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Existence and exponential stability of the damped waveequation with a dynamic boundary condition and a delay

term.

Stéphane Gerbi

LAMA, Université de Savoie, Chambéry, France

Jeudi 24 avril 2014

Joint work with Belkacem Said-Houari, Alhosn University, Abu Dhabi, UAE

S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Chambéry 2014 1 / 33

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Outline of the talk

1 Introdution

2 Well-posedness of the problem : existence and uniqueness.Setup and notationsSemigroup formulation : existence and uniqueness.

3 Asymptotic behavior

4 Some remarks


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Outline

1 Introdution



4 Some remarks


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Position of the problem

color code : Yellow : dynamic boundary conditions , red : time delayConsider the damped wave equation, with dynamic boundary conditions and timedelay :

utt −∆u − α∆ut = 0, x ∈ Ω, t > 0 ,

u(x , t) = 0, x ∈ Γ0, t > 0 ,

utt(x , t) = −(∂u∂ν

(x , t) +α∂ut

∂ν(x , t) + µ1ut(x , t) + µ2ut(x , t − τ )

)x ∈ Γ1, t > 0 ,

u(x , 0) = u0(x) x ∈ Ω ,

ut(x , 0) = u1(x) x ∈ Ω ,

ut(x , t − τ ) = f0(x , t − τ ) x ∈ Γ1, t ∈ (0, τ ) ,(1)

where u = u(x , t) , t ≥ 0 , x ∈ Ω which is a bounded regular domain of RN , (N ≥ 1),∂Ω = Γ0 ∪ Γ1, mes(Γ0) > 0, Γ0 ∩ Γ1 = ∅, α, µ1, µ2 > 0 and u0 , u1, f0 are givenfunctions. Moreover, τ > 0 represents the time delayQuestions 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?


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Dynamic boundary condition

utt −∆u − α∆ut = 0, x ∈ Ω, t > 0 ,

u(x , t) = 0, x ∈ Γ0, t > 0 ,

utt(x , t) = −(∂u∂ν

(x , t) +α∂ut

∂ν(x , t) + µ1ut(x , t))

)x ∈ Γ1, t > 0 ,

u(x , 0) = u0(x) x ∈ Ω ,

ut(x , 0) = u1(x) x ∈ Ω ,

Longitudinal vibrations in a homogeneous bar in which there are viscous effects, andspring-mass system, Pellicer and Sola-Morales, 90’s

Artificial boundary condition for unbounded domain : transparent and absorbing,and a lot of mix between these two types, Majda-Enquist 80’s,

Ω is an exterior domain of R3 in which homogeneous fluid is at rest except forsound waves. Each point of the boundary is subjected to small normaldisplacements into the obstacle. This type of dynamic boundary conditions areknown as acoustic boundary conditions, Beale , 80’s

Wentzell boundary conditions for PDE , Jérôme Goldstein, Gisèle Ruiz-Goldsteinand co workers, 2000’s


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Related works : damped waves with dynamic boundary conditions

In the absence of delay, and with a nonlinear source terms, Gerbi and Said-Houari[GS2008, GS2011] showed the local existence, an exponential decay when the initialenergy is small enough, an exponential growth when the initial energy is large enoughand a blow-up phenomenon for linear boundary conditions (m = 2)

utt −∆u − α∆ut = |u|p−2u, x ∈ Ω, t > 0

u(x , t) = 0, x ∈ Γ0, t > 0

utt(x , t) = −[∂u∂ν

(x , t) +α∂ut

∂ν(x , t) + r |ut |m−2ut(x , t)

]x ∈ Γ1, t > 0

u(x , 0) = u0(x), ut(x , 0) = u1(x) x ∈ Ω .

[GS2008] S. Gerbi and B. Said-Houari, Local existence and exponential growth for a semilinear damped wave equation withdynamic boundary conditions. Advances in Differential Equations Vol. 13, No 11-12, pp. 1051-1074, 2008.

[GS2011] S. Gerbi and B. Said-Houari, Asymptotic stability and blow up for a semilinear damped wave equation with dynamicboundary conditions. Nonlinear Analysis: Theory, Methods & Applications Vol. 74, pp. 7137-7150, 2011.


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Related works : delay term in 1D

Datko [Dat91], showed that solutions of : wtt − wxx − awxxt = 0, x ∈ (0, 1), t > 0,

w (0, t) = 0, wx (1, t) = −kwt (1, t − τ ) , t > 0,

a , k , τ > 0 become unstable for an arbitrarily small values of τ and any values of a andk. Datko et al [DLP86] treated the following one dimensional problem:

utt(x , t)− uxx(x , t) + 2aut(x , t) + a2u(x , t) = 0, 0 < x < 1, t > 0,

u(0, t) = 0, t > 0,

ux(1, t) = −kut(1, t − τ ), t > 0,

(2)

If ke2a + 1e2a − 1

< 1 then the delayed feedback system is stable for all sufficiently small

delays. If ke2a + 1e2a − 1

> 1, then there exists a dense open set D in (0,∞) such that for

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 ofwave equations. SIAM J. Control Optim., 24(1):152–156, 1986.


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Related works : wave equations and boundary feedback delay

Nicaise and Pignotti,[NP06], examined a system of wave equation with a linear boundarydamping term with a delay:

utt −∆u = 0, x ∈ Ω, t > 0 ,

u(x , t) = 0, x ∈ Γ0, t > 0 ,∂u∂ν

(x , t) = µ1ut(x , t) + µ2ut(x , t − τ ) x ∈ Γ1, t > 0 ,

u(x , 0) = u0(x), x ∈ Ω ,

ut(x , 0) = u1(x) x ∈ Ω ,

ut(x , t − τ ) = g0(x , t − τ ) x ∈ Ω, τ > 0 ,

(3)

and proved under the assumption µ2 < µ1 that null stationary state is exponentiallystable. They also proved instability if this condition fails.They also studied [NP08, NVF09], internal feedback, time-varying delay and distributeddelay.[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-varyingdelays. DCDS-S., S2(3):559–581, 2009.


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Main results

Suppose that :

Coefficient conditioncase 1: µ1 > µ2 or

case 2: µ1 ≤ µ2 and α >

(µ2

1 − µ22)

2µ1

1β?

β? < 0 defined later

then Problem (1) has aunique global solution,this solution decays exponentially to the null solution.

Remark 1If µ1 > µ2, as in the works of Nicaise and Pignotti, we can choose α = 0 so thatno strong damping is necessary.


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Outline

1 Introdution



4 Some remarks


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Setup and notations

First we reformulate the boundary delay problem, then by a semigroup approachand using the Lumer-Phillips’ theorem we will prove the global existence.

Notations

H1Γ0

(Ω) =u ∈ H1(Ω)/ uΓ0 = 0

γ1 the trace operator from H1

Γ0(Ω) on L2(Γ1)

H1/2(Γ1) = γ1(H1

Γ0(Ω)).

E (∆, L2(Ω)) =u ∈ H1(Ω) such that ∆u ∈ L2(Ω)

For u ∈ E (∆, L2(Ω)) ,

∂u∂ν∈ H−1/2(Γ1) and we have Green’s formula:

∫Ω

∇u(x)∇v(x)dx =

∫Ω

−∆u(x)v(x)dx +

⟨∂u∂ν

; v⟩

Γ1

∀v ∈ H1Γ0(Ω),

where 〈.; .〉Γ1 means the duality pairing H−1/2(Γ1) and H1/2(Γ1).


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Reformulation of the delay term

As in [NP06], we add the new variable:

z (x , ρ, t) = ut (x , t − τρ) , x ∈ Γ1, ρ ∈ (0, 1) , t > 0. (4)

Then, we have

τ zt (x , ρ, t) + zρ (x , ρ, t) = 0, in Γ1 × (0, 1)× (0,+∞) . (5)

Therefore, problem (1) is equivalent to:

utt −∆u − α∆ut = 0, x ∈ Ω, t > 0 ,

τ zt(x , ρ, t) + zρ(x , ρ, t) = 0, x ∈ Γ1, ρ ∈ (0, 1) , t > 0 ,

u(x , t) = 0, x ∈ Γ0, t > 0 ,

utt(x , t) = −(∂u∂ν

(x , t) + α∂ut

∂ν(x , t) + µ1ut(x , t) + µ2z(x , 1, t)

)x ∈ Γ1, t > 0 ,

z(x , 0, t) = ut(x , t) x ∈ Γ1, t > 0 ,

u(x , 0) = u0(x) x ∈ Ω ,

ut(x , 0) = u1(x) x ∈ Ω ,

z(x , ρ, 0) = f0(x ,−τρ) x ∈ Γ1, ρ ∈ (0, 1) .

(6)


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Outline

1 Introdution



4 Some remarks


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Semigroup formulation

Let V := (u, ut , γ1(ut), z)T ; then V satisfies the problem:V ′(t) = (ut , utt , γ1(utt), zt)T = AV (t), t > 0,V (0) = V0,

(7)

where ′ denotes the derivative with respect to time t, V0 := (u0, u1, γ1(u1), f0(.,−.τ ))T

and the operator A is defined by:

A

u

v

w

z

=

v

∆u + α∆v

−∂u∂ν− α∂v

∂ν− µ1v − µ2z (., 1)

− 1τzρ


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Domain and energy space

Energy space:

H = H1Γ0(Ω)× L2 (Ω)× L2(Γ1)× L2(Γ1)× L2(0, 1),

H is a Hilbert space with respect to the inner product⟨V , V

⟩H

=

∫Ω

∇u.∇udx +

∫Ω

v vdx +

∫Γ1

wwdσ + ξ

∫Γ1

∫ 1

0zzdρdσ

for V = (u, v ,w , z)T , V = (u, v , w , z)T and ξ defined later.The domain of A is the set of V = (u, v ,w , z)T such that:

(u, v ,w , z)T ∈ H1Γ0(Ω)× H1

Γ0(Ω)× L2(Γ1)× L2 (Γ1;H1(0, 1))

(8)

u + αv ∈ E (∆, L2(Ω)) ,∂(u + αv)

∂ν∈ L2(Γ1) (9)

w = γ1(v) = z(., 0) on Γ1 (10)


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The constant β?

For β ∈ R, define :

C (β) = infu∈H1

Γ0(Ω)

‖∇u‖22 + β‖u‖22,Γ1‖u‖22

(11)

C (β) is the first eigenvalue of the operator −∆ under the Dirichlet-Robinboundary conditions :

u(x) = 0, x ∈ Γ0

βu(x) +∂u∂ν

(x) = 0 x ∈ Γ1

From Kato’s perturbation theory, C (β) is a continuous decreasing function and asC (0) > 0, it exists β? < 0 such that

C (β?) = 0.

Definition of β?

it exists β? < 0 such that ∀β > β? , C (β) > 0


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Existence result

Suppose that :

Coefficient conditioncase 1: µ1 > µ2 or

case 2: µ1 ≤ µ2 and α >

(µ2

1 − µ22)

2µ1

1β?

β? < 0

Theorem 1

Let V0 ∈ H, then there exists a unique solution V ∈ C (R+;H) of problem (7).Moreover, if V0 ∈ D (A), then

V ∈ C (R+;D (A)) ∩ C 1 (R+;H) .


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Sketch of the proof. First step : A is dissipative 1

Let V = (u, v ,w , z)T ∈ D (A), we have:

〈AV ,V 〉H =

∫Ω

∇u.∇vdx +

∫Ω

v (∆u + α∆v) dx

+

∫Γ1

w(−∂u∂ν− α∂v

∂ν− µ1v − µ2z (σ, 1)

)dσ − ξ

τ

∫Γ1

∫ 1

0zzρdρdσ.

Since u + αv ∈ E(∆, L2(Ω)) and∂(u + αv)

∂ν∈ L2(Γ1), using Green’s formula and the

compatibility condition (10) gives:

〈AV ,V 〉H = −µ1

∫Γ1

v2dσ − µ2

∫Γ1

z (σ, 1) vdσ − α∫

Ω

|∇v |2 dx − ξ

τ

∫Γ1

∫ 1

0zρzdρdσ.

But from the compatibility condition (10), we get:

− ξ

τ

∫Γ1

∫ 1

0zρz dρ dσ =

ξ

2τ

∫Γ1

(v2 − z2(σ, 1, t)

)dσ .


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〈AV ,V 〉H = −α∫

Ω

|∇v |2 dx −(µ1 −

ξ

2τ

)∫Γ1

v2dσ − ξ

2τ

∫Γ1

∫ 1

0z2(σ, 1, t)dσ

−µ2

∫Γ1

v(σ, t)z (σ, 1) dσ

Fix δ > 0, Young’s inequality gives :

−∫

Γ1

v(σ, t)z (σ, 1) dσ ≤ δ

2

∫Γ1

z2 (σ, 1) dσ +12δ

∫Γ1

v2(σ, t)dσ

Finally

〈AV ,V 〉H +α

∫Ω

|∇v |2 dx +

(µ1 −

ξ

2τ− µ2

2δ

)∫Γ1

v2dσ+(ξ

2τ− δµ2

2

)∫Γ1

z2(σ, 1, t)dσ ≤ 0


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Fix δ and ξ

Choose δ =µ1

µ2and ξ =

µ22τ

µ1

〈AV ,V 〉H + α

∫Ω

|∇v |2 dx +µ2

1 − µ22

2µ1

∫Γ1

v2dσ ≤ 0

case 1: µ1 > µ2. For all α ≥ 0

∀V ∈ H 〈AV ,V 〉H ≤ 0.

case 2: µ1 ≤ µ2 , α > 0. Set β =µ2

1 − µ22

2αµ1.

〈AV ,V 〉H + C(β)‖u‖22 ≤ 0

Suppose : α >

(µ2

1 − µ22)

2µ1

1β?

. Thus C(β) > 0 and we get :

∀V ∈ H 〈AV ,V 〉H ≤ 0.


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λI −A is surjective for all λ > 0. Step 1 : formulation

λI −A is surjective for all λ > 0. Let F = (f1, f2, f3, f4)T ∈ H. We seekV = (u, v ,w , z)T ∈ D (A) solution of

(λI −A)V = F ,

which writes:

λu − v = f1 (12)

λv −∆(u + αv) = f2 (13)

λw +∂(u + αv)

∂ν+ µ1v + µ2z(., 1) = f3 (14)

λz +1τzρ = f4 (15)

To find V = (u, v ,w , z)T ∈ D (A) solution of the system (12), (13), (14) and (15), weproceed as in [NP06], with two major changes:

1 the dynamic condition on Γ1 which adds an unknown and an equation,2 the presence of v = ut in this dynamic boundary condition.


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λI −A is surjective for all λ > 0. Step 2: : knowing u, determine v , z ,w

Suppose u is determined with the appropriate regularity. Then from (12), we get:

v = λu − f1 . (16)

Therefore, from the compatibility condition on Γ1, (10), we determine z(., 0) by:

z(x , 0) = v(x) = λu(x)− f1(x), for x ∈ Γ1 (17)

Thus, from (15), z is solution of the linear Cauchy problem:zρ = τ

(f4(x)− λz(x , ρ)

)for x ∈ Γ1 , ρ ∈ (0, 1)

z(x , 0) = λu(x)− f1(x)(18)

The solution of the Cauchy problem (18) is given by:

z(x , ρ) = λu(x)e−λρτ − f1e−λρτ + τe−λρτ∫ ρ

0f4(x , σ)eλστdσ for x ∈ Γ1 , ρ ∈ (0, 1).

(19)So we have at the point ρ = 1 , for x ∈ Γ1,

z(x , 1) = λu(x)e−λτ + z1(x) , z1(x) = −f1e−λτ + τe−λτ∫ 1

0f4(x , σ)eλστdσ (20)

Since f1 ∈ H1Γ0(Ω) and f4 ∈ L2(Γ1)× L2(0, 1), z1 ∈ L2(Γ1).

Knowing u, we may deduce v by (16), z by (19) and using (20), we deduce w = γ1(v)

by (14).S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Chambéry 2014 23 / 33

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λI −A is surjective for all λ > 0. Step 3. u = u + αv

Set u = u + αv . From equations (13) and (14), u must satisfy:

λ2

1 + λαu −∆u = f2 +

λ

1 + λαf1 in Ω

u = 0 on Γ0

∂u∂ν

= −λ(µ2e−λτ + (λ+ µ1

)1 + λα

u + f (x) for x ∈ Γ1

(21)

with f1 ∈ L2(Ω) , f2 ∈ L2(Ω) , f ∈ L2(Γ1).The variational formulation of problem (21) is to find u ∈ H1

Γ0(Ω) such that:

∫Ω

λ2

1 + λαuω +∇u∇ωdx +

∫Γ1

λ(µ2e−λτ + (λ+ µ1

)1 + λα

u(σ)ω(σ)dσ = (22)∫Ω

(f2 +

λ

1 + λαf1)ωdx +

∫Γ1

f (σ)ω(σ)dσ ∀ω ∈ H1Γ0(Ω)


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λI −A is surjective for all λ > 0. End of proof

Since λ > 0 , µ1 > 0 , µ2 > 0, the left hand side of (22) defines a coercive bilinearform on H1

Γ0(Ω).

Thus by applying the Lax-Milgram lemma, there exists a unique u ∈ H1Γ0

(Ω)solution of (22).

Now, choosing ω ∈ C∞c , by Green’s formula u ∈ E (∆, L2(Ω)).

We recover u , v , z and finally setting w = γ1(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 Lummer-Phillips’ theorem.


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Outline

1 Introdution



4 Some remarks


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E is decreasing along trajectories

Let ξ > 0 , we define the functional energy of the solution of problem (6) as

E(t) = E(t, z , u) =12

[‖∇u(t)‖22 + ‖ut(t)‖22 + ‖ut(t)‖22,Γ1

]+

ξ

2

∫Γ1

∫ 1

0z2(σ, ρ, t) dρ dσ. (23)

E is greater than the usual one : E1(t) =12

[‖∇u(t)‖22 + ‖ut(t)‖22 + ‖ut(t)‖22,Γ1

].

Set β =µ2

1 − µ22

2αµ1.

Lemma 2

For u solution of (6), and for any t ≥ 0, we have:dE (t)

dt≤ −αC(β)‖ut‖22

Corollary 1Suppose the damping coefficient condition is fulfilled (that is β > β? , C(β) > 0). Thenthe energy E is decreasing along the trajectories.


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Sketch of the proof and asymptotic behavior

We multiply the first equation in (6) by ut and perform integration by parts to get:

12ddt

[‖∇u(t)‖22 + ‖ut(t)‖22 + ‖ut(t)‖22,Γ1

]+ α ‖∇ut(t)‖22

+µ1 ‖ut(t)‖22,Γ1 + µ2

∫Γ1

ut(σ, t)ut(σ, t − τ )dσ = 0 .(24)

By defintion of z , we have:∫

Γ1

ut(σ, t)ut(σ, t − τ )dσ =

∫Γ1

ut(σ, t)z(σ, 1, t)dσ

Fix δ > 0, Young’s inequality gives :∣∣∣∣∫Γ1

ut(σ, t)z(σ, 1, t)dσ∣∣∣∣ ≤ δ

2

∫Γ1

z2 (σ, 1) dσ +12δ

∫Γ1

u2t (σ, t)dσ

Differentiating the term∫

Γ1

∫ 1

0z2(σ, ρ, t) dρ dσ with respect to t and using the fact

that zt = − zρτ, we get

Finally

d Edt≤ −α‖∇ut‖2 −

(µ1 −

ξ

2τ− µ2

2δ

)‖ut‖22,Γ1 −

(ξ

2τ− δµ2

2

)∫Γ1

z2(σ, 1, t)dσ


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Asymptotic behavior

Fix δ and ξ

Choose δ =µ1

µ2and ξ =

µ22τ

µ1, set β =

µ21 − µ2

2

2αµ1.

dE (t)

dt≤ −αC (β)‖ut‖22

The asymptotic stability result reads as follows:

Theorem 3

Assume the damping coefficient relation is fulfiled. Then there exist two positiveconstants C and γ independent of t such that for u solution of problem (6), wehave:

E (t) ≤ Ce−γt , ∀ t ≥ 0 . (25)


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Lyapunov function

For ε > 0, to be chosen later, we define the Lyapunov function:

L(t) = E(t) + ε

∫Ω

u(x , t)ut(x , t) dx + ε

∫Γ1

u(σ, t)ut(σ, t) dσ

+εα

2

∫Ω

|∇u(x , t)|2 dx (26)

+εξ

2

∫Γ1

∫ 1

0e−2τρz2(σ, ρ, t) dρ dσ.

There exist two positive constants β1 and β2 > 0 depending on ε such that for all t ≥ 0

β1E(t) ≤ L(t) ≤ β2E(t) . (27)

By taking the time derivative of the function L defined by (26), using problem (6),performing several integration by parts, and using the previous inequality on the

derivative of E and the same Young’s inequality with δ =µ1

µ2and ξ =

µ22τ

µ1, we choose

ε > 0 such that there exist two positive constants C∗ and γ independent of t:

L(t) ≤ C∗e−γt , ∀t ≥ 0 .

Consequently, by using (27) once again, we conclude that it exists C > 0 such that:

E(t) ≤ Ce−γt , ∀t ≥ 0 .


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Outline

1 Introdution



4 Some remarks


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Some final remarks

1 Since the energy associated to (1) is less than the one associated to (6), it isobvious that the exponential stability of the solution associate to problem (6)implies the exponential stability of the one associated to (1).

2 The presence of the strong damping term −α∆ut in equation (1) plays an essentialrole in the behavior of the system. The condition µ1 < µ2 is a necessary conditionin the case α = 0, since Nicaise and Pignotti [NP06] showed an instability result ifthis condition fails.

3 Adapting the same method to the system with internal feedback:

utt −∆u − α∆ut + b (x)(µ1ut(x , t) + µ2ut

(x , t − τ

))= 0, x ∈ Ω, t > 0

u(x , t) = 0, x ∈ Γ0, t > 0

utt(x , t) = −[∂u∂ν

(x , t) +α∂ut

∂ν(x , t)

]x ∈ Γ1, t > 0

u(x , 0) = u0(x), ut(x , 0) = u1(x) x ∈ Ω ,

u(x , t − τ ) = f0(x , t − τ ) x ∈ Ω× (0, τ)

(28)

with b ∈ L∞ (Ω) is a function which satisfies

b (x) ≥ 0, a.e. in Ω and b (x) > b0 > 0 a.e. in ω

where ω ⊂ Ω is an open neighborhood of Γ1, the results are still valid.S. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Chambéry 2014 32 / 33

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Instability result?Can we show that if

µ1 < µ2 and α ≤(µ2

1 − µ22)

2µ1

1β?

we can find solution with constant energy or energy that goes to infinity?

Hint: Try to find a solution of the form:

u(t, x) = eλtφ(x) with λ ∈ C,<(λ) > 0.

[GS2012] S. Gerbi and B. Said-Houari. Existence and exponential stability of a dampedwave equation with dynamic boundary conditions and a delay term AppliedMathematics and Computations, 218(1):11900–11910, 2012

Thank you for your attentionS. Gerbi (LAMA, UdS, Chambéry) Dynamic boundary conditions and delay term Chambéry 2014 33 / 33

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