An inverse problem for the magnetic Schrödinger equation in infinite cylindrical domains
aa r X i v : . [ m a t h . A P ] M a y AN INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITECYLINDRICAL DOMAINS
M. BELLASSOUED, Y. KIAN, AND E. SOCCORSIA
BSTRACT . We study the inverse problem of determining the magnetic field and the electric potential enteringthe Schrödinger equation in an infinite 3D cylindrical domain, by Dirichlet-to-Neumann map. The cylindricaldomain we consider is a closed waveguide in the sense that the cross section is a bounded domain of theplane. We prove that the knowledge of the Dirichlet-to-Neumann map determines uniquely, and even Hölder-stably, the magnetic field induced by the magnetic potential and the electric potential. Moreover, if the maximalstrength of both the magnetic field and the electric potential, is attained in a fixed bounded subset of the domain,we extend the above results by taking finitely extended boundary observations of the solution, only.
Keywords:
Inverse problem, magnetic Schrödinger equation, Dirichlet-to-Neumann map, infinite cylindricaldomain.
1. I
NTRODUCTION
Statement of the problem.
Let ω be a bounded and simply connected domain of R with C boundary ∂ω , and set Ω := ω × R . For T > , we consider the initial boundary value problem (IBVP) ( i∂ t + ∆ A + q ) u = 0 , in Q := (0 , T ) × Ω ,u (0 , · ) = 0 , in Ω ,u = f, on Σ := (0 , T ) × Γ , (1.1)where ∆ A is the Laplace operator associated with the magnetic potential A ∈ W , ∞ (Ω) , ∆ A := X j =1 (cid:0) ∂ x j + ia j (cid:1) = ∆ + 2 iA · ∇ + i ( ∇ · A ) − | A | , (1.2)and q ∈ L ∞ (Ω) . We define the Dirichlet-to-Neumann (DN) map associated with (1.1), as Λ A,q ( f ) := ( ∂ ν + iA · ν ) u, f ∈ L (Σ) , (1.3)where ν ( x ) denotes the unit outward normal to ∂ Ω at x , and u is the solution to (1.1).In the remaining part of this text, two magnetic potentials A j ∈ W , ∞ (Ω) , j = 1 , , are said gaugeequivalent, if there exists Ψ ∈ W , ∞ (Ω) obeying Ψ | Γ = 0 , such that A = A + ∇ Ψ . (1.4)In this paper we examine the uniqueness and stability issues in the inverse problem of determining theelectric potential q and the gauge class of A , from the knowledge of Λ A,q .1.2.
Physical motivations.
The system (1.1) describes the quantum motion of a charged particle (the vari-ous physical constants are taken equal to ) constrained by the unbounded domain Ω , under the influence ofthe magnetic field generated by A , and the electric potential q . Carbon nanotubes whose length-to-diameterratio is up to / , are commonly modeled by infinite waveguides such as Ω . In this context, the inverseproblem under consideration in this paper can be rephrased as to whether the strength of the electromagnetic quantum disorder (namely, the magnetic field and the electric impurity potential q , see e.g. [21, 29]) can bedetermined by boundary measurement of the wave function u .1.3. State of the art.
Inverse coefficients problems for partial differential equations such as the Schrödingerequation, are the source of challenging mathematical problems, and have attracted many attention over thelast decades. For instance, using the Bukhgeim-Klibanov method (see [14, 35, 36]), [3] claims Lipschitzstable determination of the time-independent electric potential perturbing the dynamic (i.e. non stationary)Schrödinger equation, from a single boundary measurement of the solution. In this case, the observationis performed on a sub boundary fulfilling the geometric optics condition for the observability, derived byBardos, Lebeau and Rauch in [2]. This geometrical condition was removed by [8] for potentials which are a priori known in a neighborhood of the boundary, at the expense of weaker stability. In the same spirit,[22] Lipschitz stably determines by means of the Bughkgeim-Klibanov technique, the magnetic potentialin the Coulomb gauge class, from a finite number of boundary measurements of the solution. Uniquenessresults in inverse problems for the DN map related to the magnetic Schrödinger equation are also availablein [24], but they are based on a different approach involving geometric optics (GO) solutions. The stablerecovery of the magnetic field by the DN map of the dynamic magnetic Schrödinger equation is establishedin [9] by combining the approach used for determining the potential in hyperbolic equations (see [5, 7,11, 28, 44, 47, 49]) with the one employed for the idetification of the magnetic field in elliptic equations(see [23, 45, 50]). Notice that in the one-dimensional case, [1] proved by means of the boundary controlmethod introduced by [4], that the DN map uniquely determines the time-independent electric potential ofthe Schrödinger equation. In [10] the time-independent electric potential is stably determined by the DNmap associated with the dynamic magnetic Schrödinger equation on a Riemannian manifold. This result wasrecently extended by [6] to simultaneous determination of both the magnetic field and the electric potential.As for inverse coefficients problems of the Schrödinger equation with either Neumann, spectral, or scatteringdata, we refer to [23, 25, 26, 32, 37, 38, 45, 46, 50, 53].All the above mentioned results are obtained in a bounded domain. Actually, there is only a small numberof mathematical papers dealing with inverse coefficients problems in unbounded domains. One of them,[43], examines the problem of determining a potential appearing in the wave equation in the half-space.Assuming that the potential is known outside a fixed compact set, the author proves that it is uniquelydetermined by the DN map. Unique determination of compactly supported potentials appearing in thestationary Schrödinger equation in an infinite slab from partial DN measurements is established in [39].The same problem is addressed by [37] for the stationary magnetic Schrödinger equation, and by [54] forbi-harmonic operators with perturbations of order zero or one. The inverse problem of determining thetwisting function of an infinite twisted waveguide by the DN map, is addressed in [17]. The analysis carriedout in [28, 44, 47, 49] is adapted to unbounded cylindrical domains in [17] for the determination of time-independent potentials with prescribed behavior outside a compact set, by the hyperbolic DN map. In [34],electric potentials with suitable exponential decay along the infinite direction of the waveguide, are stablyrecovered from a single boundary measurement of the solution. This is by means of a specifically designedCarleman estimate for the dynamic Schrödinger equation in infinite cylindrical domains, derived in [33]. Thegeometrical condition satisfied by the boundary data measurements in [34] is relaxed in [12] for potentialswhich are known in a neighborhood of the boundary. In [18], time-dependent potentials that are periodic inthe translational direction of the waveguide, are stably retrieved by the DN map of the Schrödinger equation.In [30], periodic potentials are stably retrieved from the asymptotics of the boundary spectral data of theDirichlet Laplacian. Finally, we refer to [19, 20], for the analysis of the Calderón problem in a periodicwaveguide.1.4.
Well-posedness.
We start by examining the well-posedness of the IBVP (1.1) in the functional space C ([0 , T ] , H (Ω)) ∩ C ([0 , T ] , H − (Ω)) . Namely, we are aiming for sufficient conditions on the coefficients N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 3 A , q and the non-homogeneous Dirichlet data f , ensuring that (1.1) admits a unique solution in the trans-position sense. We say that u ∈ L ∞ (0 , T ; H − (Ω)) is a solution to (1.1) in the transposition sense, if theidentity h u, F i L ∞ (0 ,T ; H − (Ω)) ,L (0 ,T ; H (Ω)) = h f, ∂ ν v i L (Σ) , holds for any F ∈ L (0 , T ; H (Ω)) . Here v denotes the unique C ([0 , T ] , H (Ω)) -solution to the transposi-tion system ( i∂ t v + ∆ A + q ) v = F, in Q,v ( T, · ) = 0 , in Ω ,v = 0 , on Σ . (1.5)We refer to Subsection 2.3 for the full definition and description of transposition solutions to (1.1).Since ∂ Ω is not bounded, we introduce the following notations. First, we set H s ( ∂ Ω) := H sx ( R , L ( ∂ω )) ∩ L x ( R , H s ( ∂ω )) , s > , where x denotes the longitudinal variable of Ω . Next, we put H r,s ((0 , T ) × X ) := H r (0 , T ; L ( X )) ∩ L (0 , T ; H s ( X )) , r, s > , where X is either Ω or ∂ Ω . For the sake of shortness, we write H r,s ( Q ) (resp., H r,s (Σ) ) instead of H r,s ((0 , T ) × Ω) (resp., H r,s ((0 , T ) × ∂ Ω) ). Finally, we define H , (Σ) := { f ∈ H , (Σ); f (0 , · ) = ∂ t f (0 , · ) = 0 } , and state the existence and uniqueness result of solutions to (1.1) in the transposition sense, as follows. Theorem 1.1.
For
M > , let A ∈ W , ∞ (Ω , R ) and q ∈ W , ∞ (Ω , R ) satisfy the condition k A k W , ∞ (Ω) + k q k W , ∞ (Ω) M. (1.6) Then, for each f ∈ H , (Σ) , the IBVP (1.1) admits a unique solution in the transposition sense u ∈ H (0 , T ; H (Ω)) , and the estimate k u k H (0 ,T ; H (Ω)) C k f k H , (Σ) , (1.7) holds for some positive constant C depending only on T , ω and M . Moreover, the normal derivative ∂ ν u ∈ L (Σ) , and we have k ∂ ν u k L (Σ) C k f k H , (Σ) . (1.8)It is clear from the definition (1.3) and the continuity property (1.8), that the DN map Λ A,q belongs to B ( H , (Σ) , L (Σ)) , the set of linear bounded operators from H , (Σ) into L (Σ) .1.5. Non uniqueness.
There is a natural obstruction to the identification of A by Λ A,q , arising from theinvariance of the DN map under gauge transformation. More precisely, if Ψ ∈ W , ∞ (Ω) verifies Ψ | Γ = 0 ,then we have u A + ∇ Ψ = e − i Ψ u A , where u A (resp., u A + ∇ Ψ ) denotes the solution to (1.1) associated with themagnetic potential A (resp., A + ∇ Ψ ), q ∈ L ∞ (Ω) and f ∈ H , (Σ) . Further, as ( ∂ ν + i ( A + ∇ Ψ) · ν ) u A + ∇ Ψ = e − i Ψ ( ∂ ν + iA · ν ) u A = ( ∂ ν + iA · ν ) u A on Σ , by direct calculation, we get that Λ A,q = Λ A + ∇ Ψ ,q , despite of the fact that the two potentials A and A + ∇ Ψ do not coincide in Ω (unless ψ is uniformly zero). M. BELLASSOUED, Y. KIAN, AND E. SOCCORSI
This shows that the best we can expect from the knowledge of the DN map is to identify ( A, q ) modulogauge transformation of A . When A | ∂ Ω is known, this may be equivalently reformulated as to whether themagnetic field defined by the 2-formd A := X i,j =1 (cid:0) ∂ x j a i − ∂ x i a j (cid:1) d x j ∧ d x i , and the electric potential q , can be retrieved by Λ A,q . This is the inverse problem that we examine in theremaining part of this text.1.6.
Main results.
We define the set of admissible magnetic potentials as A := (cid:8) A = ( a i ) i ; a , a ∈ L ∞ x ( R , H ( ω )) ∩ W , ∞ (Ω) and a ∈ C (Ω) satisfies (1.9) − (1.10) (cid:9) , where sup x ∈ Ω X α ∈ N , | α | h x i d | ∂ αx a ( x ) | < ∞ for some d > , (1.9)and ∂ αx a ( x ) = 0 , x ∈ ∂ Ω , α ∈ N such that | α | . (1.10)Here H ( ω ) denotes the closure of C ∞ ( ω ) in the H ( ω ) -topology, and h x i := (1 + x ) / .The first result of this paper claims stable determination of the magnetic field d A and unique identificationof electric potential q , from the knowledge of the full data, i.e. the DN map defined by (1.3), where both theDirichlet and Neumann measurements are performed on the whole boundary Σ . Theorem 1.2.
Fix A ∗ := ( a i, ∗ ) i ∈ W , ∞ (Ω , R ) , and for j = 1 , , let q j ∈ W , ∞ (Ω) , and A j :=( a i,j ) i ∈ A ∗ + A , satisfy the condition: X i =1 ∂ x i ( ∂ x ( a i, − a i, ) − ∂ x i ( a , − a , )) = 0 , in Ω . (1.11) Then, Λ A ,q = Λ A ,q yields ( d A , q ) = ( d A , q ) .Assume moreover that the estimate X j =1 (cid:0) k A j k W , ∞ (Ω) + k q j k W , ∞ (Ω) + k e j k W , ∞ (Ω) (cid:1) + k A ∗ k W , ∞ (Ω) M, (1.12) holds for some M > , with e j ( x ′ , x ) := Z x −∞ ( a ,j ( x ′ , y ) − a , ∗ ( x ′ , y )) d y , ( x ′ , x ) ∈ Ω . Then there exist two constants µ ∈ (0 , and C > , both of them depending only on T , ω and M , suchthat we have k d A − d A k L ∞ x ( R ,L ( ω )) C k Λ A ,q − Λ A ,q k µ . (1.13)In (1.13) and in the remaining part of this text, k · k denotes the usual norm in B ( H , (Σ) , L (Σ)) . Noticethat in Theorem 1.2 we make use of the full DN map, as the magnetic field d A and the electric potential q arerecovered by observing the solution to (1.1) on the entire lateral boundary Σ . In this case we may considergeneral unknown coefficients, in the sense that the behavior of A and q with respect to the infinite variableis not prescribed (we only assume that these coefficients and their derivatives are uniformly bounded in Ω ). In order to achieve the same result by measuring on a bounded subset of Σ only, we need some extrainformation on the behavior of the unknown coefficients with respect to x . Namely, we impose that the N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 5 strength of the magnetic field generated by A = ( a i ) i , reaches its maximum in the bounded subset ( − r, r ) × ω of Ω , for some fixed r > , i.e. k ∂ x i a j − ∂ x j a i k L ∞ x ( R ,L ( ω )) = k ∂ x i a j − ∂ x j a i k L ∞ x ( − r,r ; L ( ω )) , i, j = 1 , , . (1.14)Thus, with reference to (1.14), we set Γ r := ∂ω × ( − r, r ) , introduce the space H , ((0 , T ) × Γ r ) := { f ∈ H , (Σ); f (0 , · ) = ∂ t f (0 , · ) = 0 and supp f ⊂ [0 , T ] × ∂ω × [ − r, r ] } , and define the partial DN map Λ A,q,r , by Λ A,q,r ( f ) := ( ∂ ν + iA · ν ) u | (0 ,T ) × Γ r , f ∈ H , ((0 , T ) × Γ r ) , where u denotes the solution to (1.1). The following result states for each r > , that the magnetic fieldinduced by potentials belonging (up to an additive W , ∞ (Ω , R ) -term) to A r := { A = ( a i ) i ∈ A satisfying (1.14) } , can be retrieved from the knowledge of the partial DN map Λ A,q,r ′ , provided we have r ′ > r . Theorem 1.3.
For j = 1 , , let q j ∈ W , ∞ (Ω , R ) , and let A j ∈ W , ∞ (Ω , R ) satisfy A − A ∈ A r , forsome r > . Suppose that there exists r ′ > r , such that Λ A ,q ,r ′ = Λ A ,q ,r ′ . Then, we have d A = d A .Furthermore, if k q − q k L ∞ x ( R ,H − ( ω )) = k q − q k L ∞ x ( − r,r ; H − ( ω )) , we have in addition q = q .Assume moreover that (1.11) - (1.12) hold. Then, the estimate k d A − d A k L ∞ x ( R ,L ( ω )) C k Λ A ,q ,r ′ − Λ A ,q ,r ′ k µ , (1.15) holds with two constants C > , and µ ∈ (0 , , depending only on T , ω , M , r and r ′ . We stress out that Theorem 1.3 applies not only to magnetic (resp., electric) potentials A j (resp., q j ), j = 1 , , which coincide outside ω × ( − r, r ) , but to a fairly more general class of magnetic potentials,containing, e.g., r -periodic potentials with respect to x . More generally, if g ∈ W , ∞ ( R , R + ) (resp. g ∈ W , ∞ ( R , R + ) ) is an even and non-increasing function in R + , then it is easy to see that potentials of theform g × A j (resp., g × q j ), where A j (resp., q j ) are suitable r -periodic magnetic (resp., electric) potentialswith respect to x , fulfill the conditions of Theorem 1.3.Notice that the absence of stability for the electric potential q , manifested in both Theorems 1.2 and 1.3,arises from the infinite extension of the spatial domain Ω in the x direction. Indeed, the usual derivationof a stability equality for q , from estimates such as (1.13) or (1.15), requires that the differential operatord be invertible in Ω . Such a property is true in bounded domains (see e.g. [53]), but, to the best of ourknowledge, it is not known whether it can be extended to unbounded waveguides. One way to overcomethis technical difficulty is to impose certain gauge condition on the magnetic potentials, by prescribing theirdivergence. In this case, we establish in Theorem 1.4, below, that the electric and magnetic potentials canbe simultaneously and stably determined by the DN map.1.6.1. Simultaneous stable recovery of magnetic and electric potentials.
We first introduce the set of diver-gence free transverse magnetic potentials, A := { A = ( a , a , a , a ∈ L ∞ x ( R , H ( ω )) ∩ W , ∞ (Ω) , ∂ x a + ∂ x a = 0 in Ω } , in such a way that we have ∇ · A = ∇ · A ∗ for any A ∈ A ∗ + A . Here A ∗ ∈ W , ∞ (Ω) is an arbitraryfixed magnetic potential. Since identifying A ∈ A ∗ + A from the knowledge of the DN map, amounts todetermining the magnetic field d A , we have the following result. M. BELLASSOUED, Y. KIAN, AND E. SOCCORSI
Theorem 1.4.
Let
M > , and let A ∗ ∈ W , ∞ (Ω , R ) . For j = 1 , , let q j ∈ W , ∞ (Ω , R ) , and let A j ∈ A ∗ + A satisfy (1.12) . Then, there exist two constant µ ∈ (0 , and C = C ( T, ω, M ) > , suchthat we have k A − A k L ∞ x ( R ,L ( ω )) + k q − q k L ∞ x ( R ,H − ( ω )) C k Λ A ,q − Λ A ,q k µ . (1.16) Assume moreover that the two following conditions k A − A k L ∞ x ( R ,L ( ω )) = k A − A k L ∞ x ( − r,r ; L ( ω )) , (1.17) and k q − q k L ∞ x ( R ,H − ( ω )) = k q − q k L ∞ x ( − r,r ; H − ( ω )) , (1.18) hold simultaneously for some r > . Then, for each r ′ > r , we have k A − A k L ∞ x ( R ,L ( ω )) + k q − q k L ∞ x ( R ,H − ( ω )) C k Λ A ,q ,r ′ − Λ A ,q ,r ′ k µ , (1.19) where C is a positive constant depending only on T , ω , M , r and r ′ . Comments.
The key ingredient in the analysis of the inverse problem under examination is a suitableset of GO solutions to the magnetic Schrödinger equation appearing in (1.1). These functions are specificallydesigned for the waveguide geometry of Ω , in such a way that the unknown coefficients can be recovered by aseparation of variables argument. More precisely, we seek GO solutions that are functions of x = ( x ′ , x ) ∈ Ω , but where the transverse variable x ′ ∈ ω and the translational variable x ∈ R are separated. Thisapproach was already used in [31], for determining zero order unknown coefficients of the wave equation.Since we consider first order unknown coefficients in this paper, the main issue here is to take into accountboth the cylindrical shape of Ω and the presence of the magnetic potential, in the design of the GO solutions.When the domain Ω is bounded, we know from [9] that the magnetic field d A is uniquely determinedby the DN map associated with (1.1). The main achievement of the present paper is to extend the abovestatement to unbounded cylindrical domains. Actually, we also improve the results of [9] in two directions.First, we prove simultaneous determination of the magnetic field d A and the electric potential q . Second, theregularity condition imposed on admissible magnetic potentials entering the Schrödinger equation of (1.1),is weakened from W , ∞ (Ω) to W , ∞ (Ω) .To our best knowledge, this is the first mathematical paper claiming identification by boundary mea-surements, of non-compactly supported magnetic field and electric potential. Moreover, in contrast to theother works [12, 18, 33] dealing with the stability issue of inverse problems for the Schrödinger equationin an infinite cylindrical domain, available in the mathematics literature, here we no longer require that thevarious unknown coefficients be periodic, or decay exponentially fast, in the translational direction of thewaveguide.Finally, since the conditions (1.14) and (1.17)-(1.18) are imposed in ω × ( − r, r ) only, and since thesolution to (1.1) lives in the infinitely extended cylinder (0 , T ) × Ω , we point out that the results of Theorems1.3 and 1.4 cannot be derived from similar statements derived in a bounded domain.1.7. Outlines.
The paper is organized as follows. In Section 2 we examine the forward problem associatedwith (1.1), by rigorously defining the transposition solutions to (1.1), and proving Theorem 1.1. In Section3, we build the GO solutions to the Schrödinger equation appearing in (1.1), which are the key ingredientin the analysis of the inverse problem carried out in the two last sections of this paper. In Section 4, weestimate the X-ray transform of first-order partial derivatives of the transverse magnetic potential, and theFourier transform of the aligned magnetic field, in terms of the DN map. Finally, Section 5 contains theproofs of Theorems 1.2, 1.3, and 1.4.
N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 7
2. A
NALYSIS OF THE FORWARD PROBLEM
In this section we study the forward problem associated with (1.1), that is, we prove the statement ofTheorem 1.1. Although this problem is very well documented when Ω is bounded (see e.g. [9]), to our bestknowledge, it cannot be directly derived from any published mathematical work in the framework of theunbounded waveguide Ω under consideration in this paper.The proof of Theorem 1.1, which is presented in Subsection 2.4, deals with transposition solutions to(1.1), that are rigorously defined in Subsection 2.3. As a preliminary, we start by examining in Subsection2.1, the elliptic part of the dynamic magnetic Schrödinger operator appearing in (1.1), and we establish anexistence and uniqueness result for the corresponding system in Subsection 2.2.2.1. Elliptic magnetic Schrödinger operator.
For A ∈ W , ∞ (Ω , R ) , we set ∇ A := ∇ + iA , where iA denotes the multiplier by iA , and notice for all u ∈ H (Ω) , that |∇ A u ( x ) | > (1 − ǫ ) |∇ u ( x ) | + (1 − ǫ − ) | Au ( x ) | , ǫ > , x ∈ Ω . (2.1)Next, for q ∈ L ∞ (Ω; R ) , we introduce the sesquilinear form h A,q ( u, v ) := Z Ω ∇ A u ( x ) · ∇ A v ( x ) d x − Z Ω q ( x ) u ( x ) v ( x ) d x, u, v ∈ D ( h A,q ) := H (Ω) , and consider the self-adjoint operator H A,q in L (Ω) , generated by h A,q . In light of [34, Proposition 2.5], H A,q acts on its domain D ( H A,q ) := H (Ω) ∩ H (Ω) , as the operator − (∆ A + q ) , where ∆ A := ∇ A · ∇ A is expressed by (1.2).Further, for all x ∈ Ω fixed, taking ǫ = | A ( x ) | / (1 + | A ( x ) | ) in (2.1), we get that |∇ A u ( x ) | > |∇ u ( x ) | / (1 + | A ( x ) | ) − | u ( x ) | , whence h A, ( u, u ) + k u k L (Ω) > k∇ u k L (Ω) k A k L ∞ (Ω) , u ∈ H (Ω) , where h A, stands for h A,q when q is uniformly zero. Thus, we deduce from the Poincaré inequality andLax Milgram’s theorem, that for any v ∈ H − (Ω) , there exists a unique φ v ∈ H (Ω) satisfying − ∆ A φ v + φ v = v. (2.2)Next, for u and v in H − (Ω) , we put h u, v i − := Re (cid:18)Z Ω ∇ A φ u ( x ) · ∇ A φ v ( x ) d x + Z Ω φ u ( x ) φ v ( x ) d x (cid:19) , and check that the space H − (Ω) , endowed with the above scalar product, is Hilbertian. Having said that,we may now prove the following technical result. Lemma 2.1.
For each A ∈ W , ∞ (Ω , R ) , the linear operator B A := ∆ A , with domain D ( B A ) := H (Ω) ,is self-adjoint and negative in H − (Ω) .Proof. We proceed as in the proof of [16, Proposition 2.6.14 and Corollary 2.6.15]. Namely, we pick u and v in C ∞ (Ω) , and write h B A u, v i − = h w, v i − + h u, v i − , with w := B A u − u . Taking into account that φ w = − u , we obtain that h B A u, v i − = − Re (cid:18)Z Ω ∇ A u ( x ) · ∇ A φ v ( x ) d x + Z Ω u ( x ) φ v ( x ) d x (cid:19) + h u, v i − . (2.3) M. BELLASSOUED, Y. KIAN, AND E. SOCCORSI
Next, integrating by parts, we get that − Re (cid:18)Z Ω ∇ A u ( x ) · ∇ A φ v ( x ) d x + Z Ω u ( x ) φ v ( x ) d x (cid:19) = − Re h u, − ∆ A φ v + φ v i L (Ω) = − Re h u, v i L (Ω) , so (2.3) yields h B A u, v i − = − Re h u, v i L (Ω) + h u, v i − . (2.4)Further, since h u, u i − = Re h φ u , ( − ∆ A + 1) φ u i L (Ω) = Re h φ u , u i L (Ω) and k φ u k L (Ω) h u, u i − , wesee that h u, u i − k u k L (Ω) . Therefore, we obtain h B A u, u i − = −k u k L (Ω) + h u, u i − , (2.5)by taking v = u in (2.4).By density of C ∞ (Ω) in H (Ω) , both estimates (2.4) and (2.5) remain valid for all u and v in H (Ω) .As a consequence, the operator B A is dissipative. Furthermore, − B A being surjective from H (Ω) onto H − (Ω) , by (2.2), we get that B A is m -dissipative. Moreover, as it follows readily from (2.4) that h B A u, v i − = h u, B A v i − , u, v ∈ H (Ω) , we see that the graph of B A is contained into the one of its adjoint B ∗ A . Therefore, B A is self-adjoint, invirtue of [16, Corollary 2.4.10]. (cid:3) Existence and uniqueness result.
For further use, we establish the following existence and uniquenessresult for the system ( i∂ t + ∆ A + q ) v = F, in Q,v (0 , · ) = 0 , in Ω ,v = 0 , on Σ , (2.6)with homogeneous Dirichlet boundary condition and suitable source term F . Lemma 2.2.
Let M , A and q be the same as in Theorem 1.1.(i) Assume that F ∈ L (0 , T ; H (Ω)) . Then, the system (2.6) admits a unique solution v ∈ C ([0 , T ] , H (Ω)) ,satisfying k v k C ([0 ,T ] ,H (Ω) C k F k L (0 ,T ; H (Ω)) , (2.7) for some constant C > , depending only on T , ω and M .(ii) If F ∈ W , (0 , T ; L (Ω)) , then (2.6) admits a unique solution v ∈ Z := C ([0 , T ] , L (Ω)) ∩ C ([0 , T ] , H (Ω) ∩ H (Ω)) , and there exists C = C ( T, ω, M ) > , such that k v k Z C k F k W , (0 ,T ; L (Ω)) . Proof.
The proof boils down on the following statement, borrowed from [18, Lemma 2.1].Let X be a Banach space, U be a m-dissipative operator in X with dense domain D ( U ) and B ∈C ([0 , T ] , B ( D ( U ))) . Then for all v ∈ D ( U ) and f ∈ C ([0 , T ] , X ) ∩ L (0 , T ; D ( U )) ( resp. f ∈ W , (0 , T ; X )) there is a unique solution v ∈ Z = C ([0 , T ] , D ( U )) ∩ C ([0 , T ] , X ) to the followingCauchy problem v ′ ( t ) = U v ( t ) + B ( t ) v ( t ) + f ( t ) ,v (0) = v , N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 9 such that k v k Z = k v k C ([0 ,T ] ,D ( U )) + k v k C ([0 ,T ] ,X ) C ( k v k D ( U ) + k f k ∗ ) . Here C is some positive constant depending only on T and k B k C ([0 ,T ] , B ( D ( U ))) , and k f k ∗ stands for thenorm k f k C ([0 ,T ] ,X ) ∩ L (0 ,T ; D ( U )) ( resp. k f k W , (0 ,T ; X ) ) .Notice that the operator i B A is skew-adjoint, since B A is self-adjoint in H − (Ω) . Hence i B A is m -dissipative with dense domain in H − (Ω) . Further, the multiplier by iq being bounded in C [0 , T ] , H (Ω) ,we obtain (i) by applying the above result with X = H − (Ω) , U = i B A,q , f = iF , B ( t ) = iq , and v = 0 .Similarly, as H A,q is self-adjoint in L (Ω) , then the operator − i H A,q is m -dissipative with dense domainin L (Ω) , so we derive (ii) by applying [18, Lemma 2.1] with X = L (Ω) , U = − i H A,q , f = iF , B ( t ) = 0 , and v = 0 . (cid:3) Remark 2.3.
Since w ( t, x ) := v ( T − t, x ) , for ( t, x ) ∈ Q , is solution to ( i∂ t + ∆ A + q ) w = F, in Q,w ( T, · ) = 0 , in Ω ,w = 0 on Σ , (2.8) whenever v is solution to the IBVP (2.6) , where the function ( t, x ) F ( T − t, x ) is substituted for F ,we infer from Lemma 2.2 that the transposed system (2.8) admits a unique solution w in C ([0 , T ] , H (Ω)) (resp., Z ) provided F is in L (0 , T ; H (Ω)) (resp., W , (0 , T ; L (Ω)) ). Transposition solutions.
As a preamble to the definition of transposition solutions to (1.1), we estab-lish that the normal derivative of the C ([0 , T ] , H (Ω)) -solution to (2.6) is lying in L (Σ) . Lemma 2.4.
Let M , A and q be as in Lemma 2.2. Then, the linear map F ∂ ν v , where v denotes the C ([0 , T ] , H (Ω)) -solution to (2.6) associated with F ∈ L (0 , T ; H (Ω)) , given by Lemma 2.2, is boundedfrom L (0 , T ; H (Ω)) into L (Σ) .Proof. Since k v k C ([0 ,T ] ,H (Ω)) C k F k L (0 ,T ; H (Ω)) , by (2.7), we may assume without loss of generalitythat A = 0 and q = 0 .Assume that F ∈ W , (0 , T ; L (Ω)) in such a way v ∈ Z , in virtue of Lemma 2.2. Let N ∈ C ( ω ) sat-isfy N = ν on ∂ω , where ν denotes the unit outward normal vector to ∂ω . Put N ( x ′ , x ) := ( N ( x ′ ) , for all x ′ ∈ ω and x ∈ R , so that N ∈ C (Ω) ∩ W , ∞ (Ω) verifies N = ν on ∂ Ω . Then, we have h i∂ t v + ∆ v, N · ∇ v i L ( Q ) = h F, N · ∇ v i L ( Q ) . (2.9)By integrating by parts with respect to t , we get h ∂ t v, N · ∇ v i L ( Q ) = h v ( T, · ) , N · ∇ v ( T, · ) i L (Ω) − h v, N · ∇ ∂ t v i L ( Q ) = h v ( T, · ) , N · ∇ v ( T, · ) i L (Ω) + h N · ∇ v, ∂ t v i L ( Q ) − I, (2.10)where I := R Q N ·∇ ( v∂ t v ) d x d t . Taking into account that N ·∇ = N ·∇ x ′ , where ∇ x ′ denotes the gradientoperator with respect to x ′ ∈ ω , we have I = R Q N · ∇ x ′ ( v∂ t v ) d x d t , hence I = Z Q ∇ x ′ · ( v ( t, x ) ∂ t v ( t, x ) N ( x ′ )) d x ′ d x d t − h ( ∇ · N ) v, v∂ t v i L ( Q ) = Z Σ v ( t, x ) ∂ t v ( t, x ) N ( x ′ ) · ν ( x ′ ) d x ′ d x d t − h ( ∇ · N ) v, ∂ t v i L ( Q ) = −h ( ∇ · N ) v, ∂ t v i L ( Q ) , (2.11) by Green’s formula, since v | Σ = 0 . Putting (2.10)-(2.11) together, we obtain that Re h i∂ t v, N · ∇ v i L ( Q ) = i h v ( T, · ) , N · ∇ v ( T, · ) i L (Ω) − h ( ∇ · N ) v, i∂ t v i L ( Q ) = i h v ( T, · ) , N · ∇ v ( T, · ) i L (Ω) + h ( ∇ · N ) v, ∆ v i L ( Q ) − h ( ∇ · N ) v, F i L ( Q ) . (2.12)Applying the Green formula with respect to x ′ ∈ ω and integrating by parts with respect to x ∈ R , we findthat h ( ∇ · N ) v, ∆ v i L ( Q ) = −h ( ∇ · N ) ∇ v, ∇ v i L ( Q ) − h v ∇ ( ∇ · N ) , ∇ v i L ( Q ) , so (2.12) entails Re h i∂ t v, N · ∇ v i L ( Q ) = i h v ( T, · ) , N · ∇ v ( T, · ) i L (Ω) − h ( ∇ · N ) ∇ v, ∇ v i L ( Q ) −h v ∇ ( ∇ · N ) , ∇ v i L ( Q ) − h ( ∇ · N ) v, F i L ( Q ) . This and (2.7) yield (cid:12)(cid:12) Re h i∂ t v, N · ∇ v i L ( Q ) (cid:12)(cid:12) C k v k C ([0 ,T ] ,H (Ω)) (cid:0) k v k C ([0 ,T ] ,H (Ω)) + k F k L (0 ,T ; H (Ω)) (cid:1) C k F k L (0 ,T ; H (Ω)) . From this and (2.9), it then follows that (cid:12)(cid:12) Re h ∆ v, N · ∇ v i L ( Q ) (cid:12)(cid:12) C k F k L (0 ,T ; H (Ω)) (2.13)On the other hand, we get upon applying the Green formula with respect to x ′ ∈ ω and integrating by partswith respect to x ∈ R , that h ∆ v, N · ∇ v i L ( Q ) = −h∇ v, ∇ ( N · ∇ v ) i L ( Q ) + h∇ v · ν, N · ∇ v i L (Σ) = −h∇ v, ∇ ( N · ∇ v ) i L ( Q ) + k ∂ ν v k L (Σ) . (2.14)Moreover, since Re (cid:0) ∇ v · ∇ ( N · ∇ v ) (cid:1) = Re (cid:0) ( H ∇ v ) · ∇ v (cid:1) + N · ∇ |∇ v | with H := ( ∂ x i N j ) i,j and N := ( N j ) j , we infer from (2.14) thatRe h ∆ v, N · ∇ v i L ( Q ) = k ∂ ν v k L (Σ) − Re h H ∇ v, ∇ v i L ( Q ) − Z Q N · ∇ |∇ v | d x d t. (2.15)Further, by applying once more the Green formula with respect to x ′ ∈ ω , we find for a.e. ( t, x ) ∈ (0 , T ) × R , that Z ω N ( x ′ , x ) · ∇ (cid:12)(cid:12) ∇ v ( t, x ′ , x ) (cid:12)(cid:12) d x ′ = Z ω N ( x ′ ) · ∇ x ′ (cid:12)(cid:12) ∇ v ( t, x ′ , x ) (cid:12)(cid:12) d x ′ = k∇ v ( t, · , x ) k L ( ∂ω ) − h ( ∇ · N ) ∇ v ( t, · , x ) , ∇ v ( t, · , x ) i L ( ω ) . (2.16)Bearing in mind that v | Σ = 0 , we have |∇ v | = | ∂ ν v | on Σ , so we deduce from (2.16) that Z Q N · ∇ |∇ v | d x d t = k ∂ ν v k L (Σ) − h ( ∇ · N ) ∇ v, ∇ v i L ( Q ) . From this and (2.15), it then follows that k ∂ ν v k L (Σ) = 2 Re h ∆ v, N · ∇ v i L ( Q ) + 2 Re h H ∇ v, ∇ v i L ( Q ) − h ( ∇ · N ) ∇ v, ∇ v i L ( Q ) , and hence k ∂ ν v k L (Σ)) C (cid:0) k F k L (0 ,T ; H (Ω)) + k v k C ([0 ,T ] ,H (Ω)) (cid:1) C k F k L (0 ,T ; H (Ω)) , according to (2.7) and (2.13). By density of W , (0 , T ; H (Ω)) in L (0 , T ; H (Ω)) , it is clear that theabove estimate extends to every F ∈ L (0 , T ; H (Ω)) , which proves the desired result. (cid:3) N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 11
Armed with Lemma 2.4, we now introduce the transposition solution to (1.1). For F ∈ L (0 , T ; H (Ω)) ,we denote by v ∈ C ([0 , T ] , H (Ω)) the solution to (2.8), given by Remark 2.3. Since ( t, x ) v ( T − t, x ) is solution to (2.6) associated with the source term ( t, x ) F ( T − t, x ) , we infer from Lemma 2.4, thatthe mapping F ∂ ν v is bounded from L (0 , T ; H (Ω)) into L (Σ) . Therefore, for each f ∈ L (Σ) , themapping ℓ f : F
7→ h f, ∂ ν v i L (Σ) , is an anti-linear form on L (0 , T ; H (Ω)) . Thus, there exists a unique u ∈ L ∞ (0 , T ; H − (Ω)) such that wehave h u, F i L ∞ (0 ,T ; H − (Ω)) ,L (0 ,T ; H (Ω)) = ℓ f ( F ) , F ∈ L (0 , T ; H (Ω)) , (2.17)according to Riesz’s representation theorem. The function u , characterized by (2.17), is named the solutionin the transposition sense to (1.1).2.4. Proof of Theorem 1.1.
Let w ∈ L ∞ (0 , T ; H − (Ω)) be the solution in the transposition sense to thesystem ( i∂ t + ∆ A + q ) w = 0 , in Q,w (0 , · ) = 0 , in Ω ,w = ∂ t f, on Σ . For any t ∈ (0 , T ) , put v ( t, · ) := R t w ( s, · ) d s , in such a way that v is the solution in the transposition senseto the system ( i∂ t + ∆ A + q ) v = 0 , in Q,v (0 , · ) = 0 , in Ω ,v = ∂ t f, on Σ . (2.18)We have v = ∂ t f ∈ H , / (Σ) by [41, Section 4, Proposition 2.3], and since H , / (Σ) ⊂ L (0 , T ; H / ( ∂ Ω)) ,and − ∆ A v = iw + qv in Q , from the first line of (2.18), then v ∈ L (0 , T ; H (Ω)) ∩ W , ∞ (0 , T ; H − (Ω)) .Moreover, we have the following estimate k v k L (0 ,T ; H (Ω)) C (cid:16) k w k L (0 ,T ; H − (Ω)) + k qv k L (0 ,T ; H − (Ω)) + k ∂ t f k L (0 ,T ; H / ( ∂ Ω)) (cid:17) , (2.19)where the constant C > depends only on T , ω , and M .On the other hand, from the very definition of the transposition solution w , we obtain k w k L (0 ,T ; H − (Ω)) T / k w k L ∞ (0 ,T ; H − (Ω)) C k ∂ t f k L (Σ) C k f k H , (Σ) , (2.20)with the aid of Lemma 2.4. As a consequence we have k qv k L (0 ,T ; H − (Ω)) k q k W , ∞ (Ω) T k w k L (0 ,T ; H − (Ω)) C k f k H , (Σ) . (2.21)Putting (2.19)–(2.21) together, we find that k v k L (0 ,T ; H (Ω)) C k f k H , (Σ) , (2.22)for some constant C = C ( T, ω, M ) > .Finally, as u ( t ) = R t v ( s ) ds is solution to (1.1) in the transposition sense, we have k u k H (0 ,T ; H (Ω)) (1 + T ) / k v k L (0 ,T ; H (Ω)) , hence (1.7) follows from this and (2.22). We turn now to proving (1.8). To do that, we pick f ∈ C ∞ ([0 , T ] × ∂ Ω) ∩ H , (Σ) , and proceed as inthe derivation of Lemma 2.4. We get that k ∂ ν u k L (Σ) C (cid:0) k u k H (0 ,T ; H (Ω)) + k f k H , (Σ) (cid:1) , for some constant C = C ( T, ω, M ) > , so we deduce from (1.7) that k ∂ ν u k L (Σ) C k f k H , (Σ) . The desired result follows from this by density of C ∞ ([0 , T ] × ∂ Ω) ∩ H , (Σ) in H , (Σ) .3. GO SOLUTIONS
In this section we build GO solutions to the magnetic Schrödinger equation in Ω . These functions areessential tools in the proof of Theorems 1.2, 1.3 and 1.4. As in [32], we take advantage of the translationalinvariance of Ω with respect to the longitudinal direction x , in order to adapt the method suggested byBellassoued and Choulli in [9] for building GO solutions to the magnetic Schrödinger equation in a boundeddomain, to the framework of the unbounded waveguide Ω . Moreover, as we aim to reduce the regularityassumption imposed on the magnetic potential by the GO solutions construction method, we follow thestrategy developed in [23, 37, 38, 45] for magnetic Laplace operators, and rather build GO solutions to theSchrödinger equation associated with a suitable smooth approximation of the magnetic potential.Throughout the entire section, we consider two magnetic potentials A j = ( A ♯j , a j, ) ∈ W , ∞ (Ω , R ) × W , ∞ (Ω , R ) , j = 1 , , and two electric potentials q j ∈ W , ∞ (Ω , R ) , obeying the conditions k A j k W , ∞ (Ω) + k q j k W , ∞ (Ω) M, j = 1 , , (3.1)and ∂ αx A = ∂ αx A on ∂ Ω , for all α ∈ N such that | α | . (3.2)For σ > , we denote by A ♯j,σ a suitable C ∞ ( R , R ) ∩ W ∞ , ∞ ( R , R ) -approximation of A ♯j , we shallspecify in Lemma 3.3, below. We seek solutions u j,σ to the magnetic Schrödinger equation of (1.1), where ( A j , q j ) is substituted for ( A, q ) , of the form u j,σ ( t, x ′ , x ) := Φ j (2 σt, x ) b j,σ (2 σt, x ) e iσ ( x ′ · θ − σt ) + ψ j,σ ( t, x ) , t ∈ R , x = ( x ′ , x ) ∈ ω × R . (3.3)Here, θ ∈ S := { y ∈ R : | y | = 1 } is fixed, b j,σ ( t, x ) := exp (cid:18) − i Z t θ · A ♯j,σ ( x ′ − sθ, x ) d s (cid:19) , t ∈ R , x = ( x ′ , x ) ∈ ω × R , (3.4) Φ j is a solution to the following transport equation ( ∂ t + θ · ∇ x ′ ) Φ j = 0 in R × Ω , (3.5)and we imposed that the remainder term ψ j,σ ∈ L ( Q ) scales at best like σ − / when σ is large, i.e. lim σ → + ∞ σ / k ψ j,σ k L ( Q ) = 0 . (3.6)Such a construction requires that A ♯j,σ be sufficiently close to A ♯j , as will appear in the coming subsection. N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 13
Magnetic potential mollification.
We aim to define a suitable smooth approximation A ♯j,σ ∈ C ∞ ( R , R ) ∩ W ∞ , ∞ ( R , R ) of A ♯j = ( a ,j , a ,j ) , for j = 1 , . This preliminarily requires that A ♯j be appropriately extended to a largerdomain than Ω , as follows. Lemma 3.1.
Let A ♯j , for j = 1 , , be in W , ∞ (Ω , R ) and satisfy (3.2) . Let ˜ ω be a smooth open boundedsubset of R , containing ω . Then, there exist two potentials ˜ A ♯ and ˜ A ♯ in W , ∞ ( R , R ) , both of thembeing supported in ˜Ω := ˜ ω × R , such that we have ˜ A ♯j = A ♯j in Ω , for j = 1 , , and ˜ A ♯ = ˜ A ♯ in ˜Ω \ Ω . (3.7) Moreover, the two estimates k ˜ A ♯j k W , ∞ ( R ) C max (cid:16) k A ♯ k W , ∞ (Ω) , k A ♯ k W , ∞ (Ω) (cid:17) , j = 1 , , (3.8) hold for some constant C > , depending only on ω and ˜ ω .Proof. By [48, Section 3, Theorem 5] and [34, Lemma 2.7], there exists ˜ A ♯ ∈ W , ∞ ( R , R ) , such that ˜ A ♯ = A ♯ in Ω , and (3.8) holds true for j = 1 . Then, upon possibly substituting χ ˜ A ♯ for ˜ A ♯ , where χ ∈ C ∞ ( R , R ) is supported in ˜Ω and verifies χ ( x ) = 1 for all x ∈ Ω , we may assume that ˜ A ♯ is supportedin ˜Ω as well.Next, we put ˜ A ♯ ( x ) := A ♯ ( x ) , if x ∈ Ω , ˜ A ♯ ( x ) , if x ∈ R \ Ω . (3.9)Then, it is clear from (3.2) that ˜ A ♯ ∈ W , ∞ ( R , R ) and that it satisfies (3.8) with j = 2 . (cid:3) Having seen this, we define for each σ > the smooth approximation a σ ∈ C ∞ ( R , R ) ∩ W ∞ , ∞ ( R , R ) of a function ˜ a ∈ W , ∞ ( R , R ) , supported in ˜Ω , by a σ ( x ) := Z R χ σ ( x − y ) (˜ a ( y ) + ( x − y ) · ∇ ˜ a ( y )) d y, x ∈ R . (3.10)Here we have set χ σ ( x ) := σχ ( σ / x ) for all x ∈ R , where χ ∈ C ∞ ( R , R + ) is such that supp χ ⊂ { x ∈ R ; | x | } and Z R χ ( x ) d x = 1 . This terminology is justified by the fact that a σ gets closer to ˜ a as the parameter σ becomes larger, as can beseen from the following result. Lemma 3.2.
Let ˜ a ∈ W , ∞ ( R , R ) be supported in ˜Ω and satisfy k ˜ a k W , ∞ ( R ) M , for some M > .Then, there exists a constant C > , depending only on ω , ˜ ω , and M , such that for all σ > , we have k a σ − ˜ a k W k, ∞ ( R ) Cσ ( k − / , k = 0 , , (3.11) where W , ∞ (Ω) stands for L ∞ (Ω) , and k a σ k W k, ∞ ( R ) Cσ ( k − / , k > . (3.12) Proof.
We only establish (3.11), the estimate (3.12) being obtained with similar arguments. For x ∈ R fixed, we make the change of variable η = σ / ( x − y ) in (3.10). We get a σ ( x ) = Z R χ ( η )˜ a ( x − σ − / η ) d η + σ − / Z R χ ( η ) (cid:16) η · ∇ ˜ a ( x − σ − / η ) (cid:17) d η. (3.13)On the other hand, we have Z R χ ( η )˜ a ( x − σ − / η ) d η − ˜ a ( x ) = Z R χ ( η ) (cid:16) ˜ a ( x − σ − / η ) − ˜ a ( x ) (cid:17) d η = − σ − / Z R χ ( η ) (cid:18)Z η · ∇ ˜ a ( x − sσ − / η ) d s (cid:19) d η, so we infer from (3.13) that a σ ( x ) − ˜ a ( x ) = σ − / Z R χ ( η ) (cid:18)Z η · (cid:16) ∇ ˜ a ( x − σ − / η ) − ∇ ˜ a ( x − sσ − / η ) (cid:17) d s (cid:19) d η. (3.14)By the Sobolev embedding theorem (see e.g. [27, Theorem 1.4.4.1] and [51, Lemma 3.13]), we know that ˜ a ∈ C , ( R ) satisfies the estimate k ˜ a k C , ( R ) C k ˜ a k W , ∞ ( R ) , where C > is independent of ˜ a . Fromthis and (3.14), it then follows that | a σ ( x ) − ˜ a ( x ) | C k ˜ a k W , ∞ ( R ) (cid:18)Z R χ ( η ) | η | d η (cid:19) σ − / , which, together with the estimate k ˜ a k W , ∞ ( R ) M , yields (3.11) with k = 0 . Further, upon differentiating(3.14) with respect to x i , for i = 1 , , , and upper bounding the integrand function ( η, s )
7→ ∇ ∂ i ˜ a ( x − σ − / η ) − ∇ ∂ i ˜ a ( x − sσ − / η ) by k ˜ a k W , ∞ ( R ) , uniformly over R × (0 , , we obtain (3.11) for k = 1 . (cid:3) We notice for further use from (3.10) and the expression of χ σ , that a σ ( x ) = Z R ( χ σ ( x − y ) − ∇ · (( x − y ) χ σ ( x − y ))) ˜ a ( y ) d y = Z R (cid:16) σχ ( σ / ( x − y )) + σ / ( x − y ) · ∇ χ ( σ / ( x − y )) (cid:17) ˜ a ( y ) d y, x ∈ R . Making the change of variable z = σ / ( x − y ) in the above integral, we find that a σ ( x ) = Z R (4 χ ( z ) + z · ∇ χ ( z )) ˜ a ( σ − / z − x ) d z, x ∈ R . Since χ is compactly supported in R , this entails that k a σ k L ∞ ( R ) C k ˜ a k L ∞ ( R ) , σ > , (3.15)where the constant C > depends only on χ .Let ˜ A ♯j = (˜ a ,j , ˜ a ,j ) , j = 1 , , be given by Lemma 3.1. With reference to (3.10), we define the smoothmagnetic potentials A ♯j,σ = ( a ,j,σ , a ,j,σ ) ∈ C ∞ ( R , R ) ∩ W ∞ , ∞ ( R , R ) , by setting a i,j,σ ( x ) := Z R χ σ ( x − y ) (˜ a i,j ( y ) + ( x − y ) · ∇ ˜ a i,j ( y )) d y, x ∈ R , i, j = 1 , . (3.16)Thus, applying Lemma 3.2 with ˜ a = ˜ a i,j for i, j = 1 , , we obtain the following result. N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 15
Lemma 3.3.
For j = 1 , , let A ♯j be the same as in Lemma 3.1, and satisfy (3.1) . Then, there exists aconstant C > , depending only on ω and M , such that for each j = 1 , , and all σ > , we have k A ♯j,σ − A ♯j k W k, ∞ (Ω) k A ♯j,σ − ˜ A ♯j k W k, ∞ ( R ) Cσ ( k − / , k = 0 , , (3.17) where ˜ A ♯j is given by Lemma 3.1, and k A ♯j,σ k W k, ∞ ( R ) Cσ ( k − / k ˜ A ♯j k W , ∞ ( R ) Cσ ( k − / , k > . (3.18)For further use, we notice from (3.4) and from (3.18) with k = 2 , that the following estimate k b j,σ k W , ∞ ( R × Ω) + k ∂ t b j,σ k W , ∞ ( R × Ω) C, j = 1 , , (3.19)holds uniformly in σ > , for some constant C > which is independent of σ . Moreover, it can be checkedthrough direct calculation from (3.4), that θ · ∇ x ′ b j,σ ( t, x ) = − i X m =1 θ m Z t X k =1 θ k ∂ x k a j,m,σ ( x ′ − sθ, x ) d s ! b j ( t, x )= i X k =1 θ k Z t dd s a j,k,σ ( x ′ − sθ, x ) d s ! b j ( t, x )= i (cid:16) θ · A ♯j,σ ( x ′ − tθ, x ) − θ · A ♯j,σ ( x ′ , x ) (cid:17) b j ( t, x ) , for all ( t, x ) ∈ Q , from where we see that b j,σ is solution to the transport equation ( ∂ t + θ · ∇ x ′ + iθ · A ♯j,σ ) b j,σ = 0 , in Q, σ ∈ R ∗ + , j = 1 , . (3.20)We turn now to building suitable GO solutions to the magnetic Schrödinger equation of (1.1).3.2. Building GO solutions to magnetic Schrödinger equations.
For j = 1 , , we seek GO solutionsto the magnetic Schrödinger equation of (1.1) with ( A, q ) replaced by ( A j , q j ) , obeying (3.3)–(3.6), wherethe function A ♯j,σ , appearing in (3.4), is the smooth magnetic potential described by Lemma 3.3. Thisrequires that the functions Φ j , appearing in 3.3, be preliminarily defined more explicitly. To do that, we set B (0 , r ) := { x ′ ∈ R ; | x ′ | < r } for all r > , and take R > so large that ˜ ω ⊂ B (0 , R − , where ˜ ω isthe same as in Lemma 3.1. Next, we pick φ j ∈ C ∞ ( R ) , such that supp φ j ( · , x ) ⊂ D R := B (0 , R + 1) \ B (0 , R ) , x ∈ R , (3.21)and put Φ j ( t, x ) := φ j ( x ′ − tθ, x ) , ( t, x ) ∈ R × R . (3.22)It is apparent from (3.21) and the embedding ω ⊂ B (0 , R − , that supp φ j ( · , x ) ∩ ω = ∅ , x ∈ R , (3.23)and from (3.22), that Φ is solution to the transport equation (3.5).In the sequel, we choose σ > σ ∗ := ( R + 1) /T , in such a way that supp Φ j ( ± σt, · , x ) ∩ ω = supp φ j ( · ∓ σtθ, x ) ∩ ω = ∅ , ( t, x ) ∈ [ T, + ∞ ) × R . (3.24)Notice that upon possibly enlarging R , we may assume that σ ∗ > , which will always be the case in theremaining part of this text.Next, for k ∈ N , we introduce the following subspace of H k ( R ) , H kθ := { φ ∈ H k ( R ); θ · ∇ x ′ · φ ∈ H k ( R ) and supp φ ( · , x ) ⊂ D R for a.e. x ∈ R } , endowed with the norm N k,θ ( φ ) := k φ k H k ( R ) + k θ · ∇ x ′ φ k H k ( R ) , φ ∈ H θ . (3.25)For notational simplicity, we put N θ,σ ( φ ) := N ,θ ( φ ) + σ / N ,θ ( φ ) . (3.26)The coming statement claims existence of GO solutions u j,σ , expressed by (3.3), with L (0 , T ; H k (Ω)) -norm of correction term ψ j,σ bounded by N θ,σ ( φ j ) /σ − k for k = 0 , . Proposition 3.1.
Let
M > , and let A j ∈ W , ∞ (Ω , R ) and q j ∈ W , ∞ (Ω , R ) , j = 1 , , satisfy (3.1) - (3.2) . Then, for all σ > σ ∗ , there exists u j,σ ∈ C ([0 , T ] , L (Ω)) ∩ C ([0 , T ] , H (Ω)) obeying (3.3) - (3.6) ,where Φ j is defined by (3.21) - (3.22) , such that we have (cid:0) i∂ t + ∆ A j + q j (cid:1) u j,σ = 0 in Q, and the correction term satisfies ψ j,σ = 0 on Σ , for j = 1 , , and ψ ,σ ( T, · ) = ψ ,σ (0 , · ) = 0 in Ω .Moreover, the following estimate σ k ψ j,σ k L ( Q ) + k∇ ψ j,σ k L ( Q ) C N θ,σ ( φ j ) , j = 1 , , (3.27) holds for some constant C > depending only on T , ω , and M , where the function φ j ∈ C ∞ ( R ) fulfills (3.21) .Proof. We prove the result for j = 2 , the case j = 1 being obtained in the same way.In light of (3.3)–(3.5) and the identity ( i∂ t + ∆ A + q ) u ,σ = 0 imposed on u ,σ in Q , we seek a solution ψ ,σ to the following IBVP ( i∂ t + ∆ A + q ) ψ ,σ = g σ , in Q,ψ (0 , · ) = 0 , in Ω ,ψ = 0 , on Σ , (3.28)where g σ := − ( i∂ t + ∆ A + q ) ( w σ ϕ σ ) , with w σ ( t, x ′ ) := e iσ ( x ′ · θ − σt ) and ϕ σ ( t, x ) := ϑ σ (2 σt, x ) , where ϑ σ := Φ b ,σ . (3.29)Next, taking into account that ( i∂ t + ∆ A + q ) w σ = ( i ∇ · A − | A | − σθ · A ♯ + q ) w σ , and recallingfrom (3.5) and (3.20) that i ( ∂ t + 2 σθ · ∇ x ′ ) ϕ σ = 2 σθ · A ♯ ,σ ϕ σ , we get by straightforward computations that g σ ( t, x ) = − w σ ( t, x ) X m =0 , g m,σ (2 σt, x ) , with g ,σ := (∆ A + q ) ϑ σ , g ,σ := 2 σθ · ( A ♯ ,σ − A ♯ ) ϑ σ . (3.30)As g σ ∈ W , (0 , T ; L (Ω)) , by (3.21)-(3.22), we know from Lemma 2.2 that (3.28) admits a uniquesolution ψ ,σ ∈ C ([0 , T ] , L (Ω)) ∩ C ([0 , T ] , H (Ω) ∩ H (Ω)) . Moreover, since ψ ,σ ( t, x ) = − i Z t e − i ( t − s ) H A ,q g σ ( s, x ) d s, ( t, x ) ∈ Q, where H A ,q is the self-adjoint operator acting in L (Ω) , which is defined in Subsection 2.1, we have k ψ ,σ ( t, · ) k L (Ω) Z t k e − i ( t − s ) H A ,q g σ ( s, · ) k L (Ω) d s k g σ k L (0 ,T ; L (Ω)) , N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 17 uniformly in t ∈ (0 , T ) . This entails k ψ ,σ k L ( Q ) T / k g σ k L (0 ,T ; L (Ω)) , which together with (3.30),yields k ψ ,σ k L ( Q ) T / X m =0 , Z T k g m,σ (2 σt, · ) k L (Ω) d t σ − T / X m =0 , k g m,σ k L ( R ,L (Ω)) . (3.31)We are left with the task of bounding each term k g m,σ k L ( R ,L (Ω)) , for m = 0 , , separately. We start with m = 0 , and obtain k g ,σ k L ( R ,L (Ω)) = Z R k (∆ A + q )(Φ b ,σ )( s, · ) k L (Ω) d s C k b ,σ k W , ∞ ( R × Ω) k φ k H ( R ) C k φ k H ( R ) , (3.32)by combining estimate (3.19) with definitions (3.21)-(3.22) and (3.30). Next, applying (3.17) with k = 0 ,we get that k g ,σ k L ( R ,L (Ω)) Cσ k A ♯ ,σ − A ♯ k L ∞ (Ω) k φ k L ( R ) Cσ / k φ k L ( R ) , (3.33)Putting (3.31)–(3.33) together, and recalling (3.25), we find that σ k ψ ,σ k L ( Q ) C (cid:16) k φ k H ( R ) + σ / k φ k L ( R ) (cid:17) C (cid:16) N ,θ ( φ ) + σ / N ,θ ( φ ) (cid:17) . (3.34)It remains to bound k∇ ψ ,σ k L ( R ,L (Ω)) from above. To do that, we apply [10, Lemma 3.2], which ispermitted since g σ (0 , · ) = 0 , with ε = σ − , getting k∇ ψ ,σ ( t, · ) k L (Ω) C (cid:0) σ k g σ k L (0 ,T ; L (Ω)) + σ − k ∂ t g σ k L (0 ,T ; L (Ω)) (cid:1) C X m =0 , (cid:18) σ Z T k g m,σ (2 σt, · ) k L (Ω) d t + Z T k ∂ t g m,σ (2 σt, · ) k L (Ω) d t (cid:19) C X m =0 , (cid:0) k g m,σ k L ( R ,L (Ω)) + k ∂ t g m,σ k L ( R ,L (Ω)) (cid:1) , (3.35)for every t ∈ (0 , T ) , according to (3.29)-(3.30). Further, as we have ∂ t g ,σ ( t, x ) = − (∆ A + q ) θ · (cid:16) ∇ x ′ φ + iA ♯ ,σ φ (cid:17) ( x ′ − tθ, x ) b ,σ ( t, x ) , for a.e. ( t, x ) ∈ R × Ω , by direct computation, we obtain k ∂ t g ,σ k L ( R ,L (Ω)) CN ,θ ( φ ) , (3.36)from (3.4), (3.18) with k = 2 , (3.19) with j = 2 , (3.21)-(3.22) and (3.29)-(3.30). Similarly, as ∂ t g ,σ ( t, x ) = − σθ · ( A ♯ − A ♯ ,σ )( x ) θ · (cid:16) ∇ x ′ φ + iA ♯ ,σ φ (cid:17) ( x ′ − tθ, x ) b ,σ ( t, x ) , for a.e. ( t, x ) ∈ R × Ω , we find that k ∂ t g ,σ k L ( R ,L (Ω)) Cσ k A ♯ − A ♯ ,σ k L ∞ (Ω) N ,θ ( φ ) Cσ / N ,θ ( φ ) , (3.37)according to (3.17) with j = 2 and k = 0 . Thus, we infer from (3.32)-(3.33) and (3.35)–(3.37), that k∇ ψ ,σ ( t, · ) k L (Ω) C (cid:16) N ,θ ( φ ) + σ / N ,θ ( φ ) (cid:17) , t ∈ (0 , T ) , σ > σ ∗ . This and (3.34) yield (3.27) with j = 2 , upon recalling the definition (3.26). (cid:3) Let us now establish for further use that we may substitute σ − / u j,σ for ψ j,σ in the estimate (3.27). Corollary 3.4.
For j = 1 , , let q j , A j , φ j , and u j,σ , be the same as in Proposition 3.1. Then, there exists aconstant C > , depending only on T , ω and M , such that the estimate σ k u j,σ k L ( Q ) + k∇ u j,σ k L ( Q ) Cσ / N θ,σ ( φ j ) , j = 1 , , (3.38) holds for all σ > σ ∗ .Proof. Notice from (3.22) and (3.24) that Z T k Φ j (2 σt, · ) k H k (Ω) d t = Z + ∞ k Φ j (2 σt, · ) k H k (Ω) d t = (2 σ ) − Z R k Φ j ( s, · ) k H k (Ω) d s, so we have k Φ j (2 σ · , · ) k L (0 ,T ; H k (Ω)) R / σ − / k φ j k H k ( R ) , j = 1 , , k ∈ N . (3.39)From this, (3.3), (3.19) and (3.25)–(3.27), it follows for each j = 1 , , that k u j,σ k L ( Q ) k b j,σ k L ∞ ( R × Ω) k Φ j (2 σ · , · ) k L ( Q ) + k ψ j,σ k L ( Q ) C (cid:16) σ − / k φ j k L ( R ) + σ − N θ,σ ( φ j ) (cid:17) Cσ − / N θ,σ ( φ j ) , and k∇ u j,σ k L ( Q ) k b j,σ k W , ∞ ( R × Ω) (cid:0) σ k Φ j (2 σ · , · ) k L ( Q ) + k Φ j (2 σ · , · ) k L (0 ,T ; H (Ω)) (cid:1) + k∇ ψ j,σ k L ( Q ) C (cid:16) σ / k φ j k L ( R ) + σ − / k φ j k H ( R ) + N θ,σ ( φ j ) (cid:17) Cσ / N θ,σ ( φ j ) , which yields (3.38). (cid:3) In the coming subsection we probe the medium with the GO solutions described in Proposition 3.1 inorder to upper bound the transverse magnetic potential in terms of suitable norm of the DN map.3.3.
Probing the medium with GO solutions.
Let us introduce ˜ A ♯ := ˜ A ♯ − ˜ A ♯ , and A ♯σ := A ♯ ,σ − A ♯ ,σ , σ > , (3.40)where the functions ˜ A ♯j and A ♯j,σ , j = 1 , , are defined in Lemma 3.1 and Lemma 3.3, respectively. Evi-dently, ˜ A ♯ is the function A ♯ − A ♯ , extended by zero outside Ω , and we have k A ♯σ − ˜ A ♯ k W , ∞ ( R ) X j =1 , k A ♯j,σ − ˜ A ♯j k W , ∞ ( R ) Cσ − / , σ > , (3.41)from (3.17) with k = 1 . Thus, writing A ♯σ = ( a ,σ , a ,σ ) and ˜ A ♯ = (˜ a , ˜ a ) , it follows readily from (3.16)that a i,σ ( x ) = Z R χ σ ( x − y ) (˜ a i ( y ) + ( x − y ) · ∇ ˜ a i ( y )) d y, x ∈ R , i = 1 , . (3.42)The main purpose of this subsection is the following technical result. Lemma 3.5.
Let
M > and θ ∈ S be fixed. For j = 1 , , let A j ∈ W , ∞ (Ω , R ) , q j ∈ W , ∞ (Ω , R ) ,obey (3.1) - (3.2) , and let φ j be defined by (3.21) . Then, for every σ > σ ∗ , there exists a constant C > ,depending only on T , ω , and M , such that we have σ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z (0 ,T ) × R θ · ˜ A ♯ ( x )( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x ′ d x d t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) C (cid:16) σ k Λ A ,q − Λ A ,q k + σ − / (cid:17) N θ,σ ( φ ) N θ,σ ( φ ) , (3.43) N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 19 where k · k stands for the usual norm in B ( H , (Σ) , L (Σ)) , and ˜ A ♯ is given by (3.40) .Proof. We proceed in two steps. The first step is to establish a suitable orthogonality identity for A := A − A and V := i ∇ · A − ( | A | − | A | ) + q − q , which is the key ingredient in the derivation of theestimate (3.43), presented in the second step. Step 1: Orthogonality identity.
We probe the system with the GO functions u j,σ , j = 1 , , given byProposition 3.1, and recall for further use that u j,σ ∈ C ([0 , T ] , L (Ω)) ∩ C ([0 , T ] , H (Ω)) is expressed by(3.3) and satisfies the following equation (cid:0) i∂ t + ∆ A j + q j (cid:1) u j,σ = 0 , in Q. (3.44)Since A ♯ ,σ ∈ W , ∞ (Ω) and φ ∈ C ∞ ( R ) , it follows readily from (3.3)-(3.4) and (3.22) that u ,σ − ψ ,σ ∈ C ∞ ([0 , T ] , W , ∞ (Ω)) . Thus, F := − ( i∂ t + ∆ A + q ) ( u ,σ − ψ ,σ ) ∈ W , (0 , T ; L (Ω)) , andthere is consequently a unique solution z ∈ C ([0 , T ] , L (Ω)) ∩ C ([0 , T ] , H (Ω) ∩ H (Ω)) to the IBVP ( i∂ t + ∆ A + q ) z = F, in Q,z (0 , · ) = 0 , in Ω ,z = 0 , on Σ , (3.45)in virtue of Lemma 2.2. Further, as ( u ,σ − ψ ,σ )(0 , · ) = 0 in Ω , by (3.22)-(3.23), we infer from (3.45) that v := z + u ,σ − ψ ,σ ∈ C ([0 , T ] , L (Ω)) ∩ C ([0 , T ] , H (Ω)) verifies ( i∂ t + ∆ A + q ) v = 0 , in Q,v (0 , · ) = 0 , in Ω ,v = f σ , on Σ , (3.46)where we have set f σ ( t, x ) := u ,σ ( t, x ) = u ,σ ( t, x ) − ψ ,σ ( t, x ) = (Φ b ,σ )(2 σt, x ) e iσ ( x ′ · θ − σt ) , ( t, x ) ∈ Σ . (3.47)From this and Proposition 3.1, it then follows that w := v − u ,σ is the C ([0 , T ] , L (Ω)) ∩C ([0 , T ] , H (Ω) ∩ H (Ω)) -solution to the IBVP ( i∂ t + ∆ A + q ) w = 2 iA · ∇ u ,σ + V u ,σ , in Q,w (0 , · ) = 0 , in Ω ,w = 0 , on Σ , (3.48)In light of (3.48), we deduce from (3.44) with j = 1 , upon applying the Green formula, that h iA · ∇ u ,σ + V u ,σ , u ,σ i L ( Q ) = h ( i∂ t + ∆ A + q ) w, u ,σ i L ( Q ) = h ( ∂ ν + iA · ν ) w, u ,σ i L (Σ) . (3.49)Next, taking into account that A = A on ∂ Ω , from (3.2), we see that ( ∂ ν + iA · ν ) w = ( ∂ ν + iA · ν ) v − ( ∂ ν + iA · ν ) u ,σ = ( ∂ ν + iA · ν ) v − ( ∂ ν + iA · ν ) u ,σ = (Λ A ,q − Λ A ,q ) f σ , according to (3.47) and the last line of (3.46). This and (3.49) yield the following orthogonality identity i h A · ∇ u ,σ , u ,σ i L ( Q ) + h V u ,σ , u ,σ i L ( Q ) = h (Λ A ,q − Λ A ,q ) f σ , g σ i L (Σ) , (3.50) with g σ ( t, x ) := u ,σ ( t, x ) = u ,σ ( t, x ) − ψ ,σ ( t, x ) = (Φ b ,σ )(2 σt, x ) e iσ ( x ′ · θ − σt ) , ( t, x ) ∈ Σ . (3.51)Having established (3.50), we turn now to proving the estimate (3.43). Step 2: Derivation of (3.43) . In light of (3.3), we have h A · ∇ u ,σ , u ,σ i L ( Q ) = I σ + iσ Z Q θ · A ♯ ( x )(Φ Φ )(2 σt, x )( b ,σ b ,σ )(2 σt, x ) d x d t, (3.52)with I σ := Z Q A · ∇ (Φ b ,σ )(2 σt, x ) (cid:16) (Φ b ,σ )(2 σt, x ) + e iσ ( x ′ · θ − σt ) ψ ,σ ( t, x ) (cid:17) d x d t + Z Q A · ∇ ψ ,σ ( t, x ) (cid:16) e iσ ( x ′ · θ − σt ) (Φ b ,σ )(2 σt, x ) + ψ ,σ ( t, x ) (cid:17) d x d t + iσ Z Q θ · A ♯ ( x )(Φ b ,σ )(2 σt, x ) ψ ,σ ( t, x ) e iσ ( x ′ · θ − σt ) d x d t. We infer from (3.19), (3.27), and (3.39), that | I σ | Cσ − / N θ,σ ( φ ) N θ,σ ( φ ) , σ > σ ∗ . Putting this together with (3.50) and (3.52), we find that σ (cid:12)(cid:12)(cid:12)(cid:12)Z Q θ · A ♯ ( x )( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x ′ d x d t (cid:12)(cid:12)(cid:12)(cid:12) C (cid:16)(cid:12)(cid:12) h V u ,σ , u ,σ i L ( Q ) (cid:12)(cid:12) + (cid:12)(cid:12) h (Λ A ,q − Λ A ,q ) f σ , g σ i L (Σ) (cid:12)(cid:12) + σ − / N θ,σ ( φ ) N θ,σ ( φ ) (cid:17) . (3.53)Next, we notice from (3.38) that (cid:12)(cid:12) h V u ,σ , u ,σ i L ( Q ) (cid:12)(cid:12) Cσ − / N θ,σ ( φ ) N θ,σ ( φ ) . (3.54)Moreover, in view of (3.47) and (3.51), we have (cid:12)(cid:12) h (Λ A ,q − Λ A ,q ) f σ , g σ i L (Σ) (cid:12)(cid:12) k Λ A ,q − Λ A ,q kk f σ k H , (Σ) k g σ k L (Σ) k Λ A ,q − Λ A ,q kk u ,σ − ψ ,σ k H , (Σ) k u ,σ − ψ ,σ k L (Σ) , with k u ,σ − ψ ,σ k L (Σ) k u ,σ − ψ ,σ k L (0 ,T ; H (Ω)) Cσ k Φ (2 σ · , · ) k L (0 ,T ; H (Ω)) k b ,σ k W , ∞ ( R × Ω) Cσ / N θ,σ ( φ ) , and k u ,σ − ψ ,σ k H , (Σ) C (cid:0) k u ,σ − ψ ,σ k H (0 ,T ; H (Ω)) + k u ,σ − ψ ,σ k L (0 ,T ; H (Ω) (cid:1) Cσ k Φ (2 σ · , · ) k L (0 ,T ; H (Ω)) (cid:0) k b ,σ k W , ∞ ( R × Ω) + k ∂ t b ,σ k W , ∞ ( R × Ω) (cid:1) Cσ / N θ,σ ( φ ) , according to (3.3), (3.19), (3.25), and (3.39). As a consequence, we have (cid:12)(cid:12) h (Λ A ,q − Λ A ,q ) f σ , g σ i L (Σ) (cid:12)(cid:12) Cσ k Λ A ,q − Λ A ,q kN θ,σ ( φ ) N θ,σ ( φ ) . (3.55)This and (3.53)-(3.54) yield (3.43). (cid:3) N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 21
4. P
RELIMINARY ESTIMATES X -ray transform. In this subsection we estimate the partial X -ray transform in R , of the functions ˜ ρ j ( x ′ , x ) := θ · ∂ ˜ A ♯ ∂x j ( x ) = X i =1 , θ i ∂ ˜ a i ∂x j ( x ) , x ∈ R , j = 1 , , , (4.1)in terms of the DN map. We recall that the partial X -ray transform in the direction θ ∈ S , of a function f ∈ X := { ϕ ∈ L ( R ); x ′ ϕ ( x ′ , x ) ∈ L ( R ) for a.e. x ∈ R } , (4.2)is defined as P ( f )( θ, x ′ , x ) := Z R f ( x ′ + sθ, x ) d s, x ′ ∈ R , x ∈ R . (4.3)The X -ray transform stability estimate is as follows. Lemma 4.1.
Let
M > , and let A j and q j , for j = 1 , , be as in Proposition 3.1. Then, there exists aconstant C > , depending only on T , ω and M , such that for all θ ∈ S , all ξ ′ ∈ R , and all φ ∈ C ∞ ( R ) satisfying supp φ ( · , x ) ⊂ D − R ( θ ) := { x ′ ∈ D R , x ′ · θ } for every x ∈ R , the estimate (cid:12)(cid:12)(cid:12)(cid:12)Z R φ ( x ) P (˜ ρ j )( θ, x ′ , x ) exp (cid:18) − i Z R θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) d x (cid:12)(cid:12)(cid:12)(cid:12) C (cid:16) σ k Λ A ,q − Λ A ,q k + σ − / (cid:17) N θ,σ ( φ ) N θ,σ ( ∂ x j φ ) , (4.4) holds uniformly in σ > σ ∗ and j = 1 , , .Proof. Let φ j ∈ C ∞ ( R ) , j = 1 , , be supported in D R × R . Bearing in mind that ˜ ω ⊂ B (0 , R − , we inferfrom (3.42) that ˜ A ♯ and A ♯σ are both supported in B (0 , R ) × R . Further, as | x ′ − σtθ | > σ ∗ T − R > R + 1 for all x ′ ∈ B (0 , R ) and t > T , we see that ˜ A ♯ ( x )( φ φ )( x ′ − σtθ, x ) = A ♯σ ( x )( φ φ )( x ′ − σtθ, x ) = 0 , x = ( x ′ , x ) ∈ R , t > T, As a consequence we have Z T Z R θ · (cid:16) ˜ A ♯ ( x ) − A ♯σ ( x ) (cid:17) ( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x d t = Z + ∞ Z R θ · (cid:16) ˜ A ♯ ( x ) − A ♯σ ( x ) (cid:17) ( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x d t. Next, making the substitution s = σt in the above integral, we get that σ (cid:12)(cid:12)(cid:12)(cid:12)Z T Z R θ · (cid:16) ˜ A ♯ ( x ) − A ♯σ ( x ) (cid:17) ( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x d t (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)Z + ∞ Z B (0 ,R ) × R θ · (cid:16) ˜ A ♯ ( x ) − A ♯σ ( x ) (cid:17) ( φ φ )( x ′ − sθ, x )( b ,σ b ,σ )(2 s, x ) d x d s (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k ˜ A ♯ − A ♯σ k L ∞ ( R ) Z R +10 Z B (0 ,R ) × R (cid:12)(cid:12) ( φ φ )( x ′ − sθ, x ) (cid:12)(cid:12) d x d s k ˜ A ♯ − A ♯σ k L ∞ ( R ) Z R +10 Z R (cid:12)(cid:12) ( φ φ )( x ′ − sθ, x ) (cid:12)(cid:12) d x d s ( R + 1) k ˜ A ♯ − A ♯σ k L ∞ ( R ) k φ k L ( R ) k φ k L ( R ) . From this and (3.17) with k = 0 , and (3.25), it follows that σ (cid:12)(cid:12)(cid:12)(cid:12)Z T Z R θ · (cid:16) ˜ A ♯ ( x ) − A ♯σ ( x ) (cid:17) ( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x d t (cid:12)(cid:12)(cid:12)(cid:12) Cσ − / k φ k L ( R ) k φ k L ( R ) Cσ − / N θ,σ ( φ ) N θ,σ ( φ ) . (4.5)On the other hand, since ( b ,σ b ,σ )(2 σt, x ′ + 2 σtθ, x ) = exp (cid:18) − i Z σt θ · A ♯σ ( x ′ + (2 σt − s ) θ, x ) d s (cid:19) = exp (cid:18) − i Z σt θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) , for a.e. ( t, x ) ∈ (0 , T ) × R , we have σ Z T Z R θ · A ♯σ ( x )( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x ′ d x d t = σ Z T Z R θ · A ♯σ ( x ′ + 2 σtθ, x )( φ φ )( x )( b ,σ b ,σ )(2 σt, x ′ + 2 σtθ, x ) d x ′ d x d t = Z R ( φ φ )( x ) (cid:18)Z T σθ · A ♯σ ( x ′ + 2 σtθ, x ) exp (cid:18) − i Z σt θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) d t (cid:19) d x ′ d x = i Z R ( φ φ )( x ) (cid:18)Z T dd t exp (cid:18) − i Z σt θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) d t (cid:19) d x ′ d x = i Z R ( φ φ )( x ) (cid:18) exp (cid:18) − i Z σT θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) − (cid:19) d x ′ d x . (4.6)As A ♯σ is supported in B (0 , R ) × R and | x ′ + sθ | > σ ∗ T − ( R + 1) > R , for all x ′ ∈ D R and all s > σT ,we have Z σT θ · A ♯σ ( x ′ + sθ, x ) d s = Z + ∞ θ · A ♯σ ( x ′ + sθ, x ) d s, x ′ ∈ D R , x ∈ R . (4.7)Similarly, as | x ′ + sθ | = | x ′ | + s + 2 sx ′ · θ > R , for every x ′ ∈ D − R ( θ ) and s < , we have Z −∞ θ · A ♯σ ( x ′ + sθ, x ) d s = 0 , x ′ ∈ D − R ( θ ) , x ∈ R . This and (4.7) entail Z σT θ · A ♯σ ( x ′ + sθ, x ) d s = Z R θ · A ♯σ ( x ′ + sθ, x ) d s, x ′ ∈ D − R ( θ ) , x ∈ R . (4.8)Having seen this, we take φ := ∂ x j φ , for j = 1 , , , and φ := φ , in (4.6), and find σ Z T Z R θ · A ♯σ ( x )( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x ′ d x d t = i Z R ∂ x j φ ( x ) (cid:18) exp (cid:18) − i Z σT θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) − (cid:19) d x ′ d x = − Z R φ ( x ) (cid:18)Z σT θ · ∂ x j A ♯σ ( x ′ + sθ, x ) d s (cid:19) exp (cid:18) − i Z σT θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) d x, (4.9) N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 23 upon integrating by parts. Taking into account that φ is supported in D − R ( θ ) × R , we deduce from (4.8) and(4.9), that σ Z T Z R θ · A ♯σ ( x )( φ φ )( x ′ − σtθ, x )( b ,σ b ,σ )(2 σt, x ) d x ′ d x d t = − Z R φ ( x ) (cid:18)Z R θ · ∂ x j A ♯σ ( x ′ + sθ, x ) d s (cid:19) exp (cid:18) − i Z R θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) d x ′ d x = − Z R φ ( x ) P ( ρ j,σ )( θ, x ′ , x ) exp (cid:18) − i Z R θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) d x ′ d x . (4.10)Here we used (4.3) and the notation ρ j,σ ( x ) := θ · ∂ x j A ♯σ ( x ) = X i =1 , θ i ∂ x j a i,σ ( x ) , x ∈ R , j = 1 , , . Finally, using once more that the functions A ♯σ and ˜ A ♯ are supported in B (0 , R ) , we infer from (3.41) and(4.1)-(4.3), that (cid:12)(cid:12) ( P ( ρ j,σ ) − P (˜ ρ j )) ( θ, x ′ , x ) (cid:12)(cid:12) Cσ − / , ( x ′ , x ) ∈ B (0 , R ) × R , for some positive constant C , depending only on ω and M . This entails that (cid:12)(cid:12)(cid:12)(cid:12)Z R φ ( x ) ( P ( ρ j,σ ) − P (˜ ρ j )) ( θ, x ′ , x ) exp (cid:18) − i Z R θ · A ♯σ ( x ′ + sθ, x ) d s (cid:19) d x ′ d x (cid:12)(cid:12)(cid:12)(cid:12) Cσ − / k φ k L ( R ) , which, together with (3.43),(4.5), and (4.10), yields (4.4). (cid:3) As will be seen in the coming section, the result of Lemma 4.1 is a key ingredient in the estimation of thepartial Fourier transform of the aligned magnetic field, in terms of the DN map. To this purpose, we recallfor all f ∈ X , where X is defined in (4.2), that the partial Fourier transform with respect to x ′ ∈ R of f ,expresses as b f ( ξ ′ , x ) := (2 π ) − Z R f ( x ′ , x ) e − ix ′ · ξ ′ d x ′ , ξ ′ ∈ R , x ∈ R . (4.11)Further, setting θ ⊥ := { x ′ ∈ R ; x ′ · θ = 0 } , we recall for further use from [9, Lemma 6.1], that x ′ ( f )( θ, x ′ , x ) ∈ L ( θ ⊥ ) for a.e. x ∈ R , and that [ P ( f )( θ, ξ ′ , x ) := (2 π ) − / Z θ ⊥ P ( f )( θ, x ′ , x ) e − ix ′ · ξ ′ d x ′ = (2 π ) / b f ( ξ ′ , x ) , ξ ′ ∈ θ ⊥ , x ∈ R . (4.12)4.2. Aligned magnetic field estimation.
Let us now estimate the Fourier transform of the aligned magneticfield ˜ β ( x ) := ( ∂ x ˜ a − ∂ x ˜ a ) ( x ) , x ∈ R , (4.13)with the aid of Lemma 4.1. More precisely, we aim to establish the following result. Lemma 4.2.
Let
M > , and let A j and q j , for j = 1 , , be as in Proposition 3.1. Then, there exist twoconstants ǫ ∈ (0 , and C > , both of them depending only on T , ω , and M , such that the estimates k b β ( ξ ′ , · ) k L ∞ ( R ) C h ξ ′ i (cid:0) σ k Λ A ,q − Λ A ,q k + σ − ǫ (cid:1) , (4.14) and k ∂ x b β ( ξ ′ , · ) k L ∞ ( R ) C h ξ ′ i (cid:0) σ k Λ A ,q − Λ A ,q k + σ − ǫ (cid:1) , (4.15) hold for all σ > σ ∗ and all ξ ′ ∈ R , with h ξ ′ i := (cid:0) | ξ ′ | (cid:1) / . Proof.
We shall only prove (4.14), the derivation of (4.15) being obtained in a similar fashion.Fix θ ∈ S ∩ ξ ′⊥ . We first introduce the following partition of B (0 , R ) ∩ θ ⊥ . For N ∈ N ∗ := { , , . . . } fixed, we pick x ′ , . . . , x ′ N in B (0 , R + 1 / ∩ θ ⊥ , and choose ϕ , . . . , ϕ N in C ∞ ( R , [0 , , such that supp ϕ k ⊂ B ( x ′ k , / ∩ θ ⊥ for k = 1 , . . . , N, and N X k =1 ϕ k ( x ′ ) = 1 for x ′ ∈ B (0 , R ) ∩ θ ⊥ . (4.16)Next, we set r x ′ k := (cid:16) ( R + 3 / − | x ′ k | (cid:17) / , in such a way that B ( x ′ k − r x ′ k θ, / ⊂ D − R ( θ ) , k = 1 , . . . , N. (4.17)In order to define a suitable set of test functions φ ∗ ,k , k = 1 , . . . , N , we fix x ∈ R , pick a function α ∈ C ∞ ( R , R + ) which is supported in ( − , and normalized in L ( R ) , and put α σ ( s ) := σ µ α (cid:0) σ µ ( x − s ) (cid:1) , s ∈ R , (4.18)for some positive real parameter µ , we shall make precise below. Then, the test function φ ∗ ,k is defined forall y = ( y ′ , y ) ∈ R , by φ ∗ ,k ( y ) := h (cid:16) y ′ · θ + r x ′ k (cid:17) e − i y ′ · ξ ′ ϕ / k ( y ′ − ( y ′ · θ ) θ ) exp (cid:18) i Z R θ · A ♯σ ( y ′ + sθ, y ) d s (cid:19) α σ ( y ) , (4.19)where h ∈ C ∞ ( R ) is supported in (0 , / , and normalized in L ( R ) .For every y ′ ∈ R \ B ( x ′ k − r x ′ k θ, / , it is easily seen from the basic inequality | y ′ − ( x ′ k − r x ′ k θ ) | | y ′ − ( y ′ · θ ) θ − x ′ k | + | y ′ · θ + r x ′ k | , that either of the two real numbers | y ′ − ( y ′ · θ ) θ − x ′ k | or | y ′ · θ + r x ′ k | is greater than / , and hence that h ( y ′ · θ + r x ′ k ) ϕ / k ( y ′ − ( y ′ · θ ) θ ) = 0 . As a consequence, we have supp φ ∗ ,k ( · , y ) ⊂ B (cid:16) x ′ k − r x ′ k θ, / (cid:17) ⊂ D − R ( θ ) , y ∈ R , k = 1 , . . . , N, (4.20)directly from (4.17) and (4.19). Moreover, since θ · ∇ y ′ (cid:18)Z R θ · A ♯σ ( y ′ + sθ, y ) d s (cid:19) = θ · Z R dd s A ♯σ ( y ′ + sθ, y ) d s = 0 , ( y ′ , y ) ∈ R , we derive from Lemma 3.3 for all m ∈ N , that (cid:10) ξ ′ (cid:11) k φ ∗ ,k k H m ( R ) + k θ · ∇ x ′ φ ∗ ,k k H m ( R ) C (cid:10) ξ ′ (cid:11) m +1 σ µm +max(0 , ( m − / , where C is a positive constant, independent of σ . Therefore, we have N ,θ ( φ ∗ ,k ) C h ξ ′ i and N ,θ ( φ ∗ ,k ) C h ξ ′ i σ µ , whence N θ,σ ( φ ∗ ,k ) C (cid:10) ξ ′ (cid:11) σ µ +1 / . (4.21)Similarly, we find that (cid:10) ξ ′ (cid:11) k ∂ x j φ ∗ ,k k H m ( R ) + k θ · ∇ x ′ ∂ x j φ ∗ ,k k H m ( R ) C (cid:10) ξ ′ (cid:11) m +2 σ µm +max(0 , ( m − / , j = 1 , , and (cid:10) ξ ′ (cid:11) k ∂ x φ ∗ ,k k H m ( R ) + k θ · ∇ x ′ ∂ x φ ∗ ,k k H m ( R ) C (cid:10) ξ ′ (cid:11) m +1 σ µ ( m +1)+max(0 , ( m − / . Thus, we have N ,θ ( ∂ x j φ ∗ ,k ) C h ξ ′ i σ µ and N ,θ ( ∂ x j φ ∗ ,k ) C h ξ ′ i σ µ +1 / , for j = 1 , , , andconsequently N θ,σ ( ∂ x j φ ∗ ,k ) C (cid:10) ξ ′ (cid:11) σ µ +1 / , j = 1 , , . N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 25 according to (3.25). From this and (4.21), it then follows that N θ,σ ( φ ∗ ,k ) N θ,σ ( ∂ x j φ ∗ ,k ) C (cid:10) ξ ′ (cid:11) σ µ +2 / , j = 1 , , . (4.22)Having seen this, we turn now to estimating b ˜ ρ j , where ˜ ρ j is defined by (4.1). As A ♯σ ∈ W ∞ , ∞ ( R , R ) ,we infer form (4.19) that φ ∗ ,k ∈ C ∞ ( R ) , and from (4.20) that supp φ ∗ ,k ⊂ D R × R . Thus, by performingthe change of variable y ′ = x ′ + tθ ∈ θ ⊥ ⊕ R θ , in the following integral, we deduce from (4.18)-(4.19) that Z R Z R φ ∗ ,k ( y ′ , y ) P (˜ ρ j )( θ, y ′ , y ) exp (cid:18) − i Z R θ · A ♯σ ( y ′ + sθ, y ) d s (cid:19) d y ′ d y = Z R Z R Z θ ⊥ φ ∗ ,k ( x ′ + tθ, y ) P (˜ ρ j )( θ, x ′ + tθ, y ) exp (cid:18) − i Z R θ · A ♯σ ( x ′ + sθ, y ) d s (cid:19) d x ′ d t d y = Z R Z R Z θ ⊥ h ( t + r x ′ k ) e − ix ′ · ξ ′ ϕ k ( x ′ ) α σ ( y ) P (˜ ρ j )( θ, x ′ , y ) d x ′ d t d y = Z R Z θ ⊥ e − iy ′ · ξ ′ ϕ k ( y ′ ) α σ ( y ) P (˜ ρ j )( θ, y ′ , y ) d y ′ d y . (4.23)Thus, taking µ > so small that κ := 1 / − µ > , we deduce from this, (4.4), and (4.22), that (cid:12)(cid:12)(cid:12)(cid:12)Z R Z θ ⊥ e − iy ′ · ξ ′ ϕ k ( y ′ ) α σ ( y ) P (˜ ρ j )( θ, y ′ , y ) d y ′ d y (cid:12)(cid:12)(cid:12)(cid:12) C h ξ ′ i (cid:0) σ k Λ A ,q − Λ A ,q k + σ − κ (cid:1) , x ∈ R . (4.24)Moreover, we see from (4.1) that ˜ ρ j ∈ C , ( R ) . Since supp ˜ ρ j ⊂ B (0 , R ) × R , by Lemma 3.1, then x
7→ P (˜ ρ j )( θ, x ) ∈ C , ( R ) , and we deduce from (3.18) upon making the substitution s = σ µ ( x − y ) in the following integral, that (cid:12)(cid:12)(cid:12)(cid:12)Z R Z θ ⊥ e − iy ′ · ξ ′ ϕ k ( y ′ ) P (˜ ρ j )( θ, y ′ , y ) α σ ( y ) d y ′ d y − Z θ ⊥ e − iy ′ · ξ ′ ϕ k ( y ′ ) P (˜ ρ j )( θ, y ′ , x ) d y ′ (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z R Z θ ⊥ e − iy ′ · ξ ′ ϕ k ( y ′ ) α ( s ) (cid:0) P (˜ ρ j )( θ, y ′ , x − σ − µ s ) − P (˜ ρ j )( θ, y ′ , x ) (cid:1) d y ′ d s (cid:12)(cid:12)(cid:12)(cid:12) Z − Z θ ⊥ ∩ B (0 ,R +1) α ( s ) (cid:12)(cid:12) P (˜ ρ j )( θ, y ′ , x − σ − µ s ) − P (˜ ρ j )( θ, y ′ , x ) (cid:12)(cid:12) d y ′ d s Cσ − µ , for some constant C > depending only on ω and M . Here, we used the fact that φ ∗ ,k and α are supportedin B (0 , R + 1) and ( − , , respectively. This and (4.24) yield (cid:12)(cid:12)(cid:12)(cid:12)Z θ ⊥ e − iy ′ · ξ ′ ϕ k ( y ′ ) P (˜ ρ j )( θ, y ′ , x ) d y ′ (cid:12)(cid:12)(cid:12)(cid:12) C h ξ ′ i (cid:0) σ k Λ A ,q − Λ A ,q k + σ − µ + σ − κ (cid:1) , (4.25)for all x ∈ R and k = 1 , ..., N . Further, as A ♯ is supported in B (0 , R ) × R by assumption, it holds truethat ∂ x j A ♯ ( y ′ + sθ, x ) = 0 for all s ∈ R , x ∈ R , and all y ′ ∈ θ ⊥ such that | y ′ | > R . Therefore, we have P (˜ ρ j )( θ, y ′ , x ) = 0 , y ′ ∈ θ ⊥ ∩ ( R \ B (0 , R )) , x ∈ R , in virtue of (4.1), and hence Z θ ⊥ e − iy ′ · ξ ′ P (˜ ρ j )( θ, y ′ , x ) d y ′ = Z θ ⊥ ∩ B (0 ,R ) e − iy ′ · ξ ′ P (˜ ρ j )( θ, y ′ , x ) d y ′ , x ∈ R . In light of (4.12) and (4.16), this entails that b ˜ ρ j ( ξ ′ , x ) = 12 π N X k =1 Z θ ⊥ ∩ B (0 ,R ) e − iy ′ · ξ ′ ϕ k ( y ′ ) P (˜ ρ j )( θ, y ′ , x ) d y ′ , x ∈ R . (4.26)Taking µ ∈ (0 , / , in such a way that we have κ > µ , we infer from (4.25)-(4.26) that k b ˜ ρ j ( ξ ′ , · ) k L ∞ ( R ) N X k =1 (cid:18) sup x ∈ R (cid:12)(cid:12)(cid:12)(cid:12)Z θ ⊥ e − iy ′ · ξ ′ ϕ k ( y ′ ) P (˜ ρ j )( θ, y ′ , x ) dy ′ (cid:12)(cid:12)(cid:12)(cid:12)(cid:19) C h ξ ′ i (cid:0) σ k Λ A ,q − Λ A ,q k + σ − µ (cid:1) , x ∈ R . (4.27)The last step of the proof is to notice from (4.1), (4.11), and the identity P m =1 , θ m ξ m = θ · ξ ′ = 0 , that b ˜ ρ j ( ξ ′ , x ) = i X m =1 , θ m ξ j b ˜ a m ( ξ ′ , x ) = i X m =1 , θ m (cid:16) ξ j b ˜ a m − ξ m b ˜ a j (cid:17) ( ξ ′ , x ) , x ∈ R , j = 1 , . Thus, assuming that ξ ′ = ( ξ , ξ ) ∈ R \ { } , we get from (4.13) upon choosing θ = ( ξ / | ξ ′ | , − ξ / | ξ ′ | ) ,that b ˜ ρ j ( ξ ′ , x ) = − ξ j | ξ ′ | b ˜ β ( ξ ′ , x ) , x ∈ R . From this and (4.27), it then follows that k b ˜ β ( ξ ′ , · ) k L ∞ ( R ) | ξ | + | ξ || ξ ′ | k b ˜ β ( ξ ′ , · ) k L ∞ ( R ) C h ξ ′ i (cid:0) σ k Λ A ,q − Λ A ,q k + σ − µ (cid:1) , which yields (4.14) for ξ ′ = 0 . Since b ˜ β (0 , x ) = 0 for every x ∈ R , from (4.13), then (4.14) holds for ξ ′ = 0 as well, and the proof is complete. (cid:3) Armed with Lemma 4.2, we turn now to proving the three main results of this paper.5. P
ROOF OF T HEOREMS
AND A ′ = ( a ′ i ) i ∈ A , and put A := ( a , a , , where a i ( x ′ , x ) := a ′ i ( x ′ , x ) − Z x −∞ ∂ x i a ′ ( x ′ , s ) d s, x = ( x ′ , x ) ∈ ω × R , i = 1 , . (5.1)Since a ′ ∈ C (Ω) fulfills (1.9)-(1.10), from the very definition of A , we have a ′ ∈ L x ( R , H ( ω )) ,where H ( ω ) denotes the closure of C ∞ ( ω ) in H ( ω ) . Thus, e ( x ) := R x −∞ a ′ ( x ′ , s ) d s lies in W , ∞ (Ω) ∩ L ∞ x ( R , H ( ω )) , and we deduce from the identity A = A ′ − ∇ e , arising from (5.1), thatd A ′ = d A, and Λ A ∗ + A ′ ,q = Λ A ∗ + A,q , for all A ∗ ∈ W , ∞ (Ω) and all q ∈ W , ∞ (Ω) . Moreover, it is easy to see that A obeys (1.9), in the sensethat we have ∂ αx A ( x ) = 0 , x ∈ ∂ Ω , α ∈ N , | α | . (5.2)Therefore, for each A ∗ ∈ W , ∞ (Ω , R ) and any A j ∈ A ∗ + A , j = 1 , , we may assume without loss ofgenerality, that the difference A − A reads A = ( a , a , , (5.3)and fulfills (5.2). We shall systematically do that in the sequel. For further reference, we put A ♯ := ( a , a ) ,where a j , j = 1 , , are extended by zero outside Ω . Summing up, we have N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 27
Proof of Theorem 1.2.
We establish the uniqueness result ( d A , q ) = ( d A , q ) in Subsection 5.1.1,while the proof of the stability estimate (1.13) can be found in Subsection 5.1.2.5.1.1. Uniqueness result.
For ξ ′ = ( ξ , ξ ) ∈ R \ { } , we set ξ ′⊥ := ( −| ξ ′ | − ξ , | ξ ′ | − ξ ) , and wedecompose A ♯ into the sum ( A ♯ · ξ ′ ) | ξ ′ | − ξ ′ + ( A ♯ · ξ ′⊥ ) ξ ′⊥ , in such a way that the partial Fourier transformof ∂ x A ♯ , reads ∂ x c A ♯ ( ξ ′ , x ) = (cid:18) π Z R e − ix ′ · ξ ′ ∂ x A ♯ ( x ′ , x ) · ξ ′ d x ′ (cid:19) ξ ′ | ξ ′ | + i ∂ x b β ( ξ ′ , x ) | ξ ′ | ξ ′⊥ , x ∈ R . (5.4)Next, recalling the hypothesis Λ A ,q = Λ A ,q , we get β = ∂ x a − ∂ x a = 0 in Ω , (5.5)upon sending σ to infinity in (4.14). Moreover, we have ∇ x ′ · ∂ x A ♯ = ∇ · ∂ x A = 0 , in virtue of (1.11),whence Z R e − ix ′ · ξ ′ ∂ x A ♯ ( x ′ , x ) · ξ ′ d x ′ = i Z R ∇ x ′ e − ix ′ · ξ ′ · ∂ x A ♯ ( x ′ , x )= − i Z R e − ix ′ · ξ ′ ∇ x ′ · ∂ x A ♯ ( x ′ , x ) d x ′ = 0 . (5.6)Putting this together with (5.4)-(5.5), we find that | ξ ′ | ∂ x c A ♯ ( ξ ′ , x ) = 0 for a.e. x ∈ R . Since ξ ′ is arbitraryin R \ { } , this entails that ∂ x A ♯ = 0 , and hence that ∂ x a = ∂ x a = 0 in R . From this, (5.5), and thefact that a is uniformly zero, it then follows that d A = d A .Further, taking into account that ∂ αx A = ∂ αx A = ∂ αx A ∗ on ∂ Ω , for every α ∈ N such that | α | ,we infer that A ∈ W , ∞ ( R , R ) . This and the identity d A = 0 , yield A = ∇ Ψ , where the function Ψ( x ) = R x · A ( tx ) d t lies in W , ∞ ( R , R ) . Moreover, since A vanishes in R \ Ω , we may assumeupon possibly adding a suitable constant, that the same is true for Ψ . Therefore, Ψ | ∂ Ω = 0 , and we find Λ A ,q = Λ A + ∇ Ψ ,q = Λ A ,q , by combining the identity A = A + ∇ Ψ with the gauge invarianceproperty of the DN map. From this and the assumption Λ A ,q = Λ A ,q , it then follows that Λ A ,q = Λ A ,q . (5.7)It remains to show that the function q = q − q , duly extended by zero outside Ω , is uniformly zero in R .This can be done upon applying the orthogonality identity (3.50) with A = A , i.e with A = 0 and V = q .In light of (5.7), we obtain that h qu ,σ , u ,σ i L ( Q ) = 0 , σ > σ ∗ . (5.8)Here u j,σ , for j = 1 , , is given by (3.3), and we have ( b ,σ b ,σ )( t, x ) = 1 for all ( t, x ) ∈ (0 , T ) × R , from(3.4), since A = A .Next, pick φ ∈ C ∞ ( R ) , with support in { x ∈ R ; | x | < } , and such that k φ k L ( R ) = 1 . We fix y ∈ D R ( θ ) × R , and choose δ > so small that φ ( x ) = φ ( x ) := δ − / φ ( δ − ( x − y )) is supported in D R × R . Thus, upon multiplying (5.8) by σ , and then sending σ to infinity, we find with the aid of (3.19)and (3.27), that Z + ∞ (cid:18)Z R q ( δx ′ + y ′ + sθ, δx + y ) | φ ( x ′ , x ) | d x ′ d x (cid:19) d s = 0 , δ > . (5.9) Actually, if y ′ ∈ D − R ( θ ) , then we have | y ′ + sθ | > R for any s , and hence q ( δx ′ + y ′ + sθ, δx + y ) = 0 ,uniformly in | x | < , provided δ ∈ (0 , . This and (5.9) yield that Z R (cid:18)Z R q ( δx ′ + y ′ + sθ, δx + y ) | φ ( x ′ , x ) | d x ′ d x (cid:19) d s = 0 , δ ∈ (0 , , ( y ′ , y ) ∈ D − R ( θ ) × R . (5.10)By performing the change of variable t = − s in the above integral, and then substituting ( − θ ) for θ in theresulting identity, we get that Z R (cid:18)Z R q ( δx ′ + y ′ + sθ, δx + y ) | φ ( x ′ , x ) | d x ′ d x (cid:19) d s = 0 , δ ∈ (0 , , ( y ′ , y ) ∈ D − R ( − θ ) × R . This and (5.10) yield that Z R (cid:18)Z R q ( δx ′ + y ′ + sθ, δx + y ) | φ ( x ′ , x ) | d x ′ d x (cid:19) d s = 0 , δ ∈ (0 , , ( y ′ , y ) ∈ D R × R . Next, sending δ to zero in the above identity, and taking into account that φ is normalized in L ( R ) , weobtain for each θ ∈ S , that P ( q )( θ, y ′ , y ) = Z R q ( y ′ + sθ, y ) d s = 0 , ( y ′ , y ) ∈ D R × R . This entails q = 0 , since the partial X-ray transform is injective.5.1.2. Proof of the stability estimate (1.13) . We have | ξ ′ | (cid:12)(cid:12)(cid:12) ∂ x c A ♯ ( ξ ′ , x ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) ∂ x b β ( ξ ′ , x ) (cid:12)(cid:12)(cid:12) , ξ ′ ∈ R , x ∈ R , by (5.4) and (5.6), so we infer from (4.14)-(4.15) that | b β ( ξ ′ , x ) | + | ξ ′ || ∂ x c A ♯ ( ξ ′ , x ) | C h ξ ′ i (cid:0) σ k Λ A ,q − Λ A ,q k + σ − ǫ (cid:1) , ξ ′ ∈ R , x ∈ R , (5.11)for all σ > σ ∗ , the constants C and ǫ being the same as in Lemma 4.2.Fix ρ ∈ (1 , + ∞ ) and put C ρ := { ξ ′ ∈ R ; ρ − | ξ ′ | ρ } . Then, upon applying the Planchereltheorem, we obtain k ∂ x a j ( · , x ) k L ( ω ) k ∂ x b a j ( ξ ′ , x ) k L ( B (0 ,ρ − )) + ρ − Z R \ B (0 ,ρ ) h ξ ′ i | ∂ x b a j ( ξ ′ , x ) | d ξ ′ + k ∂ x b a j ( ξ ′ , x ) k L ( C ρ ) , x ∈ R , j = 1 , . (5.12)Further, as k ∂ x b a j ( ξ ′ , x ) k L ( B (0 ,ρ − )) | ω | ρ − k a j k W , ∞ (Ω) and R R \ B (0 ,ρ ) h ξ ′ i | ∂ x b a j ( ξ ′ , x ) | d ξ ′ k a j k W , ∞ (Ω) , then there exists a constant C > , depending only on M and ω , such that we have k ∂ x b a j ( ξ ′ , x ) k L ( B (0 ,ρ − )) + ρ − Z R \ B (0 ,ρ ) h ξ ′ i | ∂ x b a j ( ξ ′ , x ) | d ξ ′ Mρ , (5.13)according to (1.12). On the other hand, we derive from (5.11) that | ∂ x b a j ( ξ ′ , x ) | Cρ ( σ δ + σ − ǫ ) , ξ ′ ∈ C ρ ∩ B (0 , ρ ) , x ∈ R , σ > σ ∗ , j = 1 , , where δ := k Λ A ,q − Λ A ,q k . Putting this and (5.12)-(5.13) together, we get for every σ > σ ∗ , that k ∂ x a j k L ∞ x ( R ,L ( ω )) C (cid:0) ρ σ δ + ρ σ − ǫ + ρ − (cid:1) , x ∈ R , j = 1 , . (5.14)Now, choosing ρ so large that ρ > σ ǫ/ ∗ , we get upon taking σ = ρ /ǫ > σ ∗ in (5.14), that k ∂ x a j ( ., x ) k L ( ω ) C (cid:0) ρ M ǫ δ + ρ − (cid:1) , x ∈ R , j = 1 , , (5.15) N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 29 where M ǫ := 16 + 96 /ǫ . Thus, if δ < δ := σ − ǫ ( M ǫ +2) / ∗ , then we have δ − / ( M ǫ +2) > σ ǫ/ ∗ , and we mayapply (5.15) with ρ = δ − / ( M ǫ +2) , getting k ∂ x a j k L ∞ x ( R ,L ( ω )) Cδ µ , with µ := 2 M ǫ + 2 ∈ (0 , , j = 1 , . (5.16)From this and the fact, arising from (1.12), that k ∂ x a j k L ∞ x ( R ,L ( ω )) (2 M δ − µ ) δ µ ∗ for all δ > δ , itthen follows that (5.16) remains valid for every δ > .Finally, arguing as before with β instead of ∂ x A ♯ , we obtain in a similar way from (5.11), that the norm k β k L ∞ x ( R ,L ( ω )) is upper bounded, up to some multiplicative constant depending only on M and ω , by δ µ ,and hence (1.13) follows from this and (5.16).5.2. Proof of Theorem 1.3.
The proof is an adaptation of the one of (1.2), where the adaptation is to takeinto account the extra information given by (1.14). Actually, since A j = ( a ,j , a ,j , a , ∗ ) and A = ( A ♯ , with A ♯ = ( a , a ) , by (5.3), then (1.14) yields k ∂ x a − ∂ x a k L ∞ x ( R ,L ( ω )) = k ∂ x a − ∂ x a k L ∞ x ( − r,r ; L ( ω )) , (5.17)and k ∂ x a j k L ∞ x ( R ,L ( ω )) = k ∂ x a j k L ∞ x ( − r,r ; L ( ω )) , j = 1 , . (5.18)More precisely, we still consider GO solutions u ,σ and u ,σ , defined by (3.3)-(3.4) and (3.22), with φ = ∂ x j φ , for j = 1 , , , and φ = φ , where φ is given by (4.18)-(4.19). The parameter x appearing in (4.18),is taken in ( − r, r ) , and we impose σ > ( r ′ − r ) − , in such a way that φ ∈ C ∞ ( D − R ( θ ) × ( − r ′ , r ′ )) .Moreover, the functions f σ ( t, x ) = Φ (2 σt, x ) b (2 σt, x ) e iσ ( x.θ − σt ) and g σ = Φ (2 σt, x ) b (2 σt, x ) e iσ ( x.θ − σt ) , ( t, x ) ∈ Σ , lie in H , ((0 , T ) × Γ r ′ ) , and we infer from (3.50) upon arguing as the derivation of Lemma 4.2, that k b β ( ξ ′ , · ) k L ∞ ( − r ′ ,r ′ ) C h ξ ′ i (cid:0) σ k Λ A ,q ,r ′ − Λ A ,q ,r ′ k + σ − ǫ (cid:1) , and that k ∂ x b β ( ξ ′ , · ) k L ∞ ( r ′ ,r ′ ) C h ξ ′ i (cid:0) σ k Λ A ,q ,r ′ − Λ A ,q ,r ′ k + σ − ǫ (cid:1) . for all ξ ′ ∈ R and some ǫ > . Here, the constant C depends only on ω , T , M , r , r ′ and ǫ . The desiredresult follows from this and (5.17)-(5.18) by arguing in the same way as in the proof of Theorem 1.2.5.3. Proof of Theorem 1.4.
We only prove (1.16), the derivation of (1.19) being quite similar to the oneof (1.15). To this end, we fix ξ ′ ∈ R , and remind that A = ( A ♯ , ∈ A , with A ♯ = ( a , a ) , satisfies ∂ x a + ∂ x a = 0 in R , so we get c A ♯ ( ξ ′ , x ) · ξ ′ = i (2 π ) − Z R A ♯ ( x ′ , x ) · ∇ x ′ e − ix ′ · ξ ′ d x ′ = − i (2 π ) − Z R e − ix ′ · ξ ′ ( ∂ x a + ∂ x a ) ( x ′ , x ) d x ′ = 0 , x ∈ R , upon integrating by parts. Thus, remembering that ξ ′⊥ = ( −| ξ ′ | − ξ , | ξ ′ | − ξ ) whenever ξ ′ = 0 , we obtain c A ♯ ( ξ ′ , x ) = ( c A ♯ ( ξ ′ , x ) · ξ ′⊥ ) ξ ′⊥ = ( − ξ b a + ξ b a )( ξ ′ , x ) ξ ′⊥ | ξ ′ | , x ∈ R , and consequently | ξ ′ | c A ♯ ( ξ ′ , x ) = − i b β ( ξ ′ , x ) , x ∈ R , (5.19) from (4.13), the above identity being valid for ξ ′ = 0 as well. Therefore, arguing as in the derivation of(1.13) from (4.14), we infer from (4.15) and (5.19) that k A k L ∞ x ( R ,L ( ω )) C k Λ A ,q − Λ A ,q k µ , (5.20)where C > and µ ∈ (0 , are two constants depending only on T , ω , and M .We turn now to estimating k q k L ∞ x ( R ,H − ( ω )) , where q = q − q − . With reference to (4.16)-(4.17), wefix x ∈ R and k ∈ { , . . . , N } , and pick a function φ ∗ ,k , expressed by (4.18)-(4.19) in the particular casewhere A ♯σ is uniformly zero, i.e. φ ∗ ,k ( y ) := h (cid:16) y ′ · θ + r x ′ k (cid:17) e − i y ′ · ξ ′ ϕ / k ( y ′ − ( y ′ · θ ) θ ) α σ ( y ) , y ′ ∈ R , y ∈ R . (5.21)In view of Proposition 3.1, we consider a GO solution u j,σ , j = 1 , , to the magnetic Schrödinger equation ( i∂ t + ∆ A j + q j ) u j,σ = 0 in Q , described by (3.3) with Φ = Φ ∗ ,k , Φ = Φ ∗ ,k , and Φ ∗ ,k ( t, x ) := φ ∗ ,k ( x ′ − tθ, x ) , t ∈ R , x ′ ∈ R , x ∈ R . (5.22)Bearing in mind that ∇ · A = 0 , we then apply (3.50) with V = q − A · ( A + A ) , getting h qu ,σ , u ,σ i L ( Q ) = h A · (( A + A ) u ,σ − i ∇ u ,σ ) , u ,σ i L ( Q ) + h (Λ A ,q − Λ A ,q ) f σ , g σ i L (Σ) , where f σ and g σ are given by (3.47) and (3.51), respectively. Thus, we have (cid:12)(cid:12) h qu ,σ , u ,σ i L ( Q ) (cid:12)(cid:12) C k A k L ∞ (Ω) k u ,σ k L ( Q ) k u ,σ k L (0 ,T ; H (Ω)) + (cid:12)(cid:12) h (Λ A ,q − Λ A ,q ) f σ , g σ i L (Σ) (cid:12)(cid:12) , and consequently (cid:12)(cid:12) h qu ,σ , u ,σ i L ( Q ) (cid:12)(cid:12) Cσ µ (cid:16) k A k L ∞ (Ω) + σ / k Λ A ,q − Λ A ,q k (cid:17) (cid:10) ξ ′ (cid:11) , (5.23)by (3.38), (3.55), and (4.21).On the other hand, it follows readily from (3.3) and (5.22), that h qu ,σ , u ,σ i L ( Q ) = Z Q q ( y )Φ ∗ ,k (2 σt, y )( b ,σ b ,σ )(2 σt, y ) d y d t + R k,σ , (5.24)where b j,σ , j = 1 , , is given by (3.4), and R k,σ := Z Q q ( y )Φ ∗ ,k (2 σt, y ) (cid:16) b ,σ (2 σt, y ) e iσ ( y ′ · θ − σt ) ψ ,σ ( t, y ) + ψ ,σ ( t, y ) b ,σ (2 σt, y ) e − iσ ( y ′ · θ − σt ) (cid:17) d y d t + Z Q q ( y )( ψ ,σ ψ ,σ )( t, y ) d y d t. Therefore, | R k,σ | is majorized by k q k L ∞ (Ω) (cid:0) k Φ ∗ ,k (2 σ · , · ) k L ( Q ) (cid:0) k ψ ,σ k L ( Q ) + k ψ ,σ k L ( Q ) (cid:1) + k ψ ,σ k L ( Q ) k ψ ,σ k L ( Q ) (cid:1) Cσ − / N θ,σ ( φ ∗ ,k ) , in virtue of (3.25)–(3.27) and (3.39), so we infer from (4.21) that | R k,σ | Cσ µ − / (cid:10) ξ ′ (cid:11) . (5.25)We turn now to examining the first term in the right hand side of (5.24). In light of (3.4), we have Z Q q ( y )Φ ∗ ,k (2 σt, y )( b ,σ b ,σ )(2 σt, y ) d y d t = Z Q q ( y )Φ ∗ ,k (2 σt, y ) d y d t + r k,σ , (5.26)with r k,σ := Z Q q ( y )Φ ∗ ,k (2 σt, y ) (cid:16) e − i R σt θ · A ♯σ ( y ′ − sθ,y ) d s − (cid:17) d y d t. (5.27) N INVERSE PROBLEM FOR THE MAGNETIC SCHRÖDINGER EQUATION IN INFINITE CYLINDRICAL DOMAINS 31
Next, as e − i R σt θ · A ♯σ ( y ′ − sθ,y ) d s − − i R σt θ · A ♯σ ( y ′ − τ θ, y ) e − i R τ θ · A ♯σ ( y ′ − sθ,y ) d s d τ , we have (cid:12)(cid:12)(cid:12) e − i R σt θ · A ♯σ ( y ′ − sθ,y ) d s − (cid:12)(cid:12)(cid:12) σT k A ♯σ k L ∞ ( R ) Cσ k A k L ∞ (Ω) , ( t, y ) ∈ Q. Here we used the fact, arising from (3.9) and (3.15)-(3.16), that for any σ > , k A ♯σ k L ∞ ( R ) is majorized,up to some multiplicative constant that is independent of σ , by k A ♯ k L ∞ (Ω) . Therefore, we infer from (1.12),(3.39), and (4.21), that | r k,σ | Cσ k A k L ∞ ( R ) k Φ ∗ ,k (2 σ · , · ) k L ( Q ) C k A k L ∞ ( R ) k φ ∗ ,k k L ( R ) C k A k L ∞ ( R ) h ξ ′ i σ µ . (5.28)We are left with the task of examining the integral Z Q q ( y )Φ ∗ ,k (2 σt, y ) d y d t = Z T Z R q ( y ) φ ∗ ,k ( y ′ − σtθ, y ) d y ′ d y d t = 12 σ Z σT Z R q ( y ′ + sθ, y ) φ ∗ ,k ( y ) d y ′ d y d s, (5.29)appearing in the right hand side of (5.26). To do that, we notice for all σ > σ ∗ , that q ( y ′ + sθ, y ) φ ∗ ,k ( y ) = 0 , s ∈ ( −∞ , ∪ (2 σT, + ∞ ) , y ′ ∈ R , y ∈ R , since q and φ ∗ ,k are supported in B (0 , R ) × R and D − R ( θ ) × R , respectively, and that | y ′ + sθ | > R whenever y ′ ∈ D − R ( θ ) and s ∈ ( −∞ , ∪ (2 σT, + ∞ ) . In view of (4.3) and (5.29), this entails that Z Q q ( y )Φ ∗ ,k (2 σt, y ) d y d t = 12 σ Z R Z R q ( y ′ + sθ, y ) φ ∗ ,k ( y ) d y ′ d y d s = 12 σ Z R P ( q )( θ, y ′ , y ) φ ∗ ,k ( y ) d y ′ d y . Thus, arguing in the same way as in the derivation of (4.23), we infer from (5.21) that (cid:12)(cid:12)b q ( ξ ′ , x ) (cid:12)(cid:12) C h ξ ′ i (cid:16) σ / k Λ A ,q − Λ A ,q k + σ µ +1 k A k L ∞ (Ω) + σ µ − / (cid:17) , (5.30)for every σ > σ ∗ .The next step of the proof is to upper bound k A k L ∞ (Ω) in terms of k Λ A ,q − Λ A ,q k . To do that, we pick p > and apply Sobolev’s embedding theorem (see e.g. [13, Corollary IX.14]), getting k A ( · , x ) k L ∞ ( ω ) C k A ( · , x ) k W ,p ( ω ) for a.e. x ∈ R , where the constant C > depends only on ω . Interpolating, we thusobtain that k A ( · , x ) k L ∞ ( ω ) C k A ( · , x ) k / W ,p ( ω ) k A ( · , x ) k / L p ( ω ) , x ∈ R . This and (5.20) yield k A k L ∞ (Ω) C k A k / L ∞ x ( R ,L p ( ω ) ) C k A k /pL ∞ x ( R ,L ( ω ) ) C k Λ A ,q − Λ A ,q k µ /p , for some constant C > depending only on ω , M and T . Then, we find by substituting the right hand sideof the above estimate for k A k L ∞ (Ω) in (5.30), that (cid:12)(cid:12)b q ( ξ ′ , x ) (cid:12)(cid:12) C (cid:10) ξ ′ (cid:11) (cid:16) σ / k Λ A ,q − Λ A ,q k + σ µ +1 k Λ A ,q − Λ A ,q k µ /p + σ µ − / (cid:17) , σ > σ ∗ . (5.31)With the notations of Subsection 5.1.2, we infer from (5.31) and the estimate Z R \ B (0 ,ρ ) h ξ ′ i − | b q ( ξ ′ , x ) | d ξ ′ Mρ , x ∈ R , which holds true for any ρ ∈ (1 , + ∞ ) , that k q k L ∞ x ( R ,H − ( ω )) C (cid:16) ρ σ / δ µ /p + σ µ − / + ρ − (cid:17) , σ > σ ∗ , (5.32)where δ = k Λ A ,q − Λ A ,q k ∈ (0 , . Thus, for µ ∈ (0 , / and δ ∈ (cid:16) , σ − (41 − µ ) p/ (6 µ ) ∗ (cid:17) , we obtain(1.16) with µ := (1 − µ ) µ / (7 p (41 − µ )) by taking ρ = δ − µ and σ = δ − µ / (1 − µ ) in (5.32).R EFERENCES [1] S. Avdonin, S. Lenhart and V. Protopopescu,
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