Comportement asymptotique de quelques problèmes dépendants d'un paramètre

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(1)´publique Alge ´rienne De ´mocratique et Populaire Re `re de l’enseignement supe ´rieur et de la Ministe recherche scientifique UNIVERSITE FERHAT ABBAS- SETIF 1. ` THESE Pr´ esent´ e de la facult´ e des Sciences D´ epartement de Math´ ematiques Pour l’obtention du diplˆ ome de. DOCTORAT EN SCIENCES Option : Math´ ematiques Par. HADI Sarra. Th` eme Comportement asymptotique de quelques probl` emes d´ ependants d’un param` etre. Soutenu le : 13/05/2018 Devant le jury : Mr M. ACHACHE. Prof. Univ. de Sétif 1. Pr´ esident. Mr A. AIBECHE. Prof. Univ. de Sétif 1. Encadreur. Mr S. GUESMIA. Dr. Univ. d’El-Qassim (Arabie Saoudite). Examinateur. Mr A. Rougirel. Prof. Univ. de Poitiers (France). Examinateur. Mr L. Chiter. Prof. Univ. de Sétif 1. Examinateur. Mr M. Zerguine. Dr. Univ. de Batna2. Examinateur.

(2) Thesis presented to obtain the doctoral degree in sciences Speciality: Differential equations Option: Mathematics By: Hadi Sarra Titled:. Asymptotic Behaviour of Some Parameter Dependents Problems Supervisors: Aibeche Aissa, Guesmia Senoussi.

(3) This work is dedicated to my father Si El Laifa,.

(4) Acknowledgements First of all, I would like to thank to my thesis director. Professor Aibeche Aissa for his advice and patience during the realization of this work. I would like to thank Dr Senoussi Guesmia who kindly accepted to be the supervisor of this thesis. My special thanks to. Dr Sengouga Abdelmouhcene for his cooperation in the published work. I wish to express my sincere thanks to Professors M. Achache, A. Rougirel, L. Chiter, and M. Zerguine who kindly agreed to serveonmy thesis committee. I express my thanks to Professor M. Achache for agreeing to help me correct this manuscript. I thank him for all his good remarks and for all the improvements that brought him to this thesis. The last thanks goes to my husband Djaafar and our sons Rahil, Anes and Mohammed Idris.. 2.

(5) Abstract In this thesis we study the asymptotic behaviour for the solution of a linear and nonlinear wave equation in a non-cylindrical domain, becoming unbounded in some direction as the time goes to infinity. If the limit of the source term is independent of this direction and t, the wave converges to the solution of an elliptic problem defined on a lower dimensional domain. The rate of convergence depends on the limit behaviour of the source term and on the coefficient of the nonlinear term.. Key words and phrases Wave equation, Asymptotic behaviour in time, non-cylindrical domains.. R´ esum´ e Dans cette thése, nous étudions le comportement asymptotique de la solution d’une équation d’onde (linéaire et non linéaire) dans un domaine non-cylindrique devenant non bornée, dans une certaine direction lorsque le temps t tend vers l’infini. Si la limite du terme source est indépendante de cette direction et t, la solution de l’équation des ondes converge vers la solution d’un problème elliptique défini sur un domaine de dimension inférieure. Le taux de convergence dépend de la limite du terme source et du coefficient du terme non linéaire.. Mots cl´ es Equation d’onde, Comportement asymptotique dans le temps, Domaines non cylindrique. 3.

(6) Contents Introduction. 6. 1 Preliminaries. 12. 1.1. Basic notions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 13. 1.2. Functions spaces on the noncylindrical domain . . . . . . . . . . . . . . . . .. 18. 2 Asymptotic behaviour of linear wave equations. 20. 2.1. Problem setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 21. 2.2. Existence and uniqueness of a weak solution . . . . . . . . . . . . . . . . . .. 21. 2.3. The limit problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 28. 2.4. Special cut-off functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 28. 2.5. Energy Estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 29. 2.5.1. A priori estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 29. Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 33. 2.6.1. Convergence theorems . . . . . . . . . . . . . . . . . . . . . . . . . .. 33. 2.6.2. Exponential convergences . . . . . . . . . . . . . . . . . . . . . . . .. 36. 2.6. 3 Asymptotic behaviour of nonlinear wave equations. 38. 3.1. Problem setting . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 39. 3.2. Existence and uniqueness of regular solutions . . . . . . . . . . . . . . . . . .. 40. 3.3. Limit problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 48. 3.4. Special cut-off functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 49. 3.5. Energy estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 50. 4.

(7) 3.5.1 3.6. A priori estimate . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 50. Main results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 56. 3.6.1. Convergence theorems . . . . . . . . . . . . . . . . . . . . . . . . . .. 56. 3.6.2. Convergence in arbitrary interior regions . . . . . . . . . . . . . . . .. 59. 3.6.3. Exponential convergence . . . . . . . . . . . . . . . . . . . . . . . . .. 60. General conclusion and perspectives. 63. Appendix. 65. 3.7. Notations for estimates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 66. 3.8. Some useful inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 66. Bibliographie. 67. 5.

(8) Introduction enerally the determination of approximate solutions for partial differential equations uses. G the resolution of simpler models, either by a general technique such as in the case of the finite element method, or by taking advantage of some particularities of our problem to consider another problem simpler. Our work is motivated by some recent works on the asymptotic behaviour of the solutions of boundary value problems of different types in a cylindrical domain Ω` , when the size of Ω` becomes unbounded, in some directions, as the parameter ` → ∞ (independently of the time), see for instance Chipot [6], Chipot and Mardare [10], Guesmia [19] for elliptic and parabolic and Stokes problems and [5], [18] for hyperbolic problems. In recent years, there is much interest in evolution problems set in time-dependent domains. These problems arise in many real world applications when the spatial domain of the considered phenomena depends strongly on time, see for instance the survey paper [24] and the references cited therein. In the present work, we study the asymptotic behaviour for the solution of linear and nonlinear wave equation in a noncylindrical domain, becoming unbounded, in some direction as the time t goes to infinity. If the limit of the source term is independent of this directions and the time t, the wave converges to the solution of an elliptic problem defined on a lower dimensional domain. The rate of convergence depends on the limit behaviour of the source term and on the coefficient of the nonlinear term. We consider a time-dependent family of bounded subset in Rn1 × Rn2 defined as Ωt := (−`0 − `t; `0 + `t)n1 × ω,. t ≥ 0,. where n1 , n2 are positive integers, ω is a bounded open subset of Rn2 with sufficiently 6.

(9) smooth boundary, `0 > 0 and the speed of expansion ` is constant and at hand, we give to `t the same role of the parameter ` in the previous works. Let us denote the points in Rn1 × Rn2 as. x = (X1 , X2 ) = x1 , ..., xn1 , x01 , ..., x0n2 ,. We also consider the following noncylindrical domain, and its lateral boundary Qt :=. [. 0<s<t. {s} × Ωs ,. Σt :=. [. 0<s<t. {s} × ∂Ωs .. More specifically, we are interested in the asymptotic behaviour, as t → ∞, of the solution of the wave equation in two cases: -The first case we study the following linear wave equation with a dissipation term    u00 (t, x) − 4u (t, x) + βu0 (t, x) = f (t, x) in Qt ,   u(t, x) = 0 on Σt ,     u(0, x) = u0 (x) , u0 (0, x) = u1 (x) in Ω ,. (1). 0. where the primes stand for the time derivative, 4 is the Laplace operator and β > 0 is a. positive constant. -The second case, we study the following nonlinear wave equation    u00 (t, x) + βu0 (t, x) − 4u(t, x) + γ(t) |u|ρ u = f (t, x) in Qt ,   u(t, x) = 0 on Σt ,     u(0, x) = u0 (x) , u0 (0, x) = u1 (x) in Ω0 ,. (2). where f, γ are functions which satisfy some appropriate conditions.. The existence of solutions for linear and nonlinear wave problems in noncylindrical domains has been studied by several authors. Lions [26] introduced the so-called penalty method to show the existence of weak solutions in expanding domains, see also Mederios [29]. Under some time likeness conditions on Qt , Cooper and Bardos [12], Cooper and Medeiros [13] obtained the existence and uniqueness of weak solutions. Clark [11], Ferreira [16], Ferriera and Lar’kin [17] also considered problems with variable coefficients. Several works dealt with the asymptotic behaviour in time for the solutions of evolution problems for different types of nonlinearity and boundary conditions in noncylindrical domains. 7.

(10) Using the multiplier method, Bardos and Chen [3] proved that the energy of the linear wave equation decays when the domain is timelike and expanding. Nakao and Narazaki [31] and Rabello [32] studied the decay of the energy for weak solutions of nonlinear wave problems in expanding domains. There idea relays on the penalization method, introduced by Lions [26]. Another method consists in considering a suitable change of variables that transforms the noncylindrical domain to a cylindrical one, establish energy estimates for the new problem, then derive the desired energy estimates for the noncylindrical problem, see for instance [22], [25]. The drawback of this method is that the differential operator of the transformed problem is, in general, more complicated. We have to mention that in this thesis we study the problem directly in the noncylindrical domains, without any change of variables. The idea is based on the use of some special cut-off functions, depending on (t; X1 ), to obtain local estimates of the difference between the wave and its limit, this techniques introduced by Guesmia [20] for a parabolic problem in noncylindrical domain, see also Chipot and Yeressian [9], Roughly speaking, if f (t; x) converges to some f∞ (X2 ) and γ(t) converges to 0 (in the nonlinear case), faster enough in a sense to be made precise later, we obtain a convergence u(t) → u∞ in interior regions of the domain Qt . Here u∞ is the solution of an elliptic problem defined on ω, then the rate of convergence u(t) → u∞ is analyzed and improved under some assumptions. The main features of this work can be summarized as follows: -In [22], [31], and [32] the size of the domain remains bounded as t → ∞, and the limit of the solution of the considered problem is zero. This situation arises when the decay in the energy of the solution, due to the expansion of the domain and damping terms, overtakes the contribution of the source term. In this work, Ωt becomes unbounded in n1 directions and the limit of the solution, in interior regions of the domain, is not necessarily zero, as t goes to infinity. To the best of our knowledge, the asymptotic behaviour of such problems has not been considered before. -In contrast with [20], the source term f in this work depends on all the variables (t; x) ∈ R+ × (−`0 − `t; `0 + `t)n1 × ω and not only on X2 ∈ ω.. 8.

(11) The rest of this thesis is divided into three chapters: In the first chapter, we present some elementary notions of functional analysis freet, used in the sequel, then we state the basic results about the existence and uniqueness of u(t). The second chapter is titled Asymptotic Behaviour of Linear Wave Equation, is reserved to show some local energy estimates for the difference u (t) − u∞ and we give some convergence results, and discuss some particular cases where the rate of convergences is exponential. Afterwards, we define of u∞ , the candidate limit u(t) as t → +∞, and the cut-off functions. The last chapter, titled Asymptotic Behaviour of Non Linear Wave Equation, in this chapter, we define u∞ , the candidate limit of u(t) as t → +∞, and the cut-off functions needed in the sequel, then we give an energy estimate for u(t) as well as a local energy estimate for the difference u(t) − u∞ . At the convergence results and discuss some particular cases where the rate of convergence increase exponentially. for the convenience of the reader, we will agree some standard notations commonly used during this work, see for instance [4], [7]. However, to make this thesis consistent and to avoid ambiguity, a list of notations for estimates and definitions is included the appendix, as well as some elementary inequalities frequently used in through this work.. 9.

(12) Notations General notations R+. : set of all positive real numbers.. Rn. : n-dimensional real Euclidean space, R = R1 .. Rn+1. : denote the point of Rn+1 as (x1 , X2 ) = (x1 , x2 , ..., xn ) .. Ω. : open domain in Rn .. ∂Ω. : is the Γ boundary of Ω.. Ω`. : denote the rectangle (−`, `) × ω, , where ω is a bounded subset in Rn .. Ωt. : {(−`0 − `t, `0 + `t) × ω,. Ωnt 1 ×n2. : {(−`0 − `t, `0 + `t)n1 × ω,. t ≥ 0} t ≥ 0} .. S⊂Ω .. S ⊂⊂ Ω : S strongly included in Ω, i.e the closure of S t Sm. : {(s, x1 ); t − m < s < t, − t0 − σ (m − t + s) < x1 < t0 + σ (m − t + s)}. :=. : equal by definition.. x. : (X1 , X2 ) ∈ Rn where X1 = (x1 , ..., xq ) ∈ Rq ,. X2 = (xq+1 , ..., xn ) ∈ Rn−q .. Functions and differential operators For a function u : Ω ⊂ Rn → R, we note x ∈ Ω.. u (x). : u (x1 , x2 , ..., xn ) ,. u(t, x). : u (t, x1 , x2 , ..., xn ) ,. ∇u. : (∂x1 u, ∂x2 u, ..., ∂xn u)T. for t ≥ 0, x ∈ Ω. gradient of u.. : ∂x21 u + ∂x22 u + ... + ∂x2n u Laplacian of u. ∂ 2u ∂u 00 0 , similarly u (t, x) = 2 . u (t, x) : ∂t T∂t ∂u ∇x,t u : , ∂x1 u, ∂x2 u, ..., ∂xn u , for t ≥ 0, x ∈ Ω. ∂t 4u. f (t, x). : the source term.. % (t, x) : special cut-off functions. χij.   1 if i = j, : the characteristic function denoted by χij =  0 if i 6= j. 10.

(13) Function spaces For a function u : Ω → R, we note C k (Ω). : u is k times continuously differentiable, k ∈ N.. D (Ω). : u is infinitely differentiable with compact support in Ω.. Lp (Ω). : space of distributions on Ω. R : u is measurable and Ω |u (x)|p dx < ∞,. D0 (Ω) L∞ (Ω). 1 ≤ p < ∞.. : u is measurable and |u (x)| ≤ C a.e. in Ω for some constant C. ∀α ∈ N, |α| ≤ k where K ∈ N.. H k (Ω). : u ∈ L2 (Ω) \ Dα u ∈ L2 (Ω) ,. H0k (Ω). : the closure of D (Ω) in H k (Ω).. W k,p (Ω). : u ∈ Lp (Ω) \ Dα u ∈ Lp (Ω) ,. C k ([a, b] ; X). : space of k times continuously differentiable functions. ∀α ∈ Nn , |α| ≤ k where K ∈ N, 1 ≤ p ≤ ∞.. from an interval [a, b] to the Banach space X. Lp ([a, b] ; X). : space of measurable functions u on [a, b] with values in X and such that |u|p is integrable, 1 ≤ p < ∞.. L∞ ([a, b] ; X) : space of measurable functions u on [a, b] such that there exists a constant C such that |u|pX ≤ C for almost every x ∈ [a, b] .. 11.

(14) Chapter 1 Preliminaries e expose in this chapter some functional analysis tools that we will need in the sequel. W of the thesis. In particular, we will recall the main results that deal with existence and uniqueness of the solution of the linear and nonlinear wave equations. Let us denote by X a Banach space (i.e., X is a complete normed vector space equipped with the norms are given by k.kX ) which will specified later.. 12.

(15) 1.1. Basic notions. The space Lp (a, b; X) Definition 1 For a, b ∈ R we denote by Lp (a, b; X) , 1 ≤ p < +∞. the space of (class of. functions) f : (a, b) → X that are measurable and such that Zb. kf (t)kpX dt < +∞.. a. Definition 2 ([6]) We denote by L∞ (a, b; X) the space of functions essentially bounded on (a, b), i.e., such that there exists M > 0 such that kf (t)kX 6 M. t ∈ (a, b) .. a.e.. Then it is easy to show Theorem 3 ([6]) Equipped with the norms . kf kLp (a,b;X) = . Zb a.  p1. kf (t)kpX dt ,. 1 ≤ p < +∞,. kf kL∞ (a,b;X) = inf M ∈ R∗+ \ kf (t)kX 6 M. the spaces Lp (a, b; X) , 1 ≤ p < +∞, are Banach spaces.. a.e.. t ∈ (a, b) ,. Proof. Let (fn ) be a Cauchy sequence in Lp (a, b; X) , it suffices to show that an extracted a converges subsequence in Lp (a, b; X) . We extract a subsequence (fnk ) such that. 1 fn −fn k ∀k ≥ 1. ≤ k k Lp (a,b;X) k+1 2. We will show that (fnk ) converges in Lp (a, b; X) . We set n X

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(17) fn gn (t) = k=1. he comes.

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(19) , (t) − f (t) n k+1 k. kgn kLp (a,b;X) ≤ 1. 13.

(20) We deduce from the monotonic convergence theorem that gn (t) converge to g(t) a.e.. with g (t) ∈ Lp .. On the other hand, we have for m ≥ n ≥ 2 |fm (t) − fn (t)| ≤ |fm (t) − fm−1 (t)| + ... + |fn+1 (t) − fn (t)| ≤ g (t) − gn−1 (t) , it follows that (fn ) (t) is from Cauchy and converges towards a limit noted by f (t) . We have |f (t) − fn (t)| ≤ g (t). for n ≥ 2,. a.e.. we deduce that f (t) ∈ Lp . Finally kfn − f kLp (a,b;X) → 0 The space D0 (a, b; X) We denote by D(a, b; X) the space of infinitely differentiable functions on (a, b) with compact support contained in X. Definition 4 ([6]) A vector valued distribution on (a, b) with values in X is a linear mapping T : D(a, b) → X ϕ 7→ hT, ϕi , which is continuous on D(a, b) in the sense that lim hT, ϕi i = hT, ϕi. i→+∞. in X,. for any sequence ϕi → ϕ in D(a, b). We can now give the definition of the derivative of a distribution. Definition 5 ([6]) We will denote by D0 (a, b; X), the space of all distributions on (a, b) with values in X.. 14.

(21) Definition 6 ([6]) Let T ∈ D0 (a, b; X), then for every integer k, we define a distribution T (k) by the formula. . (k) dk ϕ k T , ϕ = (−1) T, k . dt. T (k) is called the k th derivative of T on (a, b) and will also be denoted by. dk ϕ = T (k) . dtk. Remark 7 If X, Y are two Banach spaces such that X ,→ Y is continuously embedding. Then clearly D0 (a, b; X) ,→ D0 (a, b; Y ) and Lp (a, b; X) ,→ Lp (a, b; Y ). ∀1 ≤ p ≤ +∞.. The space H 1 (a, b; V, V 0 ) Let H, V are two separable Hilbert spaces, we suppose that V is dense in H (V ,→ H) and we identify H with his dual H 0 . So we have V ,→ H ,→ V 0 , that is, each space being dense in the next with continuous embedding. We will denote by (·, ·). the scalar product in H or the product of duality V 0 , V ,. k·kH. the norm in H,. ((·, ·)) , k·kV respectively the scalar product and the norm in V , k·kV 0. the norm in V 0 .. A suitable choice could be H01 (Ω) ,→ L2 (Ω) ,→ H −1 (Ω) , then we can define Definition 8 ([6]) We define the space H 1 (a, b; V, V 0 ) by . H 1 (a, b; V, V 0 ) = u ∈ L2 (a, b; V ), u0 ∈ L2 (a, b; V 0 ) , 15.

(22) where u0 is calculated in the sense of the distribution D0 (a, b; X). And we set 2. kuk2H 1 = kuk2V + ku0 kV 0 .. (1.1). H 1 (a, b; V, V 0 ) is the Hilbert space. Now we can state Theorem 9 Let u ∈ H 1 (a, b; V, V 0 ), then u can be identified with a continuous function on [a, b] with values in H. moreover that H 1 (a, b; V, V 0 ) ,→ C 0 ([a, b] ; H), where C 0 ([a, b] ; H) denote the space of continuous functions on [a, b] with values in H equipped with the topology of the uniform convergence on [a, b]. Density Theorem 10 ([6]) D(a, b; V ) is dense in H 1 (a, b; V, V 0 ). Extension Theorem 11 ([6]) There exists a continuous linear operator of extension that we denote by P from H 1 (a, b; V, V 0 ) into H 1 (−∞, +∞; V, V 0 ) such that P (u) = u. a.e on (a, b). ∀u ∈ H 1 (a, b; V, V 0 ).. Proof. [6] We assume that b is finite. If a = −∞ then one sets   u (t) t≤b P u (t) =  u (2b − t) t ≥ b.. It is easy to show that this solves the problem. If a is finite, then setting u = θ1 u + θ2 u, where θ1 , θ2 are two smooth functions such that θ1 = 0 on (−∞, a] ,. θ2 = 0 on [b, +∞) , 16. θ1 + θ2 = 1 for a ≤ t ≤ b..

(23) We have recall (1.1) kθ1 ukH 1 ≤ C kukH 1 , kθ2 ukH 1 ≤ C kukH 1 , and we conclude by using the previous step Integration by parts Theorem 12 ([6]) Let u, v ∈ H 1 (a, b; V, V 0 ) with is a, b finite. Then Zb. 0. hu (t), v(t)i dt +. a. Zb. hu(t), v 0 (t)i dt = ((u(b), v(b))) − ((u(a), v(a))) .. a. Proof. For u, v ∈ D(a, b; V ), we have Zb. 0. hu (t), v(t)i dt +. a. Zb. 0. hu(t), v (t)i dt =. u(t)v(t)|ba. a. −. Zb. 0. hu(t), v (t)i dt +. a. Zb. hu(t), v 0 (t)i dt. a. = u(b)v(b) − u(a)v(a) = ((u(b), v(b))) − ((u(a), v(a))) . Then the result follows by density Proposition 13 ([6]) Let u ∈ H 1 (a, b; V, V 0 ), and let v ∈ V. We have hu0 (.), vi =. d ((u(.), v)) dt. in D0 (]a, b[) .. Proof. Let ϕ ∈ D (]a, b[) . From the above proposition, we derive Zb. hu0 (t), vϕ (t)i + hvϕ0 (t) , u(t)i dt = ((u(b), vϕ(b))) − ((u(a), vϕ(a))) = 0. a. This can also be written Zb. hu0 (t), vi ϕ (t) dt = −. a. Zb. hv, u(t)i ϕ0 (t) dt = −. a. Zb. ((u(t), v)) ϕ0 (t) dt.. a. Hence the result • For more details on Lp , D0 , H 1 at vector values see, [6], [27] ... 17. ∀v ∈ V..

(24) 1.2. Functions spaces on the noncylindrical domain. We precise some functions spaces and their norms are in order on the noncylindrical domain. Let Ω be an open and bounded set of Rn with smooth boundary Γ, Q = Ω × (0, T ) an open cylindrical domain. We will use the following notations Ωs = Q ∩ {t = s}. for s > 0,. Ω0 = Q ∩ {t = 0} ,. such that Ωs 6= ∅ for all s ≥ 0,. Γs = ∂Ωs ,. Σ=. ∪ Γs. 0<s<T. ∂Q = Ω0 ∪ Σ,. where ∂Q the boundary of Q. Our assumptions on Q are    i) Ωt is monotone increasing, that is, Ω∗t ⊂ Ω∗s if t < s,       where Ω∗s is the projection of Ωt on the hyperplane t = 0.   ii) we assume the following regularity property:      For each t ∈ ]0, T [ , if u ∈ H01 (Ω) and u = 0 in Ω\Ωt      then u ∈ H 1 (Ω ) . t 0. (1.2). In order to simplify the notation we will identify Ω∗t with Ωt and we define The space Lq (0, T ; Lp (Ωt )). Definition 14 ([26]) The spaces Lp (Ωt ) (resp. H01 (Ωt )) identify, for all t in [0, T ] to closed subspaces of Lp (Ω) (respH01 (Ω)) . Definition 15 ([16]) We define the spaces Lq (0, T ; Lp (Ωt )) , Lq (0, T ; H01 (Ωt )) as Lq (0, T ; Lp (Ωt )) = {v : Lq 0, T ; H01 (Ωt ) = v :. v ∈ Lq (0, T ; Lp (Ω)) , such that v(t) ∈ Lp (Ωt ) a.e. (1 ≤ q ≤ ∞)} , . v ∈ Lq 0, T ; H01 (Ω) , such that v(t) ∈ H01 (Ωt ) a.e. (1 ≤ q ≤ ∞) .. Definition 16 ([26]) i) If 1 ≤ q < ∞, we consider the norms Z T 1q Z T 1q q q kvkLq (0,T ;Lp (Ωt )) = kv(t)kLp (Ωt ) dt , kvkLq (0,T ;H 1 (Ωt )) = kv(t)kH 1 (Ωt ) dt . 0 0 0 0. ii) If q = ∞ we consider the norms kvkL∞ (0,T ;Lp (Ωt )) = ess sup kv(t)kLp (Ωt ) ,. kvkL∞ (0,T ;H 1 (Ωt )) = ess sup kv(t)kH 1 (Ωt ) . 0. 0<t<T. 18. 0<t<T. 0.

(25) Definition 17 ([26]) The spaces Lq (0, T ; Lp (Ωt )) (resp. Lq (0, T ; H01 (Ωt ))) identify, identify, for all t in [0, T ] to the closed subspaces of Lq (0, T ; Lp (Ω)) (resp.Lq (0, T ; H01 (Ω))) . Timelikness condition Let ν = (ν1 , ν2 , ..., νn+1 , νt ) = (νx , νt ) be the unit outward normal at (t, x) on Σt , Throughout we assume that |νt | < |νx |. (timelikness condition of Σt ). on Σt for any t > 0,. (1.3). where |.| denotes the usual Euclidean norm. Since the regular solution u satisfy u = 0 on Σt , then all the tangential derivatives of u are also vanishing on Σt , so ∇x,t u =. ∂u .ν ∂ν. on Σt ,. which implies that u0 =. ∂u νt , ∂ν. ∇u =. 19. ∂u νx ∂ν. on Σt ..

(26) Chapter 2 Asymptotic behaviour of linear wave equations n this chapter, we study the asymptotic behaviour in time for the solution of a damped. I wave equation in a noncylindrical domain becoming unbounded, in some direction as t → ∞. If the limit of the source term is independent of this direction and t, the wave converges to the solution of an elliptic problem defined on a lower dimensional domain. The rate of convergence depends on the limit behaviour of the source term.. 20.

(27) 2.1. Problem setting. We consider, for an evolution problem, a bounded subset in Rn+1 defined as: Ωt := (−`0 − `t, `0 + `t) × ω,. t ≥ 0,. (2.1). where n is a positive integer, ω is a bounded open subset of Rn with sufficiently smooth boundary, `0 > 0 and ` are positive constants. We also consider the following noncylindrical domain, and its lateral boundary [ {s} × Ωs , Qt :=. Σt :=. [. 0<s<t. 0<s<t. {s} × ∂Ωs .. We denote the points in Qt as (t, x) = (t, x1 , X2 ), and we set     0 ∂ u u x1 .  , ∇x,t u =  X2 = (x01 , ..., x0n ) ∈ Rn , ∇u =  ∇u ∇X2 u. We are going to consider the asymptotic behavior, as t → +∞, of the solution of the following wave equation, with a dissipation term u00 − ∆u + βu0 = f. in Qt ,. (2.2). with a homogeneous boundary condition u (t, x) = 0. on Σt ,. (2.3). and two initial ones u (0, x) = u0 (x) ,. u0 (0, x) = u1 (x) .. (2.4). where the primes stand for the time derivatives, ∆ is the Laplacian operator and β > 0 is a positive constant.. 2.2. Existence and uniqueness of a weak solution. We consider the following problem defined on the noncylindrical domain Qt    u00 (t, x) + βu0 (t, x) − 4u(t, x) = f (t, x) in Qt   u(t, x) = 0 on Σt     u(0, .) = u0 (x) , u0 (0, .) = u1 (x) in Ω , 0. 21. (2.5).

(28) where the initial data and the source term satisfy u0 ∈ H01 (Ω0 ), u1 ∈ L2 (Ω0 ),. f ∈ L2 0, T ; L2 (Ωs ) ,. Now we are in a position to state our existence and uniqueness result. Existence. The idea is to transform the noncylinder problem in the cylinder problem, through the penalization function M ∈ L∞ (Ω× ]0, T [) that was introduced by Lions [26]. Transformation of the problem Let the domain S = Ω × (0, T ) as those introduced in [[26] Chap 3 page 414] where Qt ⊂ S = Ω × (0, T ) , where Ωt , Ω are verifying the assumptions (1.2). In S    u00 + βu0ε − 4uε + 1ε M u0ε = fe   ε uε = 0     u (0, x) = u e, u0 (0, x) = u e ε. 0. 1. ε. we consider the penalized problem ε > 0, on ∂Ω × ]0, T [ ,. (2.6). x ∈ Ω,. where M ∈ L∞ (Ω× ]0, T [), the penalty function defined by   1 in Ω× ]0, T [ − Qt , M (t, x) =  0 in Q , t. and fe, u e0 , u e1 be the extensions by zero outside of Q, Ω0 of f, u0 and u1 respectively. i.e.,     f   u in Qt , u0 in Ω0 , in Ω0 , 1 fe = , u e0 = , u e1 =  0  0  0 in S − Qt in Ω − Ω0 in Ω − Ω0 Accordingly, we first prove the following lemma. Lemma 18 For each ε > 0 there exists a function uε defined in the cylinder S = Ω× ]0, T [ such that uε ∈ L∞ 0, T ; H01 (Ω) , 22. (2.7).

(29) u0ε ∈ L∞ 0, T ; L2 (Ω) , u00ε ∈ L2 0, T ; H −1 (Ω) ,. 1 u00ε + βu0ε − 4uε + M u0ε = fe, ε uε (0, .) = u e0. Proof of lemma.. u0ε (0, .) = u e1. (2.8) (2.9). in L2 0, T ; H −1 (Ω) ,. (2.10). in Ω,. (2.11). in Ω.. (2.12). Let (wυ )υ∈N be a basis of H01 (Ω) and Vm = [w1 , ..., wm ] the subspace. generated by the m first vectors of the basis (wυ )υ∈N . For m ∈ N, consider the function uε m : [0, tε m ) → Vm as solutions of the following system  1 0 00 0  e, w ∀w ∈ Vm , , w) − (4u , w) + , w) + (βu i) (u , w) = f (M u  ε m εm εm εm  ε  Pm u0 m ∈ Vm , e0 in H01 (Ω) ii) uε m (0) = u0 m = i=1 αim wi → u   P  m  iii) u0 (0) = u e1 in L2 (Ω) u1 m ∈ Vm . 1m = εm i=1 βim wi → u. (2.13). The system (2.13) has solution in [0, tε m ) , 0 < tε m < T.. A priori estimates: Setting w = u0ε m (t) in (2.13), we obtain (u00ε m , u0ε m ) + (βu0ε m , u0ε m ) + (∇uε m , ∇u0ε m ) + or else 1 1d 0 2 1d 2 |uε m | + |∇uε m |2 + β |u0ε m | + 2 dt 2 dt ε. Z. Ω. 1 (M u0ε m , u0ε m ) = fe, u0ε m , ε |M. 2 (u0ε m )|. Integrating form 0 to t < tε m , we have. . 0 e dx ≤ f , uε m .. Z t Z Z 1 0 2 1 1 t 2 2 2 0 |u | + |∇uε m | + β |uε m | ds + |M (u0ε m )| dxds 2 εm 2 ε 0 0 Ω Z t 1 ≤ fe, u0ε m ds + |u1 m |2 + |∇u0 m |2 . 2 0. Applying Cauchy-Schwartz and Young inequalities, we deduce that Z t Z Z 2 t

(30)

(31) e

(32)

(33) 2 β t 0 2 0 e f , uε m ds ≤ |u | ds,

(34) f

(35) dxds + β 0 2 0 εm 0 since fe ∈ L2 (S), Then. Z Z t β t 0 2 0 e f , uε m ds ≤ C + |u | ds. 2 0 εm 0 23. (2.14).

(36) Then (2.14) becomes Z Z i βZ t 1h 0 2 1 t 2 2 2 0 |uε m | + |∇uε m | + |uε m | ds + |M (u0ε m )| dxds 2 2 0 ε 0 Ω 1 ≤ C + |u1 m |2 + |∇u0 m |2 , 2 or else. i β 1h 0 2 2 |uε m | + |∇uε m | + 2 2. Z. t. 0. 2 |u0ε m |. 1 ds + ε. Z tZ 0. Ω. 2. |M (u0ε m )| dxds ≤ C1 .. (2.15). where C1 is a positive constant independent of m and ε. Pass to the limit: Observe that the estimates (2.15) implies that there exists a subsequence of (uε m ) , which we still denote by (uε m ) and a function uε , such that  ∗   1) uε m * uε   ∗ 2) u0ε m * u0ε     3) √1 M u0 * ε. εm. in L∞ (0, T ; H01 (Ω)) , in L∞ (0, T ; L2 (Ω)) , √1 M u0 ε ε. (2.16). in L2 (0, T ; L2 (Ω)) .. It follows from (2.16), that we can pass to the limit in the approximate system (2.13), we obtain 1 u00ε + βu0ε − 4uε + M u0ε = fe, in L2 0, T ; H −1 (Ω) , ε There is no difficulty to verify that the initial conditions: So from (2.16)1 , ((2.16)2 and lemma 1.2 [chap 1 in [26]] we have. or. uε m (0) = u0 m → u e0 uε m (0) = u0 m * u e0. in H01 (Ω) ,. weakly in L2 (Ω). then (2.11) it holds. Furthermore, according to lemma 1.3 [chap 1 in [26]], we get ∗. (u00ε m , wj ) * (u00ε , wj ). in L∞ (0, T ). then according to lemma 1.2 [chap 1 in [26]] (u0ε m (0) , wj ) → (u0ε (0) , wj ) and according to (2.13)iii , we have (u0ε (0) , wj ) = (e u1 , wj ) , so (2.12) is satisfied 24.

(37) Theorem of existence Theorem 19 Assume that (1.2) holds, then given u0 ∈ H01 (Ω0 ) , u1 ∈ L2 (Ω0 ) and f ∈ L2 (Qt ) there exists a function u defined in Qt such that. u ∈ L∞ 0, T ; H01 (Ωt ) ,. (2.17). u0 ∈ L∞ 0, T ; L2 (Ωt ) , u00 ∈ L2 0, T ; H −1 (Ωt ) ,. (2.19). in L2 0, T ; H −1 (Ωt ) ,. u00 + βu0 − 4u = f,. u0 (0, .) = u1 ,. u(0, .) = u0 ,. (2.18). (2.20). in Ω0 .. Remark 20 We observe that initial conditions make sense since by (2.17), (2.18), (2.19). Proof of theorem. Observe that the estimates obtained are independent of ε too. Therefore, by the same argument used to obtain uε from uε m , which is the solution of (2.13) we can pass to the limit when ε → 0 in uε , or a subsequence, obtaining a function u independent of ε and m, such that.  ∗   uε * v   ∗ u0ε * v 0     u →v ε. in L∞ (0, T ; H01 (Ω)) , in L∞ (0, T ; L2 (Ω)) ,. (2.21). in L2 (Ω × ]0, T [). From estimate (2.15), we have. i β 1h 0 2 |uε | + |∇uε |2 + 2 2. Z. t. 0. 2 |u0ε |.

(38) 2 Z tZ

(39)

(40) 1

(41) 0

(42) √ M (uε )

(43) dxds ≤ C2 , ds +

(44) ε

(45) 0. (2.22). Ω. where C2 is a positive constant independent of ε. From (2.22) we obtain that Z TZ 2 |M (u0ε )| dxdt ≤ C2 ε, 0. Ω. we conclude that M v0 = 0. in Ω× ]0, T [ ,. but M = 1 in Ω× ]0, T [ − Qt , therefore v0 = 0 in Ω× ]0, T [ − Qt , but v(x, 0) = u e0 , v(x, 0) = 0 in Ω − Ω0 and when applying (1.2)i , we obtain that v=0. in Ω× ]0, T [ − Qt . 25.

(46) Then by (1.2)ii and if u be a restrecte of v in Q, then u ∈ L∞ 0, T ; H01 (Ωt ) .. Now, we consider the equation of the lemma i. e (u00ε , w) + (βu0ε , w) + (∇uε , ∇w) +. 1 (M u0ε , w) = (f, w) , ε. for all w ∈ H01 (Ω) .. Therefore, restricting the above equation on Qt , and with the functions M satisfy M u0ε = 0 we obtain in D0 (0, T ) (u00ε , w) + (βu0ε , w) + (∇uε , ∇w) = (f, w) ,. for all w ∈ H01 (Ωt ) .. (2.23). It follows from (2.21), that we can pass to the limit, ε → 0, in the equation (2.23) thus obtaining (2.20). In order to check up that u(0) = u0 it is sufficient to use (2.21)1 followed by an integration by parts. By the same argument, using (2.21)2 , it can be shown that u0 (0) = u1 . We observe that u00 = f + 4u − βu0 ,. (2.24). since 4 ∈ L (H01 (Ωt ) ; H −1 (Ωt )) , then. 4u ∈ L∞ 0, T ; H −1 (Ωt ). and (2.18), so that (2.24) leads to. u00 ∈ L2 0, T ; L2 (Ωt ) + L∞ 0, T ; H −1 (Ωt ) + L2 (Ωt ) ,. hence, in particular u00 ∈ L∞ (0, T ; H −1 (Ωt ) + L2 (Ωt )) . Thanks to Lemma 1.2 ( ch 1 in [26]),. that u0 is continuous [0, T ] → H −1 (Ωt ) + L2 (Ωt ) , such that u0 (0, x) = u1 , makes sense. This completes the proof of the theorem Existence of regular solutions Theorem 21 Let u0 ∈ H01 (Ω0 ) ∩ H 2 (Ω0 ) ,. u1 ∈ H01 (Ω0 ) and f ∈ H 1 0, T ; L2 (Ωt ). Then there exists a regular solution u satisfying u ∈ L∞ 0, T ; H01 (Ωt ) ∩ H 2 (Ωt ) ,. u0 ∈ L∞ 0, T ; H 1 (Ωt ) , 26. . (2.25). u00 ∈ L∞ 0, T ; L2 (Ωt ) ..

(47) Remark 22 The following identity holds Z Z 00 0 0 (u + βu − 4u) u (t)dx = f (t)u0 (t)dx, Ωt. Ωt. for a.e. t ∈ [0, T ] .. (2.26). So we can take u0 as a test function. Uniqueness The uniqueness of solution for problem (2.5) is based to the timelikness condition of Qt . Theorem 23 For Ωt , defined by (2.1), the solution of Problem (2.5) is unique under the following assumption for the timelikness condition of Σt 0 ≤ ` ≤ 1.. (2.27). Proof. Lets u, v be two solutions in the sense of the theorem 20, we put J = u − v, then we have J 00 − 4J + βJ 0 = 0. (2.28). J 0 (0) = 0 0, T ; H01 (Ωt ) ∩ H 2 (Ωt ) , J 0 ∈ L∞ 0, T ; H 1 (Ωt ) J(0) = 0,. J ∈ L∞. (2.29) (2.30). Formally first, multiply both members of (2.28) by J 0 , we deduce that i 1∂ h 0 2 2 2 (J ) + |∇J| + β (J 0 ) − div [J 0 ∇J] = 0 2 ∂t. Integrating over Qt , 0 ≤ t ≤ T, we obtain Z h Z h Z i i 1 1 2 2 0 2 0 2 |J | + |∇J| dx + β |J 0 | + |∇J|2 dx |J | dxds = 2 Ωt 2 Ω0 Qt Z Z 1 2 − (J 0 ) + |∇J|2 νt dσ + (J 0 ∇J) νx dσ. 2 Σt Σt Or else Z h Z Z h Z i i 1 1 1 2 2 0 2 0 2 0 2 |J | + |∇J| dx+ (J ) + |∇J| dx+β (J ) dxds = 2 Ωt 2 Ω0 2 Σt Qt. . ∂J ∂ν. 2. νt (|νx |2 − νt2 dσ. the second term is always non-positive because νt ≤ 0 for expending domains, then J = 0. This completes the proof of the theorem 27. !.

(48) 2.3. The limit problem. The candidate limit of u, as t → +∞, is the solution to the following elliptic problem defined on ω.    −4X2 u∞ = f∞  . u∞ = 0. in ω, (2.31) on ∂ω,. where 4X2 is the Laplace operator in X2 defined by 4X2 = ∂x20 + ... + ∂x20n , and 1. f∞ ∈ L2 (ω) ,. (2.32). then assuming that f → f∞ as t → +∞, in a sense to be made precise below, one expects that the limit problem becomes independent of t and x1 . We put w := u − u∞. and F := f − f∞ ,. since u∞ depends only on X2 , then u0 = w0 , ∆u − ∆X2 u∞ = ∆w and w satisfies w00 − ∆w + βw0 = F. in Qt ,. (2.33). with the following initial conditions w (0, x) = u0 (x) − u∞ (X2 ) , w0 (0, x) = u1 (x) . Remark 24 Note that if u∞ 6= 0 on Σt , then w 6= 0 on Σt . As a consequence, w is not a valid test function for equation (2.33).. 2.4. Special cut-off functions. For a fixed t > 1, let m be a positive integer with m ≤ t − 1. Consider the family of sets t := {(s, x1 ); t − m < s < t, − `0 − ` (m − t + s) < x1 < `0 + ` (m − t + s)} , Sm t t which is increasing in m (see Figure 1), i.e. Sm ⊂ Sm+1 , and satisfies t Sm ×ω ⊂. [. t−m<s<t. {s} × Ωs ⊂ ]t − m, t[ × R × ω. 28.

(49) The associated family of smooth cut-off functions is defined as %m = %m (s, x1 ) : [0, t] × R → R, such that. t %m = 0 in [0, t] × R \ Sm+1 ,. t 0 ≤ %m ≤ 1, %m = 1 in Sm ,. |∂x1 %m | , |%0m | ≤ θ,. A t Sm AA A A- 6 A A A A Sm+1 AA AA A -A A A A A A . , AA , A Ωs A , A A A A - x1 ω Ω0 t. −tA0 0 t Figure 2.1: The family of sets Sm. Figure 1:. where θ is a constant independent of t and m. We have in particular %m (0, x1 ) = 0 and %. such thatm. = 0 near the lateral boundary Σt . The supports of ∂x1 %m and %0m are included in. t t Sm+1 \Sm .. 2.5. t 0 ≤ %m ≤ 1, %m = 1 in Sm ,. t %m = 0 in [0, t] × R \ Sm+1 ,. Energy Estimates. |∂x1 %m | , |%0m | ≤ θ,. 2.5.1. A priori estimate. where θ is a constant independent of t and m. We have in particular %m (0, x1 ) = We can take the following estimation. 0 and %m = 0 near the lateral boundary Σt . The supports of ∂x1 %m and %0m are Lemma 25 Under the assumptions (2.25) and (2.27), the unique solution of Problem (2.5). t t included in for Sm+1 satisfies, a.e. \S t ∈m[0,. T ]. Z. Ωt. 0. 2. 2. |u (t)| + |∇u (t)| dx + β. Z. Qt. 0 2. |u | dxds ≤. 3. Local Energy Estimates. 29. Z. Ω0. u. 1 2.

(50)

(51) 2 1 +

(52) ∇u0

(53) dx + β. Z. Qt. f 2 dxds..

(54) Proof. We multiply the differential equation of problem (2.5) by u0 , we get i 1 ∂ h 0 2 2 (u ) + |∇u|2 + β (u0 ) − div (u0 ) ∇u = u0 f. 2 ∂s Integrating over Qt , 0 ≤ t ≤ T, we obtain Z h Z Z h Z i i 1 1 2 2 0 2 0 2 0 2 |u | + |∇u| dx + β |u | dxds ≤ |u | + |∇u| dx + (u0 ) .f dxds 2 Ωt 2 Qt Ω0 Qt Z Z 1 1 2 (u0 ) + |∇u|2 νt dσ + (u0 ∇u) νx dσ. − 2 Σt 2 Σt Or else 1 2 −. Z h Ωt. 1 2. Z. 0 2. 2. |u | + |∇u|. Σt. . ∂u ∂ν. 2. i. Z. 1 dx + β |u | dxds ≤ 2 Qt ! 0 2. Z. Ω0. h. 0 2. 2. |u | + |∇u|. i. dx +. Z. (u0 ) .f dxds. Qt. νt + (|νx |2 − νt2 dσ.. The third term in the left part is always non-positive because νt is non-positive and (1.3). Therefore we have Z Z h Z h Z i i 1 1 2 2 0 2 0 2 0 2 |u | dxds ≤ |u | + |∇u| dx + β (u0 ) .f dxds. |u | + |∇u| dx + 2 Ωt 2 Ω0 Qt Qt R Applying Cauchy-Schwartz and Young inequalities for the term Qt (u0 ) .f dxds, we get Z Z Z β 1 0 0 2 (u ) .f dxds ≤ (u ) dxds + f 2 dxds. 2 Qt 2β Qt Qt we deduce the result In order to derive local energy estimates we will make use of %m and its proprieties. So we have the following lemma. Lemma 26 Under the assumptions (2.25) and (2.32), the solutions of Problem (2.5) and Problem (2.31) satisfy Z Z 2 D (t) %m dx + Ωt. t ×ω Sm. D dxds ≤ c4. Z. t t ×ω \Sm (Sm+1 ). D dxds + c4. for a.e. t ∈ [0, T ] . where 2. D (t, x) := |w0 (t, x)| + |∇w (t, x)|2 . 30. Z. t Sm+1 ×ω. F 2 dxds,.

(55) Proof. • A local energy identity. Let us start by taking 2w%2m as a test function for (2.33), we get ∂ 2 2 2%m ww0 − 2%2m |w0 | − 4%0m .%m ww0 , ∂s ∂ 2 2 β%2m ww0 = β.%m w − 2β%0m %m w2 , ∂s 2%2m w 4w = 2∇ · %2m w∇w − 2∇ %2m w ∇w = 2∇ · %2m w∇w − 4%m w (∇%m · ∇w) − 2α.%2m |∇w|2 , 2%2m ww00. . =. so, (2.33) it comes that. ∂ [β%2m w2 + 2%2m ww0 ] − 2β%0m %m w2 − 2%2m |w0 |2 − 4%0m %m ww0 ∂s +2%2m |∇w|2 − 2∇ · (%2m w∇w) + 4%m w (∇%m · ∇w) = 2w%2m F.. (2.34). For some constant α > 0, multiplying (2.33) by 2αw0 %2m , yields ∂ 2 0 2 2 2 0 00 2α%m w w = α%m |w | − 2α%0m %m |w0 | ∂s ∂ 2 0 2α%m w 4w = 2α∇ · %2m w0 ∇w − 2α∇ · %2m w0 ∇w − %2m |∇w|2 + 2α%0m %m |∇w|2 ∂s 2 0 0 2 0 2 2αβ%m w w = 2αβ%m |w | , so, (2.33) it comes that ∂ 2 02 α%m |w | + α%2m |∇w|2 − 2α%0m %m |w0 |2 + 2αβ%2m |w0 |2 ∂s −2α%0m %m |∇w|2 − 2α∇ · (%2m w0 ∇w) + 4α%m w0 (∇%m · ∇w) = 2αw0 %2m F.. (2.35). Summing up the above identities (2.34) and (2.35), we get ∂ 2 2 β%m w + 2%2m ww0 + α%2m |w0 |2 + α%2m |∇w|2 − 2β%0m %m w2 − 2%2m |w0 |2 − 4%0m %m ww0 ∂s −2α%0m %m |w0 |2 + 2αβ%2m |w0 |2 + 2%2m |∇w|2 − 2∇ · (%2m w∇w) + 4%m w (∇%m · ∇w) −2α%0m %m |∇w|2 − 2α∇ · (%2m w0 ∇w) + 4α%m w0 (∇%m · ∇w) = 2w%2m F + 2αw0 %2m F.. Then collecting the terms with derivatives of %m in the right-hand side of the identity, we obtain ∂ 2 2 β%m w + 2%2m ww0 + α%2m |w0 |2 + α%2m |∇w|2 + 2 (αβ − 1) %2m |w0 |2 + 2%2m |∇w|2 ∂s = 2β%0m %m w2 + 4%0m %m ww0 + 2α%0m %m |w0 |2 + 2α%0m %m |∇w|2 −4%m w (∇%m · ∇w) − 4α%m w0 (∇%m · ∇w) + 2α∇ · (%2m w0 ∇w) +2∇ · (%2m w∇w) + 2w%2m F + 2αw0 %2m F.. 31.

(56) Integrating on Qt , and taking into account the fact that %m = 0 for t = 0 and on Σt , it comes Z h i 2 β%2m w2 (t) + 2%2m ww0 (t) + α%2m |w0 (t)| + α%2m |∇w (t)|2 dx Ωt Z h i 2 + 2 (αβ − 1) %2m |w0 | + 2%2m |∇w|2 dxds ZQt h i 2 2β%0m %m w2 + 4%0m %m ww0 + 2α%0m %m |w0 | + 2α%0m %m |∇w|2 dxds = Z Z Qt 0 [4%m w (∇%m · ∇w) − 4α%m w (∇%m · ∇w)] dxds + 2w%2m F + 2αw0 %2m F dxds. − Qt. Qt. To make the left-hand side as positive definite quadratic form, we use the fact that 1 02 0 2 2ww ≥ − βw + |w | , β. then choose α >. 1 , we obtain β 2. 2. β%2m w2 + 2%2m ww0 + α%2m |w0 | + α%2m |∇w|2 ≥ δ0 %2m |w0 | + α%2m |∇w|2 , 1 where δ0 = α − > 0. Integrating on Qt we obtain β Z Z 2 2 2 0 2 βδ0 |w0 | + 2 |∇w|2 %2m dxds δ0 |w (t)| + α |∇w (t)| %m dx + Ωt Qt Z 2 ≤ 2β%0m %m w2 + 4%0m %m ww0 + 2α%0m %m |w0 | − 4%m w (∇%m · ∇w) dxds Z Qt Z 2 0 0 + 2α%m %m |∇w| − 4α%m w (∇%m · ∇w) dxds + 2w%2m F + 2αw0 %2m F dxds. Qt. Qt. Using Young’s inequality in the two last integrals, we get Z Z 2 2 2 0 2 δ0 |w (t)| + α |∇w (t)| %m dx + βδ0 |w0 | + 2 |∇w|2 %2m dxds Ωt Qt Z 2 ≤ 2β%0m %m w2 + 4%0m %m ww0 + 2α%0m %m |w0 | − 4%m w (∇%m · ∇w) dxds Z Qt + 2α%0m %m |∇w|2 − 4α%m w0 (∇%m · ∇w) dxds Qt Z Z 2 0 2 2 +ε w + |w | %m dxds + c0 F 2 %2m dxds Qt. Qt. for some constant ε > 0. Here and in the sequel, ci denotes positive constants depending (at most) on θ, α and ω, but not on t. 32.

(57) . 1 2 • End of proof Again applying Young’s inequality AB ≤ δA + B several times for 4δ different choices of δ > 0, and noting that the supports of %0m and |∇%m | are included in 2. t t Sm+1 \Sm , we end up with Z Z 2 2 2 0 2 |w0 | + |∇w|2 %2m dxds |w (t)| + |∇w (t)| %m dx + Qt Ωt Z 2 ≤ c1 |w0 | + |w|2 + |∇w|2 dxds t t (S \Sm )×ω Z Z m+1 2 0 2 2 F 2 %2m dxds |w | + |w| %m dxds + c3 + c2 ε Qt. Qt. Applying the Poincaré inequality in the X2 −direction yields Z Z Z 2 2 2 2 2 2 |w (t)| %m dx ≤ cω |∇X2 w| %m dx ≤ cω |∇w|2 %2m dx, Ωt. Ωt. Ωt. where cω is the Poincaré constant. Choosing ε such that c2 ε < 1 and setting 2. D (t, x) := |w0 (t, x)| + |∇w (t, x)|2 . This completes the proof. 2.6 2.6.1. Main results Convergence theorems. To obtain the convergences, we use an iteration technique for we read the following result. Theorem 27 Under the assumptions (2.25), (2.27), (2.32), (2.39) and (2.38), we have u0 → 0,. ∂x1 u → 0,. ∇X2 u → ∇X2 u∞. in L2 S1t × ω , as t → +∞.. Moreover, if f = f∞ , the above convergence is exponential. Proof. First we note that Z. t t ×ω \Sm (Sm+1 ). D dxds =. Z. t Sm+1 ×ω. 33. D dxds −. Z. t ×ω Sm. D dxds,.

(58) then, dropping the first integral term in the inequality of Lemma 26, it comes that Z Z Z D dxds + c4 F 2 dxds, D dxds ≤ c4 (1 + c4 ) t Sm+1 ×ω. t ×ω Sm. where c4 > 0. We have then Z Z D dxds ≤ k. t Sm+1 ×ω. D dxds + k. F 2 dxds,. t ×ω Sm+1. t ×ω Sm+1. t ×ω Sm. Z. c4 < 1. Iterating the above inequality for m = 1, · · · , [t] − 1, ([·] denotes the 1 + c4 integer part), we have Z Z Z D dxds ≤ k D dxds + k F 2 dxds S1t ×ω S t ×ω S2t ×ω Z2 Z Z 2 2 2 ≤k D dxds + k F dxds + k F 2 dxds where k =. S3t ×ω. S3t ×ω. S2t ×ω. .. .. ≤k. Z. [t]−1. [t]−1. D dxds +. t ×ω S[t]. Noting that t − 1 < [t] ≤ t, we get Z. S1t ×ω. D dxds ≤ c5 e−γt. Z. X. k. j. Z. [t]−1. D dxds +. F dxds .. t S1+j ×ω. j=1. t ×ω S[t]. !. 2. X. kj. Z. !. F 2 dxds .. t S1+j ×ω. j=1. (2.36). where γ = − ln k and c5 > 0. In order to estimate the right-hand side, we write Z Z D dxds ≤ D dxds t ×ω S[t]. Qt. ≤ 2 ≤ 2. Z. Qt. Z. Qt. 2. |u0 | + |∇u|2 + |∇X2 u∞ |2 dxds 0 2. 2. |u | + |∇u| dxds +. 2 |∇X2 u∞ |2L2 (ω). Z t Z 0. `0 +`s. −`0 −`s. . dx1 ds.. One can easily show that |∇u∞ |L2 (ω) ≤ |f∞ |L2 (ω) , and taking into account Lemma 25, it follows that Z Z D dxds ≤ 2t t ×ω S[t]. Ω0. 1 2. u. Z 2 ≤ c6 t + t. Qt.

(59)

(60) 2 1 +

(61) ∇u0

(62) dx + β f 2 dxds 34. Z. Qt. f dxds + 4 |f∞ |2L2 (ω) `t2 + 2t0 t 2.

(63) for large t. Substituting this in (2.36), we obtain Z Z 2 D dxds ≤ c7 t + t S1t ×ω. 2. . f dxds e−γt + g (t). Qt. (2.37). where g is a positive real function defined by [t]−1. g (t) =. X. kj. Z. !. F 2 dxds .. t S1+j ×ω. j=1. The estimate (2.37) may ensure the convergence u → u∞ if we assume that g (t) → 0, as t → +∞, Z f 2 dxds = o eγt . t. (2.38) (2.39). Qt. This completes the proof of the theorem. Remark 28 The assumption (2.39) holds for example if time. In particular, this holds if Z Ωt. R. Qt. f 2 dxds grows polynomially in. |f − f∞ |2 dx → 0, as t → +∞.. Using Lemma 26 we have in particular, for m = 1, Z Z Z 2 2 2 0 0 2 |w (t)| + |∇w(t)| dx ≤ c4 |w | + |∇w| dxds + c4 S2t ×ω. Ω1. F 2 dxds,. S2t ×ω. then applying the precedent iteration for m = 2, · · · , [t] − 1 and arguing as above, we end up with the following corollary. Corollary 29 Under the assumptions (2.25), (2.27), (2.32), (2.39) and (2.38), we have u0 (t) → 0,. ∂x1 u (t) → 0,. ∇X2 u (t) → ∇X2 u∞. in L2 (Ω1 ) , as t → +∞.. Moreover, if f = f∞ , the above convergence is exponential. Finally, iterating m between [t]−1. X. t 2. kj. j=[ 2t ]. and [t] − 1 as above, and assuming that ! Z F 2 dxds. t S1+j ×ω. leads to the following corollary. 35. → 0, as t → +∞,. (2.40).

(64) Corollary 30 Let O be a bounded subset of R × ω, then we can choose t large enough such that O ⊂⊂ Ω 2t , and under the assumptions (2.25), (2.27), (2.32), (2.38) and (2.40), we have u0 (t) → 0,. ∇X1 u (t) → 0,. in L2 (O) , as t → +∞.. ∇X2 u (t) → ∇X2 u∞. Moreover, if f = f∞ , the above convergence is exponential. This means that in any interior bounded region of R × ω, the local energy decays to 0.. 2.6.2. Exponential convergences. The rate of convergence, in the precedent theorems, depends essentially on the convergence of f towards f∞ . We now give some examples of source terms, other than the trivial case f = f∞ , where we get an exponential rate of convergence in Theorem 27. • Assume that. Z. Ωt. |f (t) − f∞ |2 dx ≤ C1 e−γ1 t ,. for ∀t > 0,. for some positive constants C1 , γ1 such that γ1 > 0. In one hand the necessary assumption (2.39) holds since Z. Qt. 2. f dxds ≤. Z. 2. Qt. |f∞ | dxds + C1. for large t. On the other hand we have Z Z t Z 2 F dxds ≤ t S1+j ×ω. t−(1+j). ≤ C1 Then, since k j = e−γj , Z j k. t S1+j ×ω. i.e.. Z. Z. 0. t. e−γ1 s ds ≤ c9 t2. F 2 dxds. Ωs. t. t−(1+j). e−γ1 s ds ≤. C1 (1 + j)e−γ1 t × eγ1 (1+j) . γ1. F 2 dxds ≤ c10 te−γ1 t × e(γ1 −γ)j , . g (t) ≤ c10 te−γ1 t  36. [t]−1. X j=1. 1 ≤ j ≤ [t] − 1, . e(γ1 −γ)j  .. (2.41).

(65) If γ1 < γ then. P[t]−1 j=1. e(γ1 −γ)j is bounded, and if γ1 > γ then. P[t]−1 j=1. e(γ1 −γ)j ≤ c11 e(γ1 −γ)t .. Thus, due to (2.37) and (2.41), we end up with Z 2 |u0 | + |∇ (u − u∞ )|2 dxds ≤ M1 t3 e− min{γ,γ1 }t , S1t ×ω. for some positive constant M1 . • Another assumption, analogous to that made in [31], is to assume that Z t Z |f − f∞ |2 dxds ≤ C2 e−γ2 t , for ∀t > 0, t−1. Ωs. for some positive constants C2 , γ2 such that γ2 > 0. Then assumption (2.39) holds and again R we have Qt |f (t)|2 dxds ≤ c12 t2 . To estimate g, we note that Z. t S1+j ×ω. and it follows that Z. F 2 dxds ≤. t S1+j ×ω. Z. t. t−(1+j). Z. F 2 dxds =. Ωs. Z j X i=0. t−(j−i)−1. F 2 dxds ≤ C2 (j + 1) e−γ2 t ≤ C2 te−γ2 t ,. Thus. . g (t) ≤ C2 te−γ2 t . [t]−1. X j=1. t−(j−i). Z. F 2 dxds. Ωs. 1 ≤ j ≤ [t] − 1.. . e−γj  ≤ c13 t e−γ2 t ,. and due to (2.37), it follows that Z 2 |u0 | + |∇ (u − u∞ )|2 dxds ≤ M2 t3 e− min{γ,γ2 }t , S1t ×ω. for some positive constants M2 .. 37. !.

(66) Chapter 3 Asymptotic behaviour of nonlinear wave equations e study the asymptotic behaviour for the solution of nonlinear wave equations in a. W noncylindrical domain, becoming unbounded in some directions, as the time t goes to infinity. If the limit of the source term is independent of these directions and t, the wave converges to the solution of an elliptic problem defined on a lower dimensional domain. The rate of convergence depends on the limit behaviour of the source term and on the coefficient of the nonlinear term.. 38.

(67) 3.1. Problem setting. Let as denote the points in Rn1 × Rn2 as x = (X1 , X2 ) = x1 , ..., xn1 , x01 , ..., x0n2 ,. where n1 , n2 are positive integers and we set. ∇X1 u = (∂x1 u, ..., ∂xn1 u)T , ∇X2 u = (∂x01 u, ..., ∂x0n2 u)T ,     ∇X1 u u0  , ∇x,t u =  . ∇u =  ∇X2 u ∇u. Then we consider a time-dependent family of bounded subsets in Rn1 × Rn2 defined as Ωt := (−`0 − `t, `0 + `t)n1 × ω,. t ≥ 0,. where ω is a bounded open subset of Rn2 . In this chapter we study the asymptotic bihaviour of the solution of the following nonlinear wave equation when t → +∞    u00 (t, x) + βu0 (t, x) − 4u(t, x) + γ(t) |u|ρ u = f (t, x) in Qt ,   u(t, x) = 0 on Σt ,     u(0, .) = u0 (x) , u0 (0, .) = u1 (x) in Ω ,. (3.1). 0. where Qt is a noncylindrical domain of Rn1 × Rn2 with lateral boundary Σt such that Qt :=0<s<t {s} × Ωs ,. Σt :=0<s<t {s} × ∂Ωs .. The prime stands for the time derivative, 4 is the Laplace operator, β is a positive constant and γ(t) is a nonnegative function.. 39.

(68) 3.2. Existence and uniqueness of regular solutions. Transformation problem in the cylindrical domain Let us consider the subsets Ωs of Rn1 × ω, given by Ωs := {(X1 , X2 ) ∈ Rn1 × ω \ X1 = K(s)Y1 ,. Y1 = (−`0 , `0 )n1 } ,. ` s. `0 At first we will study our problem in a cylinder Q = (0, s) × Ω,. s ∈ (0, t) ,. where K(s) = 1 +. Ω = (−`0 , `0 )n1 × ω, where. domains Q and Qt are related by the diffeomorphism h defined by h : Qt → Q X1 (s, x) → s, , X2 , K(s) so h−1 defined by h−1 : Q → Qt. (3.2). (s, y) → (s, K(s)Y1 , X2 ) . For each u ∈ L2 (Qt ) ; v(s, Y1 ) = u(s, K(s)Y1 , X2 ). By change of variables X1 = K(s)Y1 , we obtain v ∈ L2 (Q) . Our object consists of changing of variables v(s, Y1 ) = u(s, K(s)Y1 , X2 ).. Under this transformation problem (3.1) in Qt is formulated in the cylindrical domain Q as follows n +n  n1P +n2 ∂ 1P 2 ∂v ∂v 0  00 0  v (t, y) + βv (t, y) − a (t, y) + b (t, y)  ij i  ∂yj ∂yi  i,j=1 ∂yi i=1   n1P +n2  ∂v   + ci (t, y) + γ (t) |v (t, y)|ρ v (t, y) = Φ (t, y)    ∂y i=1. in Q,. i. v=0       v(0) = v 0 (y) = u0 (K(0)Y1 ) = u0 (Y1 )    +n2 ∂v 0  K 0 (0) n1P    v 0 (0) = v 1 (y) = u1 (K(0)y) + yi K(0) i=1 ∂yi. 40. on Σ = ∂Ω × (0, t) , Y1 ∈ Ω, Y1 ∈ Ω..

(69) where f (t, x) = f (s, K(s)Y1 , X2 ) = Φ(t, y), aij (t, y) = χij − K 02 yi yj K −2 " −2 2 # ` ` 1+ t = χij − yi yj , χij the characteristic function, `0 `0 −1 `0 0 −1 +t yi , bi (t, y) = −2K K yi = −2 ` ci (t, Y1 ) = 1 − (n1 + n2 ) K 02 − K 002 K − βK 0 K K −2 yi −2 `0 `0 = 1 − (n1 + n2 ) − β +t +t yi . ` ` Operators Let us consider the following family of operators in L (H01 (Ω) , H −1 (Ω)) ( i.e., L (H01 (Ω) , H −1 (Ω)) = the continuous linear application space of H01 (Ω) in H −1 (Ω)). X ∂ ∂ A (s) = − aij (t, y) , t ≥ 0. ∂ ∂y y i 16i,j6n +n 1. 2. where aij = aji We suppose that. and aij ∈ W 3,∞ 0, T ; C 0 Ω for all i, j = 1, ..., n1 + n2 . X. 16i,j6n1 +n2. (3.3). aij (s, y)ζi ζj > λ kζk2. where λ is a positive constant. If we denote by a (s, ., .) the family of bilinear forms associated with A (s) , we have X Z ∂u ∂v a (s, u, v) = aij (t, y) dy, u, v ∈ H01 (Ω) . ∂ ∂ yi yj 16i,j6n Ω From the hypothesis on aij , we obtain that a (t, u, v) is symmetric and a (t, u, u) ≥ λ kuk2H 1 (Ω) 0. for all u ∈ H01 (Ω). 41. s ∈ (0, t) .. (3.4).

(70) Suppose that functions ρ, γ, f, u0 , u1 , K satisfy the following conditions • Concerning the speed of expansion, in the n1 first directions, it satisfies 0 ≤ ` ≤ 1.. (3.5). This ensures that Σt satisfies the so-called timelikness condition |νt | ≤ |νx | on Σt , for t > 0, • The nonlinear term in Problem (3.1) is subject to the following assumptions (Recall that x ∈ Rn1 +n2 ).  2  0<ρ≤ if n1 + n2 > 2, (n1 +n2 )−2  0<ρ≤∞ if n1 = n2 = 1.. γ(t) ≥ 0,. γ 0 (t) ≤ 0,. γ(t), γ 0 (t) ∈ L∞ (0, t). (3.6) (3.7). • The initial data and the source term satisfy u0 ∈ H02 (Ω0 ), u1 ∈ H01 (Ω0 ), • H.1:. f ∈ H 1 0, t; L2 (Ωs ) ,. (3.8). K ∈ C 4 (0, t) , min K(t) = α0 > 0,. 0≤s<t. sup K 0 (t) = <. max K(t) = α1 > 0,. 0≤s<t. 1 , M. M = sup {|y| , y ∈ Ω} , Rn1 +n2

(71)

(72) K 0 (s) ≥ 0, |K 00 (s)| , |K 000 (s)| ,

(73) K (4) (s)

(74) ≤ C, ∀s ∈ (0, t) , Rt Rt m1 = 0 K 0 (s)ds < ∞, m2 = 0 |K 00 (s)| ds < ∞,

(75) Rt Rt

(76) m3 = 0 |K 000 (s)| ds < ∞, m4 = 0

(77) K (4) (s)

(78) ds < ∞, Rt Rt m5 = 0 (K 0 (s))2 ds < ∞, m6 = 0 (K 00 (s))2 ds < ∞, Rt Rt m7 = 0 (K 0 (s))3 ds < ∞, m8 = 0 |K 00 (s)|3 ds < ∞. 0≤s<t. Now we are in a position to state our existence and uniqueness result.. 42. (3.9).

(79) Existence of regular solutions Theorem 31 Let t > 0. Under the assumptions eqreftlike (3.8), Then there exists a unique solution for Problem (3.1), in the sense that u ∈ L∞ 0, t; H01 (Ωs ) ∩ H 2 (Ωs ) ,. u0 ∈ L∞ 0, t; H 1 (Ωs ) ,. u00 ∈ L2 0, t; L2 (Ωs ) .. We can take u0 as a test function, i.e. the following identity holds Z Z ρ 00 0 0 (u + βu − 4u + γ(s) |u| u) u (s)dx = f (s) u0 (s)dx for a.e. s ∈ (0, t) . Ωs. (3.10). Ωs. Remark 32 Here and in the sequel we use notations of [17]. At first we will study our problem in a cylinder Q. Taking into account H.1, we can verify that • H.2:. aij = aji. and aij ∈ W 3,∞ 0, t; C 0 Ω ,. a(t, v, v) > λ kvk2. in Q. (λ > 0) .. Let f, u0 , u1 satisfy (3.8), by (3.2), we obtain v 0 ∈ H02 (Ω), v 1 ∈ H01 (Ω).. (3.11). Theorem 33 Under conditions of Theorem 31, for any f ∈ H 1 (0, t; L2 (Ω)) , there exists a unique function v(t, y) satisfying initial data (3.11) v ∈ L∞ 0, t; H01 (Ω) ∩ H 2 (Ω) ,. v 0 ∈ L∞ 0, t; H01 (Ω) ,. for t > 0, the identity holds n +n  n1P +n2 ∂ 1P 2 ∂v ∂v 0  00 0  v + βv − a + b ij i  ∂yj ∂yi i,j=1 ∂yi i=1  n1P +n2 ∂v    + ci + γ (t) |v|ρ v ∂yi i=1. where w is an arbitrary function from L2 (Ω) .. 43.       . v 00 ∈ L2 0, t; L2 (Ω) , .  , w (t)  = (Φ, w) (t). (3.12). (3.13).

(80) Proof. For small ε > 0 we consider in a cylinder Q the following nonlinear problem n +n  n1P +n2 ∂ +n2 ∂v 1P 2 ∂vε0 n1P ∂vε  0 00  − + βv b ci K v a + + + γ (t) |vε |ρ vε = Φ in Q,  i ε ij ε ε  ∂yj ∂yi ∂yi  i,j=1 ∂yi i=1 i=1     v =0 on ∂Ω × (0, t) , ε   vε (0) = v 0 (Y1 ) = u0 (K(0)Y1 ) = u0 (Y1 ) Y1 ∈ (−`0 , `0 )n1 ,    +n2 ∂v 0  ` n1P    vε0 (0) = v 1 (y) = u1 (y) + yi Y1 ∈ (−`0 , `0 )n1 . `0 i=1 ∂yi (3.14). where Kε = 1 + ε.. Let (wν )ν∈N be a basis in H02 (Ω) . For each m ∈ N we define vm,ε (t, y) =. m X. Ψrmε (t) wr (y). r=1. where unknown functions Ψrmε (t) are solutions to the following Cauchy problem for the system of ordinary differential equations 0 +n2 ∂vm,ε K 0 n1P , wr + + a(t, vm,ε , wr ) − 2 yi K i=1 ∂yi n1P +n2 ∂vm,ε , wr + (γ (t) |vm,ε |ρ vm,ε , wr ) + ci (t) ∂y i i=1. 00 (Kε vm,ε , wr ). = (Φ, wr ). 0 β(vm,ε , wr ). 1 ≤ r ≤ m,. Ψrmε (0) = v 0 , wr ,. (3.15). Ψ0rmε (0) = v 1 , wr .. This problem has solutions Ψrmε ∈ C 2 ((0, Tmε )), 0 < Tmε < T.. A priori estimate 1. In our calculations we will omit indices m, ε. Multiplying (3.15) by 2Ψ0r , summing over r, using the hypothesis H.2, (3.6) and(3.7) we get i

(81) 2 d h

(82)

(83) √ Kε v 0 (t)

(84) + a(t, v (t) , v (t)) + (γ (t) , N (u)) + 2β |v 0 (t)|2 − (γ 0 (t) , N (u)) dt 0 +n2 n1P +n2 K 0 (t) n1P ∂v 0 ∂v 0 0 −a (t, v (t) , v (t)) − 4 , v yi + 2 ci (s) ,v K(t) i=1 ∂yi ∂yi i=1. (3.16). = 2 (Φ (t) , v 0 ) , where N (u) =. Ru 0. |s|ρ sds ≥ 0. Integrating (3.16) from 0 to t, using the hypothesis H.1-H.2,. (3.6) and (3.7), and observing that Kε (v 0 )2 ≥ (v 0 )2 ≥ 0, we obtain Z t Z t 2 2 2 2 0 |v (t)| + λ kvk ≤ C + γ1 (s) |vs | + kvk (s) ds + |Φ|2 (s) ds, 0. 0. 44. (3.17).

(85) where C is a positive constant independent of m and t, γ1 ∈ L1 (0, t) . Hence, by Gronwall’s Lemma 2. 0. 2. |v (t)| + λ kvk + β. Z. 0. t. |vs |2 ds ≤ C. (3.18). A priori estimate 2. Now we differentiate equation (3.14) with respect to t, multiply the result by 2Ψ00r and sum over r to obtain i

(86) 2 d h

(87)

(88) √ Kε v 00 (t)

(89) + a(t, v 0 (t) , v 0 (t)) + 2a0 (t, v (t) , v 0 (t)) + 2β |v 00 (t)|2 − 2a00 (t, v (t) , v 0 (t)) dt 0 0 n1P +n2 n1P +n2 ∂v ∂v −3a0 (t, v 0 (t) , v 0 (t)) + 2 bi , v 00 + 2 ci , v 00 ∂yi ∂yi i=1 i=1 0 +2 (γ (t) |v|ρ v) , v 00 = 2 (Φ0 (t) , v 00 ) , (3.19) Integrating (3.19) from 0 to t, using the hypothesis H.1-H.2, (3.6) and (3.7) and observing that Kε (v 0 )2 ≥ (v 0 )2 ≥ 0, we have. Rt |v 00 (t)|2 + λ kv 0 (t)k2 + β 0 |vss |2 ds ≤ C1 + |(v 00 (0) , v 00 (0))| Rt Rt 2 2 γ (s) |v | + kv k (s) ds + |Φs |2 (s) ds, 1 ss s 0 0. Remark 34 We need an estimate for v 00 (0). Putting t = 0 in (3.14), we obtain |vtt (0)| ≤ C, where C does not depend of m and t. Now, using the above Remark, observing that γ1 ∈ L1 (0, t) , by Gronwall’s Lemma we get Z β t 2 2 00 |vss |2 ds ≤ C. (3.20) |v (t)| + λ kv(t)k + 4 0 where C is a positive constant independent of m and t. Let us now study the nonlinear term. Since γ ∈ L1 (0, t) ∩ L∞ (0, t) , we have from (3.18) and (3.20). By compactness arguments.

(90)

(91)

(92) γ (t) |vm,ε |ρ+1

(93). L2 (0,T ;L2 (Ω)). γ (t) |vm,ε |ρ vm,ε → γ (t) |v ε |ρ v ε. ≤ C.. a.e in Q,. (3.21). m → ∞.. (3.22). From (3.21), (3.22), we conclude γ (t) |vm,ε |ρ vm,ε → γ (t) |vε |ρ vε 45. weakly in L2 (Q) .. (3.23).

(94) From the a priori estimates obtained we can see that there exists a subsequence of (vm,ε ), which we still denote by ((vm,ε )m∈N , such that  ∗   vm,ε * vε in L∞ (0, t; H01 (Ω)) ,     ∗  v0 * in L∞ (0, t; H01 (Ω)) , vε0 m,ε ∗  00 00  vm,ε * vε00 in L2 (Q) , Kε vm,ε * Kε vε00      γ (t) |v |ρ v * γ (t) |v |ρ v in L2 (Q) . m,ε m,ε ε ε. in L∞ (0, t; L2 (Ω)) ,. Letting m tend to ∞, we deduce that n +n 1P 2 ∂ ∂v ε (Kε vε00 , w) + β(vε0 , w) + aij (t, y) , w (t) ∂yi i=1 ∂yi n +n n1P +n2 ∂v 1P 2 ∂v 0 ε ci , w (t) + (γ (t) |vε |ρ vε , w) (t) + bi ε , w (t) + ∂yi ∂yi i=1 i=1 = (Φ, w) (t). where w is an arbitrary function from H01 (Ω) . Obviously, initial conditions (3.14) are satisfied. Observe that estimates obtained are also independent of ε. Therefore, by the same argument we can pass to the limit when ε goes to zero in {vε }. Thus we obtain a function v ∈ L∞ 0, t; H01 (Ω) ,. v 0 ∈ L∞ 0, t; H01 (Ω) ,. v 00 ∈ L2 (Q) ,. v 00 ∈ L∞ 0, t; L2 (Ω). satisfying the identity n1P +n2 ∂ ∂v aij (t, y) , Z (t) = ∂yi ∂yi i=1 n1P +n2 ∂v ∂v 0 ρ 00 0 + ci − γ (t) |v| v , Z (t) Φ − v − βv − bi ∂yi ∂yi i=1 ≡ (P (t, y) , Z) (t). where Z is an arbitrary function from H01 (Ω) and P ∈ L2 (Ω).. It follows from the properties of a function v(t, y) that P (t, y) ∈ L∞ (0, t; L2 (Ω)). The theory of elliptic equations gives us v ∈ L∞ 0, T ; H01 (Ω) ∩ H 2 (Ω) .. This completes the existence part of Theorem 33. 46.

(95) Uniqueness Let v1 , v2 be two distinct solutions to (3.13). Putting w = 2(v1 − v2 ), we obtain  n +n 1P 2 d 0 ∂w0 0  2 2 0 0  |w (t)| + a(t, w (t) , w (t)) + 2β |w (t)| − a (t, w (t) , w (t)) + 2 bi ,w    dt ∂yi i=1   n1P +n2 ∂w   +2 ci , w0 + 2 (γ (t) |v1 |ρ v1 − γ (t) |v2 |ρ v2 , w0 ) = 0, ∂y i i=1     w=0 on Σ      w (0) = 0, w (0) = 0. t. (3.24). Green’s formula gives. nX 1 +n2. 2. i=1. and −. nX 1 +n2 i=1. nX 1 +n2 ∂bi ∂w0 0 0 2 ,w = − , (w ) bi ∂yi ∂y i i=1 . ∂bi , (w0 )2 ∂yi. . =. nX 1 +n2. 2K 0 K −1 , (w0 )2. i=1. With regard to the nonlinear term, we obtain. . 2 |(γ (t) |v1 |ρ v1 − γ (t) |v2 |ρ v2 , w0 )| Z ≤ 2γ (t) |(|v1 |ρ v1 − |v2 |ρ v2 , w0 )| dy ΩZ ≤ 2Cρ γ (t) (|v1 |ρ v1 − |v2 |ρ v2 ) |w| |w0 | dy.. (3.25). (3.26). Ω. Since injection. H01. q. (Ω) ,→ L (Ω) is continuous, if. 1 n1 +n2. +. 1 2. +. 1 q. = 1 and ρ (n1 + n2 ) ≤ q,. then |u|ρLp , |v|ρLp ∈ Ln1 +n2 (Ω) . From (3.26), we find 2 |(γ (t) |v1 |ρ v1 − γ (t) |v2 |ρ v2 , w0 )| ≤ 2Cρ γ (t) kwk |w0 | .. (3.27). Integrating (3.24) from 0 to t < ∞, using the hypothesis H.1−H.2, (3.3), (3.4), (3.24), (3.25), (3.27) and the inequality of Schwartz, we have R Rt |w0 (t)|2 + λ Ω |∇w (t)|2 dy + 2β 0 |ws (s)|2 ds RtR + 0 Ω 2 (n1 + n2 ) K 0 K −1 |ws (s)|2 dyds Rt Rt Rt ≤ Cε 0 γ1 (s) |ws (s)|2 ds + Cε 0 γ1 (s) |∇w (s)|2 ds + ε 0 |ws (s)|2 ds, 47.

(96) From here 0. 2. |w (t)| + λ. Z. 2. Ω. |∇w (t)| dy ≤ C. where γ2 (t) = max {γ (t) , γ2 (t)}. Z. t. 0. γ2 (s) |∇w (s)|2 + |ws (s)|2 ds,. ∀t ∈ (0, ∞) .. Since γ2 (t) ∈ L1 (0, T ), we have by Gronwall’s lemma ∇w (t) ≡ 0 a.e. t ∈ (0, ∞). With w − Σ = 0 we conclude that w (t) ≡ 0 in Q, hence v1 = v2 . The proof of the uniqueness part of Theorem 33 is completed. Proof of theorem 32.. Let v be the solution from Theorem 33 and u defined by (3.11).. Then u ∈ L∞ (0, t; H01 (Ωs ) ∩ ∈ H 2 (Ωs )) , u00 ∈ L2 (Qt ) ,. u0 ∈ L∞ (0, t; H 1 (Ωs )) ,. u00 ∈ L∞ (0, t; H01 (Ωs ) ,. u(0) = u0 ,. u0 (0) = u1 .. If w ∈ L2 (0, t; H01 (Ωs )), let φ (t, y) = w (t, K(t)Y1 ) for (t, y) ∈ Q. We note that (3.13) is valid. Changing the variable X1 = K(t)Y1 , we obtain (3.10) from (3.13). Let u1 , u2 be two solutions to (3.10), and v1 , v2 be the functions obtained through the isomorphism h. Then u1 , u2 are the solutions to (3.13). By the uniqueness result of Theorem 33, we have v1 = v2 , so u1 = u2 . Thus the proof of Theorem 31 is completed. 3.3. Limit problem. We assume that the source term f becomes independent of the variables (t, X1 ), i.e. f (t, X1 , X2 ) → f∞ (X2 ). as t → +∞,. for some f∞ ∈ L2 (ω) , To handle the nonlinear term, in the estimations below, we need to assume that γ (t) → 0. as t → +∞,. The sense of these two convergences will be made precise below. 48. (3.28).

(97) Passing formally to the limit in (3.1), we obtain the candidate limit of u(t), as t → +∞, is the solution of the elliptic problem defined on ω   −4 u = f in ω, X2 ∞ ∞  u∞ = 0 on ∂ω,. (3.29). where is the Laplace operator in X2 defined by 4X2 = ∂x20 + ... + ∂x20n , so the limit problem 1. 2. to become independent of (t; X1 ), as t → +∞. Remark 35 the problem (3.29) has a unique solution u ∈ H01 (ω) and one can check easily that |∇X2 u∞ |L2 (ω) ≤ |f∞ |L2 (ω). (3.30). Remark 36 By the Sobolev embedding theorem (Recall that ω ⊂ Rn2 ), we have • if n2 ∈ {1, 2} , then H 1 (ω) ⊂ Lρ+2 (ω). for 0 < ρ ≤ ∞.. • if n2 ≥ 3, then due to (3.6) we have 0 < ρ ≤ 2 , hence H 1 (ω) ⊂ Lρ+2 (ω) . n2 − 2. 2 which implies that 0 < ρ ≤ (n1 + n2 ) − 2. Therefore, under assumption (3.6), it holds that |u∞ |Lρ+2 (ω) ≤ Cs |∇X2 u∞ |L2 (ω) , for n2 ≥ 1 and some constant Cs depending only on ω. Combining this inequality with (3.30) we have |u∞ |Lρ+2 (ω) ≤ Cs |f∞ |L2 (ω) .. 3.4. (3.31). Special cut-off functions. To estimate the converge of u(t) towards u∞ , we consider the functions w (t, X1 , X2 ) := u (t, X1 , X2 ) − u∞ (X2 ). and F (t, X1 , X2 ) := f (t, X1 , X2 ) − f∞ (X2 ) ,. for (X1 , X2 ) ∈ Ωt and t ≥ 0. Since u∞ depends only on X2 , then the function w satisfies the equation w00 + βw0 − 4w + γ(t) |u|ρ u = F 49. in Qt ,. (3.32).