A geometric inverse problem for the Boussinesq system

Manuscript submitted to Website: http://AIMsciences.org AIMS’ Journals Volume X, Number 0X, XX 200X pp. X–XX A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM A. Doubova, E. Fern´ andez-Cara and M. Gonz´ alez-Burgos Universidad de Sevilla Dpto. E.D.A.N., Aptdo. 1160, 41080, Sevilla, SPAIN J.H. Ortega Universidad del B´ıo-B´ıo Dpto. de Ciencias B´asicas, Casilla 447, Fernando May, Chill´an, CHILE (Communicated by Aim Sciences) Abstract. In this work we present some results for the inverse problem of the identification of a single rigid body immersed in a fluid governed by the stationary Boussinesq equations. First, we establish a uniqueness result. Then, we show the way the observation depends on perturbations of the rigid body and we deduce some consequences. Finally, we present a new method for the partial identification of the body assuming that it can be deformed only through fields that, in some sense, are finite dimensional. In the proofs, we use various techniques, related to Carleman estimates, differentiation with respect to domains, data assimilation and controllability of PDEs. 1. Introduction and main results. Let Ω ⊂RNbe a simply connected bounded open set (N= 2 or N= 3) whose boundary ∂Ω is of class W2,∞. Let γbe a nonempty open subset of ∂Ω and let us denote by 1γthe characteristic function of γ. Let D∗be a fixed nonempty open set such that D∗⊂⊂ Ω. We will consider the following family of subsets of Ω: D={D⊂Ω : Dis a simply connected nonempty open set, ∂D is of class W2,∞,D⊂⊂ D∗}. In this paper we will deal with the following inverse problem: Given (ϕ, ψ)and (α, β)in appropriate spaces, find a set D∈ D such that a solution (u, p, θ)of the Boussinesq system        −ν∆u+ (u·∇)u+∇p=θg, ∇·u= 0 in Ω\D, −κ∆θ+u·∇θ= 0 in Ω\D, u=ϕ, θ =ψon ∂Ω, u= 0, θ = 0 on ∂D, (1) Key words and phrases. Boussinesq system, inverse problem, differentiation with respect to domains, data assimilation, controllability. Effort partially supported by D.G.E.S., Grant BFM2003–06446 and Grant CONICYTFONDECYT 1030943. 1 2 A. DOUBOVA, E. FERN´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA satisfies the additional conditions σ(u, p)·n:= (−pId. + 2ν e(u)) ·n=α, κ ∂θ ∂n =βon γ. (2) In (1), u,pand θare respectively a velocity field, a pressure distribution and a temperature θ. The constant vector gis the gravitational force and ν > 0 and κ > 0 are given constants, respectively representing the kinematic viscosity and thermal conductivity of the fluid. In (2), Id. is the identity matrix and e(u) is the linear strain tensor, given by e(u) = 1 2(∇u+t∇u). All along this paper we will assume that, among other things, (ϕ, ψ)6= (0,0). The interpretation of problem (1)–(2) is the following. We assume that a stationary Newtonian viscous fluid sensible to temperature effects fills an unknown domain Ω\Dat rest. The velocity ϕand the temperature ψon the outer boundary ∂Ω are given and we are able to measure the normal stresses σ(u, p)·nand also the normal heat flux κ∂θ ∂n on γ⊂∂Ω. Then the question is whether we can determine Dfrom Ω, ϕ,ψand these measurements. A related problem concerning a Navier-Stokes fluid was considered in [12]. A similar problem has been analyzed in [18]. There, instead of (1), one has    −∆u= 0 in Ω \D, u=ϕon ∂Ω, u= 0 on ∂D, (3) and the role of the additional information (2) is replaced by ∂u ∂n =αon γ. (4) Other problems of this kind have been analized by several authors; see for instance [1], [2], [3], [4], [5], [6], [7], [10], [11], [14], [19] and [20]. In what concerns the associated direct problem, i.e. the determination of (u, p, θ) (and then αand β) from Ω, D,ϕand ψ, we have the following standard result: Theorem 1. Assume that D∈ D and (ϕ, ψ)∈H1/2(∂Ω)N×H1/2(∂Ω) satisfies Z∂Ω ϕ·n dΓ = 0.(5) (a) For any ν > 0and any κ > 0,(1) possesses at least one solution (u, p, θ)that belongs to H1(Ω \D)N×L2(Ω \D)×H1(Ω \D)and verifies        kukH1≤C νkψkH1/2+1 κkϕkH1/2kψkH1/2+νkϕkH1/2+kϕk2 H1/2, kθkH1≤CkψkH1/2+1 κkϕkH1/2kψkH1/2+νkϕkH1/2+kϕk2 H1/2, (6) where Conly depends on Ωand D∗and σ(u, p)·n∈H−1/2(∂Ω)N,∂θ ∂n ∈H−1/2(∂Ω). (b) There exists a positive constant K0=K0(Ω, D∗)such that, if (ϕ, ψ)satisfies      K(ν, κ, ϕ, ψ) := kψkH1/2+1 κkϕkH1/2kψkH1/2+νkϕkH1/2+kϕk2 H1/2 ≤K0(Ω, D∗)ν2κ ν+κ, (7) then the solution of (1) is unique (pis unique up to a constant). A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 3 (c) If in addition r∈[2,+∞)and (ϕ, ψ)∈W2−1/r,r(∂Ω)N×W2−1/r,r(∂Ω), the previous solutions of (1) satisfy (u, p, θ)∈W2,r(Ω\D)N×W1,r(Ω\D)×W2,r(Ω\D) and σ(u, p)·n∈W1−1/r,r(∂Ω)Nand ∂θ ∂n ∈W1−1/r,r(∂Ω). For completeness, we will present the proof of this result in Section 6. Remark 1. In the sequel, we will always assume that (ϕ, ψ)∈H1/2(∂Ω)N× H1/2(∂Ω), (5) is satisfied, and (7) holds. Accordingly, for each D∈ D, we can speak of the unique solution (u, p, θ)∈H1(Ω\D)N×L2(Ω\D)×H2(Ω\D) of (1). On the other hand, it is clear from (6) that, if (7) is fulfilled, the weak solution to (1) satisfies k∇ukL2+1 κk∇θkL2≤C(Ω, D∗)ν+κ νκ K(ν, κ, ϕ, ψ)≤C(Ω, D∗)K0(Ω, D∗)ν. (8) In the sequel, we will have to consider several linear systems of the forms        −ν∆ξ+ (u·∇)ξ+ (ξ·∇)u+∇χ=ρg, ∇·ξ= 0 in Ω \D, −κ∆ρ+u·∇ρ+ξ·∇θ= 0 in Ω \D, ξ=ϕ, ρ =ψon ∂Ω, ξ= 0, ρ = 0 on ∂D (9) and        −ν∆ξ−(∇ξ)tu−(u·∇)ξ+∇χ=−ρ∇θ, ∇·ξ= 0 in Ω \D, −κ∆ρ+−u·∇ρ=ξ·gin Ω \D, ξ=ϕ, ρ =ψon ∂Ω, ξ= 0, ρ = 0 on ∂D, (10) where u∈H1(Ω \D)N,∇ · u= 0 in Ω \Dand θ∈H1(Ω \D). Under these assumptions, there exists a constant K1(Ω, D∗) such that, whenever k∇ukL2+1 κk∇θkL2≤K1(Ω, D∗)ν, (11) the systems (9) and (10) possess exactly one weak solution. In view of (8), there exists a new constant K2(Ω, D∗)≤K0(Ω, D∗) such that, if we have K(ν, κ, ϕ, ψ)≤K2(Ω, D∗)ν2κ ν+κ(12) and uand θsolve (together with some p) the nonlinear system (1), then the existence and uniqueness of weak solution is ensured for (9) and (10). In the context of the inverse problem (1)–(2), the first property we will analyze is uniqueness. Thus, let D0and D1be two sets in Dand let us consider the systems        −ν∆ui+ (ui·∇)ui+∇pi=θig, ∇·ui= 0 in Ω \Di, −κ∆θi+ui·∇θi= 0 in Ω \Di, ui=ϕ, θi=ψon ∂Ω, ui= 0, θi= 0 on ∂Di, (13) for i= 0 and i= 1. We have the following uniqueness result: Theorem 2. Assume that (ϕ, ψ)∈H1/2(∂Ω)N×H1/2(∂Ω),(ϕ, ψ)6= (0,0) and satisfies (5) and (7). Let D0and D1be two sets in D, let (ui, pi, θi)be the unique solution of (13) and let us set αi=σ(ui, pi)·nand βi=κ∂θi ∂n for i= 0,1. Then, if α0=α1and β0=β1on γ, (14) one has D0=D1. 4 A. DOUBOVA, E. FERN´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA For the proof of this result, we will an argument already used, for instance, in [6] and [13]. To this end, we need an appropriate unique continuation property, which will be proved in Section 2. We will also be concerned by the way σ(u, p)·n and κ∂θ ∂n depend on (small) perturbations of Dand some related consequences (see Theorem 3 and the remarks after it). In order to represent the deformations of a set D∈ D, let us introduce W={m∈W2,∞(RN;RN) : kmkW2,∞≤ε, m = 0 in Ω \D∗}, where ε > 0 is small enough. For each m∈ W, we define a new domain D+mby D+m={z∈RN:z=x+m(x), x ∈D}. D D+m D* D D+m D* Ω D D+m D* Figure 1. Deformations of D It is then known that, if εis small enough, for any D∈ D and any m∈ W, one has again D+m∈ D; see for instance [24]. For each m∈ W, let us consider the “perturbed” Boussinesq system        −ν∆v+ (v·∇)v+∇q=η g, ∇·v= 0 in Ω \(D+m), −κ∆η+v·∇η= 0 in Ω \(D+m), v=ϕ, η =ψon ∂Ω, v= 0, η = 0 on ∂(D+m). (15) In view of Theorem 1 and our assumptions on (ϕ, ψ), for each m∈ W the Boussinesq system (15) possesses exactly one solution (v, q, η) that belongs to H2(Ω \(D+m))N×H1(Ω \(D+m)) ×H2(Ω \(D+m)) and satisfies σ(v, q)·n∈ H1/2(∂Ω)Nand ∂θ ∂n ∈H1/2(∂Ω). Let us denote by (u, p, θ) the solution of (1), i.e. the solution to (15) for m= 0. Our second aim in this paper is to deduce identities of the form, (σ(v, q)·n−σ(u, p)·n=L1m+o(m) on γ, κ∂η ∂n −κ∂θ ∂n =L2m+o(m) on γ, where L1and L2are linear operators and o(m) kmkW2,∞→0 as kmkW2,∞→0.(16) Theorem 3. Assume that D∈ D,m∈ W and (ϕ, ψ)∈H3/2(∂Ω)N×H3/2(∂Ω) satisfies (5) and (12) and let (v, q, η)and (u, p, θ)be the solutions of (15) and (1), respectively. Then, A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 5 (a) We have σ(v, q)·n−σ(u, p)·n=σ(u′, p′)·n+o(m)on γ, (17) κ∂η ∂n −κ∂θ ∂n =κ∂θ′ ∂n +o(m)on γ, (18) where o(m)satisfies (16) and, for each m∈ W,(u′, p′, θ′)is the solution of the associated linear problem        −ν∆u′+ (u′·∇)u+ (u·∇)u′+∇p′=θ′g, ∇·u′= 0 in Ω\D, −κ∆θ′+u′·∇θ+u·∇θ′= 0 in Ω\D, u′+ (m·∇)u∈H1 0(Ω \D)N, θ′+ (m·∇)θ∈H1 0(Ω \D). (19) (b) Assume that ξ∈C2(∂Ω),supp ξ⊂⊂ γand ξ≡1on ˜γ, a relative open set of ∂Ωsuch that ˜γ⊂⊂ γ. Then, for any (¯y, ¯z)∈C2(γ)N×C2(γ)satisfying Zγ ¯yξ ·n dΓ = 0,(20) we have        Zγ (σ(v, q)·n−σ(u, p)·n)·¯yξ dΓ + κZγ∂η ∂n −∂θ ∂n¯zξ dΓ =−νZ∂D (m·n)∂u ∂n ·∂y ∂n dΓ−Z∂D (m·n)∂θ ∂n κ∂z ∂n dΓ + o(m). (21) Here, (y, π, z)is the solution of the adjoint system        −ν∆y−(∇y)tu−(u·∇)y+∇π=−z∇θ, ∇·y= 0 in Ω\D, −κ∆z−u·∇z=y·gin Ω\D, y= ¯yξ, z = ¯zξ on ∂Ω, y= 0, z = 0 on ∂D. (22) For the proof, our main tool will be the domain variation techniques introduced by F. Murat and J. Simon in [21] and [22]; see also [24] and [9]. We will present the proof of this result in Section 4. Remark 2. Notice that, in view of (17)–(19), for each m∈ W we can compute the local derivatives (u′, p′, θ′) and then the differences σ(v, q)·n−σ(u, p)·nand κ∂η ∂n −κ∂θ ∂n on γup to second-order perturbations. On the other hand, we see from (21) that the same quantity can be easily computed using (y, π, z), which is independent of m. Corollary 1. Let the assumptions of Theorem 3 be satisfied and assume that ∂D ∈ W3,∞and the perturbation mis of the form m=λn +m⊥on ∂D, where λ∈R and m⊥·n= 0. Then, if (¯y, ¯z)satisfies (20) and K:= Z∂D ν∂u ∂n ·∂y ∂n +κ∂θ ∂n ∂z ∂ndΓ6= 0, we have: λ=−1 KZγ (σ(v, q)·n−σ(u, p)·n)·¯yξ dΓ + κZγ∂η ∂n −∂θ ∂n¯zξ dΓ+o(m). 6 A. DOUBOVA, E. FERN ´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA Remark 3. Assume that ∂Ω, ϕand ψare regular enough and we have already computed a first regular approximation ˜ Dto the solution of our inverse problem. Then, the associated solution (˜u, ˜p, ˜ θ) and consequently ˜α|γ≡σ(˜u, ˜p)·n|γand ˜ β|γ=κ∂˜ θ ∂n|γ are known. Assume that we intend to compute a new (and possibly better) approximation of the form ˜ D+mwhere m=λn +m⊥on ∂˜ D,λ∈Rand m⊥·n= 0. From (21), for each ¯yand ¯zas in Corollary 1, we can write Zγ(σ(v, q)·n−˜α)·¯yξ +κ∂η ∂n −˜ β¯zξdΓ = −λ˜ K+o(λ), where ˜ K:= Z∂˜ D ν∂˜u ∂n ·∂y ∂n +κ∂˜ θ ∂n ∂z ∂n!dΓ and (y, π, z) is the solution of (22). So, the “good” strategy is to choose λ, if possible, according to the formula: λ=−1 ˜ KZγ(α−˜α)·¯yξ + (β−˜ β)¯zξdΓ. Indeed, this is a way to ensure that, at least at first order, the projections of σ(v, q)·n|γand α|γin the direction of ¯yand the projections of κ∂η ∂n |γand β|γin the direction of ¯zcoincide. Remark 4. More generally, starting from an already computed candidate ˜ Dto the solution of problem (1)–(2), let us try to determine a better candidate of the form ˜ D+m, where m·n|∂˜ D∈Mand Mis a finite dimensional space. Let {f1,...,fd} be a basis of M. Then we can write m·n|∂˜ D= d X i=1 aifi for some aito be determined. Let us introduce dlinearly independent functions (¯yi,¯zi)∈C2(γ)N×C2(γ) satisfying (20). Using again (21), we see now that Zγ(σ(v, q)·n−˜α)·¯yjξ+κ∂η ∂n −˜ β¯zjξdΓ = − d X i=1 ˜ Kijai+o(m), where ˜ Kij := Z∂˜ D fi ν∂˜u ∂n ·∂yj ∂n +κ∂˜ θ ∂n ∂zj ∂n !dΓ∀i, j = 1,...,d and, for each j, (yj, πj, zj) is the solution of (22) corresponding to the data ¯yjand ¯zj. Consequently, a strategy to compute the coefficients aiis to solve (if possible) the system of equations d X i=1 ˜ Kijai=−hα−˜α, ¯yj1γi−hβ−˜ β, ¯zj1γi,1≤j≤d. These ideas are being considered in a work in progress which will appear in the next future. A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 7 Our third aim in this paper is the (partial) identification of D. More precisely, we will analyze the following question. Let us assume that D∈ D is known and thus we can solve the direct problem (1) and compute (α, β) from (2). Let us also assume that we know the observation (αm, βm) corresponding to a modified domain D+m, with m∈ W. Then we want to know whether we are able to compute m·n|∂D from D, (α, β) and (αm, βm). Our third main result is the following: Theorem 4. Assume that (ϕ, ψ)∈H3/2(∂Ω)N×H3/2(∂Ω),(ϕ, ψ)6≡ 0and satisfies (5) and (7) and the corresponding solution of (1) satisfies  ∂u ∂n 2 + ∂θ ∂n 2 6= 0 on ∂D. (23) Also, assume that m∈ W and (m·n)|∂D belongs to a finite dimensional space M⊂W1,∞(∂D). Then (m·n)|∂D can be computed explicitly, up to second-order terms, from Ω,D,M,α,β,αmand βm. More precisely, there exists a computable function HΩ,D,M :H1/2(γ)N×H1/2(γ)7→ M such that (m·n)|∂D =HΩ,D,M (αm−α, βm−β) + o(m) for all m∈ W with (m·n)|∂D ∈M. Remark 5. The assumption (23) is reasonable. Indeed, in view of the unique continuation results we will prove below (see corollary 2), the set of points of ∂D where (23) is not satisfied has no interior point. On the other hand, (23) is satisfied whenever, for instance, we have (ϕ, ψ)∈W2−1/r,r(∂Ω)N×W2−1/r,r(∂Ω) for some r > N,ψ≥0, ψ6≡ 0 and the other assumptions on (ϕ, ψ) are fulfilled. This is a consequence of Hopf’s maximum principle applied to the equation satisfied by θ; see for example [17]. In fact, in this case we obtain ∂θ ∂n <0 on ∂D, which trivially implies (23). To our knowledge, it is unknown whether (23) is implied by the other assumptions on (ϕ, ψ) imposed in Theorem 4. For the proof of this theorem, we will use (again) domain variation techniques and also some recent results on data assimilation introduced by J.-P. Puel in [23]. For clarity, we will first present the argument in the case of the similar but simpler problem (3)–(4), which involves only the Laplace equation; see Section 5. The rest of this paper is organized as follows. In Section 2, we will prove a unique continuation property needed for the proof of Theorem 2. Theorems 2, 3 and 4 are respectively proved in sections 3, 4 and 5. Finally, Section 6 deals with the proofs of Theorem 1 as well as other technical results. 2. A unique continuation property. In this section, we will present a unique continuation property which will be used in the proof of Theorem 2. Let G⊂RN be a bounded connected open set (N= 2 or N= 3) whose boundary ∂G is of class W1,∞. In the sequel, Cdenotes a generic positive constant. We will prove the following result: 8 A. DOUBOVA, E. FERN ´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA Theorem 5. Let ω⊂Gbe a nonempty open set. Assume that a∈L∞(G)N, b∈L∞(G)N,d∈L∞(G)and ∇·a=∇·b= 0 in G. Then any solution (v, q, η)∈ H1(G)N×L2(G)×H1(G)of the linear system −ν∆v+ (a·∇)v+ (v·∇)b+∇q=ηg, ∇·v= 0 in G, −κ∆η+a·∇η+v·∇d= 0 in G, (24) satisfying v= 0 and η= 0 in ω is zero everywhere, i.e. satisfies v≡0in G,q≡Const. in Gand η≡0in G. The proof of this theorem is based on the ideas and results in [15]. It will be composed of four steps. First, we recall an appropriate local Carleman inequality (see Section 2.1). Then, in Section 2.2, using this Carleman inequality, we will prove the result of Theorem 5 but in a ball and for potentials aand bwith sufficiently small L∞norms. Next, in Section 2.3, we will show the result in small balls. Finally, in Section 2.4, we will conclude the proof. 2.1. Carleman inequality. In [15], the following result can be found: Proposition 1. Let U⊂RNbe an open set, K⊂Ua nonempty compact set, ajk ∈C∞(RN)for 1≤j≤s,1≤k≤Nand ϕ∈ D(RN). Let us set L1f= s X j=1 N X k=1 ajk∂kfj∀f= (f1,...,fs)∈L2(U)s and a0(x, ξ) = N X j=1 ξ2 j−(∂jϕ(x))2, b0(x, ξ) = 2 N X j=1 ξj∂jϕ(x)∀(x, ξ)∈U×RN and let us assume that ϕsatisfies the following property:    ∇ϕdoes not vanish in U; furthermore, ∃C0>0such that ∂ξa0(x, ξ)·∂xb0(x, ξ)−∂xa0(x, ξ)·∂ξb0(x, ξ)≥C0 for all (x, ξ)∈U×RNsuch that a0(x, ξ) = b0(x, ξ) = 0. (25) Then, there exist constants C1>0and h1>0such that, for any couple (y, F )∈ H1 0(U)×L2(U)ssatisfying supp (y)∪supp (F)⊂Kand ∆y−L1F∈L2(U)and any h∈(0, h1), one has: ZK e2ϕ/h(|y|2+h2|∇y|2)dx ≤C1ZK e2ϕ/h(h|F|2+h3|∆y−L1F|2)dx. (26) 2.2. A unique continuation property for small coefficients. In this paragraph, we will deduce the result in Theorem 5 but for potentials with sufficiently small norm. Let B(0; r) be an open ball of radius r > 0 centered at the origin. We consider system (24) in B(0; 2), i.e. −ν∆v+ (a·∇)v+ (v·∇)b+∇q=ηg, ∇·v= 0 in B(0; 2), −κ∆η+a·∇η+v·∇d= 0 in B(0; 2). (27) We have the following result: A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 9 Lemma 1. Assume that a∈L∞(B(0; 2))N,b∈L∞(B(0; 2))N,d∈L∞(B(0; 2)) and ∇·a=∇·b= 0 in B(0; 2). Then there exists ǫ > 0such that, if kak∞≤ǫand kbk∞≤ǫ, any solution (v, q, η)∈H1(B(0; 2))N×L2(B(0; 2)) ×H1(B(0; 2)) of (27) satisfying v= 0 and η= 0 in B(0; 1) is zero everywhere. Proof. Let (v, q, η)∈H1(B(0; 2))N×L2(B(0; 2)) ×H1(B(0; 2)) be a solution of (27) satisfying v= 0 and η= 0 in B(0; 1). Since q≡Const. in B(0; 1), it is not restrictive to assume that it also vanishes in B(0; 1). •Step 1: Let us first see that Proposition 1 can be applied in this context for some appropriate choices of U,K,L1and ϕ. Let us choose ε > 0 and let us set K={x∈RN:3 4≤ |x| ≤ 2−ε}and U={x∈RN:1 2<|x|<2}. Let ϕ∈ D(RN) be such that ϕ(x) = e−δ|x|2∀x∈B(0; 2),(28) where δ > 4. Notice that ∂jϕ(x) = −2δϕ(x)xj,(29) ∂j∂kϕ(x) = −2δϕ(x)δjk + 4δ2ϕ(x)xjxk,(30) where δjk is the usual Kronecker’s symbol. Assume that a0(x, ξ) = b0(x, ξ) = 0, i.e. N X j=1 ξ2 j= N X j=1 (∂jϕ(x))2and N X j=1 ξj∂jϕ(x) = 0. Then, in view of (29) and (30) we have ∂ξa0(x, ξ)·∂xb0(x, ξ)−∂xa0(x, ξ)·∂ξb0(x, ξ) = 64δ3ϕ(x)3 N X j=1 x2 j"δ N X k=1 x2 k−1#= 64δ3ϕ(x)3|x|2[δ|x|2−1] ≥16δ3e−3δ/4δ 4−1. Consequently, (25) is satisfied by this function ϕin this open set U. •Step 2: We introduce a function ζ∈ D(◦ K) such that ζ= 1 in 1−ε≤ |x| ≤ 2−2ε. We put ˜v=ζv, ˜q=ζq, ˜η=ζη, (31) where (v, q, η)∈H1(B(0; 2))N×L2(B(0; 2)) ×H1(B(0; 2)) is a solution of (27). It is then clear that (˜v, ˜q, ˜η)∈H1 0(◦ K)N×L2(◦ K)×H1 0(◦ K). From (27) we can readily see −ν∆˜v+∇˜q+∇·(˜vb) = ˜ηg −(a·∇) ˜v+H1in ◦ K,(32) where H1∈L2(◦ K) is given by H1=b(v·∇)ζ−2ν∇ζ·∇v−νv∆ζ+ (a·∇ζ)v+q∇ζ. (33) This is true because ∇·v= 0 in U. Taking the divergence in the first equation of (27), we see that ∆q=−∇·((a·∇)v+ (v·∇)b) + ∇η·g=−∇·((a·∇)v+ (∇v)b) + ∇η·g. (34) 16 A. DOUBOVA, E. FERN ´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA Lemma 3. Assume (ϕ, ψ)∈H3/2(∂Ω)N×H3/2(∂Ω) satisfies (5) and (12). Then (a) The mapping m7→ (v, q, η)◦(Id.+m), which is defined in Wand takes values in H1(Ω \D)N×L2(Ω \D)×H1(Ω \D), is differentiable at 0, with (total) derivative denoted by ( ˙u, ˙p, ˙ θ)(m). That is to say, there exists a linear continuous mapping m7→ ( ˙u(m),˙p(m),˙ θ(m)) such that (v, q, η)◦(Id. + m)−(u, p, θ) = ( ˙u, ˙p, ˙ θ)(m) + o(m),(56) where o(m)satisfies (16). (b) For each ω⊂⊂ Ω\D, the mapping m7→ (v, q, η)|ω, which is defined in Wand takes values in H1(ω)N×L2(ω)×H1(ω), is differentiable at 0. In other words, m7→ (v, q, η)is locally differentiable. The local derivative at 0in the direction mis denoted by (u′, p′, θ′)(m). (c) Furthermore, (u′, p′, θ′)(m)is the unique solution of the linear system (19) and ( ˙u, ˙p, ˙ θ)(m) = (u′, p′, θ′)(m) + (m·∇)(u, p, θ).(57) In view of (56) and (57), taking into account that m= 0 in a neighborhood of ∂Ω, we find that σ(v, q)·n−σ(u, p)·n=σ(u′, p′)·n+o(m) on γ, κ∂η ∂n −κ∂θ ∂n =κ∂θ′ ∂n +o(m) on γ. This proves (17) and (18). Since (ϕ, ψ)∈H3/2(Ω)N×H3/2(Ω), the solution (u, p, θ) of (1) satisfies (u, p, θ)∈ H2(Ω \D)N×H1(Ω \D)×H2(Ω \D), so we have u′= 0 on ∂Ω and u′=−(m·n)∂u ∂n on ∂D, θ′= 0 on ∂Ω and θ′=−(m·n)∂θ ∂n on ∂D. Let ¯y∈C2(γ)Nsatisfy (20), with ξ∈C2(∂Ω), supp ξ⊂⊂ γand ξ≡1 on ˜γ, a relative open set of ∂Ω such that ˜γ⊂⊂ γ. Let ¯z∈C2(γ) be given and let (y, π, z) be the associated solution of (22). We will justify that (21) holds. Multiplying the first equation of (19) by yand integrating by parts, we get        ZΩ\D σ(u′, p′)∇y dx +ZΩ\D ((u·∇)u′+ (u′·∇)u)·y dx =Z∂Ω (σ(u′, p′)·n)·¯yξ dΓ + ZΩ\D θ′g·y dx. (58) Here, the first term of the left hand side can also be written in the form ZΩ\D σ(u′, p′)∇y dx =−ZΩ\D ν∆y·u′dx +Z∂D (σ(y, π)·n)·u′dΓ =−ZΩ\D ν∆y·u′dx −Z∂D (m·n)∂u ∂n ·(σ(y, π)·n)dΓ. The second term of the left hand side of (58) satisfies ZΩ\D ((u·∇)u′+ (u′·∇)u)·y dx ≡ZΩ\Dui∂iu′ jyj+u′ i∂iujyjdx =ZΩ\D−ui∂iyju′ j+u′ i∂iujyjdx =ZΩ\D−(u·∇)y+ (∇u)ty·u′dx. A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 17 Therefore, we obtain from (58) that Z∂Ω (σ(u′, p′)·n)·¯yξ dΓ + ZΩ\D θ′g·y dx =−Z∂D (m·n)∂u ∂n ·(σ(y, π)·n)dΓ−ZΩ\D z∇θ·u′dx. On the boundary ∂D, since uand yvanish, we have ∂u ∂n ·n=∇·u= 0 on ∂D and σ(y, π)·n= 2νe(u)·n−pn =ν∂y ∂n +ν(∇·y)n−pn on ∂D. Consequently, ∂u ∂n ·(σ(y, π)·n) = ν∂u ∂n ·∂y ∂n on ∂D and        Z∂Ω (σ(u′, p′)·n)·¯yξ dΓ + ZΩ\D θ′g·y dx =−νZ∂D (m·n)∂u ∂n ·∂y ∂n dΓ−ZΩ\D z∇θ·u′dx. (59) On the other hand, multiplying the third equation in (19) by zand integrating by parts, we have Z∂Ω ∂θ′ ∂n ¯zξ dΓ = ZΩ\D κ∇θ′·∇z dx +ZΩ\D (u·∇θ′+u′·∇θ)z dx. (60) The first term in the right hand side of this equality is as follows: ZΩ\D κ∇θ′·∇z dx =−ZΩ\D κθ′∆z dx +Z∂D κθ′∂z ∂n dΓ =−ZΩ\D κθ′∆z dx −Z∂D (m·n)∂θ ∂n κ∂z ∂n dΓ.(61) The second term in the right hand side of (60) reads ZΩ\D (ui∂iθ′+u′ i∂iθ)z dx =−ZΩ\D ui∂iz θ′dx +ZΩ\D u′ i∂iθz dx ≡ −ZΩ\D (u·∇z)θ′+ZΩ\D z∇θ·u′dx. (62) From (60), (61) and (62), we deduce that Z∂Ω ∂θ′ ∂n ¯zξ dΓ−ZΩ\D θ′g·y dx =−Z∂D (m·n)∂θ ∂n κ∂z ∂n dΓ + ZΩ\D z∇θ·u′dx. (63) Finally, adding (59) and (63) we obtain that        Zγ (σ(u′, p′)·n)·¯yξ dΓ + Zγ ∂θ′ ∂n ¯zξ dΓ =−νZ∂D (m·n)∂u ∂n ·∂y ∂n dΓ−Z∂D (m·n)∂θ ∂n κ∂z ∂n dΓ. (64) Now, using (17) and (18) in (64) we get (21). This ends the proof of Theorem 3. 5. Proof of Theorem 4. To clarify the situation, we start presenting a sketch of the proof of Theorem 4 in the much more simple case of the Laplace equation. 18 A. DOUBOVA, E. FERN ´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA 5.1. A simple case: the Laplace equation. Given ϕin an appropriate space, D∈ D and m∈ W, we consider the following problems:    −∆u= 0 in Ω \D, u=ϕon ∂Ω, u= 0 on ∂D (65) and    −∆v= 0 in Ω \D+m, v=ϕon ∂Ω, v= 0 on ∂(D+m). (66) We now set α=∂u ∂n |γand αm=∂v ∂n |γ. Then we have the following result: Theorem 6. Assume that ϕ∈H3/2(∂Ω) does not vanish identically. Also, assume that m∈ W and (m·n)|∂D belongs to a finite dimensional space M⊂W1,∞(∂D). Then (m·n)|∂D can be computed explicitly, up to second-order terms, from Ω,D, M,αand αm. More precisely, there exists a computable function GΩ,D,M such that (m·n)|∂D =GΩ,D,M (αm−α) + o(m) (67) for all small mwith (m·n)|∂D ∈M. Sketch of the proof: We will proceed in three steps: •Step 1: Domain variations techniques. Using domain variation techniques, we can write that ∂v ∂n −∂u ∂n =∂u′ ∂n +o(m) on γ, where o(m) satisfies (16) and u′is the solution of      −∆u′= 0 in Ω \D, u′= 0 on ∂Ω, u′=−(m·n)∂u ∂n on ∂D. (68) Therefore, we have ∂u′ ∂n =αm−α+o(m) on γ and the proof is reduced to compute (m·n)|∂D from ∂u′ ∂n |γup to second-order perturbations. •Step 2: A (non standard) data assimilation approach. At this point, our approach is inspired by the techniques introduced by J.-P. Puel in [23]. Thus, assume that m∈ W and (m·n)|∂D ∈M, where M⊂W1,∞(∂D) is a finite dimensional space. Then ∂u′ ∂n |∂D belongs to a suitable finite dimensional space E⊂H−1/2(∂D). Notice that ∂u ∂n ∈C0(∂D). For simplicity, let us assume that ∂u ∂n 6= 0 on ∂D. Then, to determine (m·n)|∂D, it suffices to compute the integrals Z∂D (m·n)∂u ∂n h dΓ, h ∈M. Let PE:L2(∂D)7→ Ebe the usual orthogonal projector and let us assume that, for each h∈M, we can solve the following control problem: Find w∈L2(γ) such that the solution θhof      −∆θh= 0 in Ω \D, θh=w1γon ∂Ω, ∂θh ∂n =hon ∂D (69) A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 19 satisfies PE(θh|∂D) = 0.(70) Then, using (68) and making some integrations by parts, we find: −Z∂D (m·n)∂u ∂n h dΓ = Z∂D∪∂Ω u′∂θh ∂n dΓ =Z∂D ∂u′ ∂n PE(θh|∂D)dΓ + Z∂Ω ∂u′ ∂n θhdΓ = Zγ ∂u′ ∂n w dΓ. That is to say, we get the equalities −Z∂D (m·n)∂u ∂n h dΓ = Zγ ∂u′ ∂n w dΓ, where ∂u′ ∂n is known (up to second-order terms; this is the consequence of step 1) and wcan be computed solving (69)–(70). This shows that (m·n)|∂D can be computed explicitly by solving as many control problems of the previous kind as dim M. •Step 3: Resolution of the exact finite dimensional control problem (69)–(70). Finally, it can be seen that, for any h∈L2(∂D), there exist wand θhsuch that (69)–(70) hold. Indeed, it is sufficient to apply a classical unique continuation property for the Laplace equation in combination with the arguments in [25]; see the proof of Lemma 4 for more details. This ends the proof of Theorem 6. 5.2. The general case: a Boussinesq system. We will now follow the steps of the previous proof in order to deduce Theorem 4. We have (ϕ, ψ)∈W2−1/r,r(∂Ω)N× W2−1/r,r(∂Ω)Nfor some r > N and thus (u, p, θ)∈W2,r(Ω \D)N×W1,r(Ω \D)× W2,r(Ω \D). Let us assume that m∈ W, with (m·n)|∂D ∈M. Let us also assume that D∈ D is known, we have computed (α, β) from (2) solving the direct problem (1) and, also, that we know the observation (αm, βm) corresponding to the modified domain D+m, that is to say, αm=σ(v, q)·n|γand βm=κ∂η ∂n|γ, where (v, q, η) is the solution of (15). We recall that our goal is to compute explicitly (m·n)|∂D from (α, β) and (αm, βm). •Step 1: Domain variations. Thanks to Theorem 3, using domain variation techniques, we get the following identities: αm−α≡σ(v, q)·n−σ(u, p)·n=σ(u′, p′)·n+o(m) on γ, βm−β≡κ∂η ∂n −κ∂θ ∂n =κ∂θ′ ∂n +o(m) on γ, where (u′, p′, θ′)∈H1(Ω \D)N×L2(Ω \D)×H1(Ω \D) is the unique solution to          −ν∆u′+ (u′·∇)u+ (u·∇)u′+∇p′=θ′g, ∇·u′= 0 in Ω \D, −κ∆θ′+u′·∇θ+u·∇θ′= 0 in Ω \D, u′= 0, θ′= 0 on ∂Ω, u′=−(m·n)∂u ∂n, θ′=−(m·n)∂θ ∂n on ∂D (71) and o(m) satisfies (16). Therefore, the proof of Theorem 4 is reduced to compute (m·n)|∂D from σ(u′, p′)· n|γand κ∂θ′ ∂n |γup to second-order terms. This will be done in the next step. 20 A. DOUBOVA, E. FERN ´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA Notice that, up to now, we have not used the fact that (m·n)|∂D belongs to a finite dimensional space. •Step 2: A (non standard) data assimilation approach. Let us now assume that (m·n)|∂D ∈M⊂W1,∞(∂D), with dim M < +∞. Then, in view of (71), we have (σ(u′, p′)·n|∂D , κ∂θ′ ∂n |∂D)∈E, where E⊂H−1/2(∂D)N×H−1/2(∂D) is another finite dimensional space. As before, we use an argument inspired by the techniques in [23]. We will use the fact that the quantities Z∂D (m·n)  ∂u ∂n 2 + ∂θ ∂n 2!h dΓ, h ∈M, determine (m·n)|∂D . Indeed, under the hypothesis (23), the bilinear form (ℓ, h)7→ R∂D ℓ∂u ∂n 2+∂θ ∂n 2h dΓ is a scalar product in M. Therefore, our goal will be to write these integrals in terms of σ(u′, p′)·n|γand κ∂θ′ ∂n |γ. To this end, we will argue as follows. For the moment, let us assume that for each h∈Mwe are able to solve the following exact finite dimensional control problem: find a control (w1, w2) such that (w11γ, w21γ)∈H1/2(∂Ω)N×H1/2(∂Ω) and the corresponding weak solution (y, q, z)∈H1(Ω \D)N×L2(Ω \D)×H1(Ω \D) of          −ν∆y−(∇y)tu−(u·∇)y+∇q=−z∇θ, ∇·y= 0 in Ω \D, −κ∆z−u·∇z=g·yin Ω \D, y=w11γ, z =w21γon ∂Ω, σ(y, q)·n=∂u ∂nh, κ ∂z ∂n =∂θ ∂nhon ∂D (72) satisfies h(Φ,Ψ),(y|∂D, z|∂D)i∂D = 0 ∀(Φ,Ψ) ∈E, (73) where h·,·i∂D stands for the usual duality coupling for H−1/2(∂D)N×H−1/2(∂D) and H1/2(∂D)N×H1/2(∂D). Then, using that (u′, p′, θ′) is the solution of (71) and (73), we see that −Z∂D (m·n)  ∂u ∂n 2 + ∂θ ∂n 2!h dΓ = hσ(y, q)·n, u′i∂D∪∂Ω+κh∂z ∂n, θ′i∂D∪∂Ω =hσ(u′, p′)·n, yi∂D∪∂Ω+κh∂θ′ ∂n , zi∂D∪∂Ω =hσ(u′, p′)·n, w11γi∂Ω+κh∂θ′ ∂n , w21γi∂Ω, where h·,·iΣstands for the usual duality product in H−1/2(Σ)Nand H−1/2(Σ). Notice that this allows us compute (m·n)|∂D (up to second-order perturbations) from σ(u′, p′)·n|γand κ∂θ′ ∂n |γ. As we have seen in the previous step, this also allows us compute (m·n)|∂D from the known observations (αm, βm) and (α, β). The conclusion of this step is that the proof of Theorem 4 will be achieved if we are able to solve (72)–(73). A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 21 Remark 6. From the practical viewpoint, what we have to do is the following. Let {ℓ,...,ℓI}be a basis of Mand let us put (m·n)|∂D = I X i=1 aiℓi. Let (w1 i, w2 i) be, for each i= 1,...,I a control solving the problem (72)–(73) with h=ℓi. Then the coefficients aiare given by the unique solution of the following linear system: I X i=1 Z∂D ℓi  ∂u ∂n 2 + ∂θ ∂n 2!ℓjdΓ!ai=qj,1≤j≤I, where we have set qj=−hσ(u′, p′)·n, w1 j1γi∂Ω−κh∂θ′ ∂n , w2 j1γi∂Ω. •Step 3: Resolution of the exact finite dimensional control problem (72)–(73). We have the following result: Lemma 4. Let E⊂H−1/2(∂D)N×H−1/2(∂D)be a finite dimensional space. Let us assume that h∈L∞(∂D)and (u, θ)∈H2(Ω\D)N×H2(Ω\D)satisfies ∇·u= 0 in Ω\Dand (11). Then, there exist controls (w1, w2)such that (w11γ, w21γ)∈ H1/2(∂Ω)N×H1/2(∂Ω) and the associated solution (y, q, z)of (72) satisfies (73). Furthermore, for each ε > 0, we can choose w1and w2such that k(y|∂D, z|∂D)kL2(∂D)≤ε. (74) Proof. This result is a consequence of the unique continuation properties we have presented in Section 2 and the fact that Ehas finite dimension. There are several ways to prove it. Here, we follow the approach in [25] that, for each ε > 0, provides a control satisfying (74). Let Gbe a bounded open set with boundary ∂G of class C2such that Ω ⊂G and ∂Ω∩G=γ. Let ωbe a nonempty open subset of G\Ω. We consider the following distributed control problem: find controls (f, k)∈L2(ω)N×L2(ω) such that the corresponding solution (y, q, z)∈H1(G\D)N×L2(G\D)×H1(G\D) of          −ν∆y−(∇y)tu−(u·∇)y+∇q=−z∇θ+f1ω,∇·y= 0 in G\D, −κ∆z−u·∇z=g·y+k1ωin G\D, y= 0, z = 0 on ∂G, σ(y, q)·n=∂u ∂nh, κ ∂z ∂n =∂θ ∂nhon ∂D (75) satisfies (73). For any (a, b)∈H−1/2(∂D)N×H−1/2(∂D), let us consider the adjoint system          −ν∆ξ+ (u·∇)ξ+ (ξ·∇)u+∇χ=ρg, ∇·ξ= 0 in G\D, −κ∆ρ+u·∇ρ+ξ·∇θ= 0 in G\D, ξ= 0, ρ = 0 on ∂G, σ(ξ, χ)·n=a, κ ∂ρ ∂n =bon ∂D. (76) As a consequence of (11), for each (a, b)∈H−1/2(∂D)N×H−1/2(∂D) system (76) possesses a unique weak solution (ξ, χ, ρ)∈H1(G\D)N×L2(G\D)×H1(G\D) that satisfies kξkH1(G\D)+kχkL2(G\D)+kρkH1(G\D)≤Ck(a, b)kH−1/2(∂D), 22 A. DOUBOVA, E. FERN ´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA where Cis a positive constant only depending on G,D, and u. In addition, we have σ(ξ, χ)·n∈H−1/2(∂G)Nand ∂ρ ∂n ∈H−1/2(∂G), with similar estimates. On the other hand, if we multiply the first equation of (72) by ξand the second one by zand we make appropriate integrations by parts, we get: Zω (f·ξ+kρ)dx =Z∂D h∂u ∂n ·ξ+∂θ ∂nρdΓ−h(a, b),(y|∂D , z|∂D)i∂D .(77) Let L:H−1/2(∂D)7→ H1/2(∂D) be the canonical identification operator, with kLηkH1/2(∂D)=kηkH−1/2(∂D)∀η∈H−1/2(∂D), (Lη , v)H1/2(∂D)=hη , vi∂D ∀η∈H−1/2(∂D),∀v∈H1/2(∂D). Let e Ebe the finite dimensional space e E=LE⊂H1/2(∂D)N×H1/2(∂D) and let Pe Ebe the associated orthogonal projector in H1/2(∂D)N×H1/2(∂D). For any ε > 0, let us consider the functional Jε(a, b)        Jε(a, b) = 1 2Zω|ξ|2+|ρ|2dx +εk(I−Pe E)L(a, b)kH1/2(∂D) −Z∂D h∂u ∂n ·ξ+∂θ ∂nρdΓ∀(a, b)∈H−1/2(∂D)N×H−1/2(∂D), where (ξ, χ, ρ) is the solution of (76). Let us assume that there exists a minimizer (ˆa,ˆ b)∈H−1/2(∂D)N×H−1/2(∂D) of Jε(we will justify this temporary assumption at the end of this proof). Let us denote by (ˆ ξ, ˆχ, ˆρ) the associated solution of (76). Then Zωˆ ξ·ξ+ ˆρρdx +ε k(I−Pe E)L(ˆa,ˆ b)kH1/2(∂D)h(I−Pe E)L(ˆa,ˆ b),(a, b)i∂D −Z∂D h∂u ∂n ·ξ+∂θ ∂nρdΓ = 0 (78) for any (a, b)∈H−1/2(∂D)N×H−1/2(∂D). Let us take the controls fand kgiven by f=ˆ ξ1ω, k = ˆρ1ω. Then, we deduce from (78) and (77) that the associate state (y, q, z) satisfies (y|∂D, z|∂D) = ε k(I−Pe E)L(ˆa,ˆ b)kH1/2(∂D) (I−Pe E)L(ˆa,ˆ b) for all (a, b)∈L2(∂D)N×L2(∂D). Consequently, the controls (w1, w2) = (y|∂Ω, z|∂Ω) fulfill the statement of the lemma. In order to end the proof, let us see that there exists a unique minimizer (ˆa,ˆ b) of Jε. But this is a trivial consequence of the following properties of Jε: •Jεis lower semi-continuous and strictly convex. •Jεis coercive. More precisely, lim inf k(a,b)kH−1/2(∂D)→∞ Jε(a, b) k(a, b)kH−1/2(∂D)≥ε. This is a consequence of the following unique continuation property (given in Theorem 5): If the solution (ξ, χ, ρ) of (76) verifies ξ= 0 and ρ= 0 in ω, then (ξ, χ, ρ)≡(0,0,0) in G\D. This ends the proof. A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 23 6. Some technical results. For completeness, in this section we will present a sketch of the proof of Theorem 1, which provides existence, uniqueness and regularity properties of the solution of (1). For the proof we will use the standard Galerkin’s method and some properties of Sobolev spaces. 1 – Existence: Assume that D∈ D and (ϕ, ψ)∈H1/2(∂Ω)N×H1/2(∂Ω) is such that R∂Ωϕ·n ds = 0. For simplicity, the usual norms in the spaces L2(Ω \D), H1(Ω \D), . . . will be respectively denoted by k·kL2,k·kH1, . . . Let us set V(O) = {v∈H1 0(O)N:∇·v= 0 }, where O ⊂ RNis a given regular domain. Then, for every α1, α2>0, there exists (Φ∗ α1,Ψ∗ α2)∈H1(Ω \D∗)N×H1(Ω \D∗), that satisfies    ∇·Φ∗ α1= 0 in Ω \D∗, Φ∗ α1=ϕ, Ψ∗ α2=ψon ∂Ω, Φ∗ α1= 0,Ψ∗ α2= 0 on ∂D∗, k(Φ∗ α1,Ψ∗ α2)kH1(Ω\D∗)≤C(Ω, D∗)k(ϕ, ψ)kH1/2(∂Ω) and            ZΩ\D∗ (u·∇)Φ∗ α1·u dx≤α1kuk2 H1(Ω\D∗), ZΩ\D∗ (v·∇Ψ∗ α2)w dx≤α2kvkH1(Ω\D∗)kwkH1(Ω\D∗), for every u, v ∈V(Ω\D∗) and every w∈H1 0(Ω\D∗) (see lemmas III.6.2 and VIII.4.2 in [16]). Let us denote by Φα1and Ψα2the extensions by zero of the functions Φ∗ α1 and Ψ∗ α2to the whole set Ω. Then the couple (Φα1,Ψα2) belongs to H1(Ω \D)N× H1(Ω \D) and satisfies        ZΩ (u·∇)Φα1·u dx≤α1kuk2 H1∀u∈V(Ω), ZΩ (v·∇Ψα1)w dx≤α2kvkH1kwkH1∀v∈V(Ω),∀w∈H1 0(Ω). (79) and k(Φα1,Ψα2)kH1≤C(Ω, D∗)k(ϕ, ψ)kH1/2(∂Ω) . Let us introduce Fand G, with F= Ψα2g+ν∆Φα1−(Φα1·∇)Φα1, G =κ∆Ψα2−Φα1·∇Ψα2. Then we have kFkH−1(Ω\D)≤C(Ω, D∗)kψkH1/2+νkϕkH1/2+kϕk2 H1/2, kGkH−1(Ω\D)≤C(Ω, D∗) (κkψkH1/2+kϕkH1/2kψkH1/2).(80) We will look for a solution (u, p, θ) of (1). Let us put u=w+Φα1and θ=η+Ψα2. Then (w, p, η) must satisfy        −ν∆w+ (w·∇)Φα1+ (Φα1·∇)w+ (w·∇)w+∇p=ηg +Fin Ω \D, ∇·w= 0 in Ω \D, −κ∆η+ Φα1·∇η+w·∇η+w·∇Ψα2=Gin Ω \D, (w, η) = (0,0) on ∂Ω∪∂D. (81) It will be sufficient to show that there exist positive constants α1, α2such that the nonlinear system (81) possesses at least one weak solution, more precisely, a 24 A. DOUBOVA, E. FERN ´ ANDEZ-CARA, M. GONZ´ ALEZ-BURGOS, J. H. ORTEGA couple (w, p, η) that belongs to V(Ω \D)N×L2(Ω \D)×H1 0(Ω \D) and satisfies the previous partial differential equations in the weak or distributional sense. To this end, a standard Galerkin’s method can be used. As usual, in order to obtain the existence result, the key point is to prove appropriate “a priori” estimates on the approximated solutions {(wn, ηn)}n≥1. We have:        νk∇wnk2 L2=−ZΩ\D (wn·∇) Φα1·wn+ZΩ\D ηng·wn+hF , wniH−1, κk∇ηnk2 L2=−ZΩ\D (wn·∇Ψα2)ηn+hG , ηniH−1. Taking into account (79), we deduce that            νk∇wnk2 L2≤α1k∇wnk2 L2+C νk∇ηnk2 L2+ν 2k∇wnk2 L2+1 νkFk2 H−1, κk∇ηnk2 L2≤α2k∇wnkL2k∇ηnkL2+κ 4k∇ηnk2 L2+1 κkGk2 H−1 ≤α2 2 κk∇wnk2 L2+κ 2k∇ηnk2 L2+1 κkGk2 H−1, where Cis a positive constant only depending on Ω and g. Now, let us take α1=ν/4. It is then easy to deduce that      k∇wnk2 L2≤4C ν2k∇ηnk2 L2+4 ν2kFk2 H−1, k∇ηnk2 L2≤2α2 2 κ2k∇wnk2 L2+2 κ2kGk2 H−1. On the other hand, we can set α2 2=ν2κ2 16C.From this last inequality, we see that      k∇wnk2 L2≤16C ν2κ2kGk2 H−1+8 ν2kFk2 H−1, k∇ηnk2 L2≤4 κ2kGk2 H−1+1 CkFk2 H−1 (82) and, therefore, (wn, ηn) is uniformly bounded in V(Ω \D)×H1 0(Ω \D). In a classical way, this can be used to prove the existence of a weak solution (w, p, η) of (81) that belongs to V(Ω \D)×L2(Ω \D)×H1 0(Ω \D). Finally, from (80) and (82) we deduce that k∇wkL2≤C(Ω, D∗) νK(ν, κ, ϕ, ψ) and k∇ηkL2≤C(Ω, D∗)K(ν, κ, ϕ, ψ), where K(ν, κ, ϕ, ψ) is given by (7). Obviously, this proves (6). Therefore, (81) possesses at least one weak solution with the desired estimates. Let us now see that σ(u, p)·n∈H−1/2(∂Ω)N. In view of well known results, it suffices to prove that σ(u, p)∈L2(Ω \D)N×Nand ∇ · σ(u, p)∈Lr(Ω \D)N for some r > 1 if N= 2 and r= 6/5 if N= 3. But this is very easy to check. Indeed, we have (u, p, θ)∈H1(Ω \D)N×L2(Ω \D)×H1(Ω\D) and, consequently, σ(u, p)∈L2(Ω \D)N×N. On the other hand, ∇·σ(u, p) = (u·∇)u+θ g, whence we also have ∇·σ(u, p)∈Lβ(Ω \D)Nfor all β < 2 if N= 2 and β= 3/2 if N= 3. In a very similar way, it can be proved that ∂θ ∂n ∈H−1/2(∂Ω). 2 – Uniqueness: Let us assume that there exist two solutions (u1, p1, θ1) and (u2, p2, θ2) of (1) that verify (6) and let us set u=u1−u2, p =p1−p2and A GEOMETRIC INVERSE PROBLEM FOR THE BOUSSINESQ SYSTEM 25 θ=θ1−θ2.We assume now that (7) is satisfied. We have that (u, p, θ) satisfies    −ν∆u+ (u·∇)u1+ (u2·∇)u+∇p=θg, ∇·u= 0 in Ω \D, −κ∆θ+u1·∇θ+u·∇θ2= 0 in Ω \D, u= 0, θ = 0 on ∂Ω∪∂D, (83) By multiplying the first equation of (83) by uand integrating in Ω \D, we get νk∇uk2 L2=ZΩ\D θg ·u dx −ZΩ\D (u·∇)u1·u dx. Consequently, νk∇uk2 L2≤C(Ω)k∇θkL2k∇ukL2+k∇u1kL2kukL3kukL6 ≤C(Ω) k∇θkL2+k∇u1kL2k∇ukL2k∇ukL2.(84) We multiply the second equation of (83) by θand we integrate in Ω \D. We obtain κk∇θk2 L2=−ZΩ\D u·∇θ2θ dx ≤C(Ω, D∗)k∇ukL2k∇θkL2k∇θ2kL2. Thus, we have κk∇θkL2≤C(Ω)k∇ukL2k∇θ2kL2. Coming back to (84), we find that νk∇uk2 L2≤C(Ω) 1 κk∇θ2kL2+k∇u1kL2k∇uk2 L2. But, in view of the estimates (6) satisfied by uiand θi, we have k∇u1kL2+1 κk∇θ2kL2≤C(Ω, D∗)ν+κ νκ K(ν, κ, ϕ, ψ), whence we also obtain νk∇uk2 L2≤C(Ω, D∗)ν+κ νκ K(ν, κ, ϕ, ψ)νk∇uk2 L2. Thus, there exists K1(Ω, D∗) such that, if the couple (ϕ, ψ) satisfies (7), then C(Ω, D∗)ν+κ νκ K(ν, κ, ϕ, ψ)< ν and we necessary have u≡0. This proves the uniqueness of (u, p, θ) (pis unique up to a constant). 3 – Regularity of u,pand θ:For simplicity, let us sketch the argument for r= 2. Let us now assume that ϕ∈H3/2(∂Ω)Nand ψ∈H3/2(∂Ω). Then the weak solutions of (1) are in fact strong solutions, that is to say, they satisfy (u, p, θ)∈ H2(Ω \D)N×H1(Ω \D)×H2(Ω \D), with appropriate estimates. This fact can be deduced as a consequence of the W2,r-regularity theory for the Poisson equation and the Stokes problems (see for instance [8] and the references therein). Acknowledgements. The authors are grateful to the anonymous referee for several comments that have helped to correct some mistakes and improve considerable the paper.