Uniqueness for an hyperbolic inverse problem with time-dependent coefficient Mourad Bellassoued * — Ibtissem Ben Aïcha** * ENIT-LAMSIN University of Tunis El Manar, National Engineering School of Tunis B.P.37, 1002 Tunis TUNISIA mourad.bellassoued@enit.utm.tn ** ENIT-LAMSIN University of Aix-Marseille 85 Boulevard Charles Livon, 13284 Marseille FRANCE & University of Carthage, Faculty of sciences of Bizerte 7021, Jarzouna Bizerte TUNISIA ibtissem.benaicha.enit@utm.tn ABSTRACT. This paper deals with an hyperbolic inverse problem of determining a time-dependent coefficient a appearing in a dissipative wave equation, from boundary observations. We prove in dimension n greater than two, that a can be uniquely determined in a precise subset of the domain, from the knowledge of the Dirichlet-to-Neumann map. RÉSUMÉ. Dans ce travail, on étudie le probl ème inverse de la détermination d’un coefficient dépendant de la variable d’espace et du temps apparaissant dans une équation d’onde dissipative, à partir des mesures faites sur tout le bord du domaine. On démontre que ce coefficient peut être déterminé d’une manière unique dans une partie précise du domaine à partir des mesures de type Neumann. KEYWORDS : Inverse problems, Dissipative wave equation, Time-dependent coefficient, Uniqueness . MOTS-CLÉS : Problèmes inverses, équation d’onde dissipative, coefficient qui dépendent du temps, Unicité Special issue, LEM2I Mourad Bellassoued, Nabil Gmati, Mohamed Jaoua, Gilles Lebeau, Editors. ARIMA Journal, vol. 23, pp. 65-78 (2016) 66 ARIMA - volume 23 - 2016 1. Introduction 1.1. Statement of the problem The present paper is devoted to the study of the following hyperbolic inverse problem: Given T > 0 and a bounded domain Ω ⊂ Rn , n ≥ 2, with C ∞ boundary Γ = ❇ Ω, determine the absorbing coefficient a present in the following initial boundary value problem for the wave equation from boundary observations ⎧ ❇t2 u − ∆u + a(x, t)❇t u = 0 in Q = Ω × (0, T ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ u(x, 0) = 0, ❇t u(x, 0) = 0 in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ on Σ = Γ × (0, T ), ⎩ u(x, t) = f (x, t) (1) where f ∈ H 1 (Σ), and the coefficient a ∈ C 2 (Q) is assumed to be real valued. It is well known (see [10] and [9] ) that if the compatibility condition is satisfied, that is f (⋅, 0) = 0, then, there exists a unique solution u to the equation (1) that belongs to the following space u ∈ C([0, T ], H 1 (Ω)) ∩ C 1 ([0, T ], L2 (Ω)). Moreover, there exists a constant C > 0 such that we have ∥❇ν u∥L2 (Σ) ≤ ∥f ∥H 1 (Σ) , (2) where ν denotes the unit outward normal to Γ at x and ❇ν u stands for ∇u⋅ν. In the present paper, we focus on the uniqueness issue in the study of the inverse problem of determining the time-dependent absorbing coefficient a from the knowledge of the Dirichlet-toNeumann map. From a physical view point, the inverse problem under consideration consists in recovering the absorbing coefficient a in an homogeneous medium by probing it with disturbances generated on the boundary. The data are the responses of the medium to these disturbances measured on all the boundary. Here the coefficient a can be seen as one of the medium properties and we aim to recover it in a specific subset of the domain from boundary measurements, after probing the medium by a Dirichlet data f . The medium is assumed to be quiet initially. The problem of recovering coefficients that depend only on the spatial variable is considered by many authors. In [16] Rakesh and Symes proved a uniqueness result in recovering a time-independent potential appearing in a wave equation from measurements made on the whole boundary. The main tools in the derivation of this result are first, the construction of geometric optics solutions and second, the relation linking the hyperbolic Dirichlet-to-Neumann map to the X-ray transform. As for the uniqueness from local Neumann measurements, we refer to Eskin [7]. One can also see the paper of Isakov [11], in which a uniqueness result was proved in the determination of two time-independent coefficients appearing in a dissipative wave equation. The stability in the case where the Neumann data are observed on a subdomain of the boundary was considered by Bellassoued, Choulli and Yamamoto [1], where a stability estimate of log-type was proved in recovering a time-independent coefficient appearing in a wave equation. In [13], Isakov and Sun established a stability result of Hölder type in determining a coefficient in a subdomain from local Neumann data. As for the stability in Uniqueness for an hyperbolic inverse problem with time-dependent coefficient 67 the case where Neumann data are observed on the whole boundary, we refer to Sun [21], Cipolatti and Lopez [6]. When the coefficients depend also on the time variable, there is a uniqueness result proved by Ramm and Rakesh [17], in which they proved that a time-dependent potential appearing in a wave equation can be uniquely determined in a precise subset made of lines making an angle of 45○ with the t-axis and meeting the planes t = 0 and t = T outside Q, from global Neumann-data. It’s clear from [10] that this coefficient can not be recovered over the whole domain Q and this is actually due to the homogeneous initial conditions imposed in the system. However, Isakov proved in [12], that the time-dependent coefficient may be uniquely determined over the whole domain Q, but he needed to know much more information about the solution of the wave equation. We can also refer to [8, 18, 19, 20]. Inspired by the work of Bellassoued and Dos Santos [2], Waters [22] proved recently that one can stably recover the X-ray transform of a time-dependent lower order term present in a wave equation from the knowledge of the Dirichlet-to-Neumann map, in the Riemmanian case. In the euclidian case, Ben Aïcha [5] and Kian [14, 15], showed by taking inspiration from the work of Bellassoued-Jellali-Yamammoto [3, 4], stability results in the recovery of a zeroth order time-dependent coefficient appearing in a wave equation. In this paper, we prove that the time-dependent absorbing coefficient a can be uniquely determined with respect to the Dirichlet-to-Neumann map in a specific subset of the domain Q, provided that a is known outside this subset. 1.2. Main results In order to state our main result we first introduce the following notations. Let r > 0 be such that T > 2r and Ω ⊆ B(0, r/2) = {x ∈ Rn , ∣x∣ < r/2}. We set Qr = B(0, r/2) × (0, T ). We consider the annular region around the domain Ω, Ar = {x ∈ Rn , r r < ∣x∣ < T − } , 2 2 and the forward and backward light cones: r r Cr+ = {(x, t) ∈ Qr , ∣x∣ < t − , t > } , 2 2 r r − t, T − > t} , 2 2 r r Cr = {(x, t) ∈ Qr , ∣x∣ ≤ − t, 0 ≤ t ≤ }. 2 2 Cr− = {(x, t) ∈ Qr , ∣x∣ < T − Finally, we denote Q∗r = Cr+ ∩ Cr− and Qr,∗ = Q ∩ Q∗r . We remark that the open subset Qr,∗ is made of lines making an angle of 45○ with the t-axis and meeting the planes t = 0 and t = T outside Qr . We notice that Qr,∗ ⊂ Q. Note, that in the particular case where Ω = B(0, r/2), we have Qr,∗ = Q∗r (see Figure 1 in [5]). Our set of data will be given by the the Dirichlet-to-Neumann map Λa defined as follows Λa ∶ H 1 (Σ) Ð→ L2 (Σ) f z→ ❇ν u, 68 ARIMA - volume 23 - 2016 By (2) we have that Λa is continuous from H 1 (Σ) to L2 (Σ). We denote by ∥Λa ∥ its norm in L(H 1 (Σ), L2 (Σ)). Let us now introduce the admissible set of the absorbing coefficients a. Given a0 ∈ C 2 (Qr ) and M > 0 we set A(a0 , M ) = {a ∈ C 2 (Qr ), a = a0 in Qr ∖ Qr,∗ , ∥a∥C 2 (Q) ≤ M }. Having said that we may state the main results of this paper Theorem 1.1 (Non uniqueness) For any a ∈ A(a0 , M ) such that Supp(a) ⊂ Cr , we have Λa = Λ 0 . Theorem 1.2 (Uniqueness) Let T > 2Diam(Ω) and ai ∈ A(a0 , M ), i = 1, 2. Then, we have Λa2 = Λa1 implies a2 = a1 on Qr,∗ . The outline of this paper is as follows. In Section 2, we develop the proof of Theorem 1.1. Section 3 is devoted to the construction of geometric optics solutions to the equation (1). Using these particular solutions, we prove in Section 3 Theorem 1.2. 2. Non uniqueness in determining the time-dependent coefficient In this section we aim to show that it is hopeless to recover the time-dependent coefficient a over the whole domain in the case where the initial conditions are zero. 2.1. Preliminary This section is devoted to the proof of a fundamental result which is borrowed from [10]. Let us first introduce the following notations. We define V = ⋃ D(τ ) = ⋃ (Cr ∩ {t = τ }), 0≤τ ≤t′ 0≤τ ≤t′ where 0 < t′ < r/2. Moreover, we denote by S = ❇ Cr ∩ (Ω×]0, t′ [), and ❇ V = S ∪ D(t′ ) ∪ D(0). Lemma 2.1 Let us denote by u the solution of the dissipative wave equation ⎧ (❇t2 − ∆ + a(x, t)❇t ) u(x, t) = 0 in Q, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ u(x, 0) = 0 = ❇t u(x, 0) in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ on Σ. ⎩ u(x, t) = f (x, t) Then, u(x, t) = 0 on the set Cr . We denote by P = ❇t2 − ∆ + a(x, t)❇t . A simple calculation gives us ∫ 2P u(x, t) ❇t u(x, t) dx dt V = ∫ 2❇t2 u(x, t)❇t u(x, t) dx dt − ∫ 2∆u(x, t) ❇t u(x, t) dx dt V V + ∫ 2a(x, t)∣❇t u(x, t)∣2 dx dt V n = ∫ ❇t (∣❇t ∣2 + ∣∇u∣2 ) dx dt + ∫ ∑ ❇j (❇t u ❇j u) dx dt V j=1 V + ∫ 2a(x, t)∣❇t u(x, t)∣ dx dt. 2 V Uniqueness for an hyperbolic inverse problem with time-dependent coefficient 69 Then, using the above identity, we see that n 2 ∫ 2P u(x, t)❇t u(x, t) dx dt = ∫ ❇t e(x, t) dx dt+ ∫ ∑ ❇j Xj (x, t) dx dt+ ∫ 2a(x, t)∣❇t u(x, t)∣ dx dt, V j=1 V V V where e(x, t) = ∣❇t u(x, t)∣2 + ∣∇u(x, t)∣2 and Xj (x, t) = −2❇t u(x, t)❇j u(x, t). Next, by applying the divergence theorem, one gets n ∫ 2P u(x, t)❇t u(x, t) dx dt = ∫ (e(x, t)η + ∑ Xj (x, t)µj ) dσ + ∫ S V D(t′ ) j=1 + ∫ 2a(x, t)∣❇t u(x, t)∣2 dx dt, e(x, t′ ) dx − ∫ e(x, 0) dx D(0) (3) V where dσ denotes the surface element of S and the vector (η, µ1 , µ2 , ..., µn ) ∈ Rn+1 is the outward unit normal vector at (x, t) ∈ S such that ⎛n ⎞ η = ∑ µ2j ⎝j=1 ⎠ 1/2 (4) . On the other hand, from Cauchy-Schwarz inequality and (4), we can see that n ∫ (e(x, t)η + ∑ Xj (x, t)µj ) dσ S j=1 ≥ ∫ (∣❇t u(x, t)∣2 + ∣∇u(x, t)∣2 ) η − 2∣❇t u(x, t)∣∣∇u(x, t)∣ η dσ S ≥ 0. (5) Then, since e(x, 0) = 0 we get from (3) and (5) this estimation ∫ D(t′ ) e(x, t′ ) dx ≤ ∫ 2P u ❇t u(x, t) dx dt − ∫ 2a(x, t)∣❇t u(x, t)∣2 dx dt V V Now, using the fact that P u(x, t) = 0 for any (x, t) ∈ V , we get ∫ D(t′ ) t′ e(x, t′ ) dx ≤ C ∫ 0 ∫ D(t) (e(x, t) + ∣u(x, t)∣2 ) dx dt, (6) where, the positive constant C is depending on M . Now bearing in mind that ∣u(x, t′ )∣2 = ∣u(x, 0)∣2 + ∫ 0 t′ ❇t (∣u(x, t)∣2 ) dt ≤ ∫ t′ e(x, t) dx. (7) 0 Thus, from (6) and (7) we deduce that ∫ D(t′ ) (e(x, t′ ) + ∣u(x, t′ )∣2 ) dx ≤ ∫ 0 t′ ∫ D(t) (e(x, t) + ∣u(x, t)∣2 ) dx dt. In view of Gronwall’s Lemma we end up deducing that u(x, t) = 0 for any x ∈ D(t′ ) and t′ ∈ (0, r/2). This completes the proof of the lemma. 70 ARIMA - volume 23 - 2016 2.2. Proof of Theorem 1.1 Let a ∈ A(a0 , M ) such that Supp(a) ⊂ Cr . Let f ∈ H 1 (Σ) and u satisfy ⎧ ❇t2 u − ∆u + a(x, t)❇t u = 0 in Q, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ u(x, 0) = 0, ❇t u(x, 0) = 0 in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ on Σ, ⎩ u=f Since from Lemma 2.1, we have u = 0 in the conic set Cr and using the fact that Supp(a) ⊂ Cr , we deduce that u solves also the following hyperbolic boundary-value problem ⎧ ❇t2 v − ∆v = 0 in Q, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ v(x, 0) = 0, ❇t v(x, 0) = 0 in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ on Σ. ⎩ v=f Then, we conclude that Λa (f ) = Λ0 (f ) for all f ∈ H 1 (Σ). 3. Construction of geometric optics solutions In this section, we construct suitable geometrical optics solutions for the dissipative wave equation (1), which are key ingredients to the proof of our main result. We first state the following lemma that will be used in order to prove the main statement of this section. Lemma 3.1 (see [10]) Let T, M1 , M2 > 0, a ∈ L∞ (Q) and b ∈ L∞ (Q), such that ∥a∥L∞ (Q) ≤ M1 and ∥b∥L∞ (Q) ≤ M2 . Assume that F ∈ L1 (0, T ; L2 (Ω)). Then, there exists a unique solution u to the following equation such that ⎧ ❇t2 u − ∆u + a(x, t)❇t u + b(x, t)u(x, t) = F (x, t) in Q, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ in Ω, ⎨ u(x, 0) = 0 = ❇t u(x, 0) ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ on Σ, ⎩ u(x, t) = 0 (8) u ∈ C([0, T ]; H01 (Ω)) ∩ C 1 ([0, T ]; L2 (Ω)). Moreover, there exists a constant C > 0 such that for t ∈ (0, T ) we have ∥❇t u(⋅, t)∥L2 (Ω) + ∥∇u(⋅, t)∥L2 (Ω) ≤ C∥F ∥L1 (0,T ;L2 (Ω)) . (9) φ(x, t) = ϕ(x + tω), (10) By the use of the above lemma, we may construct suitable geometrical optics solutions to the equation (1) and to the retrograde problem. We shall first consider a function ϕ ∈ C0∞ (Rn ). Notice that for all ω ∈ S n−1 the function φ given by solves the following transport equation (❇t − ω ⋅ ∇)φ(x, t) = 0. Let us now prove the following Lemma. (11) Uniqueness for an hyperbolic inverse problem with time-dependent coefficient 71 Lemma 3.2 Let M1 , M2 > 0, a ∈ A(a0 , M1 ) and b ∈ W 1,∞ (Q) such that ∥b∥W 1,∞ (Q) ≤ M2 . Given ω ∈ S n−1 and ϕ ∈ C0∞ (Rn ), we consider the function φ defined by (10). Then, for any λ > 0, the following equation ❇t2 u − ∆u + a(x, t)❇t u + b(x, t)u = 0 in Q, (12) admits a unique solution u+ ∈ C([0, T ]; H 1 (Ω)) ∩ C 1 ([0, T ]; L2 (Ω)), of the following form u+ (x, t) = φ(x, t)̃ a+ (x, t)eiλ(x⋅ω+t) + r+ (x, t), (13) 1 t ∫ a(x + (t − s)ω, s) ds), 2 0 (14) where ̃ a+ (x, t) is given by ̃ a+ (x, t) = exp ( − and r+ (x, t) satisfies r+ (x, 0) = ❇t r+ (x, 0) = 0, in Ω, r+ (x, t) = 0 on Σ. (15) Moreover, there exists a positive constant C > 0 such that λ∥r+ ∥L2 (Q) + ∥❇t r+ ∥L2 (Q) ≤ C∥ϕ∥H 3 (Rn ) . (16) In order to prove this lemma, it will be enough to prove the existence of r+ satisfying ⎧ ⎪ (❇t2 − ∆ + a(x, t)❇t + b(x, t))r+ = g(x, t), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ r+ (x, 0) = ❇t r+ (x, 0) = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ + ⎪ ⎪ ⎩ r (x, t) = 0, (17) and obeying the estimate (16), where g(x, t) is given by g(x, t) = −(❇t2 − ∆ + a(x, t)❇t + b(x, t))(φ(x, t)̃ a+ (x, t)eiλ(x⋅ω+t) ). Bearing in mind that ̃ a+ (x, t) solves the following equation 2(❇t − ω ⋅ ∇)̃ a+ (x, t) = −a(x, t)̃ a+ (x, t), one can see from (11) that g(x, t) = −eiλ(x⋅ω+t) (❇t2 −∆+a(x, t)❇t +b(x, t))(φ(x, t)̃ a+ (x, t)) = −eiλ(x⋅ω+t) g0 (x, t), where g0 ∈ L1 (0, T, L2 (Ω)). Hence, in light of Lemma 3.1, we deduce the existence of a unique solution r+ such that r+ ∈ C([0, T ]; H01 (Ω)) ∩ C 1 ([0, T ]; L2 (Ω)), 72 ARIMA - volume 23 - 2016 and satisfying (17). We set t w(x, t) = ∫ 0 r+ (x, s) ds. (18) Then, in light of (17) and (18), we get (❇t2 − ∆ + a(x, t)❇t + b(x, t))w(x, t) = ∫ 0 t +∫ g(x, s) ds + ∫ t 0 t 0 (b(x, t) − b(x, s))r+ (x, s) ds ❇s a(x, s)r+ (x, s) ds, Then, the function w is a solution to the following equation ⎧ ⎪ (❇t2 − ∆ + a(x, t)❇t + b(x, t))w(x, t) = F1 (x, t) + F2 (x, t) in Q, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ w(x, 0) = 0 = ❇t w(x, 0) in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ on Σ. ⎩ w(x, t) = 0 Here F1 and F2 are given by F1 (x, t) = ∫ t (19) g(x, s) ds, 0 and F2 (x, t) = ∫ t 0 (b(x, t) − b(x, s))r+ (x, s) ds + ∫ t ❇s a(x, s)r+ (x, s) ds. 0 Considering τ ∈ [0, T ] and applying Lemma 3.1 on the interval [0, τ ], we obtain ∥❇t w(., τ )∥2L2 (Ω) ≤ C(∥F1 ∥2L2 (Q) + T (M12 + 4M22 ) ∫ τ 0 ∫ ∫ Ω 0 t ∣r+ (x, s)∣2 ds dx dt). Therefore, in view of (18), we deduce that ∥❇t w(., τ )∥2L2 (Ω) ≤ C(∥F1 ∥2L2 (Q) + ∫ 0 τ ≤ C(∥F1 ∥2L2 (Q) + T ∫ ∫ 0 t τ0 ∥❇s w(., s)∥2L2 (Ω) ds dt) ∥❇s w(., s)∥2L2 (Ω) ds). As a consequence, we find out from Gronwall’s Lemma that ∥❇t w(., τ )∥2L2 (Ω) ≤ C∥F1 ∥2L2 (Q) . Hence, from (18), one deduce that ∥r+ ∥L2 (Q) ≤ C∥F1 ∥L2 (Q) . In view of (19), one can easily see that F1 can be written as follows t t 1 iλ(x⋅ω+s) ) ds. ∫ g0 (x, s)❇s (e iλ 0 0 Therefore, by integrating by parts with respect to s, we get F1 (x, t) = ∫ g(x, s) ds = ∥r+ ∥L2 (Q) ≤ C ∥ϕ∥H 3 (Rn ) , λ for some C > 0. Since ∥g∥L2 (Q) ≤ C∥ϕ∥H 3 (Rn ) and using the energy estimate (9) associated to the problem (17) we obtain the following estimation ∥❇t r+ ∥L2 (Q) + ∥∇r+ ∥L2 (Q) ≤ C∥ϕ∥H 3 (Rn ) . This completes the proof of the lemma. Uniqueness for an hyperbolic inverse problem with time-dependent coefficient 73 Lemma 3.3 Let M1 , M2 > 0, a ∈ A(a0 , M1 ) , and b ∈ W 1,∞ (Q) such that ∥b∥W 1,∞ (Q) ≤ M2 . Given ω ∈ S n−1 and ϕ ∈ C0∞ (Rn ), we consider the function φ defined by (10). Then, the following equation ❇t2 u − ∆u − a(x, t)❇t u + b(x, t)u = 0 in Q, (20) admits a unique solution of the following form u− ∈ C([0, T ]; H 1 (Ω)) ∩ C 1 ([0, T ]; L2 (Ω)), u− (x, t) = ϕ(x + tω)̃ a− (x, t)e−iλ(x⋅ω+t) + r− (x, t), (21) 1 t ̃ a− (x, t) = exp ( ∫ a(x + (t − s)ω, s) ds), 2 0 (22) where ̃ a (x, t) is given by − and r− (x, t) satisfies r− (x, T ) = ❇t r− (x, T ) = 0, in Ω, Moreover, there exists a constant C > 0 such that r− (x, t) = 0 on Σ. λ∥r− ∥L2 (Q) + ∥❇t r− ∥L2 (Q) ≤ C∥ϕ∥H 3 (Rn ) . (23) (24) We prove this result by proceeding as in the proof of Lemma 3.2. Putting ̃ g (x, t) = −(❇t2 − ∆ − a(x, t)❇t + b(x, t))(φ(x, t)̃ a− (x, t)e−iλ(x⋅ω+t) ). Then, it is easy to see that if r− (x, t) is solution to the following system ⎧ ⎪ g (x, t) in Q, (❇t2 − ∆ − a(x, t)❇t + b(x, t))r− (x, t) = ̃ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ r− (x, T ) = 0 = ❇t r− (x, T ) in Ω, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − ⎪ ⎪ on Σ, ⎩ r (x, t) = 0 + − then, r (x, t) = r (x, T − t) is a solution to (17) with g(x, t) = ̃ g (x, T − t) and a(x, t), b(x, t) are replaced by a(x, T − t) and = b(x, T − t). 4. Proof of Theorem 1.2 This section is devoted to the proof of Theorem 1.2. The proof is based on the geometric optics solutions constructed in Section 3 and the following preliminary identity. We need first to introduce the following notations. Let ω ∈ S n−1 , a1 , a2 ∈ A(a0 , M ). We set 1 t ̃ a(x, t) = (̃ a− ̃ a+ )(x, t) = exp ( − ∫ a(x + (t − s)ω, s) ds), 2 0 where ̃ a−1 and ̃ a+2 are given by 1 t 1 t ̃ a+2 (x, t) = exp (− ∫ a2 (x+(t−s)ω, s) ds). a−1 (x, t) = exp ( ∫ a1 (x+(t−s)ω, s) ds), ̃ 2 0 2 0 Moreover, we define a in Rn+1 by a = a2 − a1 in Qr and a = 0 on Rn+1 ∖ Qr . 74 ARIMA - volume 23 - 2016 4.1. An identity for the absorbing coefficient The main purpose of this section is to give a preliminary identity for the absorbing coefficient a . Lemma 4.1 Let ϕ ∈ C0∞ (Ar ) and a1 , a2 ∈ A(a0 , M ). Assume that Λa2 = Λa1 , then, the following identity holds 2 a(x, t) dx dt = 0. ∫ a(x, t)ϕ (x + tω)̃ (25) Q In light of Lemma 3.2, there exists a geometrical optics solution u+ to the equation in the following form ⎧ ❇ 2 u+ − ∆u+ + a2 (x, t)❇t u+ = 0 in Q, ⎪ ⎪ ⎪ t ⎨ ⎪ + + ⎪ ⎪ in Ω, ⎩ u (x, 0) = ❇t u (x, 0) = 0 u+ (x, t) = ϕ(x + tω)̃ a+2 (x, t)eiλ(x⋅ω+t) + r+ (x, t), (26) corresponding to the coefficients a2 , where r+ (x, t) satisfies (15), (16). We denote by fλ the function fλ (x, t) = u+ (x, t)∣Σ = ϕ(x + tω)̃ a+2 (x, t)eiλ(x⋅ω+t) . We denote by u1 the solution of ⎧ ❇t2 u1 − ∆u1 + a1 (x, t)❇t u1 = 0 in Q, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ in Ω, ⎨ u1 (x, 0) = ❇t u1 (x, 0) = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ on Σ. ⎩ u1 (x, t) = fλ (x, t) Putting u = u1 − u+ . Then, u is a solution to the following system ⎧ ❇t2 u − ∆u + a1 (x, t)❇t u = a(x, t)❇t u+ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ u(x, 0) = ❇t u(x, 0) = 0 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ u(x, t) = 0, in Q, in Ω, (27) on Σ. where a = a2 −a1 . On the other hand Lemma 3.3 guarantees the existence of a geometrical optic solution u− to the adjoint problem of (1) ⎧ ❇ 2 u− − ∆u− − a1 (x, t)❇t u− − ❇t a1 (x, t)u− = 0 in Q, ⎪ ⎪ ⎪ t ⎨ ⎪ − − ⎪ ⎪ in Ω, ⎩ u (x, T ) = ❇t u (x, T ) = 0 corresponding to the coefficients a1 and −❇t a1 , in the form u− (x, t) = ϕ(x + tω)e−iλ(x⋅ω+t)̃ a1 − (x, t) + r− (x, t), (28) where r− (x, t) satisfies (23), (24). Multiplying the first equation of (27) by u− , integrating by parts and using Green’s formula, we obtain − + − ∫ a(x, t)❇t u u dx dt = ∫ (Λa2 − Λa1 )(fλ ) u dσ dt. Q Σ (29) Uniqueness for an hyperbolic inverse problem with time-dependent coefficient 75 On the other hand, by replacing u+ and u− by their expressions, we get iλ(x⋅ω+t) + − + − ̃ a2 r dx dt ∫ a(x, t)❇t u u dx dt = ∫ a(x, t)❇t ϕ(x + tω)e Q Q a−1 )dx dt a+2 r− dx dt + ∫ a(x, t)❇t ϕ(x + tω)ϕ(x + tω)(̃ a+2 ̃ + ∫ a(x, t)ϕ(x + tω)eiλ(x⋅ω+t) ❇t̃ Q + ∫ a(x, t)ϕ Q 2 a−1 dx dt + a+2 ̃ (x + tω)❇t̃ iλ ∫ Q T 0 iλ(x⋅ω+t) + − ̃ a2 r dx dt ∫ a(x, t)ϕ(x + tω)e Ω + ∫ a(x, t)ϕ(x + tω)e−iλ(x⋅ω+t)̃ a−1 ) dx dt a−1 ❇t r+ dx dt + iλ ∫ a(x, t)ϕ2 (x + tω)(̃ a+2 ̃ Q Q a dx dt + I(λ), + ∫ a(x, t)❇t r+ r− dx dt = iλ ∫ a(x, t)ϕ2 (x + tω)̃ Q Q a−1 . Then, in light of (29), we have where ̃ a=̃ a+2 ̃ a(x, t) dx dt = ∫ (Λa2 − Λa1 )(fλ ) u− dσ dt − I(λ). iλ ∫ a(x, t)ϕ2 (x + tω)̃ (30) Σ Q Note that for λ sufficiently large, we have ∣Iλ ∣ ≤ C∥ϕ∥2H 3 (Rn ) . (31) Hence, using the fact that Λa2 = Λa1 , we deduce from (30), (31) and by taking λ → +∞ the desired result. 4.2. End of the proof In this section we complete the proof of Theorem 1.2 by the use of the results we have already obtained in the previous sections. Let us first consider the following set E = {(ξ, τ ) ∈ Rn ∖ {ORn } × R, ∣τ ∣ < ∣ξ∣}, and denote by F̂ the Fourier transform of F ∈ L1 (Rn+1 ) as follows: F̂(ξ, τ ) = ∫ ∫ R Rn F (x, t)e−ix⋅ξ e−itτ dx dt. In light of (25), we have as λ goes to +∞, the following identity 2 ∫ a(x, t)ϕ (x + tω) exp ( − Q 1 t ∫ a(x + (t − s)ω, s) ds) dx dt = 0. 2 0 (32) Then, using the fact a(x, t) = 0 outside Qr,∗ and making this change of variables y = x + tω, one gets ∫ T 0 ∫ Rn a(y − tω, t)ϕ2 (y) exp ( − 1 t ∫ a(y − sω, s) ds) dy dt = 0. 2 0 Bearing in mind that ∫ 0 T ∫ Rn 1 t ∫ a(y − sω, s) ds) dy dt 2 0 T 1 t d = −2 ∫ ∫ ϕ2 (y) [ exp ( − ∫ a(y − sω, s) ds)] dy dt dt 2 0 0 Rn 1 T 2 = −2 ∫ ϕ (y)[ exp ( − ∫ a(y − sω, s) ds) − 1] dy. 2 0 Rn a(y − tω, t) ϕ2 (y) exp ( − 76 ARIMA - volume 23 - 2016 we conclude that ∫ Rn ϕ2 (y)[ exp ( − 1 T ∫ a(y − sω, s) ds) − 1] dy = 0. 2 0 (33) Now, we consider a nonnegative function ψ ∈ C0∞ (Rn ) supported in the unit ball B(0, 1) such that ∥ψ∥L2 (Rn ) = 1. We define ϕh (x) = h−n/2 ψ( x−y ), h (34) where y ∈ Ar . Then, for h > 0 sufficiently small one can see that Supp ϕh ⊂ C0∞ (Ar ) and satisfies Supp ϕh ∩ Ω = ∅, and Supp ϕh ± T ω ∩ Ω = ∅. Then, as h goes to 0 we deduce from (33) with ϕ = ϕh that exp ( − 1 T ∫ a(y − sω, s) ds) − 1 = 0. 2 0 Since a = a2 − a1 = 0 outside Qr,∗ , we conclude that ∫ a(y − tω, t) dt = 0, R a.e y ∈ Ar , ω ∈ S n−1 . (35) r On the other hand, if ∣y∣ ≤ , we notice that 2 a(y − tω, t) = 0, ∀ t ∈ R. Indeed, we have (36) r (37) ∣y − tω∣ ≥ ∣t∣ − ∣y∣ ≥ t − , 2 hence, (y − tω, t) ∉ Cr+ if t > r/2, from (37). As (y − tω, t) ∉ Cr+ if t ≤ r/2, then we have (y − tω, t) ∉ Cr+ ⊃ Qr,∗ for t ∈ R. This and the fact that a = a2 − a1 = 0 outside Qr,∗ , yield (36), and consequently, ∫ a(y − tω, t) dt = 0, R r ∣y∣ ≤ . 2 By a similar way, we prove for ∣y∣ ≥ T − r/2, that (y − tω, t) ∉ Cr− ⊃ Qr,∗ , for t ∈ R. Then we obtain n−1 (38) ∫ a(y − tω, t) dt = 0, a.e. y ∉ Ar , ω ∈ S . R Thus, by (35) and (38) we find n n−1 ∫ a(y − tω, t) dt = 0, a.e y ∈ R , ω ∈ S . R We now turn our attention to the fourier transform of a. Let ξ ∈ Rn . In light of (38) and by the use of Fubini’s Theorem, we get ∫ ∫ R Rn a(x − tω, t)e−ix⋅ξ dx dt = 0. Making the change of variables y = x − tω, one gets ∫ ∫ R Rn a(y, t)e−iy⋅ξ e−it(ω⋅ξ) dy dt = 0. Uniqueness for an hyperbolic inverse problem with time-dependent coefficient 77 Let us now consider ξ ′ ∈ S n−1 such that ξ ⋅ ξ ′ = 0. Setting ¿ 2 Á τ À1 − τ ⋅ ξ ′ ∈ S n−1 , ω = 2 ⋅ξ+Á ∣ξ∣ ∣ξ∣2 then (ξ, τ ) = (ξ, ω ⋅ ξ) ∈ E. We then deduce that ̂ a(ξ, τ ) = 0 in the set E. By an argument of analyticity, we extend this result to Rn+1 . Hence, by the injectivity of the Fourier transform we get the desired result. This completes the proof of Theorem 1.2. 5. References [1] M. B ELLASSOUED, , M. C HOULLI, , M. YAMAMOTO, Stability estimate for an inverse wave equation and a multidimensional Borg-Levinson theorem, J. Diff. Equat., vol. 247, 2,num. 465-494, 2009. [2] M. B ELLASSOUED, , D. 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