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A remark on observability of the wave equation with moving boundary

  • Kaïs Ammari , Ahmed Bchatnia EMAIL logo and Karim El Mufti
Published/Copyright: May 25, 2017

Abstract

We deal with the wave equation with assigned moving boundary (0<x<a(t)) upon which Dirichlet or mixed boundary conditions are specified. Here a(t) is assumed to move slower than light and periodically. Moreover, a is continuous, piecewise linear with two independent parameters. Our major concern will be an observation problem which is based measuring, at each t>0, of the transverse velocity at a(t). The key to the results is the use of a reduction theorem by Yoccoz [14].

A Appendix

In this section, we treat the Dirichlet observability. We assume the following.

Assumption A.1.

The function

b(t):=H(a(t)+t)-H(-a(t)+t)H(a(t)+t)+H(-a(t)+t)

satisfies

c1b(t)c2for all t,c1,c2>0.

Note that for a(t) given by (1.7), we have

b(t)=a(t)t+l2l1-l2

This function satisfies

(A.1)2aminα-β1-β2a(t)α-β1-βb(t)2a(t)α-β1+β2amaxα-β1+βfor all t,

and Assumption A.1 is satisfied.

Theorem A.2 (Dirichlet observability).

Under Assumptions 1.4 and A.1, we have that for all Tρ(F), there exists a constant C*>0 such that for all solution u of system (1.1) with the mixed boundary condition (1.3) and initial data (ϕ,ψ)Hl1(0,a(0))×L2(0,a(0)), we have

(A.2)0T|ut(a(t),t)|2dtC*(ϕHl1(0,a(0))2+ψL2(0,a(0))2).

Remark A.3.

The observability time is optimal here for the same reason as in Remark 1.6 and Theorem 1.5.

Remark A.4.

Using Φ given by (2.1), we transform system (1.1), (1.3) into

(A.3){τ2V-ξ2V=0for 0<ξ<ρ(F)/2,τ,V(0,τ)=0,Vξ(ρ(F)/2,τ)+b(t(τ))Vτ(ρ(F)/2,τ)=0for τ,V(ξ,0)=ϕ2(ξ),Vτ(ξ,0)=ψ2(ξ)for ξ(0,ω/2).

For the proof of Theorem A.2, we need the following lemmas.

Lemma A.5.

There exist positive constants C and ω such that

(A.4)EV(τ)Ce-ωτEV(0).

Proof.

Define the Lyapunov function

E1(τ)=120ρ(F)[Vξ2(ξ,τ)+Vτ2(ξ,τ)]𝑑ξ+δ0ρ(F)ξVξ(ξ,τ)Vτ(ξ,τ)𝑑ξ.

For δ<1ρ(F), we obtain

(A.5)0<(1-δρ(F))EV(τ)E1(τ)(1+δρ(F))EV(τ).

We derive E1 with respect to τ and get

E1(τ)=[VξVτ]ξ=0ξ=ρ(F)-δ20ρ(F)[Vξ2(ξ,τ)+Vτ2(ξ,τ)]𝑑ξ+δ2[ξ(Vξ2+Vτ2)]ξ=0ξ=ρ(F)
=[δ2(1+b(t(τ))2)-b(t(τ))]Vτ2(ρ(F),τ)-δ20ρ(F)[Vξ2(ξ,τ)+Vτ2(ξ,τ)]𝑑ξ.

We choose δ small enough and, taking into account (A.1) and (A.5), we get E1(τ)-ωE1(τ). The proof is complete. ∎

Lemma A.6.

If Tρ(F), then there exists C(T)>0 such that for all (ϕ2,ψ2)Hl1(0,ρ(F)/2)×L2(0,ρ(F)/2), we have

(A.6)C(T)0T|Vξ(ρ(F)/2,τ)|2dτϕ2Hl1(0,ρ(F)/2)2+ψ2L2(0,ρ(F)/2)2

and

(A.7)C(T)0T|Vτ(ρ(F)/2,τ)|2dτϕ2Hl1(0,ρ(F)/2)2+ψ2L2(0,ρ(F)/2)2.

Proof.

The energy identity for system (A.3) gives

EV(T)-EV(0)=-oTb(t(τ))|Vτ(ρ(F)/2,τ)|2𝑑τ.

Using (A.1) and (A.4), we obtain

0T|Vτ(ρ(F)/2,τ)|2dτC0Tb(t(τ))|Vτ(ρ(F)/2,τ)|2dτC(EV(0)-EV(T))CEV(0)(1-e-ωT).

This permit to conclude the second inequality in Lemma A.6.

For the first inequality, it suffices to use (A.3) and (A.1). ∎

Proof of Theorem A.2.

For the proof of (A.2), we state as above

tu=ξVtξ+τVtτ.

Next we have

|tu(a(t),t)|2=14{|ξV(ρ(F)/2,τ)[H(x+t)-H(-x+t)]+τV(ρ(F)/2,τ)[H(x+t)+H(-x+t)]}2.

Making use of Young’s inequalities, Lemma 2.4, (A.6), (A.7) and (1.8), we obtain the desired result. ∎

Acknowledgements

The authors are very grateful to the anonymous referees for their helpful comments and suggestions, that improved the manuscript.

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Received: 2015-7-23
Revised: 2016-9-6
Accepted: 2017-3-23
Published Online: 2017-5-25
Published in Print: 2017-6-1

© 2017 Walter de Gruyter GmbH, Berlin/Boston

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