hcos(ξ) ≥ 2π
2 − hcos(π
2ǫ)
≥ π cos(π
2ǫ)
.
Now we take the sequence jN ∈ N such that
xjN = jNh→ ξ =p
q, when h→ 0,
defined by
jN =
[
p(2N + 1)
2q
]
∈ N, (5.30)
where [x] means the integral part of x.In the sequel, we need the following lemma which is proved in Lemma 3.1 of [103] by
the multiplier technique.
Lemma 5.5.3. For every T > 2, h > 0 and every solution uh of (5.8), we have
(T − 2)EjN (uh, 0) ≤ h3
4
N∑
j=0
∫ T
0
(
u′j+1 − u′jh
)2
dt+2 − h
4
∫ T
0
∣
∣u′N+1
∣
∣
2dt.
We also use the following lemma proved in Lemma 2.9 of [97].
196
Lemma 5.5.4. s is a rational number with an irreducible fraction
s =p
q, where p is odd
if and only if there exists α > 0 such that∣
∣
∣
∣
sin
(
(k +1
2)πs
)∣
∣
∣
∣
> α, ∀k ∈ N.
We are ready to prove the following uniform interior observability of the discrete waveequation :
Proposition 5.5.5. Assume that ǫ ∈ (0, 1) is small enough such that
α > tan(πǫ
2
)
,
where α verifies∣
∣
∣
∣
sin
(
(k +1
2)πξ
)∣
∣
∣
∣
> α, ∀k ∈ N
(which is possible, see Lemma 5.5.4). Then there exist T = T (γ) > 2 and C = C(γ, T ) > 0such that for every solution uh of (5.8) in the class Ch(γ), we have
(T − 2)EjN (uh, 0) ≤ C
∫ T
0
∣
∣u′jN (t)∣
∣
2dt.
Proof: We rewrite the solution uh ∈ Ch(γ) of (5.8) as
uh =∑
|µk, h|h≤√γ
ckeiµk, htϕk, h,
where
µ−k, h = −µk, h, µk, h =√
λk, h, ϕ−k, h = ϕk, h, ck =
ak − ibk2
, c−k = ck.
Then, applying Ingham’s Theorem (see Lemma 5.5.1), there exist C > 0 and T > 2cos(πǫ
2)
such that∫ T
0
∣
∣u′jN (t)∣
∣
2dt ≥ C
∑
|µk, h|h≤√γ
|ck|2 |µk, h|2∣
∣
∣ϕk, hjN
∣
∣
∣
2
= C∑
|µk, h|h≤√γ
|ck|2 λk, h∣
∣
∣ϕk, hjN
∣
∣
∣
2
.(5.31)
Now, we look for a condition that guarantees
∣
∣
∣ϕk, hjN
∣
∣
∣
2
≥ α, ∀k ∈ 0, ..., N − 1 : λk, hh2 ≤ γ.
197
Recalling that jN =[
p(2N+1)2q
]
∈ N, we may write
xjN = jNh =p
q
2 − h
2± η, with η → 0 as h→ 0.
Then
ϕk, hjN= sin
(
(k +1
2)π
2jNh
2 − h
)
= sin
(
(k +1
2)π
(
p
q± η
2
2 − h
))
= sin
(
(k +1
2)πp
q
)
cos
(
(k +1
2)π
2η
2 − h
)
± cos
(
(k +1
2)πp
q
)
sin
(
(k +1
2)π
2η
2 − h
)
.
Thus, by Lemma 5.5.4, we obtain
∣
∣
∣ϕk, hjN
∣
∣
∣≥
∣
∣
∣
∣
sin
(
(k +1
2)πp
q
)∣
∣
∣
∣
∣
∣
∣
∣
cos
(
(k +1
2)π
2η
2 − h
)∣
∣
∣
∣
−∣
∣
∣
∣
cos
(
(k +1
2)πp
q
)∣
∣
∣
∣
∣
∣
∣
∣
sin
(
(k +1
2)π
2η
2 − h
)∣
∣
∣
∣
≥ α
∣
∣
∣
∣
cos
(
(k +1
2)π
2η
2 − h
)∣
∣
∣
∣
−∣
∣
∣
∣
sin
(
(k +1
2)π
2η
2 − h
)∣
∣
∣
∣
.
Now, by the choice of γ, we have
λk, hh2 ≤ γ ⇔ (2k + 1)πh
2(2 − h)≤ πǫ
2
⇔ 0≤ (k + 12)2πη
2 − h≤ πǫη ≤ π
2, because η → 0, so η <
1
2.
Since the function sinus is increasing on [0, π2], we may write
sin
(
(k +1
2)π
2η
2 − h
)
≤ sin(πǫη) ≤ sin(πǫ
2
)
,
and then∣
∣
∣
∣
cos
(
(k +1
2)π
2η
2 − h
)∣
∣
∣
∣
≥ cos(πǫ
2
)
.
Thus, by the assumption α > tan(πǫ2), we get
∣
∣
∣ϕk, hjN
∣
∣
∣≥ α cos
(πǫ
2
)
− sin(πǫ
2
)
> C > 0.
Now, we can return to the inequality (5.31) and deduce that
∫ T
0
∣
∣u′jN (t)∣
∣
2dt ≥ C
∑
|µk, h|h≤√γ
|ck|2 λk, h. (5.32)
198
Then, we recall that Lemma 5.5.3 states that, for T > 2, we have
(T − 2)EjN (uh, 0) ≤ h
4
N∑
j=0
∫ T
0
(u′j+1 − u′j)2dt+
2 − h
4
∫ T
0
∣
∣u′N+1
∣
∣
2dt.
Moreover, applying Ingham’s inequality to the two terms of this right hand side, we get
h
4
N∑
j=0
∫ T
0
(u′j+1 − u′j)2dt ≤ Ch
N∑
j=0
∑
|µk, h|h≤√γ
|ck|2λk, h(ϕk, hj+1 − ϕk, hj )2
= Ch3∑
|µk, h|h≤√γ
|ck|2λk, hN∑
j=0
(
ϕk, hj+1 − ϕk, hjh
)2
= Ch3∑
|µk, h|h≤√γ
|ck|2λ2k, h
N∑
j=0
(ϕk, hj )2 by Lemma 5.2.2
≤ Cγ∑
|µk, h|h≤√γ
|ck|2λk, hhN∑
j=0
(ϕk, hj )2 because u ∈ Ch(γ),
and
2 − h
4
∫ T
0
∣
∣u′N+1
∣
∣
2dt ≤ C
∑
|µk, h|h≤√γ
|ck|2λk, h(ϕk, hN+1)2
≤ C∑
|µk, h|h≤√γ
|ck|2λk, h because ϕk, hN+1 is bounded by 1.
Therefore, since hN∑
j=0
(ϕk, hj )2 ∼ 1, for T > 2cos(πǫ
2), there exists C > 0 such that
(T − 2)EjN (uh, 0) ≤ C∑
|µk, h|h≤√γ
|ck|2λk, h.
This estimate and (5.32) lead to the conclusion.
5.5.2 Some estimates
Lemma 5.5.6. There exist T > 0 and C(T ) > 0 such that the solution wh of (5.18) verifies
∫ 2T
0
w′jNw′jN−1(2jNh− h)(2T − t)dt ≤ C(T )
∫ 2T
0
Eh(wh, t)dt+h
4
∫ 2T
0
(w′jN
)2(2T − t)dt.
199
Proof: We multiply the first identity of (5.18) by (jh)(2T − t)wj+1−wj−1
h(which is the
discrete version of the multiplier x(2T − t)wx(x, t)), we add on j = 1, ..., jN − 1 andintegrate between 0 and 2T. This yields
I1 − I2 = 0, (5.33)
where
I1 = h
jN−1∑
j=1
∫ 2T
0
w′′j (jh)(2T − t)
wj+1 − wj−1
hdt,
I2 = h
jN−1∑
j=1
∫ 2T
0
wj+1 − 2wj + wj−1
h2(jh)(2T − t)
wj+1 − wj−1
hdt.
For the first term I1, an integration by parts yields (recalling that w′j(t = 0) = 0)
I1 = −hjN−1∑
j=1
∫ 2T
0
w′j(jh)
d
dt((2T − t)
wj+1 − wj−1
h)dt
= h
jN−1∑
j=1
∫ 2T
0
w′j(jh)
wj+1 − wj−1
hdt− I3,
where
I3 = h
jN−1∑
j=1
∫ 2T
0
w′j(jh)(2T − t)
w′j+1 − w′
j−1
hdt.
Successive changes of variables lead to
I3 =
jN∑
j=2
∫ 2T
0
w′j−1w
′j(j − 1)h(2T − t)dt−
jN−2∑
j=0
∫ 2T
0
w′j+1w
′j(j + 1)h(2T − t)dt
=
jN∑
j=1
∫ 2T
0
w′j−1w
′j(j − 1)h(2T − t)dt−
jN−2∑
j=1
∫ 2T
0
w′j+1w
′j(j + 1)h(2T − t)dt becausew0 = 0
= −jN−1∑
j=1
∫ 2T
0
w′j(w
′j+1 − w′
j−1)jh(2T − t)dt− h
jN∑
j=1
∫ 2T
0
w′j−1w
′j(2T − t)dt
+
∫ 2T
0
w′jN−1w
′jN
(jN − 1)h(2T − t)dt− h
jN−2∑
j=1
∫ 2T
0
w′j+1w
′j(2T − t)dt
+
∫ 2T
0
w′jNw′jN−1jNh(2T − t)dt
= −I3 − h
jN∑
j=1
∫ 2T
0
w′j−1w
′j(2T − t)dt− h
jN−2∑
j=1
∫ 2T
0
w′j+1w
′j(2T − t)dt
+
∫ 2T
0
w′jNw′jN−1(2jNh− h)(2T − t)dt.
200
This implies that
I3 = −h2
jN∑
j=1
∫ 2T
0
w′j−1w
′j(2T − t)dt− h
2
jN−2∑
j=1
∫ 2T
0
w′j+1w
′j(2T − t)dt
+1
2
∫ 2T
0
w′jNw′jN−1(2jNh− h)(2T − t)dt,
and therefore
I1 = h
jN−1∑
j=1
∫ 2T
0
w′j(jh)
wj+1 − wj−1
hdt+
h
2
jN∑
j=1
∫ 2T
0
w′j−1w
′j(2T − t)dt
+h
2
jN−2∑
j=1
∫ 2T
0
w′j+1w
′j(2T − t)dt− 1
2
∫ 2T
0
w′jNw′jN−1(2jNh− h)(2T − t)dt.
(5.34)
Again performing some elementary changes of variables, we get
I2 = −hjN−1∑
j=0
∫ 2T
0
(
wj+1 − wjh
)2
(2T −t)dt+∫ 2T
0
(
wjN − wjN−1
h
)2
jNh(2T −t)dt. (5.35)
Inserting (5.34) and (5.35) into the identity (5.33), we have obtained that
1
2
∫ 2T
0
w′jNw′jN−1(2jNh− h)(2T − t)dt = h
jN−1∑
j=1
∫ 2T
0
w′j(jh)
wj+1 − wj−1
hdt
+h
2
jN∑
j=1
∫ 2T
0
w′j−1w
′j(2T − t)dt+
h
2
jN−2∑
j=1
∫ 2T
0
w′j+1w
′j(2T − t)dt
+h
jN−1∑
j=0
∫ 2T
0
(wj+1 − wj
h)2(2T − t)dt−
∫ 2T
0
(wjN − wjN−1
h)2jNh(2T − t)dt.
We conclude by the inequality ab ≤ a2
2+ b2
2, by (5.19) and by the fact that jh ≤ 1.
In the same way, we have the following lemma
Lemma 5.5.7. There exist T > 0 and C(T ) > 0 such that the solution wh of (5.18) verify
∫ 2T
0
w′jNw′jN+1(2T − t)(2−2jNh−h)dt ≤ C(T )
∫ 2T
0
Eh(wh, t)dt+h
4
∫ 2T
0
(w′jN
)2(2T − t)dt.
Proof: The proof is similar to the one of the previous lemma. Here, we multiply (5.18) by(1 − jh)(2T − t)
wj+1−wj−1
h, we add this time on j = jN + 1, ..., N and integrate between 0
and 2T.
201
Remark 5.5.8. Lemmas 5.5.6 and 5.5.7 imply that there exist T > 0, a constant C(T ) > 0and another constant C > 0 (independent of T ) such that
∫ 2T
0
(w′jN+1 + w′
jN−1)w′jN
(2T − t)dt ≤ C(T )
∫ 2T
0
Eh(wh, t)dt+ Ch
∫ 2T
0
(w′jN
)2(2T − t)dt,
because h→ 0 and xjN = jNh→ pq∈ (0, 1).
Now, we are ready to prove the main estimate of this subsection.
Proposition 5.5.9. There exist T > 0 and C(T ) > 0 such that the solution wh of (5.18)verifies
∫ T
0
(w′jN
)2dt ≤ C(T )
(∫ 2T
0
(y′jN )2dt+ h2
∫ 2T
0
|(Ahuh)jN (t)|2 dt+ h4
∫ 2T
0
|(Ahu′h)jN (t)|2 dt)
,
where yh (respectively uh) is solution of (5.4) (respectively (5.8)).
Proof: By (5.18), we have
wjN+1 − 2wjN + wjN−1
h= αy′jN − h(Ahuh)jN ,
which implies that
2(w′jN
)2 = (w′jN+1 + w′
jN−1)w′jN
− αhy′′jNw′jN
+ h2(Ahu′h)jNw
′jN.
We multiply this identity by 2T − t and integrate between 0 and 2T to find
2
∫ 2T
0
(2T − t)(w′jN
)2dt =
∫ 2T
0
(w′jN+1 + w′
jN−1)w′jN
(2T − t)dt
−αh∫ 2T
0
y′′jNw′jN
(2T − t)dt+ h2
∫ 2T
0
(Ahu′h)jNw
′jN
(2T − t)dt.
By Remark 5.5.8, we obtain
2
∫ 2T
0
(2T − t)(w′jN
)2dt ≤ C(T )
∫ 2T
0
Eh(wh, t)dt+ Ch
∫ 2T
0
(w′jN
)2(2T − t)dt
−αh∫ 2T
0
y′′jNw′jN
(2T − t)dt+ h2
∫ 2T
0
(Ahu′h)jNw
′jN
(2T − t)dt.
(5.36)Moreover, by (5.17)
−αh∫ 2T
0
y′′jNw′jN
(2T − t)dt = −αh∫ 2T
0
u′′jNw′jN
(2T − t)dt− αh
∫ 2T
0
w′′jNw′jN
(2T − t)dt.
202
Now, observe that
∫ 2T
0
w′′jNw′jN
(2T − t)dt =1
2
∫ 2T
0
d
dt
(
(w′jN
)2)
(2T − t)dt
=1
2
∫ 2T
0
(w′jN
)2dt,
by integration by parts and because w′j(t = 0) = 0, ∀j. This identity in the previous one
yields
−αh∫ 2T
0
y′′jNw′jN
(2T − t)dt = −αh∫ 2T
0
u′′jNw′jN
(2T − t)dt− αh
2
∫ 2T
0
(w′jN
)2dt. (5.37)
On the other hand, by Young’s inequality, we have for all ǫ > 0 :
−αh∫ 2T
0
u′′jNw′jN
(2T − t)dt ≤ ǫ
∫ 2T
0
(w′jN
)2(2T − t)dt+α2h2T
2ǫ
∫ 2T
0
(u′′jN )2dt.
Then (5.37) becomes
−αh∫ 2T
0
y′′jNw′jN
(2T−t)dt ≤ ǫ
∫ 2T
0
(w′jN
)2(2T−t)dt+α2h2T
2ǫ
∫ 2T
0
(u′′jN )2dt−αh2
∫ 2T
0
(w′jN
)2dt.
(5.38)For the last term of the right hand side of (5.36), we still use Young’s inequality and forany ǫ > 0, we get
h2
∫ 2T
0
(Ahu′h)jNw
′jN
(2T − t)dt ≤ h4
4ǫ
∫ 2T
0
((Ahu′h)jN )2(2T − t)dt+ ǫ
∫ 2T
0
(w′jN
)2(2T − t)dt
≤ h4T
2ǫ
∫ 2T
0
((Ahu′h)jN )2dt+ ǫ
∫ 2T
0
(w′jN
)2(2T − t)dt.
(5.39)Inserting (5.38) and (5.39) into the estimate (5.36), we have obtained that
2
∫ 2T
0
(2T − t)(w′jN
)2dt ≤ C(T )
∫ 2T
0
Eh(wh, t)dt+ Ch
∫ 2T
0
(w′jN
)2(2T − t)dt
+ǫ
∫ 2T
0
(w′jN
)2(2T − t)dt+α2h2T
2ǫ
∫ 2T
0
(u′′jN )2dt− αh
2
∫ 2T
0
(w′jN
)2dt
+h4T
2ǫ
∫ 2T
0
((Ahu′h)jN )2dt+ ǫ
∫ 2T
0
(w′jN
)2(2T − t)dt,
which implies that
∫ 2T
0
(2T − t)(2 − Ch− 2ǫ)(w′jN
)2dt ≤ C(T )
∫ 2T
0
Eh(wh, t)dt+α2h2T
2ǫ
∫ 2T
0
(u′′jN )2dt
+h4T
2ǫ
∫ 2T
0
((Ahu′h)jN )2dt.
203
Taking ǫ = 14
and h small enough such that Ch ≤ 1, we arrive at
∫ 2T
0
(w′jN
)2(2T−t)dt ≤ C(T )
∫ 2T
0
Eh(wh, t)dt+4α2h2T
∫ 2T
0
(u′′jN )2dt+4h4T
∫ 2T
0
((Ahu′h)jN )2dt.
(5.40)Now, by (5.20), we have
Eh(wh, t) = Eh(wh, t) − Eh(wh, 0) = −∫ t
0
w′jN
(αy′jN − h(Ahuh)jN )ds,
which implies by Young’s inequality, for all ǫ > 0 and t ≤ 2T,
Eh(wh, t) ≤ (1 + α)ǫ
∫ t
0
(w′jN
)2ds+α
4ǫ
∫ 2T
0
(y′jN )2ds+h2
4ǫ
∫ 2T
0
(Ahuh)2jNds.
Fubini’s theorem yields
∫ 2T
0
∫ t
0
(w′jN
(s))2dsdt =
∫ 2T
0
∫ 2T
s
(w′jN
(s))2dtds =
∫ 2T
0
(w′jN
(s))2(2T − s)ds.
Thus,
∫ 2T
0
Eh(wh, t)dt ≤ (1 + α)ǫ
∫ 2T
0
(w′jN
)2(2T − t)dt+αT
2ǫ
∫ 2T
0
(y′jN )2dt
+h2T
2ǫ
∫ 2T
0
(Ahuh)2jNdt.
(5.41)
Inserting (5.41) into (5.40) and taking ǫ small enough, we obtain
∫ 2T
0
(w′jN
)2(2T − t)dt ≤ C(T )[
∫ 2T
0
(y′jN )2dt+ h2
∫ 2T
0
(Ahuh)2jNdt
+h2
∫ 2T
0
(u′′jN )2dt+ h4
∫ 2T
0
((Ahu′h)jN )2dt]
= C(T )
(∫ 2T
0
(y′jN )2dt+ 2h2
∫ 2T
0
(Ahuh)2jNdt+ h4
∫ 2T
0
((Ahu′h)jN )2dt
)
.
As∫ 2T
0
(w′jN
)2(2T − t)dt ≥∫ T
0
(w′jN
)2(2T − t)dt ≥ T
∫ T
0
(w′jN
)2dt,
we have proved the requested estimate.
204
5.5.3 “Quasi” exponential decay
We are ready to state the main result of this chapter
Theorem 5.5.10. Under the assumptions of Proposition 5.5.5, there exist T > 0, K >0, ω > 0 and β ∈ (0, 1) such that any solution of (5.4) with initial data in the class Ch(γ)verifies, for all t > 0,
Eh(yh, t) ≤ Ke−ωtEh(yh, 0)+
n−2∑
i=0
βn−i(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNds+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNds
)
,
(5.42)where n is such that 2(n− 1)T < t ≤ 2nT and uh is solution of the system (5.8).
Proof: Lemma 5.2.1 implies that
Eh(0) − Eh(2T ) = α
∫ 2T
0
(y′jN )2dt. (5.43)
Writing, like in the previous subsection, yh = uh + wh, yields
∣
∣u′jN∣
∣
2 ≤ 2∣
∣y′jN∣
∣
2+ 2
∣
∣w′jN
∣
∣
2.
By Proposition 5.5.5, for filtered solution of (5.8), we have
(T − 2)EjN (uh, 0) ≤ C
∫ T
0
∣
∣u′jN (t)∣
∣
2dt. (5.44)
By the definition (5.6) of Eh and (5.11) of EjN , we have
Eh(yh, 0) = Eh(uh, 0) ≤ EjN (uh, 0).
Therefore
(T − 2)Eh(yh, 0) ≤ C
∫ T
0
(∣
∣y′jN (t)∣
∣
2+∣
∣w′jN
(t)∣
∣
2)dt.
Then, using Proposition 5.5.9, we obtain
(T − 2)Eh(yh, 0) ≤ C(T )
(∫ 2T
0
(y′jN )2dt+ h2
∫ 2T
0
(Ahuh)2jNdt+ h4
∫ 2T
0
((Ahu′h)jN )2dt
)
.
So, for T > 2, and since the energy Eh is decreasing, we have
Eh(yh, 2T ) ≤ Eh(yh, 0) ≤ C
∫ 2T
0
(y′jN )2dt+C
(
h2
∫ 2T
0
(Ahuh)2jNdt+ h4
∫ 2T
0
((Ahu′h)jN )2dt
)
.
The identity (5.43) yields
(1 + C)Eh(yh, 2T ) ≤ CEh(yh, 0) + C
(
h2
∫ 2T
0
(Ahuh)2jNdt+ h4
∫ 2T
0
((Ahu′h)jN )2dt
)
,
205
and therefore
Eh(yh, 2T ) ≤ C
1 + CEh(yh, 0) +
C
1 + C
(
h2
∫ 2T
0
(Ahuh)2jNdt+ h4
∫ 2T
0
((Ahu′h)jN )2dt
)
.
We set β = C1+C
< 1. Since the system (5.4) is invariant by translation, by iteration, wefind for all n ∈ N,
Eh(yh, 2nT ) ≤ βnEh(yh, 0) +
n−1∑
i=0
βn−i
(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNdt+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNdt
)
.
Setting ω = 1T
ln( 1β), we have obtained
Eh(yh, 2nT ) ≤ e−ωnTEh(yh, 0)+
n−1∑
i=0
βn−i(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNdt+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNdt
)
.
For arbitary t > 0, there exists n ∈ N such that 2(n− 1)T < t ≤ 2nT . By the decreasingproperty of Eh, we conclude that
Eh(yh, t) ≤ Eh(yh, 2(n− 1)T )
≤ e−ω(n−1)TEh(yh, 0) +n−2∑
i=0
βn−i(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNds+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNds
)
≤ 1
βe−ωnTEh(yh, 0) +
n−2∑
i=0
βn−i(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNds+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNds
)
,
which finishes the proof.
In fact, we have the equivalence between the “quasi” exponential decay and the obser-vability inequality :
Proposition 5.5.11. Let the assumptions of Proposition 5.5.5 be satisfied, then the follo-wing assertions are equivalent
(i) There exist T > 0, K > 0, ω > 0 and β ∈ (0, 1) such that every solution of (5.4)with initial data in the class Ch(γ) verifies, for all t > 0
Eh(yh, t) ≤ Ke−ωtEh(yh, 0)+
n−2∑
i=0
βn−i(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNds+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNds
)
,
where n is such that 2(n− 1)T < t ≤ 2nT and uh is solution of the system (5.8).(ii) There exist positive constants T0 and C0 such that every solution uh of (5.8) verifies
EjN (uh, 0) ≤ C0
∫ T0
0
∣
∣u′jN∣
∣
2dt+ C0
(
h2
∫ T0
0
(Ahuh)2jNdt+ h4
∫ T0
0
((Ahu′h)jN )2dt
)
.
206
Proof: The implication (ii)⇒(i) is due to the proof of Theorem 5.5.10 just by adding the
term C0(h2∫ 2T
0(Ahuh)
2jNdt+ h4
∫ 2T
0((Ahu
′h)jN )2dt) in the right hand side of (5.44).
The proof of the other implication (i)⇒(ii) follows the proof of Lemma 5.4.1. For T0
large enough (T0 ≥ ln(4M3
)/ω), by (i), we have
Eh(T0) ≤3
4Eh(0) +
n−2∑
i=0
βn−i
(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNdt+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNdt
)
,
where 2(n− 1)T < T0 ≤ 2nT, and then, by (5.15), we have
Eh(uh, 0) = Eh(0) ≤ 4α
∫ T0
0
(y′jN )2dt+4
n−2∑
i=0
βn−i(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNdt+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNdt
)
.
Then, (5.17) and (5.21) imply
Eh(uh, 0) ≤ C
∫ T0
0
(u′jN )2dt+ Ch2
∫ T0
0
(Ahuh)2jNdt
+4
n−2∑
i=0
βn−i(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNdt+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNdt
)
.
Consequently, by (5.25), we obtain
EjN (uh, 0) ≤ C
∫ T0
0
(u′jN )2dt+ Ch2
∫ T0
0
(Ahuh)2jNdt
+4
n−2∑
i=0
βn−i(
h2
∫ 2(i+1)T
2iT
(Ahuh)2jNdt+ h4
∫ 2(i+1)T
2iT
(Ahu′h)
2jNdt
)
≤ C
∫ T0
0
(u′jN )2dt+ Ch2
∫ T0
0
(Ahuh)2jNdt
+4
1 − β
(
h2
∫ T0
0
(Ahuh)2jNdt+ h4
∫ T0
0
((Ahu′h)jN )2dt
)
,
because 2(i+ 1)T ≤ 2(n− 1)T < T0, which finish the proof.
We have seen in Theorem 5.4.3 that without filter, the energy does not decay uniformlyexponentially. Therefore, a natural question is : without filter, do we have a uniform ”quasi”exponential decay of the energy Eh ? The following proposition gives a negative result,which shows the advantage of the filtering technique.
Proposition 5.5.12. Assume that N is a multiple of q and then jN = N pq. Then for all
T > 0 there exist a positive constant C(T ) and initial data such that the solution uh of(5.8) with these initial data satisfies
EjN (uh, 0) ≥ C(T )
h2
(∫ T
0
∣
∣u′jN (t)∣
∣
2dt+ h2
∫ T
0
|(Ahuh)jN |2 dt+ h4
∫ T0
0
((Ahu′h)jN )2dt
)
.
207
Therefore the decay of Eh is not uniformly quasi exponential with respect to h by Proposition5.5.11.
Proof: We follow the proof of Lemma 5.4.2. We assume that N is a multiple of q and,thus, we can easily prove that jN given in (5.30) verify jN = N p
q. Choose
y(0)j = ϕN−q, h
j , y(1)j = 0.
It is easy to check thatuh = cos(
√
λN−q, ht)ϕN−q, h
solves (5.8). We have seen in Lemma 5.4.2 that
EjN (uh, 0) =2 − h
8λN−q, h,
and there exists C > 0 such that
∫ T
0
∣
∣u′jN (t)∣
∣
2dt ≤ CλN−q, hT (πp)2h
2
4,
and
h2
∫ T
0
(|(Ahuh)jN |2 dt ≤ Ch2λN−q, hT.
Moreover(Ahu
′h)jN (t) = λ
3/2N−q, hϕ
k, hjN
(− sin(√
λk, ht)).
Thus
h4
∫ T
0
((Ahu′h)jN )2dt ≤ h4λ3
N−q, h
∣
∣
∣ϕk, hjN
∣
∣
∣
2
T ≤ CλN−q, hh2T,
by (5.27) and (5.28).Therefore, there exists C(T ) > 0 such that for h small enough,
∫ T
0
∣
∣u′jN (t)∣
∣
2dt+ h2
∫ T
0
|(Ahuh)jN |2 dt+ h4
∫ T
0
((Ahu′h)jN )2dt ≤ C(T )λN−q, hh
2
≤ C(T )h2EjN (uh, 0).
Remark 5.5.13. We can easily verify that (5.42) implies that there exist T > 0, K >0, ω > 0 and C > 0 such that every solution of (5.4) with initial data in the class Ch(γ)verifies, for all t > 0
Eh(yh, t) ≤ Ke−ωtEh(yh, 0) + C(hEh(uh, 0) + h3Fh(uh, 0)), (5.45)
208
where Eh is a constant energy for the problem without damping (5.8) defined by
Eh(uh, t) =h
2
N∑
j=0
∣
∣
∣
∣
u′j+1(t) − u′j(t)
h
∣
∣
∣
∣
2
+h
2
N∑
j=1
|(Ahuh)j(t)|2 , (5.46)
and Fh is given by
Fh(uh, t) =h
2
N∑
j=0
∣
∣
∣
∣
u′′j+1(t) − u′′j (t)
h
∣
∣
∣
∣
2
+h
2
N∑
j=1
|(Ahu′h)j(t)|2
which is also constant and which is obtained by substituting uh by u′h in the energy Eh (weobserve that u′h is a solution of (5.8) because (5.8) is linear).
In the general situation, the right hand side of (5.45) behaves as a non increasingexponential function for t between 0 and − lnh
ω+ c and is constant (proportional to h) for
t large, see Figure 5.1. But for smooth enough initial data y(0) and y(1), and choosingappropriately y
(0)h and y
(1)h , we have
Eh(uh, 0) → E(u, 0) and Fh(uh, 0) → F (u, 0), when h→ 0.
Therefore, the two last terms of the right hand side of (5.45) tends to zero as h→ 0, whichmeans that the estimate (5.45) tends to the exponential stability estimate
E(y, t) ≤ Ke−ωtE(y, 0).
In that sense our estimate (5.45) is quasi optimal.
κ h
Eh(0)
t
Eh
−lnhω
........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ .................................... .................. .................. .................. .................. .................................... .................. .................. .................. ........................... .................. ........................... .................. ........................... .................. .................. ...................................................................................................................................................................................................................................................................................................................................................................................................................................
......
......
......
......
..
....
0.00
Fig. 5.1 – Behaviour of the right hand side of (5.45)
In order to illustrate our results from Theorem 5.5.10 and Proposition 5.5.12 (or Theo-rem 5.4.3), let us present the following numerical results. We consider the semi-discrete
209
problem (5.4) with N = 26 and appropriate jN such that jNh approaches ξ = 1/2. Asinitial data, we take either y(0) = ϕN−2,h, y(1) = 0 or y(0) = ϕ1,h, y(1) = 0 ; correspondingto high or low frequency data respectively. The logarithms of the discrete energies arepresented in Figure 5.2. The top curve of the figure is the logarithm of the energy of thehigh frequency datum, while the bottom curve of the figure is the logarithm of the energyof the low frequency datum. Hence we see that without filtering the decay of the energy isnot exponential (actually the energy decays very slowly from e8.3386 to e8.3358). For the lowfrequency datum a quasi-exponential decay is detected (because the curve of the logarithmof the energy is mainly a line up to t ≃ 15 and mainly constant afterwards). These twotests confirm the advantage of the filtering technique.
0 5 10 15 20 25 30 35 40−20
−15
−10
−5
0
5
10
Fig. 5.2 – The energy for a low (bottom curve) and high (top curve) frequency data
210
Conclusion
Dans cette these nous avons etudie la stabilisation de quelques equations d’evolution parretro-action (feedback) ; plus particulierement notre attention s’est portee sur les equationsdes ondes, de la chaleur et des poutres. Nous nous sommes interesses a trois axes principaux.
Tout d’abord, nous avons considere la stabilisation de l’equation des ondes sur desreseaux 1-d par des feedbacks situes aux noeuds en mettant en oeuvre deux methodesdifferentes. Dans le premier chapitre, nous nous sommes ramenes a l’etude d’une inegalited’observabilite pour le probleme conservatif, ou l’on impose la condition de Neumanna la place des conditions de dissipation, et nous avons donne des conditions spectralesfournissant la decroissance exponentielle ou polynomiale.
Dans le second chapitre, nous avons transfere des inegalites d’observabilite a poids, dejaexistantes pour le probleme conservatif (ou la condition de dissipation est remplacee par lacondition de Dirichlet), en inegalites d’observabilite faibles pour le systeme dissipe. Gracea une inegalite d’interpolation, nous avons obtenu des taux de decroissance explicites quidependent des proprietes geometriques et topologiques du reseau.
Le second axe d’etude est la stabilisation de systemes avec un terme de retard dans lesfeedbacks.
En supposant que le poids du feedback avec retard est plus petit que celui sans retard,nous avons donne des conditions pour obtenir la stabilite forte, exponentielle ou polyno-miale de l’equation des ondes sur des reseaux 1-d.
Nous avons ensuite developpe une theorie abstraite pour les equations d’evolution dusecond ordre generalisant ces resultats.
Enfin nous avons etudie le cas ou le retard depend du temps pour les equations desondes et de la chaleur. En emettant certaines hypotheses sur ce retard τ(t) (en particulierla derivee de τ doit etre majoree par 1 et τ borne) et en utilisant une fonctionnelle deLyapounov appropriee, nous avons prouve que l’energie est exponentiellement decroissanteet nous avons donne explicitement son taux de decroissance.
Le dernier axe de notre etude porte sur l’aspect numerique. Nous avons montre qu’unetechnique de filtrage permet d’obtenir une decroissance quasi-exponentielle de l’equationdes ondes discretisee en espace par differences finies avec un amortissement interne. Sans cefiltrage, le taux de decroissance ne serait pas uniforme par rapport au pas de discretisation.
Une perspective de recherche interessante porterait sur l’etude de la stabilite des equations
211
d’evolution abstraites du second ordre avec des feedbacks avec retard dependant du temps.Dans un tel cas, comme le systeme n’est pas invariant par translation, la methode du cha-pitre 3 n’est plus utilisable et une fonctionnelle de Lyapounov appropriee, comme dans lechapitre 4, pourrait repondre au probleme.
Nous pourrons etendre la stabilisation de l’equation des ondes dans un domaine de Rn
avec un terme de retard dependant du temps sur une partie du bord au cas non lineaire.Nous pourrons egalement effectuer l’etude de la stabilisation de l’approximation par
elements finis des equations d’evolution du second ordre pour des feedbacks non bornes ennous basant sur [96].
212
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