Existence and exponential stability of the damped wave equation with a dynamic boundary condition and a delay term.

Existence and exponential stability of the damped wave equation with a dynamic boundary condition and a delay term. Stéphane Gerbi LAMA, Université de Savoie, Chambéry, France Fachbereich Mathematik und Statistik, Universität Konstanz Donnerstag, 09.12.2010 Joint work with Belkacem Said-Houari actually at Konstanz University S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 1 / 32

Outline of the talk 1 Introdution 2 Well-posedness of the problem : existence and uniqueness. Setup and notations Semigroup formulation : existence and uniqueness. 3 Asymptotic behavior Exponential stability for µ 2 < µ 1 Exponential stability for µ 2 > µ 1 and α > (µ 2 µ 1 )B 2 4 Some remarks S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 2 / 32

Position of the problem Consider the damped wave equation, with delay: dynamic boundary conditions and time u tt u α u t = 0, x Ω, t > 0, u(x, t) = 0, x Γ 0, t > 0, ( ) u α ut u tt(x, t) = (x, t) + (x, t) + µ1ut(x, t) + µ2ut(x, t τ) ν ν x, t > 0, u(x, 0) = u 0(x) x Ω, u t(x, 0) = u 1(x) x Ω, u t(x, t τ) = f 0(x, t τ) x, t (0, τ), (1) where u = u(x, t), t 0, x Ω which is a bounded regular domain of R N, (N 1), Ω = Γ 0, mes(γ 0) > 0, Γ 0 =, α, µ 1, µ 2 > 0 and u 0, u 1, f 0 are given functions. Moreover, τ > 0 represents the time delay Questions to be asked : 1 Existence, uniqueness and global existence? 2 Is the stationary solution u = 0 stable and what is the rate of the decay? S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 3 / 32

Dynamic boundary condition u tt u α u t = 0, x Ω, t > 0, u(x, t) = 0, x Γ 0, t > 0, ( ) u α ut u tt(x, t) = (x, t) + (x, t) + µ1ut(x, t)) ν ν x, t > 0, u(x, 0) = u 0(x) x Ω, u t(x, 0) = u 1(x) x Ω, Longitudinal vibrations in a homogeneous bar in which there are viscous effects, Artificial boundary condition for unbounded domain : transparent and absorbing, and a lot of mix between these two types, Ω is an exterior domain of R 3 in which homogeneous fluid is at rest except for sound waves. Each point of the boundary is subjected to small normal displacements into the obstacl. This type of dynamic boundary conditions are known as acoustic boundary conditions. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 4 / 32

Related works : damped waves with dynamic boundary conditions In the absence of delay, and with a nonlinear source terms, Gerbi and Said-Houari [GS2008, GS2010] showed the local existence, an exponential decay when the initial energy is small enough, an exponential growth when the initial energy is large enough and a blow-up phenomenon for linear boundary conditions (m = 2) u tt u α u t = u p 2 u, x Ω, t > 0 u(x, t) = 0, x Γ 0, t > 0 [ ] u α ut u tt(x, t) = (x, t) + ν ν (x, t) + r ut m 2 u t(x, t) x, t > 0 u(x, 0) = u 0(x), u t(x, 0) = u 1(x) x Ω. [GS2008] S. Gerbi and B. Said-Houari, Local existence and exponential growth for a semilinear damped wave equation with dynamic boundary conditions. Advances in Differential Equations Vol. 13, No 11-12, pp. 1051-1074, 2008. [GS2010] S. Gerbi and B. Said-Houari, Asymptotic stability and blow up for a semilinear damped wave equation with dynamic boundary conditions. Mathematical Methods in the Applied Sciences in press (2010). S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 5 / 32

Related works : delay term in 1D Datko [Dat91], showed that solutions of : w tt w xx aw xxt = 0, x (0, 1), t > 0, w (0, t) = 0, w x (1, t) = kw t (1, t τ), t > 0, a, k, τ > 0 become unstable for an arbitrarily small values of τ and any values of a and k. Datko et al [DLP86] treated the following one dimensional problem: u tt(x, t) u xx(x, t) + 2au t(x, t) + a 2 u(x, t) = 0, 0 < x < 1, t > 0, u(0, t) = 0, t > 0, (2) u x(1, t) = ku t(1, t τ), t > 0, If k e2a + 1 < 1 then the delayed feedback system is stable for all sufficiently small e 2a 1 delays. If k e2a + 1 > 1, then there exists a dense open set D in (0, ) such that for e 2a 1 each τ D, system (2) admits exponentially unstable solutions. [Dat91] R. Datko. Two questions concerning the boundary control of certain elastic systems. J. Differential Equations, 92(1):27 44, 1991. [DLP86] R. Datko, J. Lagnese, and M. P. Polis. An example on the effect of time delays in boundary feedback stabilization of wave equations. SIAM J. Control Optim., 24(1):152 156, 1986. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 6 / 32

Related works : wave equations and boundary feedback delay Nicaise and Pignotti,[NP06], examined a system of wave equation with a linear boundary damping term with a delay: u tt u = 0, x Ω, t > 0, u(x, t) = 0, x Γ 0, t > 0, u (x, t) = µ1ut(x, t) + µ2ut(x, t τ) x Γ1, t > 0, ν u(x, 0) = u 0(x), x Ω, u t(x, 0) = u 1(x) x Ω, u t(x, t τ) = g 0(x, t τ) x Ω, τ (0, 1), (3) and proved under the assumption µ 2 < µ 1 that null stationary state is exponentially stable. They also proved instability if this condition fails. They also studied [NP08, NVF09], internal feedback, time-varying delay and distributed delay. [NP06] S. Nicaise and C. Pignotti. Stability and instability results of the wave equation with a delay term in the boundary or internal feedbacks. SIAM J. Control Optim., 45(5):1561 1585, 2006. [NP08] S. Nicaise and C. Pignotti. Stabilization of the wave equation with boundary or internal distributed delay. Diff. Int. Equs., 21(9-10):935 958, 2008. [NVF09] S. Nicaise, J. Valein, and E. Fridman. Stabilization of the heat and the wave equations with boundary time-varying delays. DCDS-S., S2(3):559 581, 2009. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 7 / 32

Main results We show, [GS11] that if Coefficient condition (B defined later) µ 2 < µ 1 or µ 2 µ 1 and α > (µ 2 µ 1)B 2 Problem (1) has a [GS11] unique global solution, this solution decays exponentially to the null solution. S. Gerbi and B. Said-Houari. Existence and exponential stability of the damped wave equation with a dynamic boundary condition and a delay term. Submitted S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 8 / 32

Outline 1 Introdution 2 Well-posedness of the problem : existence and uniqueness. Setup and notations Semigroup formulation : existence and uniqueness. 3 Asymptotic behavior Exponential stability for µ 2 < µ 1 Exponential stability for µ 2 > µ 1 and α > (µ 2 µ 1 )B 2 4 Some remarks S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 9 / 32

Setup and notations First we reformulate the boundary delay problem, then by a semigroup approach and the Hille-Yosida theorem we will the global existence. H 0 (Ω) = { u H 1 (Ω)/ u Γ0 = 0 }, γ 1 the trace operator from H 0 (Ω) on L 2 () and ( H 1/2 () = γ 1 H 1 Γ0 (Ω) ). E(, L 2 (Ω)) = { u H 1 (Ω) such that u L 2 (Ω) } For u E(, L 2 (Ω)), u ν H 1/2 () and we have Green s formula: Ω u(x) v(x)dx = Ω u(x)v(x)dx + u ν ; v v H 0 (Ω), where.;. Γ1 means the duality pairing H 1/2 () and H 1/2 (). We denote by the same constant B the best constant of the embedding H 1 0 (Ω) L 2 (Ω), the Poincaré s inequality and the trace inequality: u H 1 Γ 0 (Ω), u 2 B u 2 u H 1 Γ 0 (Ω), u 2,Γ1 B u 2. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 10 / 32

Relations between the coefficient Consider the following two cases : case 1: µ 2 < µ 1. We may define a positive real number ξ such that: τµ 2 ξ τ (2µ 1 µ 2 ). (4) case 2: µ 2 µ 1. We will suppose that the damping parameter α verifies: α > (µ 2 µ 1)B 2. (5) In this case, we may define a positive real number ξ satisfying the two inequalities: ξ τµ 2, (6) ( µ2 α > 2 + ξ ) 2τ µ1 B 2 > 0. (7) S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 11 / 32

Reformulation of the delay term As in [NP06], we add the new variable: z (x, ρ, t) = u t (x, t τρ), x, ρ (0, 1), t > 0. (8) Then, we have τz t (x, ρ, t) + z ρ (x, ρ, t) = 0, in (0, 1) (0, + ). (9) Therefore, problem (1) is equivalent to: u tt u α u t = 0, x Ω, t > 0, τz t (x, ρ, t) + z ρ(x, ρ, t) = 0, x, ρ (0, 1), t > 0, u(x, t) = 0, x Γ 0, t > 0, ( ) u ut u tt (x, t) = (x, t) + α (x, t) + µ1ut (x, t) + µ2z(x, 1, t) ν ν x, t > 0, z(x, 0, t) = u t (x, t) x, t > 0, u(x, 0) = u 0(x) x Ω, u t (x, 0) = u 1(x) x Ω, z(x, ρ, 0) = f 0(x, τρ) x, ρ (0, 1). (10) S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 12 / 32

Semigroup formulation Let V := (u, u t, γ 1(u t), z) T ; then V satisfies the problem: { V (t) = (u t, u tt, γ 1(u tt), z t ) T = AV (t), t > 0, V (0) = V 0, (11) where denotes the derivative with respect to time t, V 0 := (u 0, u 1, γ 1(u 1), f 0(.,.τ)) T and the operator A is defined by: v u u + α v v A = w u ν α v µ1v µ2z (., 1) ν z 1 τ zρ S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 14 / 32

Existence and uniqueness Energy space: H = H 1 Γ 0 (Ω) L 2 (Ω) L 2 () L 2 () L 2 (0, 1), H is a Hilbert space with respect to the inner product V, Ṽ = u. ũdx + vṽdx + w wdσ + ξ H Ω Ω for V = (u, v, w, z) T, Ṽ = (ũ, ṽ, w, z) T and ξ is defined by (4) or (6). The domain of A is the set of V = (u, v, w, z) T such that: Theorem 1 1 0 z zdρdσ (u, v, w, z) T H 1 Γ 0 (Ω) H 1 Γ 0 (Ω) L 2 () L 2 ( ; H 1 (0, 1) ) (12) u + αv E(, L 2 (u + αv) (Ω)), L 2 () ν (13) w = γ 1(v) = z(., 0) on (14) Suppose that µ 2 µ 1 and α > (µ 2 µ 1)B 2 or µ 2 < µ 1. Let V 0 H, then there exists a unique solution V C (R +; H) of problem (11). Moreover, if V 0 D (A), then V C (R +; D (A)) C 1 (R +; H). S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 15 / 32

Sketch of the proof. First step : A is dissipative Let V = (u, v, w, z) T D (A), we have: AV, V H = u. vdx + v ( u + α v) dx Ω ( Ω + w u ν α v ) µ1v µ2z (σ, 1) dσ ξ 1 zz ρdρdσ. ν τ 0 Since u + αv E(, L 2 (u + αv) (Ω)) and L 2 (), Green s formula gives: ν AV, V H = µ 1 w Γ 2 dσ µ 2 z (σ, 1) wdσ α v 2 dx ξ 1 z ρzdρdx. (15) 1 Ω τ 0 Case 1: µ 2 < µ 1. Choose then ξ that satisfies inequality (4). Using Young s inequality, (15) can be rewritten as : AV, V H + α ( ξ + 2τ µ2 2 Consequently, by using (4), we deduce that Ω ) ( v 2 dx + µ 1 ξ ) 2τ µ2 w 2 dσ 2 z 2 (σ, 1) dσ 0. AV, V H 0, (16) Case 2: µ 2 µ 1 and α > (µ 2 µ 1)B 2. Choose then ξ that satisfies the two inequalities (6) and (7). Using Young s inequality and the definition of the constant B, we can again prove that the inequality (16) holds. This means that in both cases A is dissipative. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 16 / 32

λi A is surjective for all λ > 0. Step 1 λi A is surjective for all λ > 0. Let F = (f 1, f 2, f 3, f 4) T H. We seek V = (u, v, w, z) T D (A) solution of which writes: λw + (u + αv) ν (λi A) V = F, λu v = f 1 (17) λv (u + αv) = f 2 (18) + µ 1v + µ 2z(., 1) = f 3 (19) λz + 1 zρ = f4 (20) τ To find V = (u, v, w, z) T D (A) solution of the system (17), (18), (19) and (20), we proceed as in [NP06], with two major changes: 1 the dynamic condition on which adds an unknown and an equation, 2 the presence of v = u t in this dynamic boundary condition. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 17 / 32

λi A is surjective for all λ > 0. Step 2. Suppose u is determined with the appropriate regularity. Then from (17), we get: v = λu f 1. (21) Therefore, from the compatibility condition on, (14), we determine z(., 0) by: z(x, 0) = v(x) = λu(x) f 1(x), for x (22) Thus, from (20), z is solution of the linear Cauchy problem: { ( ) z ρ = τ f 4(x) λz(x, ρ) for x, ρ (0, 1) z(x, 0) = λu(x) f 1(x) (23) The solution of the Cauchy problem (23) is given by: ρ z(x, ρ) = λu(x)e λρτ f 1e λρτ + τe λρτ f 4(x, σ)e λστ dσ for x, ρ (0, 1). (24) 0 So we have at the point ρ = 1, z(x, 1) = λu(x)e λτ + z 1(x) for x (25) with, 1 z 1(x) = f 1e λτ + τe λτ f 4(x, σ)e λστ dσ for x. 0 Since f 1 H (Ω) and f 4 L 2 (Γ 0 1) L 2 (0, 1), z 1 L 2 (). Consequently, knowing u, we may deduce v by (21), z by (24) and using (25), we deduce w = γ 1(v) by (19). S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 18 / 32

λi A is surjective for all λ > 0. Step 3. u = u + αv Set u = u + αv. From equations (18) and (19), u must satisfy: λ 2 1 + λα u u = f2 + λ f1 in Ω 1 + λα u = 0 on Γ 0 ) u λ (µ 2e λτ + (λ + µ 1 ν = u + f (x) for x 1 + λα (26) with f 1 L 2 (Ω), f 2 L 2 (Ω), f L 2 (). The variational formulation of problem (26) is to find u H 1 Γ 0 (Ω) such that: Ω λ 2 uω + u ωdx + 1 + λα ) λ (µ 2e Γ1 λτ + (λ + µ 1 u(σ)ω(σ)dσ = (27) Ω ( f 2 + 1 + λα ) λ 1 + λα f1 ωdx + f (σ)ω(σ)dσ ω H 0 (Ω) Since λ > 0, µ 1 > 0, µ 2 > 0, the left hand side of (27) defines a coercive bilinear form on H 1 Γ 0 (Ω). Thus by applying the Lax-Milgram lemma, there exists a unique u H 1 Γ 0 (Ω) solution of (27). Now, choosing ω C c, by Green s formula u E(, L2 (Ω)). We recover u, v, z and finally setting w = γ 1(v)), we have found V = (u, v, w, z) T D (A) solution of (λid A) V = F. The well-posedness result, Theroem 1, follows from the Hille-Yosida theorem S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 19 / 32

E is decreasing along trajectories For a positive constant ξ satisfying the strict inequality (4), we define the functional energy of the solution of problem (10) as E(t) = E(t, z, u) = 1 2 + ξ 2 E is greater than the usual one : E 1(t) = [ u(t) 22 + ut(t) 22 + ut(t) 22,Γ1 ] 1 0 z 2 (σ, ρ, t) dρ dσ. (28) [ u(t) 22 + ut(t) 22 + ut(t) 22,Γ1 ]. We will prove that the above energy E (t) is a decreasing function along the trajectories. Lemma 2 Assume that µ 1 > µ 2, then the energy defined by (28) is a non-increasing positive function and there exists a positive constant C such that for u solution of (10), and for any t 0, we have: de (t) dt [ ] C ut 2 (σ, t) dσ + z 2 (σ, 1, t) dσ α u t(x, t) 2 dx. (29) Ω S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 21 / 32

Sketch of the proof and asymptotic behavior We multiply the first equation in (10) by u t and perform integration by parts to get: 1 d ] [ u(t) 22 2 dt + ut(t) 22 + ut(t) 22,Γ1 + α u t(t) 2 2 (30) +µ 1 u t(t) 2 2, + µ 2 u t(σ, t)u t(σ, t τ)dσ = 0. Multiply the third equation in (10) by ξz, integrate the result over (0, 1): ξ 1 z ρz(σ, ρ, t) dρ dσ = ξ 1 τ 2τ ρ z 2 (σ, ρ, t) dρ dσ 0 = ξ 2τ 0 ( z 2 (σ, 1, t) z 2 (σ, 0, t) ) dσ. (31) Using the definition (8) of z in the equality (30) and using the same technique as in the first step of the proof of Theorem 1, inequality (29) holds. The asymptotic stability result reads as follows: Theorem 3 Assume µ 2 < µ 1. Then there exist two positive constants C and γ independent of t such that for u solution of problem (10), we have: E(t) Ce γt, t 0. (32) S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 22 / 32

Sketch of the proof. Step 1: Lyapunov function For ε > 0, to be chosen later, we define the Lyapunov function: L(t) = E(t) + ε u(x, t)u t (x, t) dx + ε u(σ, t)u t (σ, t) dσ Ω + εα u(x, t) 2 dx (33) 2 Ω + εξ 1 e 2tρ z 2 (σ, ρ, t) dρ dσ. 2 0 There exist two positive constants β 1 and β 2 > 0 depending on ε such that for all t 0 β 1E(t) L(t) β 2E(t). (34) By taking the time derivative of the function L defined by (33), using problem (10), performing several integration by parts, and using (29) we get: dl(t) dt [ ] C ut 2 (σ, t) dσ + z 2 (σ, 1, t) dσ α u t 2 2 ε u 2 2 + ε ut 2 2 + ε ut 2 2, εµ 1 u t (σ, t)u(σ, t) dσ εµ 2 z(σ, 1, t)u(σ, t)dσ (35) ( + εξ ) d 1 e 2tρ z 2 (σ, ρ, t) dρ dσ. 2 dt 0 S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 23 / 32

Sketch of the proof. Step 2: introduce δ By using the second equation in (10), the last term in (35) can be treated as follows: ( ) εξ d 1 e 2tρ z 2 (σ, ρ, t) dρ dσ 2 dt 0 1 = εξ ρe 2tρ z 2 (σ, ρ, t) dρ dσ εξ 1 e 2tρ z(σ, ρ, t)z ρ(σ, ρ, t) dρ dσ 0 τ 0 1 = εξ ρe 2tρ z 2 (σ, ρ, t) dρ dσ εξ 1 e 2tρ 0 τ2 0 ρ z 2 (σ, ρ, t) dρ dσ. Then, by using an integration by parts, the above formula can be rewritten as: ( εξ d ) 1 e 2tρ z 2 (σ, ρ, t) dρ dσ dt 0 = εξ e 2t z 2 (σ, ρ, t) dρ dσ + εξ ut 2 (σ, t) dσ (36) τ τ ( ) 2 1 εξ τ + 1 ρe 2tρ z 2 (σ, ρ, t) dρ dσ. 0 Using Young s inequality, and the trace inequality, we obtain, for any δ > 0: u t udσ δ u 2 2,Γ 1 + 1 4δ ut 2 2,. δb 2 u 2 2 + 1 4δ ut 2 2,. (37) S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 24 / 32

Sketch of the proof. Step 2 Similarly we have: z(σ, 1, t)u(σ, t) dσ 1 z 2 (σ, 1, t) dσ + δb 2 u 2 2. (38) 4δ Inserting (36), (37) and (38) into (35) and using Poincaré s inequality we have: [ ( C ε 1 + ξ τ + µ1 dl(t) dt [ ( ξ C ε τ e 2t + µ2 4δ )] u t 2 2, 4δ )] z 2 (σ, 1, t) dσ ( α εb 2) ( ) u t 2 2 ε 1 B 2 δ (µ 1 + µ 2 ) u 2 2 ( ) 2 1 εξ τ + 1 ρe 2tρ z 2 (σ, ρ, t) dρ dσ. (39) 0 S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 25 / 32

Sketch of the proof. Step 3: choose δ and ε We choose now δ small enough in (39) such that δ < 1 B 2 (µ 1 + µ 2). Once δ is fixed, using once again Poincaré s inequality in (39), we may pick ε small enough to obtain the existence of η > 0, such that: dl(t) ηεe(t), t 0. (40) dt On the other hand, by virtue of (34), setting γ = ηε/β 2, the last inequality becomes: dl(t) dt γl(t), t 0. (41) By integrating the previous differential inequality (41) between 0 and t, it exists C > 0 such that we have the following estimate for the function L: L(t) C e γt, t 0. Consequently, by using (34) once again, we conclude that it exists C > 0 such that: E(t) Ce γt, t 0. This completes the proof of Theorem 3. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 26 / 32

E is decreasing along trajectories Lemma 4 Assume that µ 2 > µ 1 and α > (µ 2 µ 1)B 2. For any ξ satisfying (6)-(7), the energy defined by (28) is a non-increasing positive function and there exists a positive constant κ such that for u solution of (10), and for any t 0, we have: de(t) dt [ κ z 2 (σ, 1, t) dσ + Ω ] u t(x, t) 2 dx. (42) S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 28 / 32

Sketch of the proof. Derivative of the energy: de(t) dt = α ξ 2τ Ω ( u t 2 dx µ 1 ξ 2τ z 2 (σ, 1, t) dσ µ 2 ) ut 2 (σ, t) dσ u t (σ, t)z(σ, 1, t) dσ. (43) Now, using Young s inequality, then (43) takes the form: de(t) dt α u t 2 Ω ( ) ξ 2τ µ2 2 ( dx µ 1 ξ Since, ( µ 1 ξ ) 2τ µ2 < 0, 2 using the trace inequality, we obtain the inequality: de(t) dt ) 2τ µ2 ut 2 (σ, t) dσ 2 z 2 (σ, 1, t) dσ. (44) ( ( α B 2 µ 1 ξ )) 2τ µ2 u t 2 dx 2 Ω ( ) ξ 2τ µ2 z 2 (σ, 1, t) dσ. 2 Using the two inequalities (6)-(7) that satisfy ξ, we may find κ > 0 such that the inequality (42) holds. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 29 / 32

Asymptotic behavior Theorem 5 Assume that µ 2 > µ 1 and α > (µ 2 µ 1)B 2. For any ξ satisfying (6)-(7), there exist two positive constants C and γ independent of t such that for u solution of problem (10), we have: E(t) Ce γt, t 0. (45) dl(t) dt [ ( ξ κ ε ( κ ε )] τ e 2t + µ2 z 2 (σ, 1, t) dσ 4δ Γ ( 1 B 2 + 1 + ξ )) τ + µ1 u t 2 2 4δ ε ( 1 B 2 δ (µ 1 + µ 2 ) ) u 2 2 (46) ( ) 2 1 εξ τ + 1 e 2tρ z 2 (σ, ρ, t) dρ dσ. By choosing firstly δ and then ε, we may find γ > 0 independent of t such that: dl(t) γl(t), t 0. dt This inequality permits us to conclude the proof of Theorem 5. S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 30 / 32 0

Some remarks 1 Since the energy associated to (1) is less than the one associated to (10), it is obvious that the exponential stability of the solution associate to problem (10) implies the exponential stability of the one associated to (1). 2 The presence of the strong damping term α u t in equation (1) plays an essential role in the behavior of the system. The condition µ 1 < µ 2 is a necessary condition in the case α = 0, since Nicaise and Pignotti [NP06] showed an instability result if this condition fails. 3 Adapting the same method to the system with internal feedback: ( ( ) ) u tt u α u t + b (x) µ 1u t(x, t) + µ 2u t x, t τ = 0, x Ω, t > 0 u(x, t) = 0, x Γ 0, t > 0 [ ] u α ut u tt(x, t) = (x, t) + (x, t) ν ν x, t > 0 u(x, 0) = u 0(x), u t(x, 0) = u 1(x) x Ω, u(x, t τ) = f 0(x, t τ) x Ω (0, τ) with b L (Ω) is a function which satisfies b (x) 0, a.e. in Ω and b (x) > b 0 > 0 a.e. in ω where ω Ω is an open neighborhood of, the results are still valid. (47) S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 31 / 32

Thank you for your attention S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Konstanz 2010 32 / 32