Stabilization of two strongly coupled hyperbolic equations in exterior domains
aa r X i v : . [ m a t h . A P ] M a y STABILIZATION OF TWO STRONGLY COUPLED HYPERBOLICEQUATIONS IN EXTERIOR DOMAINS
L.ALOUI , AND H.AZAZA LAMMDA-ESSTHS, UNIVERSIT´E DE SOUSSE, TUNISIA. UNIVERSIT´E DE TUNIS EL MANAR, TUNISIA.
Abstract.
In this paper we study the behavior of the total energy and the L -norm ofsolutions of two coupled hyperbolic equations by velocities in exterior domains. Only one ofthe two equations is directly damped by a localized damping term. We show that, when thedamping set contains the coupling one and the coupling term is effective at infinity and oncaptive region, then the total energy decays uniformly and the L -norm of smooth solutionsis bounded. In the case of two Klein-Gordon equations with equal speeds we deduce anexponential decay of the energy. Introduction and statement of the results
Let Ω be a domain of R d , d >
2. We denote by ∆ the Laplace operator on Ω with Dirichletboundary condition. We consider the following hyperbolic equation with localized lineardamping ∂ t u − ∆ u + mu + a ( x ) ∂ t u = 0 in R + × Ω ,u = 0 on R + × Γ , ( u (0 , . ) , ∂ t u (0 , . )) = ( u , u ) in Ω , (1.1)where a ∈ L ∞ (Ω) is a nonnegative smooth function and m ∈ R + . It is easy to verify that theenergy given by E u ( t ) = 12 Z Ω | ∂ t u ( t, x ) | + |∇ u ( t, x ) | + m | u ( t, x ) | dx, (1.2)is non-increasing and E u (0) = Z t Z Ω a ( x ) | ∂ t u ( t, x ) | dxdt + E u ( t ) , t > . When m = 0, the stabilization problem for the linear damped wave equation has beenstudied by several authors. More precisely, when Ω is bounded, the uniform decay of thetotal energy is equivalent to the geometric control condition of Bardos et al [7]. On the otherhand, if Ω is not bounded then, in general, the decay rate of the total energy cannot beuniform. Indeed, in the whole space,i.e. Ω = R d , Matsumura [19] obtained a precise L p − L q type decay estimate for solutions of (1.1), when a ( x ) = 1, E u ( t ) C (1 + t ) − − d ( i − ) I i , (1.3) Key words and phrases.
Damped wave equation, Klein-Gordon equation, Energy decay, exterior domain,observability, Stability.Email : [email protected]@gmail.com. k u ( t, . ) k L C (1 + t ) d ( i − ) I i , (1.4)where C is a positive constant, i ∈ [1 ,
2] and I i = k u k H + k u k L + k u k L i + k u k L i . Theproof in [19] is based on a Fourier transform method. In the case of exterior domains andwhen a ( x ) > a − > u of the system (1.1)satisfies E u ( t ) C (1 + t ) − I and k u ( t ) k L CI , f or all t > . (1.5)In [20], Nakao obtained the estimate (1.5) for a damper which is positive near infinity andnear a part of the boundary (Lions’s condition). Daoulatli in [11] generalized this result byassuming that each trapped ray meets the damping region which is also effective at infinity.Recently, Aloui et al [6] established the uniform stabilization of the total energy for the sys-tem (1.1) when the initial data are compactly supported. They proved that the rate of decayturns out to be the same as those of the heat equation, which shows that the effective damperat space infinity strengthens the parabolic structure in the equation.In the case m >
0, the energy (1.2) contains the L norm. Then, using the semi-groupproperty, the type of decay (1.5) implies the expnential one E u ( t ) Ce − δt E u (0) , f or all t > , (1.6)where C, δ positive constants. In [23] Zuazua considered the nonlinear Klein-gordon equationswith dissipative term and he proved the exponential decay of energy through the weightedenergy method. This result has been generalized by Aloui et al [5] for more general nonlinear-ities. We refer the reader to the works of Dehman et al [9] and Laurent et al [14] for relatedresults.In this paper we will study the stabilization problem for a system of two coupled hyperbolicequations on exterior domain. More precisely, let O be a compact domain of R d with C ∞ boundary Γ = ∂O and Ω = R d \ O ∂ t u − ∆ u + m u + b ( x ) ∂ t v + a ( x ) ∂ t u = 0 in R + × Ω ,∂ t v − γ ∆ v + m v − b ( x ) ∂ t u = 0 in R + × Ω ,u = v = 0 on R + × Γ , ( u (0 , . ) , ∂ t u (0 , . )) = ( u , u ) in Ω , ( v (0 , . ) , ∂ t v (0 , . )) = ( v , v ) in Ω , (1.7)where b ∈ L ∞ (Ω) is a smooth function, m , m ∈ R + and γ is a positive constant.We associate to the system (1.7) the energy functional given by E u,v ( t ) = 12 Z Ω |∇ u ( t, x ) | + | ∂ t u ( t, x ) | + m | u ( t, x ) | dx + 12 Z Ω γ |∇ v ( t, x ) | + | ∂ t v ( t, x ) | + m | v ( t, x ) | dx. Let H = (cid:16) H D (Ω) × L (Ω) (cid:17) be the completion of ( C ∞ (Ω)) with respect to the norm k ( w , w , w , w ) k H = (cid:16) Z Ω |∇ w | + γ |∇ w | + m | w | + m | w | + | w | + | w | dx (cid:17) . The linear evolution equation (1.7) can be rewritten under the form (cid:26) U t + AU = 0 , U (0) = U ∈ H , (1.8)where U = u∂ t uv∂ t v , U = u u v v and the unbounded operator A on H with domain D ( A ) = {U ∈ H , AU ∈ H} is defined by A = − Id − ∆ + m Id a b − Id − b − γ ∆ + m Id . From the linear semi-group theory, we can infer that for U ∈ H the problem (1.8) admits aunique solution U ∈ C ([0 , + ∞ [ , H ).In addition, if U ∈ D ( A n ), for n ∈ N , then the solution U ∈ n \ i =0 C n − i ( R + , D ( A i )).It is easy to verify that ddt E u,v ( t ) = − Z Ω a ( x ) | ∂ t u ( t, x ) | dx. (1.9)Thus E u,v ( t ) is decreasing with respect to time.In bounded domain and under some geometric conditions, Kapitonov [13] considered the caseof equal speeds ( γ = 1) and proved the uniform decay E u,v ( t ) M e − βt E u,v (0) , f or all t > , (1.10)where M, β >
0. In [3], Ammar et al studied the indirect stability of system (1.7) in thecase of one-dimensional space and when a and b have disjoint supports. More precisely, theyestablished that the ”classical” internal damping applied to only one of the equations nevergives exponential stability if γ = 1 and for the case γ = 1 they gave an explicit necessaryand sufficient conditions for the stability to occur. In [22], Toufayli generalized this result fordifferent speeds and established, under some geometric conditions, a polynomial stability.The problem of the indirect stabilization has been also studied for coupled wave equationsby displacements (weakly coupled). Indeed Alabau et al [1] considered the following system ∂ t u ( t, x ) − ∆ u ( t, x ) + b ( x ) v ( t, x ) + a ( x ) ∂ t u ( t, x ) = 0 in R + × Ω ,∂ t v ( t, x ) − ∆ v ( t, x ) + b ( x ) u ( t, x ) = 0 in R + × Ω ,u = v = 0 on R + × Γ , ( u (0 , . ) , ∂ t u (0 , . )) = ( u , u ) in Ω , ( v (0 , . ) , ∂ t v (0 , . )) = ( v , v ) in Ω , (1.11)where Ω is a bounded domain. They proved that the system (1.11) can not be exponentiallystable and when the coupling term is constant they established a polynomial decay. In [2] L.ALOUI AND H.AZAZA
Alabau et al improved this result by assuming that the regions { a > } and { b > } bothverify GCC and the coupling term satisfies a smallness assumption. This result has beengeneralized by Aloui et al [4], for more natural smallness condition on the infinity norm of thecoupling term. Recently, Daoulatli [10] showed that the rate of energy decay for solutions tothe system on a compact manifold with a boundary is determined from a first order differen-tial equation when the coupling zone and the damping zone verify the GCC.In the sequel, we fix a constant R > O ⊂ B = { x ∈ R d , | x | < R } . Suppose that there exist two positive constants a − and b − such that the damping set ω a := { a ( x ) > a − > } and the coupling set ω b := { b ( x ) > b − > } are non-empty open subsetsof Ω. As usual for damped wave (resp. Klein-Gordon) equations, we have to make somegeometric assumptions on the sets ω a and ω b so that the energy of a single wave decayssufficiently rapidly at infinity. Here, we shall use the Geometric control condition. Definition 1.1. (see [7, 15] ) We say that a set ω of Ω satisfies the geometric control condition GCC if there exists
T > such that from every point in Ω the generalized geodesic meets theset ω in a time t < T . If ω satisfies GCC , we set T ω = inf { T > , ( ω, T ) satisfies GCC } . We need also the following assumptions( A ) supp ( b ) ⊂ supp ( a ).( A ) There exists R > R such that • B cR ⊂ ω a ∩ ω b , if ( m , m ) ∈ R + × R ∗ + , • B cR ⊂ ω b and a ( x ) = βb ( x ), | x | > R , for some β >
0, if m = m = 0.For γ ∈ R ∗ + , we set I γ = E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + k ( u, v )(0) k L (Ω) and H γ = ( H ∩ ( L (Ω)) , if γ = 1 ,D ( A ) ∩ ( L (Ω)) , if γ = 1 . With this notation, we can state the stability result for the system (1.7).
Theorem 1.1.
Let γ ∈ R ∗ + and ( m , m ) ∈ { (0 , } ∪ R + × R ∗ + . We assume that ω b satisfiesthe GCC and that the assumptions ( A ) and ( A ) hold. Then for any solution ( u, v ) of thesystem (1.7) with initial data ( u , u , v , v ) ∈ H γ , we have E u,v ( t ) C (1 + t ) − I γ and k ( u, v )( t ) k L CI γ , for all t > , (1.12) where C is positive constant. In addition for ( u , u , v , v ) ∈ H , E u,v ( t ) converges to zero as t goes to infinity. In the case of Klein-Gordon-type systems we obtain the following uniform decay.
Corollary 1.
Let m , m , γ ∈ R ∗ + . Assume that ω b satisfies the GCC and the assumptions ( A ) and ( A ) hold. ⊲ If γ = 1 , then there exist positive constants C and α such that E u,v ( t ) Ce − αt E u,v (0) , for all t > , (1.13) for all solution ( u, v ) of the system (1.7) with initial data ( u , u , v , v ) ∈ H . ⊲ If γ = 1 , then there exists a positive constant C such that E u,v ( t ) Ct n n X k =0 E ∂ kt u,∂ kt v (0) , for all t > , (1.14) for all solution ( u, v ) of the system (1.7) with initial data ( u , u , v , v ) ∈ D ( A n ) . Remark 1. • To our best knowledge, our result is new for the indirect stabilizationproblem in exterior domains. • Remark that, when γ = 1 , the energy of the system (1.7) decays as fast as that of thecorresponding scalar damped equation. So the coupling through velocities, in this case,allows a full transmission of the damping effects, quite different from the couplingthrough the displacements. • To prove our main result we study the energy first at infinity ( Section 2) and thenin bounded regions (Section 3). Keeping, only the second step, we can obtain the exp-nential energy decay for the system (1.7) in bounded domains with Dirichly boundarycondition. • Due to technical difficulties we did not cover the Klein-Gordon-Wave case ( m > , m = 0 ); we will be interested in the forthcoming work. We conclude this introduction with an outline of the rest of this paper. In Section 2 weestimate the total energy at infinity by multiplier arguments. Section 3 is devoted to thestudy of the energy in bounded domain. The proof of this result is based on observabilityestimate for scalar wave equation. In order to control the compact terms, we prove in section4 a weak observability estimate that is based on a unique continuation result. Finally, inSection 5 we combine the results of the previous sections to established our main results.We denote by Ω R := Ω ∩ B R , C R,R ′ = Ω ∩ ( B R ′ \ B R ) , when 0 < R < R ′ , E R ( u, v, t ) = 12 Z | x | >R | ∂ t u ( t, x ) | + |∇ u ( t, x ) | + m | u ( t, x ) | dx + 12 Z | x | >R | ∂ t v ( t, x ) | + γ |∇ v ( t, x ) | + m | v ( t, x ) | dx,E R ( u, v, t ) = 12 Z Ω R | ∂ t u ( t, x ) | + |∇ u ( t, x ) | + m | u ( t, x ) | dx + 12 Z Ω R | ∂ t v ( t, x ) | + γ |∇ v ( t, x ) | + m | v ( t, x ) | dx, and A . B means A CB for some positive constante C . L.ALOUI AND H.AZAZA Estimate of energy near infinity
The main result of this section is as follows.
Proposition 2.1.
Let γ ∈ R ∗ + and ( m , m ) ∈ { (0 , } ∪ R + × R ∗ + . Let R > be such that ( A ) is satisfied and R > R . Then for every ε > , there exists C ε > such that for allsolution ( u, v ) of (1.7) with initial data ( u , u , v , v ) ∈ H γ , we have k ( u, v )( t ) k L ( | x | >R ) + Z t E R ( u, v, s ) ds . C ε ( E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0))+ ε Z t E u,v ( s ) ds + C ε Z t Z Ω R | u | + | v | dxds + k ( u, v )(0) k L (Ω) , (2.1) for all t > . Let ϕ ∈ C ∞ ( R d ) be a function satisfying 0 ϕ ϕ ( x ) = ( | x | > R | x | R . To prove Proposition 2 .
1, we need the following Lemma.
Lemma 2.1.
We assume the hypothesis of Proposition 2.1 and we consider ϕ as above. Thenfor every ε > , there exist C ε > such that for all solution ( u, v ) of (1.7) with initial data ( u , u , v , v ) ∈ H γ , we have Z t Z Ω b ( x ) ϕ | ∂ t v | dxds . C ε ( E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0))+ C ε Z t Z Ω R | v | dxds + ε Z t E u,v ( s ) ds, (2.2) for all t > .Proof of Lemma . . Multiplying the first and the second equation of (1.7) respectively by ϕ∂ t v and γ ϕ∂ t u and integrating the sum of these results on [0 , t ] × Ω, we obtain h Z Ω γ ϕ∂ t u∂ t v + m ϕuv dx i t + Z t Z Ω b ( x ) ϕ | ∂ t v | dxds = Z t Z Ω γ a ( x ) ϕ | ∂ t u | − ϕ∂ t u∂ t v + ϕ ∆ u∂ t v + ( m − m γ ) ϕv∂ t u + ϕ ∆ v∂ t u − (1 − γ ) ϕ∂ t v∂ t u dxds. Note that Z t Z Ω ϕ ∆ u∂ t v dxds = h Z Ω ϕ ∆ uv dx i t − Z t Z Ω ϕ ∆ ∂ t uv dxds = − h Z Ω ∇ u ( ∇ ϕv + ϕ ∇ v ) dx i t − Z t Z Ω ∆( ϕv ) ∂ t u dxds = − Z t Z Ω (∆ ϕv + ∆ vϕ + 2 ∇ v ∇ ϕ ) ∂ t u dxds − h Z Ω ∇ u ( ∇ ϕv + ϕ ∇ v ) dx i t . (2.3)Then using Young’s inequality, we get h F γ i t + Z t Z Ω b ( x ) ϕ | ∂ t v | dxds . Z t Z Ω (( 1 γ a ( x ) + 2) ϕ + C ε |∇ ϕ | ) | ∂ t u | + C ε ϕ (1 − γ ) | ∂ t u | + | ∆ ϕ | | v | dxds + ε Z t Z Ω |∇ v | + ( m − m γ ) k ϕ k ∞ | v | + | ∂ t u | + k ϕ k ∞ | ∂ t v | dxds, where F γ = Z Ω ϕ ( 1 γ ∂ t u∂ t v + m uv ) + ∇ u ( ∇ ϕv + ϕ ∇ v ) dx. By hypothesis supp ( ϕ ) ⊂ { x ∈ Ω , a ( x ) > a − } , (2.4)so, we deduce that h F γ i t + Z t Z Ω b ( x ) ϕ | ∂ t v | dxds . C ε Z t Z Ω a ( x )( | ∂ t u | + (1 − γ ) | ∂ t u | ) dxds + Z t Z Ω R | v | dxds + ε Z t E u,v ( s ) ds. (2.5)Using the energy decay (1.9) and the fact that ( m , m ) ∈ { (0 , } ∪ R + × R ∗ + , we can see that (cid:12)(cid:12)(cid:12) F γ ( s ) (cid:12)(cid:12)(cid:12) . E u,v ( s ) . E u,v (0) , ∀ s > . (2.6)Combining (1.9), (2.5) and (2.6), we obtain (2.2). (cid:3) Lemma 2.2.
Let γ ∈ R ∗ + and ( m , m ) = (0 , . Let R > be such that ( A ) is satisfiedand R > R . Then for every ε > , there exists C ε > such that for all solution ( u, v ) of (1.7) with initial data ( u , u , v , v ) ∈ H γ , we have k ( u, v )( t ) k L ( | x | >R ) + Z t E R ( v, s ) ds . C ε ( E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0))+ ε Z t E u,v ( s ) ds + C ε ( Z t Z Ω R | u | + | v | dxds + k ( u, v )(0) k L (Ω) ) , (2.7) for all t > . Where E R ( v, t ) = R | x | >R | ∂ t v ( t, x ) | + |∇ v ( t, x ) | dx. Proof of Lemma . . We write the system (1.7) in the form ∂ t u − ∆ u + a ( x ) b ( x ) ∂ t v − a ( x ) b ( x ) γ ∆ v + b ( x ) ∂ t v = 0 in R + × Ω R c , − ∂ t v + γ ∆ v + b ( x ) ∂ t u = 0 in R + × Ω R c . (2.8) L.ALOUI AND H.AZAZA
Multiplying the first equation of (2.8) by ϕv and the second one by γ ϕu and integrating thesum of these results on [0 , t ] × Ω, we obtain Z Ω ϕb ( x )2 ( 1 γ | u ( t ) | + | v ( t ) | ) dx + β Z t Z Ω ϕ ( | ∂ t v | + |∇ v | ) dxds = Z t Z Ω ϕβ | ∂ t v | + γ β ∆ ϕ | v | − ∇ u ( ∇ ϕv + ϕ ∇ v )+ ∇ v ( ∇ ϕu + ϕ ∇ u ) + (1 − γ ) ϕ∂ t u∂ t v dxds + Z Ω ϕb ( x )2 ( 1 γ | u (0) | + | v (0) | ) dx − h G γ i t dx, where G γ = Z Ω ϕ ( ∂ t uv + ∂ t vv − γ ∂ t vu ) dx. According to Lemma 2 .
1, hypothesis ( A ) and using Young’s inequality, we deduce that Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx + Z t Z Ω ϕ ( | ∂ t v | + |∇ v | ) dxds . E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + k ( u, v )(0) k L + Z t Z Ω R | v | + | u | dxds + ε Z t E u,v ( s ) ds − h G γ i t . (2.9)But we have (cid:12)(cid:12)(cid:12) G γ ( t ) (cid:12)(cid:12)(cid:12) . E u,v ( t ) + ε Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx . E u,v (0) + ε Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx. So, for ε small enough we get Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx + Z t Z Ω ϕ ( | ∂ t v | + |∇ v | ) dx ds . E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + k ( u, v )(0) k L + Z t Z Ω R | v | + | u | dxds + ε Z t E u,v ( s ) ds. (2.10)Since ϕ ≡ f or | x | > R (2.11)we deduce that Z | x | >R | u ( t ) | + | v ( t ) | dx + Z t E R ( v, s ) ds Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx + Z t Z Ω ϕ ( | ∂ t v | + |∇ v | ) dxds. Combining this estimate with (2.10), we conclude (2.7). This finishes the proof of Lemma2 . (cid:3) Now we give the proof of Proposition 2 . Proof of Proposition . . We distinguish the case m = m = 0 and the case where m ∈ R + and m ∈ R ∗ + . First case m = m = 0. Multiplying the first equation of (1.7) by ϕu and integrating on[0 , t ] × Ω, we obtain h Z Ω ϕ ( ∂ t uu + a ( x ) | u | b ( x ) uv ) dx i t + Z t Z Ω ϕ ( |∇ u | + | ∂ t u | ) dxds = Z t Z Ω ϕ | ∂ t u | + ∆ ϕ | u | + ϕb ( x ) v∂ t u dxds. (2.12)Note that we have Z t Z Ω ϕb ( x ) v∂ t u dxds = Z t Z Ω ϕv ( ∂ t v − γ ∆ v ) dxds = h Z Ω ϕ∂ t vv dx i t + Z t Z Ω ϕ ( γ |∇ v | − | ∂ t v | ) − γ ∆ ϕ | v | dxds. (2.13)So, combining this identity with (2.12) and using (2.4), we get Z t Z Ω ϕ ( | ∂ t u | + |∇ u | ) dxds . Z t Z Ω a ( x ) | ∂ t u | + Z t Z Ω ϕ ( | ∂ t v | + |∇ v | ) dxds + Z t Z Ω R | u | + | v | dxds − h Z Ω ϕ ( ∂ t uu + b ( x ) uv + a ( x ) | u | − ∂ t vv ) dx i t . (2.14)Using that, (cid:12)(cid:12)(cid:12) Z Ω ϕ ( ∂ t uu + b ( x ) uv + a ( x ) | u | − ∂ t vv )( t ) dx (cid:12)(cid:12)(cid:12) . C ε E u,v (0) + Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx (cid:12)(cid:12)(cid:12) Z Ω ϕ ( ∂ t uu + b ( x ) uv + a ( x ) | u | − ∂ t vv )(0) dx (cid:12)(cid:12)(cid:12) . E u,v (0) + k ( u, v )(0) k L , we obtain Z t Z Ω ϕ ( | ∂ t u | + |∇ u | ) dxds . C ε E u,v (0) + Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx + Z t Z Ω ϕ ( | ∂ t v | + |∇ v | ) dx ds + Z t Z Ω R | u | + | v | dxds + k ( u, v )(0) k L . (2.15)According to (2.10) and using (2.11), we get Z t E R ( u, s ) ds . C ε E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + ε Z t E u,v ( s ) ds + Z t Z Ω R | u | + | v | dxds + k ( u, v )(0) k L , (2.16)where E R ( u, t ) = R | x | >R | ∂ t u ( t, x ) | + |∇ u ( t, x ) | dx .Combining (2.7) and (2.16), we conclude (2.1). Second case m ∈ R + and m ∈ R ∗ + . Multiplying the first and the second equation of (1.7)respectively by ϕu and ϕv and integrating the sum of these results on [0 , t ] × Ω, we obtain Z Ω ϕ a ( x ) | u ( t ) | dx + Z t Z Ω ϕ ( | ∂ t u | + |∇ u | + m | u | + | ∂ t v | + |∇ v | + m | v | ) dxds = Z t Z Ω ϕ ( | ∂ t u | + | ∂ t v | ) dxds + Z t Z Ω ∆ ϕ | u | + γ | v | ) + 2 ϕb ( x ) v∂ t u dxds − h Z Ω ϕ ( ∂ t uu + ∂ t vv + b ( x ) uv ) dx i t + Z Ω ϕ a ( x ) | u (0) | dx . Z t Z Ω a ( x ) | ∂ t u | + ϕ | ∂ t v | + ε k ϕ k ∞ | v | dxds − h Z Ω ϕ ( ∂ t uu + ∂ t vv + b ( x ) uv ) dx i t + Z Ω ϕ a ( x ) | u (0) | dx + Z t Z Ω R | u | + | v | dxds. (2.17)Using the following estimates for ε small enough (cid:12)(cid:12)(cid:12) Z Ω ϕ (( ∂ t uu + ∂ t vv + b ( x ) uv )( t )) dx (cid:12)(cid:12)(cid:12) . E u,v (0) + ε Z Ω ϕ | u ( t ) | dx, (cid:12)(cid:12)(cid:12) Z Ω ϕ (( ∂ t uu + ∂ t vv + b ( x ) uv )(0)) dx (cid:12)(cid:12)(cid:12) . E u,v (0) + k u (0) k L , and according to Lemma 2 .
1, we infer (2.1). The proof of proposition 2 . (cid:3) Estimate of energy in bounded region
In this section, we will study the energy in bounded domain. For this aim, we consider afunction ψ ∈ C ∞ ( R d ) such that 0 ψ ψ ( x ) = ( | x | R | x | > R . where R > R > R and R > A ) is satisfied.It is easy to verify that ( u i , v i ) = ( ψu, ψv ) satisfies the following system ∂ t u i − ∆ u i + m u i + b ( x ) ∂ t v i + a ( x ) ∂ t u i = − ∇ ψ ∇ u − u ∆ ψ in R + × Ω R ∂ t v i − γ ∆ v i + m v i − b ( x ) ∂ t u i = − γ ∇ ψ ∇ v − γ v ∆ ψ in R + × Ω R u i = v i = 0 on R + × ∂ Ω R ( u i , u i , v i , v i ) = ( ψu , ψu , ψv , ψv ) . (3.1) Proposition 3.1.
Let γ ∈ R ∗ + , ( m , m ) ∈ { (0 , } ∪ R + × R ∗ + and ψ be as above. Assumethat the assumption ( A ) holds and that ( ω b , T ) geometrically controls Ω for some T > .Then for every ε > , there exist C ε > such that for all solution ( u, v ) of (1.7) with initialdata ( u , u , v , v ) ∈ H γ , we have Z t + Tt E R ( u, v, s ) ds . C ε Z t + Tt Z Ω a ( x )( | ∂ t u | + (1 − γ ) | ∂ t u | ) dxds + ε Z t + Tt E u,v ( s ) ds + C ε Z t + Tt Z Ω R | u | + | v | dxds + C ε Z t + Tt E R ( u, v, s ) ds + h K γ i t + Tt (3.2) for all t > . Where K γ = − Z Ω b ( x ) γ ∂ t u i ∂ t v i + ∇ u i ∇ (( b ( x ) v i ) + m ab ( x ) u i v i dx. In order to prove proposition 3 . Lemma 3.1.
Assume that the hypothesis of Proposition 3.1 hold. Then for every ε > , thereexists C ε > such that for all solution ( u, v ) of (1.7) with initial data ( u , u , v , v ) ∈ H γ ,we have Z t + Tt Z Ω b ( x ) | ∂ t v i | dxds . C ε Z t + Tt Z Ω a ( x )( | ∂ t u | + (1 − γ ) | ∂ t u | ) dxds + ε Z t + Tt E u,v ( s ) ds + C ε Z t + Tt Z Ω R | v | + | u | dxds + C ε Z t + Tt Z C R ,R |∇ u | + |∇ v | dxds + h K γ i t + Tt , (3.3) for all t > .proof of Lemma . . We multiply the first and the second equation of (3.1) respectively by b ( x ) ∂ t v i and b ( x ) γ ∂ t u i and we integrate the sum of these results on [ t, t + T ] × Ω, we get h K γ i t + Tt + Z t + Tt Z Ω b ( x ) | ∂ t v i | dxds = Z t + Tt Z Ω b ( x ) γ | ∂ t u i | − ab ( x ) ∂ t u i ∂ t v i + ( m − m γ ) b ( x ) v i ∂ t u i ) dxds − Z t + Tt Z Ω b ( x )(2 ∇ u ∇ ψ + ∆ ψu ) ∂ t v i + b ( x ) γ (2 ∇ v ∇ ψ + ∆ ψv ) ∂ t u i dxds + Z t + Tt Z Ω ( 1 γ − b ( x ) ∂ t u i ∂ t v i dxds − Z t + Tt Z Ω ∂ t u i (∆ b ( x ) v i + 2 ∇ b ( x ) ∇ v i ) dxds. From Young’s inequality and using hypothesis ( A ), we infer that h K γ i t + Tt + Z t + Tt Z Ω b ( x ) | ∂ t v i | dxds . C ε Z t + Tt Z Ω a ( x )( | ∂ t u | + (1 − γ ) | ∂ t u | ) dxds + ε Z t + Tt Z Ω ( m − m γ ) | v | + | ∂ t u | + | ∂ t v | + |∇ v | dxds + C ε Z t + Tt Z Ω R | u | + | v | dxds + C ε Z t + Tt Z C R ,R |∇ u | + |∇ v | dxds. (3.4)This implies (3.3). (cid:3) Proof of proposition . . First, we recall the following observability estimate for the waveequation ( see proposition 3, [11]).
Lemma 3.2.
Let γ, T > and O a bounded domain. Let φ be a nonnegative function on O and setting V = { φ ( x ) > } . We assume that ( V , T ) satisifies the GCC . There exists C T > , such that for all ( u , u ) ∈ H ( O ) × L ( O ) , f ∈ L loc ( R + , L ( O )) , and all t > the solution of ∂ t u − γ ∆ u + mu = f in R + × O ,u = 0 on R + × ∂ O , ( u (0 , x ) , ∂ t u (0 , x )) = ( u , u ) ∀ x ∈ O . (3.5) where m > , satisfies with E u ( t ) = 12 Z O | ∂ t u ( t, x ) | + m | u ( t, x ) | + γ |∇ u ( t, x ) | dx, the inequality Z t + Tt E u ( s ) ds C T Z t + Tt Z O φ ( x ) | ∂ t u | + | f | dxds. (3.6)Let ω b, = ω b ∩ B R = { x ∈ Ω ∩ B R , b ( x ) > b − > } . Since ( ω b , T ) satisfies the GCC , B R c ⊂ ω b and R > R , we conclude that ( ω b, , T ) geometrically controls Ω R .So, according to Lemma 3 . A ), we have Z t + Tt E v i ( s ) ds . Z t + Tt Z ω b, | ∂ t v i | dxds + Z t + Tt Z Ω b ( x ) | ∂ t u i | dxds + Z t + Tt Z C R ,R |∇ v | dxds + Z t + Tt Z Ω R | v | dxds . Z t + Tt Z Ω b ( x ) | ∂ t v i | dxds + Z t + Tt Z Ω a ( x ) | ∂ t u | dxds + Z t + Tt Z C R ,R |∇ v | dxds + Z t + Tt Z Ω R | v | dxds, t > , (3.7) where E v i ( t ) = 12 Z Ω |∇ v i ( t, x ) | + | ∂ t v i ( t, x ) | + m | v i ( t, x ) | dx. We have also Z t + Tt E u i ( s ) ds . Z t + Tt Z Ω a ( x ) | ∂ t u | + b ( x ) | ∂ t v i | dxds + Z t + Tt Z C R ,R |∇ u | dxds + Z t + Tt Z Ω R | u | dxds, t > , (3.8)where E u i ( t ) = 12 Z Ω |∇ u i ( t, x ) | + | ∂ t u i ( t, x ) | + m | u i ( t, x ) | dx. Adding the two estimates above and using (3.3), we deduce that Z t + Tt E u i ,v i ( s ) ds . C ε Z t + Tt Z Ω a ( x )( | ∂ t u | + (1 − γ ) | ∂ t u | ) dxds + ε Z t + Tt E u,v ( s ) ds + C ε Z t + Tt E R ( u, v, s ) ds + C ε Z t + Tt Z Ω R | u | + | v | dxds + h K γ i t + Tt . (3.9)Since ψ ≡ | x | R , we get Z t + Tt E R ( u, v, s ) ds Z t + Tt E u i ,v i ( s ) ds Combining this estimate with (3.9), we conclude (3.2). (cid:3) Weak observability estimate
In this section, we prove the following proposition.
Proposition 4.1.
Let γ ∈ R ∗ + and m , m ∈ R + . Let R > be such that ( A ) is satisfiedand R > R . We assume that the assumption ( A ) holds. Then for every T > T ω b and α > , there exists C T,α > , such that for all ( u , u , v , v ) ∈ ( H (Ω) × L (Ω)) , and all t > , the solution of the system (1.7) satisfies the following inequality Z t + Tt Z Ω R | v | + | u | dxds C T,α Z t + Tt Z Ω a ( x ) | ∂ t u | dxds + α Z t + Tt E u,v ( s ) ds. (4.1) Proof of Proposition . . We note that for each ( u , u , v , v ) ∈ ( H (Ω) × L (Ω)) , the solu-tion ( u, v ) are given as the limit of smooth solutions ( u n , v n )( t ) with ( u n , v n )(0) = ( u n, , v n, ) ∈ ( C ∞ (Ω)) and ( ∂ t u n , ∂ t v n )(0) = ( u n, , v n,t ) ∈ ( C ∞ (Ω)) such that ( u n, , v n, ) → ( u , v ) ∈ ( H (Ω)) and ( u n, , v n, ) → ( u , v ) ∈ ( L (Ω)) . Note that k u n ( t, . ) − u ( t, . ) k H + k ∂ t u n ( t, . ) − ∂ t u ( t, . ) k L −−−−−→ n → + ∞ , k v n ( t, . ) − v ( t, . ) k H + k ∂ t v n ( t, . ) − ∂ t v ( t, . ) k L −−−−−→ n → + ∞ , uniformly on the each closed interval [0 , T ] for any T >
0. Therefore we may assume that( u, v ) is smooth.To prove the estimate (4.1), we argue by contradiction. We assume that there exist apositive sequence ( t n ) and a sequence U n = ( u n , ∂ t u n , v n , ∂ t v n )of solution of the system (1.7) with initial data ( u n, , u n, , v n, , v n, ) ∈ ( H (Ω) × L (Ω)) , suchthat Z t n + Tt n Z Ω R | u n | + | v n | dxds > n Z t n + Tt n Z Ω a ( x ) | ∂ t u n | dxdt + α Z t n + Tt n E u n ,v n ds Set β n = Z t n + Tt n Z Ω R | u n | + | v n | dxds and ( y n , ∂ t y n , z n , ∂ t z n )( t ) := U n ( t + t n ) β n . We infer that Z T Z Ω R | y n | + | z n | dxds = 1 , (4.2) Z T Z Ω a ( x ) | ∂ t y n | dxds n , (4.3) Z T E y n ,z n ( s ) ds α . (4.4)Therefore ( y n , z n ) ⇀ ( y, z ) in L ((0 , T ) , H (Ω)) ∩ W , ((0 , T ) , L (Ω)) , with respect to the weak topology. By Rellich’s lemma, we can assume that( y n , z n ) → ( y, z ) in ( L ((0 , T ) × Ω R )) . It is easy to see that the limit ( y, z ) satisfies the system ∂ t y − ∆ y + m y + b ( x ) ∂ t z = 0 in (0 , T ) × Ω ,∂ t z − γ ∆ z + m z = 0 in (0 , T ) × Ω ,y = z = 0 on (0 , T ) × Γ ,a ( x ) ∂ t y = 0 on (0 , T ) × Ω (4.5)and Z T Z Ω R | y | + | z | dxds = 1 . (4.6) It is clear that ( ∂ t y, ∂ t z ) satisfies the following system ∂ t ( ∂ t y ) − ∆( ∂ t y ) + m ∂ t y + b ( x ) ∂ t ( ∂ t z ) = 0 in (0 , T ) × Ω ,∂ t ( ∂ t z ) − γ ∆( ∂ t z ) + m ∂ t z = 0 in (0 , T ) × Ω ,∂ t y = ∂ t z = 0 on (0 , T ) × ∂ Ω ,a ( x ) ∂ t y = 0 on (0 , T ) × Ω . (4.7)From the first and previous equations in (4.7), we deduce that b ( x ) ∂ t z = 0 on supp ( a ). But supp ( b ) ⊂ supp ( a ), so ∂ t z = 0 on supp ( b ). Setting w = ∂ t z , we have ∂ t w = 0 in (0 , T ) × ω b ,∂ t w − γ ∆ w + m w = 0 in (0 , T ) × Ω ,w = 0 on (0 , T ) × ∂ Ω ,w ∈ L ((0 , T ) × Ω) . (4.8)Using the first and second equations in (4.8), we can see that W F ( w ) ∩ (0 , T ) × ω b × R × R n is a subset of { ( t, x, τ, ξ ) ∈ (0 , T ) × Ω × R × R n ; τ − γ | ξ | = τ = 0 } = (0 , T ) × Ω × { } × { } . where W F ( w ) denotes the H -wavefront set of w . Since B cR ⊂ ω b , we deduce that w ∈ H loc ((0 , T ) × B cR ) . Next, we will show that w ∈ H loc ([0 , T ] × R R ). Let ρ = ( t , x , τ , ξ ) ∈ T ∗ ([0 , T ] × B R ) and Γ be the generalized bicharacteristic issued from ρ . Set { ρ :=(0 , x , τ , ξ ) } = Γ ∩ { t = 0 } and { ρ := ( T, x , τ , ρ ) } = Γ ∩ { t = T } , so we distin-guish two cases,1 st case: x or x / ∈ B R . In this case ρ or ρ / ∈ W F ( w )). Since T > T ω b , then usingthe propagation of regularity along the bicharacteristic flow of the operator ∂ t − γ ∆ (see[17, 18]), we obtain ρ / ∈ W F ( w ).2 nd case: x , x ∈ B R . Since ρ , ρ ∈ T ∗ ([0 , T ] × B R ) and ω b controls geometrically[0 , T ] × Ω, then Γ intersects the region [0 , T ] × ( ω b ∩ Ω R ). But w ∈ H loc ([0 , T ] × ( ω b ∩ Ω R )),then applying again the regularity propagation theorem, we deduce that ρ / ∈ W F ( w ).Therefore, we conclude that w ∈ H loc ((0 , T ) × Ω). Now, set ˜ w = ∂ t w . Since R n \ Ω R ⊂ ω b ,so ˜ w = 0 on R n \ Ω R and satisfies ∂ t ˜ w − γ ∆ ˜ w + m ˜ w = 0 in (0 , T ) × Ω R , ˜ w = 0 on (0 , T ) × ∂ Ω R , ˜ w = 0 in (0 , T ) × ( ω b ∩ Ω R ) , ˜ w ∈ L ((0 , T ) × Ω R ) (4.9)Since ω b ∩ Ω R controls geometrically Ω R , then using the classical unique continuation result(see [7, 8] ), we infer that ˜ w ≡ , T ) × Ω R . Therefore, the function z satisfies ( − γ ∆ z + m z = 0 in (0 , T ) × Ω ,z = 0 in (0 , T ) × ∂ Ω . (4.10) This implies that z = 0 on (0 , T ) × Ω. Now, from (4.5) we obtain ∂ t y − ∆ y + m y = 0 in (0 , T ) × Ω ,a ( x ) ∂ t y = 0 in (0 , T ) × Ω ,y = 0 on (0 , T ) × ∂ Ω ,y ∈ H ((0 , T ) × Ω) (4.11)Arguing as for z , we can prove that y = 0. This is in contradiction with (4.6). (cid:3) Proof of Theorem . R > R . According to (2.1) for t = nT , n ∈ N ∗ , we have Z nT E R ( u, v, s ) ds . C ε E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + Z nT Z Ω R | u | + | v | dxds ! + ε Z nT E u,v ( s ) ds + k ( u, v )(0) k L . (5.1)Next, using (3.2) with R = 2 R and R = 3 R , we get Z ( k +1) TkT E R ( u, v, s ) ds . C ε Z ( k +1) TkT Z Ω a ( x )( | ∂ t u | + (1 − γ ) | ∂ t u | ) dxds + ε Z ( k +1) TkT E u,v ( s ) ds + C ε Z ( k +1) TkT E R ( u, v, s ) ds + C ε Z ( k +1) TkT Z Ω R | u | + | v | dxds − h K γ i ( k +1) TkT , ∀ k ∈ N . (5.2)Thus n − X k =0 Z ( k +1) TkT E R ( u, v, s ) ds . n − X k =0 C ε Z ( k +1) TkT Z Ω a ( x )( | ∂ t u | + (1 − γ ) | ∂ t u | ) dxds + ε Z ( k +1) TkT E u,v ( s ) ds − h K γ i ( k +1) TkT + C ε (cid:16) Z ( k +1) TkT E R ( u, v, s ) ds + Z ( k +1) TkT Z Ω R | u | + | v | dxds (cid:17)! , ∀ k ∈ N . (5.3)This gives Z nT E R ( u, v, s ) ds . C ε Z nT Z Ω a ( x )( | ∂ t u | + (1 − γ ) | ∂ t u | ) dxds + ε Z nT E u,v ( s ) ds + C ε Z nT E R ( u, v, s ) ds + C ε Z nT Z Ω R | u | + | v | dxds − h K γ i nT , ∀ n ∈ N ∗ . (5.4) From the following estimate (cid:12)(cid:12)(cid:12) K γ ( s ) (cid:12)(cid:12)(cid:12) . E u,v (0) , ∀ s > , and using (1.9) and (5.1), we deduce that Z nT E R ( u, v, s ) ds . C ε ( E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0))+ ε Z nT E u,v ( s ) ds + C ε Z nT Z Ω R | u | + | v | dxds, ∀ n ∈ N ∗ . (5.5)So, combining (5.5) and (5.1), we conclude for small enough ε the following estimate Z nT E u,v ( s ) ds . C ε ( E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0)) k ( u, v )(0) k L + C ε Z nT Z Ω R ( | v | + | u | ) dxds. (5.6)Next, From (4.1) with R = 3 R we have n − X k =0 Z ( k +1) TkT Z Ω R | v | + | u | dxds . n − X k =0 (cid:16) Z ( k +1) TkT Z Ω a ( x ) | ∂ t u | dxds + α Z ( k +1) TkT E u,v ( s ) ds (cid:17) . Thus Z nT Z Ω R | v | + | u | dxds . E u,v (0) + α Z nT E u,v ( s ) ds. (5.7)Finally, using (5.7) for α small enough in (5.6), we find Z nT E u,v ( s ) ds . C ε ( E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0)) + k ( u, v )(0) k L (Ω) , (5.8)Therefore Z + ∞ E u,v ( s ) ds . E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + k ( u, v )(0) k L (Ω) . As the energy is decreasing then(1 + t ) E u,v ( t ) Z + ∞ E u,v ( s ) ds + E u,v (0) . E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0)+ k ( u, v )(0) k L (Ω) , for all t > . (5.9)On the other hand, using (2.1), (5.7) and (5.8), we deduce that Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx . E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + k ( u, v )(0) k L (Ω) . (5.10) Since ϕ ≡ | x | > R , Z Ω ϕ ( | u ( t ) | + | v ( t ) | ) dx > Z Ω cR | u ( t ) | + | v ( t ) | dx, (5.11)therefore Z Ω cR | u ( t ) | + | v ( t ) | dx . E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + k ( u, v )(0) k L (Ω) . (5.12)Poincare’s inequality and the fact that the energie of ( u, v ) is decreasing gives Z Ω R | u ( t ) | + | v ( t ) | dx C Ω Z Ω R |∇ u ( t ) | + |∇ v ( t ) | dx . E u,v (0) (5.13)for all t > Z Ω | u ( t ) | + | v ( t ) | dx . E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0) + k ( u, v )(0) k L (Ω) , (5.14)for all t > Proof of Corollary . From (5.9), we deduce if γ = 1 E u,v ( t ) Ct E u,v (0) , for all t > , we choose t such that Ct < γ = 1, E u,v ( t ) Ct ( E u,v (0) + (1 − γ ) E ∂ t u,∂ t v (0)) , for all t > , according to [Theoreme 2 .
1, 1] we infer that (1.14). (cid:3)
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