Application of the boundary control method to partial data Borg-Levinson inverse spectral problem

We consider the multidimensional Borg-Levinson problem of determining a potential $q$, appearing in the Dirichlet realization of the Schr\"odinger operator $A_q=-\Delta+q$ on a bounded domain $\Omega\subset \mathbb{R}^n$, $n\geq2$, from the boundary spectral data of $A_q$ on an arbitrary portion of $\partial\Omega$. More precisely, for $\gamma$ an open and non-empty subset of $\partial\Omega$, we consider the boundary spectral data on $\gamma$ given by $\mathrm{BSD}(q,\gamma):=\{(\lambda_{k},{\partial_\nu \phi_{k}}_{|\overline{\gamma}}):\ k \geq1\}$, where $\{ \lambda_k:\ k \geq1\}$ is the non-decreasing sequence of eigenvalues of $A_q$, $\{ \phi_k:\ k \geq1 \}$ an associated Hilbertian basis of eigenfunctions, and $\nu$ is the unit outward normal vector to $\partial\Omega$. We prove that the data $\mathrm{BSD}(q,\gamma)$ uniquely determine a bounded potential $q\in L^\infty(\Omega)$. Previous uniqueness results, with arbitrarily small $\gamma$, 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.
1.2.Previous literature.Our problem is a generalization to the multidimensional case of the pioneering work of Borg [4], Levinson [21], Gel'fand and Levitan [7] 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 [24] who applied the result of [28] to prove that B(q, ∂Ω) determines uniquely q.Päivärinta and Serov [25] extended this result to q ∈ L p , and Canuto and Kavian [5] to more general perturbations of the Laplacian.Isozaki [9] proved that the uniqueness still holds if finitely many eigenpairs remain unknown and [6,12,13] 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 [2].In [10], 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 [11] 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 [3] 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 [15] and, for the case of equations taking values on Hermitian vector bundles, [17].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 [16] assume that γ satisfies the geometric control condition [1].
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 [17,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 [19,20], 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 A proof of this classical result can be found e.g. in [11,Theorem 2.47].Let us now reformulate Lemma 2.1 by using the notion of domain of influence.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 [29] 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 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 .
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.
Under the assumptions of Lemma 3.1, it holds in particular that 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 Proof.By Corollary 2.1, there is a sequence 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 .
and also that for any .
From this result, we deduce the following corollary.
The proof of Lemma 3.2 can be iterated as follows.
Proof.By the proof of Lemma 3.2, there is a sequence Then by Lemma 3.2, .
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 This hypothesis as well as the set N (γ) introduced in the following lemma are illustrated in Fig. 1.
We show next that inner products on domains of influence can be used to determine the following pointwise products 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 , where t, s ∈ [0, T ] and f, g ∈ C ∞ 0 ((0, T ) × γ).Combining this with Corollary 3.1, we obtain , and therefore, denoting the volume of Ãε,x by .
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 (3.10) The construction of the set A x,ε is illustrated in Fig. 2.
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, .
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 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 (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 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 We have u g j = v j on (0, s 2 ) × Ω.Up to a reduction of δ we can assume that a(t, Integrating both sides of this expression and sending σ → +∞, we get After the change of variables z = x − β(s 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.
This together with Corollary 2.

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 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.
We extend the notion of domain of influence for any r > 0 and any open set B ⊂ Ω by setting 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 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: 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 is dense in L 2 (Ω(B, r)).
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 the family (A x,ε ) ε>0 is of bounded eccentricity and Combining the density of the set (4.7) with the arguments used in Lemma 3.4, we obtain for all F, G ∈ C ∞ 0 ((τ, T ) × B).After taking the limit ε → 0, we obtain 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).
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 On the other hand, we have and we deduce that Thus, condition (3.1) implies that, for all (t, x) and, in virtue of (3.1), we deduce that Combining this with (4.12), we deduce that the restriction of R j,σ , j = 1, 2, to The uniqueness of solutions for this IBVP implies that 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 Integrating both sides of this expression, we get Then, in view of (4.13), sending σ → +∞ and using the fact that β = 0 on [0, τ ], we get 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 and making the substitution s = t − T + δ , we find to be arbitrary, we deduce that for almost every z ∈ B(0, δ) we have which, by fixing s = 0, implies that Then, using the fact that x 1 = x 0 + (ε 0 + s 1 )ω, we deduce that 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.
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 Theorem A.1.Let (t, x) ∈ T .Assume that there exists δ > 0 such that q ∈ L ∞ (B (x, δ)) and such that This theorem is a special case of [29, Theorem 1] (see also [26] for related results).We refer also to [11,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.
The rest of the appendix concerns the proofs of the two approximate controllability results that we need.
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.
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).

Corollary 2 . 1 .
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.