Y Kian

M Morancey

L Oksanen

APPLICATION OF THE BOUNDARY CONTROL METHOD TO PARTIAL DATA BORG-LEVINSON INVERSE SPECTRAL PROBLEM

Keywords: Inverse problems, inverse spectral problem, wave equation, Boundary Control method, uniqueness, partial data, unique continuation Mathematics subject classification 2010 : 35R30, 35J10, 35L05

We consider the multidimensional Borg-Levinson problem of determining a potential q, appearing in the Dirichlet realization of the Schrödinger operator Aq = -∆ + q on a bounded domain

2, from the boundary spectral data of Aq on an arbitrary portion of ∂Ω. More precisely, for γ an open and non-empty subset of ∂Ω, we consider the boundary spectral data on γ given by BSD(q, γ) := {(λ k , ∂ν ϕ k |γ ) : k 1}, where {λ k : k 1} is the non-decreasing sequence of eigenvalues of Aq, {ϕ k : k 1} an associated Hilbertian basis of eigenfunctions, and ν is the unit outward normal vector to ∂Ω. We prove that the data BSD(q, γ) uniquely determine a bounded potential q ∈ L ∞ (Ω). Previous uniqueness results, with arbitrarily small γ, assume that q is smooth. Our approach is based on the Boundary Control method, and we give a self-contained presentation of the method, focusing on the analytic rather than geometric aspects of the method.

1. Introduction 1.1. Statement of the results. We fix Ω a C 2 bounded and connected domain of R n , n 2, and γ a non empty open set of Γ = ∂Ω. We consider the Schrödinger operator A q = -∆ x + q acting on L 2 (Ω) with Dirichlet boundary condition and q ∈ L ∞ (Ω) real valued. The spectrum of A q consists of a non decreasing sequence of eigenvalues {λ k : k ∈ N * }, with N * := {1, 2, . . .}, to which we associate a Hilbertian basis of eigenfunctions {ϕ k : k ∈ N * }. Then, we introduce the boundary spectral data restricted to the portion γ given by BSD(q, γ) := (λ k , ∂ ν ϕ k|γ ) : k ∈ N * , with ν the outward unit normal vector to Γ and ∂ ν the normal derivative. The main goal of the present paper is to prove uniqueness in the recovery of q from the data BSD(q, γ). Theorem 1.1. Assume that Ω is convex, γ ⊂ Γ is open and non empty, and q j ∈ L ∞ (Ω), j = 1, 2. Then BSD(q 1 , γ) = BSD(q 2 , γ) implies q 1 = q 2 . This result will be proved by applying the so called Boundary Control method that we adapt to the present setting with a convex domain and a bounded potential.

Let us also formulate a dynamic variant of Theorem 1.1. Fix Σ = (0, T ) × ∂Ω, Q = (0, T ) × Ω with 0 < T < ∞, and consider the initial boundary value problem (IBVP in short)

   ∂ 2 t u -∆ x u + q(x)u = 0, in Q, u(0, •) = 0, ∂ t u(0, •) = 0,
in Ω, u = f, on Σ.

(1.1)

1 Aix Marseille Univ, Université de Toulon, CNRS, CPT, Marseille, France 2 Aix Marseille Univ, CNRS, Centrale Marseille, I2M, Marseille, France 3 Department of Mathematics, University College London, London, UK

According to [START_REF] Lasiecka | Non homogeneous boundary value problems for second order hyperbolic operators[END_REF]Theorem 2.1], for f ∈ H 1 (Σ), the problem (1.1) admits a unique solution

u ∈ C([0, T ]; H 1 (Ω)) ∩ C 1 ([0, T ]; L 2 (Ω))
which satisfies ∂ ν u ∈ L 2 (Σ). Thus we can define the partial Dirichlet-to-Neumann map Λ(q, γ, T ) :

C ∞ 0 ((0, T ) × γ) f → ∂ ν u |(0,T )×γ .
We define also diam(Ω) = max{|x -y| : x, y ∈ Ω}. The dynamic variant can be stated in the following way Theorem 1.2. Assume that Ω is convex, γ ⊂ Γ is open and non empty, T > 2 diam(Ω), and q j ∈ L ∞ (Ω), j = 1, 2. Then Λ(q 1 , γ, T ) = Λ(q 2 , γ, T ) implies that q 1 = q 2 .

1.2. Previous literature. Our problem is a generalization to the multidimensional case of the pioneering work of Borg [START_REF] Borg | Eine Umkehrung der Sturm-Liouvilleschen Eigenwertaufgabe[END_REF], Levinson [21], Gel'fand and Levitan [START_REF] Gel | On the determination of a differential equation from its spectral function[END_REF] stated in an interval of R, also called Borg-Levinson inverse spectral problem. The first multidimensional formulation of this problem is given by Nachman, Sylvester and Uhlmann [START_REF] Nachman | An n-dimensional Borg-Levinson theorem[END_REF] who applied the result of [START_REF] Sylvester | A global uniqueness theorem for an inverse boundary value problem[END_REF] to prove that B(q, ∂Ω) determines uniquely q. Päivärinta and Serov [START_REF] Päivärinta | An n-dimensional Borg-Levinson theorem for singular potentials[END_REF] extended this result to q ∈ L p , and Canuto and Kavian [START_REF] Canuto | Determining Two Coefficients in Elliptic Operators via Boundary Spectral Data: a Uniqueness Result, Bolletino Unione Mat[END_REF] to more general perturbations of the Laplacian. Isozaki [START_REF] Isozaki | Some remarks on the multi-dimensional Borg-Levinson theorem[END_REF] proved that the uniqueness still holds if finitely many eigenpairs remain unknown and [START_REF] Choulli | Stability for the multi-dimensional Borg-Levinson theorem with partial spectral data[END_REF][START_REF] Kavian | Uniqueness and stability results for an inverse spectral problem in a periodic waveguide[END_REF][START_REF] Kian | A multidimensional Borg-Levinson theorem for magnetic Schrödinger operators with partial spectral data[END_REF] proved that only some asymptotic knowledge of B(q, ∂Ω) is enough for the recovery of q as well as more general coefficients.

Let us now turn to partial data results. For arbitrarily small γ, the known uniqueness results are based on the Boundary Control method introduced by Belishev [START_REF] Belishev | An approach to multidimensional inverse problems for the wave equation[END_REF]. In [START_REF] Katchalov | Multidimensional inverse problem with incomplete boundary spectral data[END_REF], under the assumption that q is smooth, Katchalov and Kurylev proved that the data B(q, ∂Ω), with the exception of finitely many eigenpairs, determines q, and [START_REF] Katchalov | Inverse boundary spectral problems[END_REF] proved that the uniqueness remains true when knowing only the partial boundary spectral data B(q, γ), with γ an arbitrary portion of the boundary. The novelty of the present paper is to consider non-smooth q.

Let us remark that more general operators than the Schrödinger operator have been considered. It was proved in [START_REF] Belishev | To the reconstruction of a Riemannian manifold via its spectral data (BC-method)[END_REF] that, when Ω is a smooth Riemanian manifold the boundary, the spectral data B(0, ∂Ω) determines the Riemanian manifold up to an isometry. Moreover, arbitrary smooth and symmetric lower order perturbations of the Laplace-Beltrami operator can be determined up to natural gauge transformations, see [START_REF] Kurylev | An inverse boundary problem for the Schrödinger operator with magnetic field[END_REF] and, for the case of equations taking values on Hermitian vector bundles, [START_REF] Kurylev | Inverse problems for the connection Laplacian[END_REF]. These results allow γ to be arbitrarily small. It is an open question, however, if the recovery of non-symmetric lower order perturbations is possible without further geometric assumptions. The known results [START_REF] Kurylev | Gelf 'and inverse problem for a quadratic operator pencil[END_REF] assume that γ satisfies the geometric control condition [START_REF] Bardos | Sharp sufficient conditions for the observation, control, and stabilization of waves from the boundary[END_REF].

All the results of the present paper can be extended to the recovery of more general coefficients on a smooth Riemannian manifold, by changing some intermediate tools and by replacing the last part of the proof, that is, the global recovery step, with the iterative process described in [START_REF] Kurylev | Inverse problems for the connection Laplacian[END_REF]Section 4.2]. The assumption of convexity allows us to simplify in various way the exposition in order to emphasize the main idea, and analytic aspects, of the Boundary Control method. The geometric aspects are mostly avoided, since for a pair points on a convex domain, the shortest path between the points is simply a line segment. For these reasons the present paper can also be considered as an introduction to the Boundary Control method.

The dynamic variant in Theorem 1.2 allows for a more fine grained notion partial data where f is supported on a part of boundary, disjoint from the part on which ∂ ν u is restricted. Such disjoint data questions have been studied in [START_REF] Lassas | An inverse problem for a wave equation with sources and observations on disjoint sets[END_REF][START_REF] Lassas | Inverse problem for the Riemannian wave equation with Dirichlet data and Neumann data on disjoint sets[END_REF], however, the techniques used the present paper do not readily extend to disjoint data cases. 1.3. Outline. In Section 2 we recall some properties of solutions of (1.1) that will be used in the proof of Theorem 1.1. In Section 3 we describe the Boundary Control method and use it to show that q can be recovered locally near γ. Building on the local recovery step, we show in Section 4 the global recovery as stated in Theorem 1.1. In Section 5 we show how to prove Theorem 1.2 by adapting the proof of Theorem 1.1. For the convenience of the reader, we prove in the appendix some well-known facts formulated in Section 2.

Finite speed of propagation and unique continuation

The Boundary Control method is based on two complementary properties of the wave equation (1.1): the finite speed of propagation and unique continuation. Loosely speaking, they give respectively the maximum and minimum speeds at which waves can propagate. It is essential for the Boundary Control method that these two speeds are the same in the case of a scalar valued wave equations such as (1.1). All the results recalled in this section are well-known, however, for the convenience of the reader, we give their proofs in the appendix.

2.1. Finite speed of propagation and domains of influence. We make the standing assumption that Ω is convex and define dist(x, S) = inf{|x -y| : y ∈ S}, x ∈ Ω, S ⊂ Ω.

We write also B(x, r) = {y ∈ Ω : |y -x| < r}, x ∈ Ω, r > 0. A typical formulation of the finite speed of propagation is as follows Lemma 2.1. Let q ∈ L ∞ (Ω), let x ∈ Ω, and let τ > 0. Define the cone

D = {(t, y) ∈ [0, τ ] × Ω : |y -x| < τ -t},
and consider u ∈ C([0, τ ]; H 1 (Ω)) ∩ C 1 ([0, τ ]; L 2 (Ω)) satisfying (∂ 2 t -∆ x + q)u = 0 in (0, τ ) × Ω. Then u |{0}×B(x,τ ) = ∂ t u |{0}×B(x,τ ) = 0, u |D∩Σ = 0, imply u |D = 0.
A proof of this classical result can be found e.g. in [START_REF] Katchalov | Inverse boundary spectral problems[END_REF]Theorem 2.47]. Let us now reformulate Lemma 2.1 by using the notion of domain of influence. Definition 2.1. For every τ > 0 and every open subset S of Γ we define the subset Ω(S, τ ) of Ω given by Ω(S, τ ) = {x ∈ Ω : dist(x, S) τ }.

The set Ω(S, τ ) is called the domain of influence of S at time τ . We give a proof of this theorem in the appendix. The proof is a short reduction to Lemma 2.1.

2.2.

Unique continuation and approximate controllability. The local unique continuation result [START_REF] Tataru | Unique continuation for solutions to PDE; between Hörmander's theorem and Holmgren's theorem[END_REF] by Tataru implies the following global Holmgren-John type unique continuation Theorem 2.2. Let q ∈ L ∞ (Ω), let S be an open subset of Γ, and let τ > 0. Consider

u ∈ C([0, 2τ ]; H 1 (Ω)) ∩ C 1 ([0, 2τ ]; L 2 (Ω)) satisfying (∂ 2 t -∆ x + q)u = 0 in (0, 2τ ) × Ω. Then u |[0,2τ ]×S = ∂ ν u |[0,2τ ]×S = 0 (2.1) implies u(τ, x) = ∂ t u(τ, x) = 0, x ∈ Ω(S, τ ).
We give a proof of this theorem and the following corollary in the appendix. The corollary is often called approximate controllability. We denote by u f = u the solution of (1.1) when emphasizing the dependence on the boundary source f .

{u f (τ, •) : f ∈ C ∞ 0 ((0, τ ) × S)} (2.2)
is dense in L 2 (Ω(S, τ )).

Here L 2 (Ω(S, τ )) is considered as the subspace of L 2 (Ω) consisting of functions vanishing outside Ω(S, τ ). Note that Theorem 2.1 implies that the set (2.2) is indeed contained in the subspace L 2 (Ω(S, τ )), and in this sense, Corollary 2.1 relates the finite speed of propagation and unique continuation. The Boundary Control method depends heavily on this relation, as described in the next section.

Local recovery of the potential

We make the following standing assumption (A) Ω is convex, γ ⊂ Γ is open and non-empty, and q j ∈ L ∞ (Ω), j = 1, 2, and write B qj = BSD(q j , γ). In this section we prove the following Theorem 3.1. Suppose that B q1 = B q2 . Then there are τ ∈ (0, T ) and a non-empty open set γ ⊂ γ such that q 1 (x) = q 2 (x), x ∈ Ω(γ , τ ).

(3.1)

3.1. From boundary spectral data to inner products of solutions. We write u f j for the solution of (1.1) with q = q j , j = 1, 2, and T = 2 diam(Ω) + 1. Moreover, we denote by ϕ j,k , k ∈ N * , a fixed Hilbertian basis of eigenfunctions of A qj , j = 1, 2. Let us begin by showing that the Fourier coefficients of u f j (t) := u f j (t, •), with respect to the bases ϕ j,k , k ∈ N * , coincide for j = 1 and j = 2. Lemma 3.1. Suppose that

B q1 = B q2 . Let f ∈ H 1 (Σ) satisfy supp(f ) ⊂ (0, T ] × γ. Then u f 1 (t), ϕ 1,k L 2 (Ω) = u f 2 (t), ϕ 2,k L 2 (Ω) , k ∈ N * , t ∈ [0, T ]. (3.2)
Proof. Write B qj = {(λ k , ψ k ) : k ∈ N * }, j = 1, 2. We start by assuming that f ∈ C ∞ 0 ((0, T ] × γ). Then u f j ∈ H 2 (Q). Setting v j,k (t) := u f j (t), ϕ j,k L 2 (Ω)
and integrating by parts we find

v j,k (t) = ∂ 2 t u f j (t), ϕ j,k L 2 (Ω) = -(-∆ x + q j )u f j (t), ϕ j,k L 2 (Ω) = -λ k v j,k (t) + ∂Ω f (t, x)∂ ν ϕ j,k (x)dσ(x).
As supp(f ) ⊂ (0, T ] × γ, we deduce that both v j,k , j = 1, 2, solve the same differential equation

v k (t) + λ k v k (t) = γ f (t, x)ψ k (x)dσ(x), v k (0) = v k (0) = 0, (3.3)
which implies (3.2). By density, (3.2) holds also for

f ∈ H 1 (Σ) satisfying supp(f ) ⊂ (0, T ] × γ.
Under the assumptions of Lemma 3.1, it holds in particular that

u f 1 (t) 2 L 2 (Ω) = u f 2 (t) 2 L 2 (Ω) = ∞ k=1 |v f k (t)| 2 , t ∈ [0, T ], (3.4)
u f 1 (t), u g 1 (s) L 2 (Ω) = u f 2 (t), u g 2 (s) L 2 (Ω) = ∞ k=1 v f k (t)v g k (s), t, s ∈ [0, T ], (3.5)
where

v f k = v k is the solution of (3.3). 3.2.
Inner products on domains of influence. We denote by 1 S the indicator function of a set S, that is, 1 S (x) = 1 if x ∈ S and 1 S (x) = 0 otherwise. Let us show that (3.5) holds when Ω is replaced by a domain of influence Ω(γ , τ ) in the following sense

Lemma 3.2. Suppose that B q1 = B q2 . Let f, g ∈ H 1 (Σ) supported in (0, T ] × γ, let γ ⊂ γ be open, and let τ ∈ (0, T ]. Then 1 Ω(γ ,τ ) u f 1 (t), u g 1 (s) L 2 (Ω) = 1 Ω(γ ,τ ) u f 2 (t), u g 2 (s) L 2 (Ω) , t, s ∈ [0, T ].
Proof. By Corollary 2.1, there is a sequence

(f k ) k∈N * in C ∞ 0 ((0, τ ) × γ ) such that u f k 1 (τ ) converges to 1 Ω(γ ,τ ) u f 1 (t) in L 2 (Ω) as k → +∞. As 1 Ω(γ ,τ ) u f 1 (t)
is the orthogonal projection of u f 1 (t) into the subspace L 2 (Ω(γ , τ )), it holds, using again the density in Corollary 2.1, that

lim k→+∞ u f k 1 (τ ) -u f 1 (t) L 2 (Ω) = inf h∈C ∞ 0 ((0,τ )×γ ) u h 1 (τ ) -u f 1 (t) L 2 (Ω)
.

Now (3.5) implies that lim k→+∞ u f k 2 (τ ) -u f 2 (t) L 2 (Ω) = lim k→+∞ u f k 1 (τ ) -u f 1 (t) L 2 (Ω) = inf h∈C ∞ 0 ((0,τ )×γ ) u h 1 (τ ) -u f 1 (t) L 2 (Ω)
and also that for any

h ∈ C ∞ 0 ((0, τ ) × γ ) u h 1 (τ ) -u f 1 (t) L 2 (Ω) = u h 2 (τ ) -u f 2 (t) L 2 (Ω)
.

The last two equalities imply

lim k→+∞ u f k 2 (τ ) -u f 2 (t) L 2 (Ω) = inf h∈C ∞ 0 ((0,τ )×γ ) u h 2 (τ ) -u f 2 (t) L 2 (Ω) . (3.6)
The convergence of (u f k 1 (τ )) k∈N * and (3.4) imply that (u f k 2 (τ )) k∈N * is a Cauchy sequence, and therefore it converges. Corollary 2.1 and (3.6) imply that u f k 2 (τ ) converges to 1 Ω(γ ,τ ) u f 2 (t). Finally, by (3.5),

1 Ω(γ ,τ ) u f 1 (t), u g 1 (s) L 2 (Ω) = lim k→+∞ u f k 1 (τ ), u g 1 (s) L 2 (Ω) = lim k→+∞ u f k 2 (τ ), u g 2 (s) L 2 (Ω) = 1 Ω(γ ,τ ) u f 2 (t), u g 2 (s) L 2 (Ω)
.

From this result, we deduce the following corollary.

Corollary 3.1. Suppose that B q1 = B q2 . Let f, g ∈ H 1 (Σ) be supported in (0, T ] × γ, and let x ∈ γ and τ ∈ (0, T ]. Then

1 B(x,τ ) u f 1 (t), u g 1 (s) L 2 (Ω) = 1 B(x,τ ) u f 2 (t), u g 2 (s) L 2 (Ω) , t, s ∈ [0, T ]. (3.7)
Proof. For ε > 0, we fix

γ ε := {y ∈ γ : |x -y| < ε}, Ω ε = Ω(γ ε , τ ) \ B(x, τ ). For j = 1, 2, since u f j ∈ C([0, T ]; H 1 (Ω))
, by the Sobolev embedding theorem for

p := 3 for n = 2 2n n-2
for n 3

we have u f j ∈ C([0, T ]; L p (Ω)).
Thus, an application of the Hölder inequality yields

1 Ωε u f j (t), u g j (s) L 2 (Ω) |Ω ε | p-2 2p u f j C([0,T ];L p (Ω)) u g j C([0,T ];L 2 (Ω)) .
Thus, for j = 1, 2, we have

lim ε→0 1 Ω(γε,τ ) u f j (t), u g j (s) L 2 (Ω) -1 B(x,τ ) u f 1 (t), u g 1 (s) L 2 (Ω) = lim ε→0 1 Ωε u f j (t), u g j (s) L 2 (Ω) = 0. (3.8)
On the other hand, Lemma 3.2 implies

1 Ω(γε,τ ) u f 1 (t), u g 1 (s) L 2 (Ω) = 1 Ω(γε,τ ) u f 2 (t), u g 2 (s) L 2 (Ω) , t, s ∈ [0, T ], ε > 0.
Combining this with (3.8) and sending ε → 0, we deduce (3.7).

The proof of Lemma 3.2 can be iterated as follows.

Lemma 3.3. Suppose that B q1 = B q2 . Let f, g ∈ H 1 (Σ) be supported in (0, T ] × γ, let γ , γ ⊂ γ be open, and let τ , τ ∈ (0, T ]. Then

1 Ω(γ ,τ ) 1 Ω(γ ,τ ) u f 1 (t), u g 1 (s) L 2 (Ω) = 1 Ω(γ ,τ ) 1 Ω(γ ,τ ) u f 2 (t), u g 2 (s) L 2 (Ω) , t, s ∈ [0, T ].
Proof. By the proof of Lemma 3.2, there is a sequence

(f k ) k∈N * in C ∞ 0 ((0, τ ) × γ ) such that for both j = 1 and j = 2, the functions u f k j (τ ) converge to 1 Ω(γ ,τ ) u f j (t) in L 2 (Ω) as k → +∞.
Then by Lemma 3.2,

1 Ω(γ ,τ ) 1 Ω(γ ,τ ) u f 1 (t), u g 1 (s) L 2 (Ω) = lim k→+∞ 1 Ω(γ ,τ ) u f k 1 (t), u g 1 (s) L 2 (Ω) = lim k→+∞ 1 Ω(γ ,τ ) u f k 2 (t), u g 2 (s) L 2 (Ω) = 1 Ω(γ ,τ ) 1 Ω(γ ,τ ) u f 2 (t), u g 2 (s) L 2 (Ω)
.

3.3. Recovery of internal data near γ. Let x ∈ Ω and let y be one of the closest point in ∂Ω to x. Then the line through x and y must intersect ∂Ω perpendicularly. Conversely, a point y ∈ ∂Ω is the closest point in ∂Ω to x = y -rν(y) for small r > 0. Here ν(y) is the outward unit normal vector at y. Furthermore, there is τ 0 > 0 such that r = dist(y -rν(y), ∂Ω), r ∈ [0, τ 0 ], y ∈ ∂Ω. We will show that Theorem 3.1 holds with any choice of τ > 0 and γ ⊂ γ satisfying

Ω(γ , τ ) ⊂ {y -rν(y) : r ∈ [0, τ 0 ), y ∈ γ}. (3.9)
This hypothesis as well as the set N (γ) introduced in the following lemma are illustrated in Fig. 1.

Figure 1. Geometric condition (3.9)
We show next that inner products on domains of influence can be used to determine the following pointwise products

Lemma 3.4. Suppose that B q1 = B q2 . Let f, g ∈ H 1 (Σ) be supported in (0, T ] × γ. Then u f 1 (t, x)u g 1 (s, x) = u f 2 (t, x)u g 2 (s, x), t, s ∈ [0, T ], x ∈ N (γ), where N (γ) = {y -rν(y) : r ∈ (0, τ 0 ), y ∈ γ}.
Proof. To illustrate the idea of the proof, let us suppose for a moment that q j , j = 1, 2, and Ω are smooth. Then for smooth f and g, also the functions u f j and u g j are smooth. Let y ∈ γ, r ∈ (0, τ 0 ), and set x = y -rν(y), Ãε,x = B(y, r + ε) \ Ω(γ, r -ε), ε > 0. Then Ãε,x → {x} as ε → 0. By taking a limit analogous to that in Corollary 3.1, it follows from Lemma 3.3 that

1 B(y,r+ε) 1 Ω(γ,r-ε) u f 1 (t), u g 1 (s) L 2 (Ω) = 1 B(y,r+ε) 1 Ω(γ,r-ε) u f 2 (t), u g 2 (s) L 2 (Ω)
, where t, s ∈ [0, T ] and f, g ∈ C ∞ 0 ((0, T ) × γ). Combining this with Corollary 3.1, we obtain

1 B(y,r+ε) (1 -1 Ω(γ,r-ε) )u f 1 (t), u g 1 (s) L 2 (Ω) = 1 B(y,r+ε) (1 -1 Ω(γ,r-ε) )u f 2 (t), u g 2 (s) L 2 (Ω)
, and therefore, denoting the volume of Ãε,x by

| Ãε,x |, | Ãε,x | -1 1 Ãε,x u f 1 (t), u g 1 (s) L 2 (Ω) = | Ãε,x | -1 1 Ãε,x u f 2 (t), u g 2 (s) L 2 (Ω)
.

Letting ε → 0, we obtain u f 1 (t, x)u g 1 (t, x) = u f 2 (t, x)u g 2 (t, x).
Let us now turn to the case of bounded q j , j = 1, 2. The above argument does not generalize immediately, since the limit with respect to ε might not exist in the non-smooth case. Our remedy is to replace the sets Ãε,x with sets of bounded eccentricity.

Let x be as above. Choose unit vectors ξ 1 , . . . , ξ n ∈ R n , that form a basis of R n , and that are small enough perturbations of -ν(y) so that the lines s → x + sξ j intersect ∂Ω in γ near y. Denote the points of intersection by z j , and consider the sets

A x,ε = B x,ε \ Ω(γ, r -ε), B x,ε = n j=1 B(z j , |x -z j | + ε), ε > 0.
(3.10)

The construction of the set A x,ε is illustrated in Fig. 2.

Figure 2. Sets A x,ε

For small ε > 0, the set A x,ε is approximated in the first order by the simplex with outward normals -ν(y) and ξ j , j = 1, . . . , n, and all the faces having distance ε to x. Thus A x,ε is of bounded eccentricity and A ε,x → {x} as ε → 0.

By repeating the proofs of Lemma 3.3 and Corollary 3.1 several times, we obtain analogously to the smooth case,

|A ε,x | -1 1 Aε,x u f 1 (t), u g 1 (s) L 2 (Ω) = |A ε,x | -1 1 Aε,x u f 2 (t), u g 2 (s) L 2 (Ω)
.

The Lebesgue differentiation theorem, see e.g. [27, Chapter 7, Theorem 7.14], implies the claim. Note that the products in the claim are interpreted as L 1 -functions.

Lemma 3.5. Suppose that

B q1 = B q2 . Let f, g ∈ H 1 (Σ) be supported in (0, T ] × γ. Then u f 1 (t, x) = u f 2 (t, x), t ∈ [0, T ], x ∈ N (γ).
Proof. We will choose u g j in Lemma 3.4 to be a suitable geometric optics solution, and begin by constructing such solutions. Let y ∈ γ, r ∈ (0, τ 0 ), and set x 0 = y -rν(y). Let δ > 0 be small and set s 1 = r + δ and

s 2 = r + 2δ. The line β(t) = x 0 + (s 1 -t)ν(y) satisfies β(s 1 ) = x 0 , β(s 1 -r) = y ∈ γ, and β(0) ∈ R n \ Ω. Hence if δ is small enough and χ ∈ C ∞ 0 (R n ) has small enough support, then the function a(t, x) = χ(x -β(t)) satisfies supp(a) ∩ ([0, s 2 ] × Ω) ⊂ (0, s 2 ] × N (γ), supp(a) ∩ ([0, s 2 ] × ∂Ω) ⊂ (0, s 2 ) × γ.
(3.11)

In particular, supp[a(0, •)] ⊂ R n \ Ω. The support of this particular solution is illustrated in Fig. 3.

Figure 3. Support of the geometric optics solution

To simplify the notation, we suppose that χ is real valued, and write ω = -ν(y). Then we consider

v j (t, x) = a(t, x)e iσ(t-x•ω) + R j,σ (t, x), (t, x) ∈ (0, T ) × Ω, σ > 1, j = 1, 2, where R j,σ solves    ∂ 2 t R j -∆ x R j + q j (x)R j = -e iσ(t-x•ω) (∂ 2 t -∆ x + q j )a, in (0, T ) × Ω, R j = 0, on (0, T ) × ∂Ω, R j (0, •) = 0, ∂ t R j (0, •) = 0, in Ω.
It follows that ∂ 2 t v j -∆ x v j + q j (x)v j = 0 in (0, T ) × Ω and, analogously to [14, Lemma 2.2] one can check that R j,σ L 2 ((0,T )×Ω) → 0, σ → +∞.

(3.12)

Moreover, (3.11) implies that

∂ k t v j (0, x) = ∂ k t a(t, x)e iσ(t-x•ω) |t=0 = 0, x ∈ Ω, k = 0, 1. (3.13) Define g(t, x) = 1 (0,s2) (t)a(t, x)e iσ(t-x•ω) , t ∈ [0, T ], x ∈ ∂Ω. Then (3.11) implies that supp(g) ⊂ (0, T ) × γ.
We have u g j = v j on (0, s 2 ) × Ω. Up to a reduction of δ we can assume that a(t,

•) |∂Ω = 0 for t s 1 , thus u g j = v j on (0, T ) × Ω. Therefore Lemma 3.4 implies that for all f ∈ C ∞ 0 ((0, T ] × γ), ψ ∈ C ∞ 0 (-δ, δ), t ∈ [0, T -δ], s ∈ R, and x ∈ N (γ), u f 1 (t + s, x)v 1 (s + s 1 , x)e iσ(s+s1-x•ω) ψ(s) = u f 2 (t + s, x)v 2 (s + s 1 , x)e iσ(s+s1-x•ω) ψ(s).
Integrating both sides of this expression and sending σ → +∞, we get

R N (γ) u f 1 (t + s, x)a(s + s 1 , x)ψ(s)dxds = R N (γ) u f 2 (t + s, x)a(s + s 1 , x)ψ(s)dxds. (3.14)
After the change of variables z = x -β(s

+ s 1 ), R R n u f 1 (t + s, z + β(s + s 1 ))χ(z)ψ(s)dzds = R R n u f 2 (t + s, z + β(s + s 1 ))χ(z)ψ(s)dzds.
As χ and ψ are arbitrary cutoff functions with small supports, it holds that u f 1 (t, x) = u f 2 (t, x) for t ∈ [0, T -δ] and x near x 0 . As δ > 0 can be taken arbitrarily small and x 0 ∈ N (γ) can be chosen arbitrarily, the claim follows.

3.4. Proof of Theorem 3.1. Choose τ > 0 and open and non-empty γ ⊂ γ satisfying (3.9). Lemmas 3.1, 3.5 and the finite speed of propagation imply that for any

f ∈ C ∞ 0 ((0, τ ] × γ ), ϕ 1,k , u f 2 (τ, •) L 2 (Ω(γ ,τ )) = ϕ 1,k , u f 1 (τ, •) L 2 (Ω) = ϕ 2,k , u f 2 (τ, •) L 2 (Ω) = ϕ 2,k , u f 2 (τ, •) L 2 (Ω(γ ,τ ))
.

This together with Corollary 2.1 implies that ϕ 1,k = ϕ 2,k in Ω(γ , τ ). Thus in Ω(γ , τ ) it holds that

0 = (A q1 -λ k )ϕ 1,k -(A q2 -λ k )ϕ 2,k = (q 1 -q 2 )ϕ 2,k , k ∈ N * .
Integrating this expression on Ω(γ , τ ) we get (q 1 -q 2 )1 Ω(γ ,τ ) , ϕ 2,k L 2 (Ω) = 0, k ∈ N * . This proves (3.1).

Global recovery

The goal of this section is to get global recovery of the potential from the local determination. More precisely, we will complete the proof of Theorem 1.1. We start by fixing the notation. From now on we consider τ ∈ (0, T ) and γ ⊂ γ such that condition (3.1) is fulfilled and we assume that T > diam(Ω) + 2τ . For any open connected set B of Ω we define the operator

K j,B h := u h j |[0,2T +τ ]×B , h ∈ C ∞ 0 ((0, 2T + τ ] × γ ). Moreover, for B ⊂ Ω, we define L j,B F := v j,F |[0,2T ]×B , F ∈ C ∞ 0 ((τ, T ) × B) with v j,F solving    ∂ 2 t v j -∆ x v j + q j (x)v j = F, in (0, ∞) × Ω, v j (0, •) = 0, ∂ t v j (0, •) = 0,
in Ω, v j = 0, on (0, ∞) × ∂Ω. We write also B(x, r) := {y ∈ R n : |y -x| < r}, x ∈ R n , r > 0.

Lemma 4.1. Fix x ∈ Ω(γ , τ ) and consider for ε > 0 the set B = B(x, ε) ∩ Ω(γ , τ ). Then, we have

K 1,B = K 2,B . (4.2)
Moreover, if B ⊂ Ω ∩ Ω(γ , τ ) we have L 1,B = L 2,B . (4.3)
Proof. The equation (4.2) follows immediately from Lemma 3.5. Let h ∈ C ∞ 0 ((0, 2T + τ ) × γ ) and set h(t, x) = h(2T + τ -t, x). Then integrating by parts, we find

2T +τ 0 B u h j (2T + τ -t, x)F (t, x)dxdt = 2T +τ 0 Ω u h j (2T + τ -t, x)F (t, x)dxdt = 2T +τ 0 Ω u h j (2T + τ -t, x)(∂ 2 t -∆ x + q j )v j (t, x)dxdt = - 2T +τ 0 γ h(2T + τ -t, x)∂ ν v j dxdt.
Then (4.2) implies

2T +τ 0 γ h∂ ν v 1 dxdt = 2T +τ 0 γ h∂ ν v 2 dxdt. As h ∈ C ∞ 0 ((0, 2T + τ ) × γ ) is arbitrary we deduce that ∂ ν v 1 = ∂ ν v 2 on (0, 2T + τ ) × γ . Thus, fixing v = v 1 -v 2 and using (3.1), we deduce that ∂ 2 t v -∆ x v + q 1 v = 0 on (0, 2T + τ ) × Ω(γ , τ ), v |(0,2T +τ )×γ = ∂ ν v |(0,2T +τ )×γ = 0.
In particular, for any t ∈ (τ, 2T ), we have

∂ 2 t v -∆ x v + q 1 v = 0 on (t -τ, t + τ ) × Ω(γ , τ ), v |[t-τ,t+τ ]×γ = ∂ ν v |[t-τ,
t+τ ]×γ = 0. Thus, the unique continuation property of Theorem 2.2 implies that, for any t ∈ (τ, 2T ), we have v 1 (t, •) = v 2 (t, •) on Ω(γ , τ ) ⊃ B. On the other hand, using the fact that supp(F ) ⊂ (τ, T ) × B ⊂ (τ, T ) × Ω, one can check that the restriction of v j to (0, τ ) × Ω solves the problem

   ∂ 2 t v j -∆ x v j + q j (x)v j = 0, in (0, τ ) × Ω, v j (0, •) = 0, ∂ t v j (0, •) = 0, in Ω, v j = 0, on (0, τ ) × ∂Ω. which implies that v 1 (t, x) = 0 = v 2 (t, x), (t, x) ∈ (0, τ ) × Ω.
Combining these two identities we deduce (4.3).

We extend the notion of domain of influence for any r > 0 and any open set B ⊂ Ω by setting

Ω(B, r) = x ∈ Ω : dist(x, B) r .
From now on, we fix ε 0 ∈ (0, τ /7), x 0 ∈ Ω(γ , τ -3ε 0 ), dist(x 0 , Γ) > 3ε 0 , B = B(x 0 , ε 0 ). Note that Ω(B, ε 0 ) ⊂ int(Ω(γ , τ )). In the remaining of this text we will prove that (4.3) implies that

q 1 = q 2 in Ω \ B.
Note first that according to the finite speed of propagation we have

supp(v j,F (T, •)) ⊂ Ω(B, r), F ∈ C ∞ 0 ((T -r, T ) × B). (4.4)
This is the analogue of Theorem 2.1 for solutions of (4.1). We will also need to use global unique continuation in the domain Ω \ B. In the appendix, we discuss unique continuation only under assumptions that allow us to avoid certain arguments of geometric nature. For this purpose let Ω ⊂ R n be a domain with smooth boundary. We fix S an open subset of ∂Ω and for every x ∈ Ω , we consider the set Z x (S) = {y ∈ S : |x -y| = dist(x, S)}. Then, we introduce the following condition on S:

(H) Let x ∈ Ω . If y ∈ Z x (S) ∩ S then [x, y] := {tx + (1 -t)y : t ∈ [0, 1]} ⊂ (Ω ∪ S). Furthermore, if y ∈ Z x (S) \ S then for every neighborhood V of y in S there exists z ∈ V ∩ S such that [x, z] ⊂ (Ω ∪ S).
As Ω is convex, this condition holds for any S ⊂ ∂Ω. Now, let us recall (as illustrated in Fig. 4) that for any x ∈ Ω \ B, y = x 0 + ε 0

{v j,F (T, •) : F ∈ C ∞ 0 ((T -r, T ) × B)} (4.5)
is dense in L 2 (Ω(B, r)).

For convenience of the reader, we have included a proof of this lemma in the appendix. The proof is analogous with the proof of Corollary 2.1. Let us also show that the norm of solution of (4.1) with

F ∈ C ∞ 0 ((τ, T ) × B) is determined by condition L 1,Ω(B,ε0) = L 2,Ω(B,ε0) , (4.6)
which follows from (4.3) and the fact that Ω(B, ε 0 ) ⊂ int(Ω(γ , τ )). We have the following analogue of (3.5)

Lemma 4.3. Condition (4.6) implies that, for all t, s ∈ [0, T ] and all

F 1 , F 2 ∈ C ∞ 0 ((τ, T ) × B), we have v 1,F1 (t, •), v 1,F2 (s, •) L 2 (Ω) = v 2,F1 (t, •), v 2,F2 (s, •) L 2 (Ω) .
(4.7)

Proof. Consider for j = 1, 2 and for all t, s ∈ (0, 2T ] the function

w j (t, s) = v j,F1 (t, •), v j,F2 (s, •) L 2 (Ω) .
Integrating by parts, for all t, s ∈ (0, 2T ), we find

(∂ 2 t -∂ 2 s )w j (t, s) = ∂ 2 t v j,F1 (t, •), v j,F2 (s, •) L 2 (Ω) -v j,F1 (t, •), ∂ 2 s v j,F2 (s, •) L 2 (Ω) = (∆ x -q j )v j,F1 (t, •) + F 1 (t, •), v j,F2 (s, •) L 2 (Ω) -v j,F1 (t, •), (∆ x -q j )v j,F2 (s, •) + F 2 (s, •) L 2 (Ω) = F 1 (t, •), L j,Ω(B,ε) F 2 (s, •) L 2 (B) -L j,Ω(B,ε) F 1 (t, •), F 2 (s, •) L 2 (B) .
Then, applying (4.6), we deduce that, for all t, s ∈ (0, 2T ), we have

(∂ 2 t -∂ 2 s )w 1 (t, s) = F 1 (t, •), L 1,Ω(B,ε) F 2 (s, •) L 2 (B) -L 1,Ω(B,ε) F 1 (t, •), F 2 (s, •) L 2 (B) = (∂ 2 t -∂ 2 s )w 2 (t, s).
(4.8) Moreover, for j = 1, 2, we have w j (0, s) = ∂ t w j (0, s) = w j (t, 0) = ∂ s w j (t, 0) = 0. Thus, applying (4.8), we deduce that w = w 1 -w 2 solves the system associated with the 1 + 1 wave equation

   ∂ 2 t w -∂ 2 s w = 0, in (0, 2T ) × (0, 2T ), w(0, •) = 0, ∂ t w(0, •) = 0, in (0, 2T ), w(•, 0) = 0, ∂ s w(•, 0) = 0, in (0, 2T ).
From unique continuation we can deduce that w(T, s) = ∂ t w(T, s) = 0, ∀s ∈ Ω({0}, T ) = (0, T ).

Therefore, w solves    ∂ 2 s w -∂ 2 t w = 0, in (0, T ) × (0, T ), w(0, s) = w(T, s) = 0, s ∈ (0, T ), w(t, 0) = ∂ s w(t, 0) = 0, t ∈ (0, T ).
Then, the uniqueness of this initial boundary value problem implies that w |(0,T )×(0,T ) = 0 which, according to the continuity of w with respect to (t, s) ∈ [0, T ] 2 , implies that for all t, s ∈ [0, T ] we have w 1 (t, s) = w 2 (t, s). This proves (4.7).

Proof of Theorem 1.1. In a similar way as in the previous section, for any x ∈ int(Ω(B, t) \ B), t ∈ (0, T ], we consider y ∈ ∂B the unique element of B such that dist(x, B) = |x -y| = s ∈ (0, τ ). One can check that there exist z 1 , . . . , z n ∈ B such that (x -z 1 , x -z 2 , . . . , x -z n ) is a basis of R n . Moreover, we can choose z 1 , . . . , z n ∈ B such that for

B x,ε := n j=1 B(z j , |x -z j | + ε), A x,ε := B x,ε \ Ω(B, s -ε), (4.9)
the family (A x,ε ) ε>0 is of bounded eccentricity and

lim ε→0 A x,ε = {x}. (4.10)
Combining the density of the set (4.7) with the arguments used in Lemma 3.4, we obtain

A x,ε -1 1 A x,ε v 1,F (τ, •), v 1,G (τ, •) L 2 (Ω) . = A x,ε -1 1 A x,ε v 2,F (τ, •), v 2,G (τ, •) L 2 (Ω)
for all F, G ∈ C ∞ 0 ((τ, T ) × B). After taking the limit ε → 0, we obtain

v 1,F (t, x)v 1,G (t, x) = v 2,F (t, x)v 2,G (t, x), t ∈ (τ, T ], x ∈ Ω, F, G ∈ C ∞ 0 ((τ, T ) × B). (4.11)
From now on our goal will be to use this identity to conclude. For this purpose, in a similar way to Theorem 3.1 we will use special solutions that we will introduce next in order to recover v 1,F (T, •) for any F ∈ C ∞ 0 ((τ, T )×B). Then we will complete the proof. Let us first fix x 1 ∈ Ω\B and consider

s 1 =dist(x 1 , B). Since T -τ > diam(Ω) we know that s 1 + ε 0 = |x 1 -x 0 | ∈ (0, T -τ ). Now consider δ 1 ∈ (0, s 1 /2) ∩ (0, ε 0 /2) ∩ (0, dist(x 1 , ∂Ω)/2)(0, (T -s 1 -ε 0 -τ )/2). We fix also ω = x 1 -x 0 |x 1 -x 0 | , δ 2 := inf t∈[0,s1+ε0+δ1] dist (x 0 + tω, ∂Ω)
and we consider δ = min(δ1,δ2)

4

. Note that according to the definition of δ 1 we have

δ 2 > 0. Now let χ ∈ C ∞ 0 (B(0, δ)), δ ∈ (0, δ) and define a(t, x) := χ (x -x 0 -(t -T + s 1 + ε 0 + δ )ω) .
Then, we consider,

u j (t, x) = a(t, x)e iσ(t-x•ω) + R j,σ (t, x), (t, x) ∈ (0, τ ) × Ω, j = 1, 2, σ > 1,
where R j,σ solves

   ∂ 2 t R j -∆ x R j + q j (x)R j = -e iσ(t-x•ω) (∂ 2 t -∆ x + q j )a, in (0, T ) × Ω, R j (T -s 1 -ε 0 , •) = 0, ∂ t R j (T -s 1 -ε 0 , •) = 0,
in Ω, R j = 0, on (0, T ) × ∂Ω.

(4.12)

It follows that ∂ 2 t u j -∆ x u j + q j (x)u j = 0 in (0, T ) × Ω and one can check that R j,σ L 2 ((0,τ )×Ω) Cσ -1 , (

with C independent of σ. Then, we consider

β ∈ C ∞ (R; [0, 1]) satisfying β(t) = 0 for t ∈ (-∞, T -s 1 -ε 0 ] and β(t) = 1 for t ∈ [T -s 1 -ε 0 + δ, +∞).
Then, we introduce

w j (t, x) := β(t)u j (t, x).
It is clear that

w j (t, x) = 0, (t, x) ∈ [0, T -s 1 -ε 0 ] × Ω. (4.14)
Moreover, in view of (4.12), we know that

w j (t, x) = a(t, x) = χ (x -x 0 -(t -T + s 1 + ε 0 + δ )ω) e iσ(t-x•ω) , (t, x) ∈ (T -s 1 -ε 0 , T ) × ∂Ω. Using the fact that dist(x 0 + (t -T + s 1 + ε 0 + δ )ω, ∂Ω) δ 2 4δ, t ∈ [T -s 1 -ε 0 , T ] (4.15)
we deduce that a(t, x) = 0, (t, x) ∈ [T -s 1 -ε 0 , T ] × ∂Ω. Combining this with, (4.14) we deduce that w j (t, x) = 0, (t, x) ∈ (0, T ) × ∂Ω.

In the same way, (4.14) implies that w j (0, x) = ∂ t w j (0, x) = 0, x ∈ Ω and fixing G j = 2β (t)∂ t u j + β (t)u j , we deduce that w j = v j,Gj . Now let us show that G 1 = G 2 = G with supp(G) ⊂ (τ, T ) × B. For this purpose note first that

G 1 -G 2 = 2β (t)∂ t (R 1,σ -R 2,σ ) + β (t)(R 1,σ -R 2,σ ). (4.16)
On the other hand, we have

|(t -T + s 1 + ε 0 + δ )ω + y| 2δ + |t -T + s 1 + ε 0 | 3δ, y ∈ B(0, δ), t ∈ [T -s 1 -ε 0 , T -s 1 -ε 0 + δ]
and we deduce that

supp(a) ∩ [T -s 1 -ε 0 , T -s 1 -ε 0 + δ] × R n ⊂ [T -s 1 -ε 0 , T -s -ε 0 + δ] × B(x 0 , 3δ).
Thus, condition (3.1) implies that, for all (t, x)

∈ (T -s 1 -ε 0 , T -s 1 -ε 0 + δ) × Ω, we get (∂ 2 t -∆ x + q j )R j,σ (t, x) = -e iσ(t-x•ω) (∂ 2 t -∆ x + q j )a(t, x) = -e iσ(t-x•ω) (∂ 2 t -∆ x + q 1 )a(t, x). Moreover, condition (4.4) implies supp(R j,σ )∩[T -s 1 -ε 0 , T -s 1 -ε 0 +δ]×Ω ⊂ [T -s 1 -ε 0 , T -s 1 -ε 0 +δ]×Ω(B(x 0 , 3δ), δ) ⊂ (τ, T )×B (4.17)
and, in virtue of (3.1), we deduce that

(∂ 2 t -∆ x + q j )R j,σ (t, x) = (∂ 2 t -∆ x + q 1 )R j,σ (t, x), (t, x) ∈ (T -s 1 -ε 0 , T -s 1 -ε 0 + δ) × Ω, j = 1, 2.
Combining this with (4.12), we deduce that the restriction of R j,σ , j = 1, 2, to

(T -s -ε 0 , T -s -ε 0 + δ) × Ω solves    ∂ 2 t R -∆ x R + q 1 (x)R = -e iσ(t-x•ω) (∂ 2 t -∆ x + q 1 )a, in (T -s -ε 0 , T -s -ε 0 + δ) × Ω, R(T -s 1 -ε 0 , •) = 0, ∂ t R(T -s 1 -ε 0 , •) = 0, in Ω, R = 0, on (T -s -ε 0 , T -s -ε 0 + δ) × ∂Ω.
The uniqueness of solutions for this IBVP implies that

R 1,σ (t, x) = R 2,σ (t, x), (t, x) ∈ (T -s 1 -ε 0 , T -s 1 -ε 0 + δ) × Ω.
Combining this with (4.16), we deduce that G 1 = G 2 = G and (4.17) implies that supp(G) ⊂ (τ, T ) × B. Applying (4.11), one can check that, for all F ∈ C ∞ 0 ((τ, T ) × B) and ψ ∈ C ∞ 0 ((τ, T )), we have

v 1,F (t, x)w 1 (t, x)e iσ(t+x•ω) ψ(t) = v 2,F (t, x)w 2 (t, x)e iσ(t-x•ω) ψ(t), t ∈ [τ, T ], x ∈ Ω.
Integrating both sides of this expression, we get

T τ Ω v 1,F β(t)aψ(t)dxdt + T τ Ω v 1,F β(t)R 1,σ e iσ(t-x•ω) ψ(t)dxdt = T τ Ω v 2,F aβ(t)ψ(t)dxdt + T τ Ω v 2,F R 2,σ e iσ(t-x•ω) β(t)ψ(t)dxdt.
Then, in view of (4.13), sending σ → +∞ and using the fact that β = 0 on [0, τ ], we get

T 0 Ω v 1,F (t, x)a(t, x)ψ(t)β(t)dxdt = T 0 Ω v 2,F (t, x)a(t, x)ψ(t)β(t)dxdt. (4.18)
Consider θ ∈ C ∞ 0 ((-δ , δ )) and fix ψ(t) := θ (t -T + δ ). Note that supp(ψ) ⊂ (T -2δ , T ) and (4.15) implies that, for all t ∈ (0, T ), supp(β(t)a(t, •)ψ(t)) ⊂ Ω. Thus, (4.18) becomes

R R n β(t)v 1,F (t, x)a(t, x)ψ(t)dxdt = R R n β(t)v 2,F (t, x)a(t, x)ψ(t)dxdt.

Making the substitution

z = x -x 0 -(t -T + s 1 + ε 0 + δ )ω, we obtain R R n β(t)v 1,F (t, z + x 0 + (t -T + s 1 + ε 0 + δ )ω) χ(z)ψ(t)dzdt = R R n β(t)v 2,F (t, z + x 0 + (t -T + s 1 + ε 0 + δ )ω) χ(z)ψ(t)dzdt
Proof. Consider for j = 1, 2 and for all t, s ∈ [0, T ] the function w j (t, s) = u h j (t, •), u f j (s, •)

L 2 (Ω)
. Then, integrating by parts, we find

∂ 2 t w j -∂ 2 s w j = ∂ 2 t u h j (t, •), u f j (s, •) L 2 (Ω) -u h j (t, •), ∂ 2 s u f j (s, •) L 2 (Ω) = (∆ x -q j )u h j (t, •), u f j (s, •) L 2 (Ω) -u h j (t, •), (∆ x -q j )u f j (s, •) L 2 (Ω) = Λ qj h(t, •), f (s, •) L 2 (γ) -h(t, •), Λ qj f (s, •) L 2 (γ) .
Moreover, for j = 1, 2, we have w j (0, s) = ∂ t w j (0, s) = w j (t, 0) = ∂ s w j (t, 0) = 0. Thus, applying (4.6), we deduce that w = w 1 -w 2 solves the system associated with the 1 + 1 wave equation

   ∂ 2 t w -∂ 2 s w = 0, in (0, T ) × (0, T ), w(0, •) = 0, ∂ t w(0, •) = 0, in (0, T ), w(•, 0) = 0, ∂ s w(•, 0) = 0, in (0, T ).
Let us now turn to the unique continuation result formulated in Section 2.2. Recall that Theorem 2.2 follows from a local Holmgren-John unique continuation. Consider a smooth surface T := {(t, x) ∈ R 1+n : ψ(t, x) = 0}. We say that the differential operator ∂ 2 t -∆ x + q is non-characteristic at a point (t, x) ∈ T if the outward unit normal vector n = (n 0 , n ) with respect to ∂{(t, x) ∈ R 1+n : ψ(t, x) 0} at (t, x), with n 0 ∈ R, n ∈ R n , satisfies n 2 0 = |n | 2 . For all r > 0 and all t ∈ R, x ∈ R n we fix B (x, r) = {y ∈ R n : |y -x| r}, B((t, x), r) = {(s, y) ∈ R n : |(s -t, y -x)| r}.

Theorem A.1. Let (t, x) ∈ T . Assume that there exists δ > 0 such that q ∈ L ∞ (B (x, δ)) and such that

∂ 2 t -∆ x + q is non-characteristic at T ∩ B((t, x), δ). Then, if u ∈ H 1 (B((t, x), δ)) solves (∂ 2 t -∆ x + q)u = 0 in B((t, x), δ) and satisfies supp(u) ⊂ {(s, y) ∈ B((t, x), δ) : ψ(s, y) 0} then supp(u) ∩ T = ∅.
This theorem is a special case of [29, Theorem 1] (see also [START_REF] Robbiano | Uniqueness in the Cauchy problem for operators with partially holomorphic coefficients[END_REF] for related results). We refer also to [START_REF] Katchalov | Inverse boundary spectral problems[END_REF]Theorem 2.66] for a proof without microlocal analysis. In order to prove Theorem 2.2, we fix Ω 1 ⊃ Ω and we consider two intermediate results.

Lemma A.1. Let x 0 ∈ R n , ρ 0 , δ 0 > 0 and q ∈ L ∞ (B (x 0 , ρ 0 + δ 0 )). Assume that there exists u ∈ H 1 ((-δ 0 , δ 0 ) × B (x 0 , ρ 0 + δ 0 )) solving ∂ 2 t u -∆ x u + qu = 0 in (-δ 0 , δ 0 ) × B (x 0 , ρ 0 + δ 0 ) and satisfying u(t, x) = 0, (t, x) ∈ [-δ 0 , δ 0 ] × B (x 0 , ρ 0 ). (A.1)
Then, we have

u(t, x) = 0, (t, x) ∈ K x0,δ0 (A.2) with K x0,δ0 = {(t, x) ∈ [-δ 0 , δ 0 ] × B (x 0 , ρ 0 + δ 0 ) : |x -x 0 | δ 0 -|t|}.
Proof. We start by introducing the set

K s = {(t, x) ∈ [-δ 0 , δ 0 ] × Ω 1 : (t 2 + s 2 ) 1/2 + |x -x 0 | δ 0 }, s ∈ [0, δ 0 ]
and we remark that K 0 = K x0,δ0 and K δ0 = {(0, x 0 )}. The surface ∂K s are smooth with the exception of the end points (t, x) = (±(δ 2 0 -s 2 ) 1/2 , x 0 ). In addition, for s ∈ (0, δ 0 ], the points ∂K s \ {(±(δ 2 0 -s 2 ) 1/2 , x 0 )} are non-characteristic with respect to ∂ 2 t -∆ x + q. Indeed, for every (t, x) ∈ ∂K s \ {(±(δ 2 0 -s 2 ) 1/2 , x 0 )} the normal derivative (modulo multiplication by -1) n(t, x) of ∂K s is given by

n(t, x) = t 2 + s 2 2t 2 + s 2 1 2 t (t 2 + s 2 ) 1 2 , x -x 0 |x -x 0 |
and we clearly have

t 2 t 2 + s 2 < 1 = x -x 0 |x -x 0 | 2 , s ∈ (0, δ 0 ]. This proves that, for s ∈ (0, δ 0 ], ∂ 2 t -∆ x + q is non-characteristic at any (t, x) ∈ ∂K s \ {(±(δ 2 0 -s 2 ) 1/2 , x 0 )}. We fix s 0 = inf{s ∈ [0, δ 0 ] : u |Ks = 0}.
Let us show ad absurdio that s 0 = 0. For this purpose let us assume that s 0 > 0. According to (A.1), we have s 0 < δ 0 and it is clear that for any (t, x) ∈ K x0,δ0 such that

(t 2 + s 0 2 ) 1/2 + |x -x 0 | < δ 0
there exists s ∈ (s 0 , δ 0 ] such that (t, x) ∈ K s . Then, the definition of s 0 implies that u = 0 on a neighborhood of (t, x).

According to the local unique continuation result of Theorem A.1, since ∂K s0 \ {(±(δ 2 0 -s 2 0 ) 1/2 , x 0 )} is non-characteristic and since supp(u) ∩ {(t, x) ∈ K x0,δ0 : (t 2 + s 0 2 ) 1/2 + |x -x 0 | < δ 0 } = ∅, we deduce that u = 0 on a neighborhood of every point of ∂K s0 \ (R × {x 0 }). Moreover, if (t, x) ∈ ∂K s0 ∩ (R × {x 0 }) we have (t, x) = (±(δ 2 0 -s 2 ) 1/2 , x 0 ) and (A.1) implies that u = 0 on a neighborhood of (t, x). Thus, there is and we have u(t, x) = 0, (t, x) ∈ [t 0 -ε, t 0 + ε] × B (x 0 , ρ).

This proves (A.8) and Theorem 2.2.

The rest of the appendix concerns the proofs of the two approximate controllability results that we need.

Proof of Corollary 2.1. In order to prove the density result we fix h ∈ L 2 (Ω(γ , τ )), extended by zero to h ∈ L 2 (Ω), such that

u f j (τ, •), h L 2 (Ω) = u f j (τ, •), h L 2 (Ω(γ ,τ ))
= Ω(γ ,τ ) u f j (τ, •)hdx = 0, f ∈ C ∞ 0 ((0, T )×γ ), t ∈ [0, τ ] (A.9) and we will prove that h = 0. For this purpose, let e j ∈ C([0, τ ]; H 1 (Ω)) ∩ C 1 ([0, τ ]; L 2 (Ω)) solves    ∂ 2 t e j -∆ x e j + q j (x)e j = 0, in (0, τ ) × Ω, e j (τ, •) = 0, ∂ t e j (τ, •) = h, in Ω, e j = 0, on (0, τ ) × ∂Ω.

(A.10)

In light of [18, ,Theorem 2.1], we have ∂ ν e j |(0,τ )×∂Ω ∈ L 2 ((0, τ ) × ∂Ω). Thus, for all f ∈ C ∞ 0 ((0, T ) × γ ), integrating by parts and applying (1.1) and (A.9), we find 0 = τ 0 Ω u f j (∂ 2 t e j -∆ x e j + q j (x)e j )dxdt = Ω u f j (τ, •)hdx + (∂ 2 t u f j -∆ x u f j + q j (x)u f j )e j dxdt = τ 0 γ f (t, x)∂ ν e j dσ(x)dt.

Allowing f ∈ C ∞ 0 ((0, T ) × γ ) be arbitrary we deduce that ∂ ν e j |(0,τ )×γ = 0. Now fixing E j defined by E j (t, x) := e j (t, x) for t τ, e j (2τ -t, x) for t > τ, we deduce that E j ∈ H 1 ((0, 2τ ) × Ω) satisfies ∂ 2 t E j -∆ x E j + q j (x)E j = 0, in (0, 2τ ) × Ω, E j = ∂ ν E j = 0, on (0, 2τ ) × γ .

Thus, in view of Theorem 2.2, we have h = ∂ t E j (τ, •) |Ω(γ ,τ ) = 0. This proves the density of (2.2).

Proof of Lemma 4.2. In order to prove the density result we fix h ∈ L 2 (Ω(B, r)), extended by zero to h ∈ L 2 (Ω), such that (∂ 2 t v j,F -∆ x v j,F + q j (x)v j,F )e j dxdt = -T 0 Ω F (t, x)e j dxdt.

Allowing F ∈ C ∞ 0 ((T -r, T ) × B) be arbitrary we deduce that e j |(T -r,T )×B = 0. Now fixing E j defined by E j (t, x) := e j (t, x) for t ∈ [0, T ), e j (2T -t, x) for t ∈ [T, 2T ), we deduce that E j ∈ H 1 ((0, 2T ) × Ω) satisfies ∂ 2 t E j -∆ x E j + q j (x)E j = 0, in (0, 2T ) × Ω, E j = 0, on (T -r, T + r) × B.

Thus, in view of the proof of Theorem 2.2, we have h = E j (T, •) |Ω(B,r) = 0. This proves the density of (4.5).

Theorem 2 . 1 .

 21 Let S be an open subset of Γ and τ ∈ (0, T ]. Let u solve (1.1) with f ∈ H 1 (Σ) satisfying supp(f ) ⊂ (0, T ] × S. Then supp[u(τ, •)] ⊂ Ω(S, τ ).

Corollary 2 . 1 .

 21 Let S be an open subset of Γ and τ ∈ (0, T ]. Then the set

  x-x0 |x-x0| is the unique element of B satisfying dist(x, B) = |y -x|. Moreover, since Ω is convex we have [x, y] ⊂ Ω and [x, y] can not meet B since y is the unique element of ∂B satisfying dist(x, ∂B) = |y -x|. Therefore, we have [x, y] ⊂ (Ω \ B) ∪ ∂B and ∂B satisfies condition (H). Theorem 2.2 with Ω replaced by Ω \ B and S = ∂B implies the following analogue of Corollary 2.1.

Figure 4 .

 4 Figure 4.

Lemma 4 . 2 .

 42 For j = 1, 2, the set

f

  (t, x)∂ ν e j dσ(x)dt + τ 0 Ω

vF

  j,F (T, •), h L 2 (Ω) = v j,F (T, •), h L 2 (Ω(B,r)) = Ω(γ ,τ ) v j,F (T, •)hdx = 0, F ∈ C ∞ 0 ((T -r, T ) × B) (A.11)and we will prove that h = 0. For this purpose, let e j ∈ C([0, τ ];H 1 (Ω)) ∩ C 1 ([0, τ ]; L 2 (Ω)) solves    ∂ 2 t e j -∆ x e j + q j (x)e j = 0, in Q, e j (T, •) = 0, ∂ t e j (T, •) = h,in Ω, e j = 0, on (0, T ) × ∂Ω (A.12)Then, for all F ∈ C ∞ 0 ((T -r, T ) × B), integrating by parts and applying (A.11), we find F (∂ 2 t e j -∆ x e j + q j (x)e j )dxdt = Ω v j,F (T, •)hdx -(t, x)e j dxdt + T 0 Ω

and making the substitution s = t -T + δ , we find

Allowing χ ∈ C ∞ 0 (B(0, δ)), θ ∈ C ∞ 0 ((-δ , δ )) to be arbitrary, we deduce that for almost every z ∈ B(0, δ) we have v 1,F (s + T -δ , z + x 0 + (s + s 1 + ε 0 ) ω) = v 2,F (s + T -δ , z + x 0 + (s + s 1 + ε 0 ) ω) , s ∈ (-δ , δ ) which, by fixing s = 0, implies that

Then, using the fact that x 1 = x 0 + (ε 0 + s 1 )ω, we deduce that v 1,F (T -δ , z + x 1 ) = v 2,F (T -δ , z + x 1 ) , a.e. z ∈ B(0, δ).

Sending δ → 0, we prove that v 1,F (T, z) = v 2,F (T, z) , a.e. z ∈ B(x 1 , δ). Now allowing x 1 ∈ Ω \ B to be arbitrary we deduce that

) Using this identity we will complete the proof of the theorem. For this purpose, note first that repeating the arguments of Lemma 3.1 one can check that

Combining this with arguments similar to the end of the proof of Theorem 3.1 we deduce that q 1 = q 2 .

Recovery from the Dirichlet-to-Neumann map

This section is devoted to proof of Theorem 1.2. The proof of this result is similar to the one of Theorem 1.1 and the boundary spectral data B(q, γ) can be replaced by the Dirichlet-to-Neumann map Λ q as far as T > 2 diam(Ω). The only point that we need to check is the following. Lemma 5.1. Let q 1 , q 2 ∈ L ∞ (Ω). For f ∈ H 1 (Σ) satisfy f |t=0 = 0, supp(f ) ⊂ [0, T ] × γ and, for j = 1, 2, we fix u f j be the solution of (1.1) with q = q j . Condition Λ q1 = Λ q2 implies that, for all t, s ∈ (0, T /2] and all h, f ∈ C ∞ 0 ((0, T ] × γ), we have

.

(5.1)

The proof is similar with that of Lemma 4.3, however, we give it for the convenience of the reader.

Repeating the argument developed in the proof of Lemma 4.3 leads to w (0,T /2)×(0,T /2) = 0. This proves (5.1).

Using (5.1) and repeating the arguments used for Theorem 3.1, we can show that

Moreover, we get

Proof. Note first that according to (3.1),

Thus, from the unique continuation property of Theorem 2.2 we deduce (5.2). In view of (5.2), we deduce (5.3) by mimicking the proof of statement (4.3) in Lemma 4.1 .

Armed with this lemma and the arguments used for the global recovery in Section 4 we can complete the proof of Theorem 1.2.

Appendix A. Proofs of classical results on wave equations

We begin by proving the reformulation of the finite speed of progation as stated in Section 2.1. an open neighborhood V of K s0 such that u |V = 0. As ∂K s0 is compact, there exists ε ∈ (0, s 0 ) such that u |Ks 0 -ε = 0. This contradicts the definition of s 0 and we deduce that s 0 = 0. This completes the proof of the lemma.

From the previous result we can deduce the following.

Then, we have

Proof. Without loss of generality we assume that τ ρ + δ. Then our goal is to show that for any t ∈ [-τ, τ ] and any x ∈ B (x 0 , ρ + δ) such that |x -x 0 | τ -|t| we have u(t, x) = 0. We divide this proof into two steps.

First step:

In particular we have u(t, x) = v(0, x) = 0, x ∈ B (x 0 , τ -|t|). This shows that

Second step: let t ∈ [ρ + δ -τ, τ -δ -ρ] and consider w defined by

Then, for any δ 0 ∈ [δ, ρ + δ) and ρ 0 = ρ + δ -δ 0 applying again Lemma A.1, we deduce that

In particular, we have

Therefore, we have

Finally, combining (A.6)-(A.7), we deduce (A.5).

We are now in the position to complete the proof of Theorem 2.2.

Proof of Theorem 2.2 under the assumption (H). Replacing u(t, x) by u(t + τ, x), we can without loss of generality assume that (∂ 2 t -∆ x + q)u = 0 on (-τ, τ ) × Ω and ∂ ν u = u = 0 on (-τ, τ ) × S. Then, fixing the cone K S,τ := {(t, x) ∈ (-τ, τ ) × Ω : dist(x, S) < τ -|t|} the proof will be completed if we show that u(t, x) = 0, (t, x) ∈ K S,τ .

(A.8)

Here we use the fact that K S,τ ∩ ({0} × Ω) = {0} × Ω(S, τ ). Fix (t 0 , x 0 ) ∈ (-τ, τ ) × Ω such that dist(x 0 , S) < τ -|t 0 |. We consider ε 1 > 0 arbitrary small. Then, according to condition (H), there exists z 0 ∈ S such that |z 0 -

and we clearly have (∂Ω \ ∂Ω 1 ) ⊂ S. By eventually reducing the size of ε 1 , we can assume that the element z ∈ R n given by

Moreover, fixing ε = 4ε 1 , which can be arbitrary small since ε 1 > 0 can be arbitrary small, we deduce that |x 0 -z| = := dist(x 0 , S) + ε. We extend u by 0 to (-τ, τ ) × Ω 1 . Since ∂ ν u = u = 0 on (-τ, τ ) × S, we deduce that u ∈ H 1 ((-τ, τ ) × Ω 1 ). Therefore extending q to Ω 1 we deduce that u solves ∂ 2 t -∆ x u + qu = 0 in (-τ, τ ) × Ω 1 . Using the fact that [x 0 , z] ⊂ Ω 1 , we consider the path

which is lying in Ω 1 . Since µ([0, ]) is compact there exists δ > 0 such that dist(µ([0, ]), ∂Ω 1 ) > 2δ and dist(µ(0), ∂Ω) > 2δ. Now let N ∈ N be such that N δ = + ε. Here we can eventually reduce the size of δ in order to have +ε δ ∈ N. Choose s j ∈ [0, ], j = 0, . . . , N, such that 0 < s j+1 -s j < δ, s 0 = 0, s N = and denote y j = µ(s j ). We can now apply Lemma A.2 to complete the proof of the theorem. Indeed, we can choose ρ ∈ (0, δ), which can be arbitrary small, such that u = 0 on (-τ, τ ) × B (y 0 , ρ). Hence, Lemma A.2 implies that u = 0 in

On the other hand, using the fact that |y 1 -y 0 | < δ, for all (t, x) ∈ [-τ +|y 1 -y 0 |+ρ, τ -|y 1 -y 0 |-ρ]×B (y 1 , ρ), we have |x -

and u(t, x) = 0, (t, x) ∈ [-τ + s 1 + ρ, τ -s 1 -ρ] × B (y 1 , ρ). Repeating this process and by eventually reducing the size of ρ, we find u(t, x) = 0, (t, x) ∈ [-τ + s j + jρ, τ -s j -jρ] × B (y j , ρ), j = 0, . . . , N.

Note that here we use the fact that |y j+1 -y j | + s j = s j+1 , j = 0, . . . , N -1.

Using the fact that s N = = dist(x 0 , S) + ε, we get [-τ + s N + N ρ, τ -s N -N ρ] × B (y N , ρ) = [-τ + + N ρ, τ --N ρ] × B (x 0 , ρ).

Therefore, using the fact that dist(x 0 , S) < τ -|t 0 | and the fact that ε and ρ are arbitrary and N is independent of ρ, we can choose ε and ρ in such a way that τ -|t 0 |-dist(x 0 , S) > 2ε + N ρ. It follows that [t 0 -ε, t 0 + ε] × B (x 0 , ρ) ⊂ [-τ + s N + N ρ, τ -s N -N ρ] × B (y N , ρ)