Linearized Calderón problem and exponentially accurate quasimodes for analytic manifolds
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This is a self-archived version of an original article. This version may differ from the original in pagination and typographic details. Author(s): Title: Year: Version: Copyright: Rights: Rights url: Please cite the original version: CC BY 4.0 https://creativecommons.org/licenses/by/4.0/ Linearized Calderón problem and exponentially accurate quasimodes for analytic manifolds © 2022 the Authors Published version Krupchyk, Katya; Liimatainen, Tony; Salo, Mikko Krupchyk, K., Liimatainen, T., & Salo, M. (2022). Linearized Calderón problem and exponentially accurate quasimodes for analytic manifolds. Advances in Mathematics, 403, Article 108362. https://doi.org/10.1016/j.aim.2022.108362 2022
Advances in Mathematics 403 (2022) 108362 Contents lists available at ScienceDirect Advances in Mathematics www.elsevier.com/locate/aim Linearized Calderón problem and exponentially accurate quasimodes for analytic manifolds Katya Krupchyk a, Tony Liimatainen b,∗, Mikko Salo c aDepartment of Mathematics, University of California, Irvine, CA 92697-3875, USA bDepartment of Mathematics and Statistics, University of Helsinki, Helsinki, PL 68, FI-00014, Finland cDepartment of Mathematics and Statistics, University of Jyväskylä, Jyväskylä, FI-40014, Finland a r t i c l e i n f o a b s t r a c t Article history: Received 2 October 2020 Received in revised form 7 March 2022 Accepted 7 March 2022 Available online 5 April 2022 Communicated by C. Fefferman MSC: primary 35R30 secondary 35A18, 35A20, 35J25 Keywords: Inverse problems Riemannian manifold Conformally transversally anisotropic Gaussian quasimodes WKB construction Wave front set In this article we study the linearized anisotropic Calderón problem on a compact Riemannian manifold with boundary. This problem amounts to showing that products of pairs of harmonic functions of the manifold form a complete set. We assume that the manifold is transversally anisotropic and that the transversal manifold is real analytic and satisfies a geometric condition related to the geometry of pairs of intersecting geodesics. In this case, we solve the linearized anisotropic Calderón problem. The geometric condition does not involve the injectivity of the geodesic Xray transform. Crucial ingredients in the proof of our result are the construction of Gaussian beam quasimodes on the transversal manifold, with exponentially small errors, as well as the FBI transform characterization of the analytic wave front set. © 2022 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). *Corresponding author. E-mail addresses: katya.krupch[email protected] (K. Krupchyk), tony.liimatainen@helsinki.fi (T. Liimatainen), mikko.j.salo@jyu.fi (M. Salo). https://doi.org/10.1016/j.aim.2022.108362 0001-8708/© 2022 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/).
2K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 1. Introduction and statement of results The inverse conductivity problem posed by Calderón [5]asks to determine the electrical conductivity of a medium from voltage and current measurements on its boundary. This problem is the mathematical model of Electrical Impedance (or Resistivity) Tomography, an imaging method with applications in seismic and medical imaging. It is also one of the most fundamental models of inverse boundary value problems for elliptic partial differential equations. For these reasons both the theoretical and applied aspects of the Calderón problem have been under intense study. We refer to the survey [53]for more information and references. In this article we are interested in the case where the electrical conductivity of the medium is anisotropic, i.e. depends on direction. This can be modeled by a matrix conductivity coefficient, or in geometric terms by having a resistivity coefficient given by a Riemannian metric gon a compact manifold Mwith smooth boundary. There are many variants of this problem. One of them is the (geometric) Calderón problem for a Schrödinger equation: given a known compact Riemannian manifold (M, g)with smooth boundary and an unknown potential q∈C∞(M), determine qfrom the knowledge of the Cauchy data on ∂M of solutions of the Schrödinger equation (−Δg+q)u=0inM. Here −Δgis the Laplace-Beltrami operator. This geometric Calderón problem is solved in [21]when dim(M) =2. The problem is open in general when dim(M) ≥3with only partial results available. In particular, the unique determination of qwas obtained in [52]in the Euclidean setting, in [27]for hyperbolic manifolds, and in [40], [29]in the real analytic setting. Going beyond these settings, the geometric Calderón problem was only solved in the case when (M, g)is CTA (conformally transversally anisotropic, see Definition 1.1 below) and under the assumption that the geodesic X-ray transform on the transversal manifold is injective [10,12]. The linearized version (at q=0) of the above problem is also of interest, since methods for the linearized problem often give insight to the original problem. In our case, the linearized problem reduces to the following simple question asking whether products of pairs of harmonic functions form a complete set in L1(M): Question 1. Let (M, g)be a compact oriented Riemannian manifold with smooth boundary. If f∈L∞(M)satisfies M fu1u2dVg=0 for all uj∈L2(M)with Δguj=0in M, j=1, 2, is it true that f≡0?
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 3 The methods of [21,10,12]give a positive answer to Question 1when dim(M) =2, or when dim(M) ≥3and (M, g)is CTA with the transversal manifold having injective geodesic X-ray transform. There have been recent attempts to improve these results when dim(M) ≥3. In [20], it is proved that Question 1has a positive answer when (M, g)is a complex Kähler manifold with sufficiently many holomorphic functions. The article [13] establishes a recovery of singularities result: if (M, g)is transversally anisotropic and the transversal manifold satisfies a certain geometric condition, one can recover transversal singularities of f. See also [11], [50]for the linearized Calderón problem with partial data in the Euclidean setting. The recent related works [14,15,31,32,34,41,47] regarded inverse problems for nonlinear elliptic equations on CTA manifolds as well as some complex manifolds, equipped with a Kähler metric. The linearized problem in Question 1plays an important role in these results. Moreover, it was proven in [34]that on a general transversally anisotropic manifold, the products of four (instead of pairs of) harmonic functions form a complete set in L1(M), and this result was used crucially to solve the corresponding inverse problems for semilinear Schrödinger equations. We also mention the related works [22,36], which solve inverse problems for nonlinear wave equations by showing that sets of products of four waves are dense. In this article we extend the result of [13]and show that if the transversal manifold is additionally real-analytic, Question 1has a positive answer (i.e. one can recover f∈ L∞(M) completely, not just some of its singularities). Let us proceed to state our results. To that end, let us first recall the following definitions, see [10], [12]. Definition 1.1. Let (M, g)be a smooth compact oriented Riemannian manifold of dimension n ≥3with smooth boundary ∂M. (i) (M, g)is called transversally anisotropic if (M, g) ⊂⊂ (T, g) where T=R ×Mint 0, g=e ⊕g0, (R, e)is the Euclidean real line, and (M0, g0)is a smooth compact (n −1)–dimensional manifold with smooth boundary, called the transversal manifold. (ii) (M, g)is called conformally transversally anisotropic (CTA) if (M, cg)is transversally anisotropic, for some positive function c ∈C∞(M). Here and in what follows Mint 0=M0\∂M0stands for the interior of M0. By choosing local coordinates xfor M0and denoting by x1the coordinate on R, the metric g=e ⊕g0 has the form dx2 1+gαβ(x)dxαdxβ, where the Einstein summation convention is used for α, β=1, ..., n −1. In this work we require that all manifolds are oriented. We mention the recent work [4], which studies the Calderón problem on non-oriented Riemannian surfaces.
4K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 Let (M, g)be transversally anisotropic of dimension n ≥3with a transversal manifold (M0, g0). Next we need some definitions related to the transversal manifold (M0, g0). Following [12], we say that a geodesic γ:[−T1, T2] →M0, 0 <T 1, T2<∞, is nontangential if γ(−T1), γ(T2) ∈∂M0, γ(t) ∈Mint 0for all −T1<t <T 2, and ˙γ(−T1), ˙γ(T2)are nontangential vectors on ∂M0. Following [13], we have the following definition. Definition 1.2. We say that (x 0, ξ 0) ∈S∗Mint 0is generated by an admissible pair of geodesics, if there are two nontangential unit speed geodesics γ1:[−T1,T 2]→M0,γ 2:[−S1,S 2]→M0, 0 <T 1, T2, S1, S2<∞, such that (i) γ1(0) =γ2(0) =x 0, (ii) ˙γ1(0) +˙γ2(0) =t0ξ 0, for some 0 <t 0<2, where ξ 0is understood as an element of Tx0Mint 0by the Riemannian duality, (iii) γ1, γ2do not have self-intersections at the point x 0, and x 0is the only point of their intersections, i.e. γ1(t)=x 0⇔t=0,γ 2(s)=x 0⇔s=0, γ1(t)=γ2(s)⇒γ1(t)=γ2(s)=x 0. Let f∈L∞(M)and let us extend f∈L∞(M)by zero to (R ×M0) \M. Writing x =(x1, x) where x1∈R, and xare local coordinates M0, we let f(λ, x)= ∞ −∞ e−iλx1f(x1,x )dx1,λ∈R,(1.1) be the partial Fourier transform of fwith respect to x1. We have for each λ ∈Rthat f(λ, · ) ∈L∞(M0) ∩E(Mint 0). When Xis a real analytic open manifold and u ∈D (X), we let WFa(u) ⊂T∗X\{0} stand for the analytic wave front set of u, see [49, Definition 6.1], [25, Sections 8.5, 9.3]. The set WFa(u) ⊂T∗X\{0}is closed conic and we have π(WFa(u)) = singsuppa(u), where π:T∗X→X, (x, ξ) → x, is the natural projection and singsuppa(u)is the analytic singular support of u, i.e. the smallest closed set such that uis real analytic in the complement [25, Theorem 8.4.5]. In particular, WFa(u) =∅if and only if uis real analytic on X. We have the following analytic microlocal result, which is an analog of Theorem 1.1 in [13], established in the C∞–case.
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 5 Theorem 1.3. Let (M, g)be a transversally anisotropic manifold of dimension n ≥3 with transversal manifold (M0, g0), and assume that Mint 0and g0|Mint 0are real analytic. Assume furthermore that f∈L∞(M)satisfies M fu1u2dVg=0,(1.2) for all uj∈L2(M)with −Δguj=0in Mint. Let (x 0, ξ 0) ∈S∗Mint 0be generated by an admissible pair of geodesics. Then for any λ ∈R, one has (x 0,ξ 0)/∈WFa( f(λ, ·)) ⊂T∗Mint 0\{0}. Here f(λ, · )refers to the partial Fourier transformation given by (1.1). Theorem 1.3 implies the following global result, which gives a positive answer to Question 1under suitable geometric assumptions. Theorem 1.4. Let (M, g)be a transversally anisotropic manifold of dimension n ≥3and assume that the transversal manifold (M0, g0)is connected, Mint 0as well as g0in Mint 0 are real analytic. Assume that every point (x 0, ξ 0) ∈S∗Mint 0is generated by an admissible pair of geodesics. Moreover, assume that f∈L∞(M)satisfies (1.2)for all uj∈L2(M) with −Δguj=0in Mint. Then f=0in M. Remark 1.5. Note that while (Mint 0, g0)is real analytic, Theorem 1.4 does not follow from the existing results in the real analytic setting, as it corresponds to deforming the zero potential by an L∞perturbation. Remark 1.6. In Theorems 1.3 and 1.4 while Mint is real analytic, the boundary ∂M need not be real analytic. As the following example shows, there exist transversally anisotropic manifolds (M, g) with a transversal manifold (M0, g0) satisfying the geometric conditions of Theorem 1.4 and with a non-invertible geodesic X-ray transform. Therefore, the geometric Calderón problem is still open on such manifolds while our Theorem 1.4 gives a positive solution to the corresponding linearized problem. Example 1.7. Let M0=S1×[0, a], a >0, be a cylinder with its usual flat metric g0. The geodesics on M0are straight lines, circular cross sections, and helices that wind around the cylinder. The geodesic X-ray transform is not invertible, since the kernel contains functions of the form f(eit, s) =h(s) where h ∈C∞ 0((0, a)) integrates to zero over [0, a]. However, it is shown in Appendix Athat every point (x 0, ξ 0) ∈S∗Mint 0is generated by an admissible pair of geodesics.
6K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 It is established in [13, Lemma 3.1] that if (M0, g0)satisfies the strict Stefanov– Uhlmann regularity condition at (x 0, ξ 0) ∈S∗Mint 0, which we now proceed to recall, then (x 0, ξ 0)is generated by an admissible pair of geodesics. Definition 1.8. The transversal manifold (M0, g0)satisfies the strict Stefanov–Uhlmann regularity condition at (x 0, ξ 0) ∈S∗Mint 0if there exists η∈S∗ x 0Mint 0such that g0(ξ 0, η) =0and such that the following holds: let γx 0,η:[−T1, T2] →M0, 0 <T 1, T2< ∞, be the geodesic with γx 0,η(0) =x 0, ˙γx 0,η=η. We have (i) γx 0,ηis nontangential, (ii) γx 0,ηcontains no points conjugate to x 0, (iii) γx 0,ηdoes not self-intersect for any time t ∈[−T1, T2]. Hence, if a transversally anisotropic manifold (M, g)is such that the transversal manifold (M0, g0)satisfies the strict Stefanov–Uhlmann regularity condition at every point of S∗Mint 0with Mint 0and g0|Mint 0real analytic, and (M0, g0)is connected, then Theorem 1.4 holds. As the following examples demonstrate, there are transversally anisotropic manifolds (M, g)with a transversal manifold (M0, g0) satisfying the geometric condition of Theorem 1.4, and with an invertible geodesic X-ray transform. Thus, for such manifolds (M, g), Theorem 1.4 also follows from [10], [12]. Example 1.9. Let (M0, g0)be a simple manifold, i.e. a compact simply connected manifold with strictly convex boundary so that no geodesic has conjugate points. Then (M0, g0) satisfies the strict Stefanov–Uhlmann regularity condition at any point of S∗Mint 0and thus also the geometric condition in Theorem 1.4. Note that in this case (M, g)is admissible in the sense of [10], and Theorem 1.4 would also follow from [10]. Example 1.10. Let S3⊂R4be the unit sphere and let μbe a geodesic arc from the north pole to the south pole of the sphere. Let M0be the closure of a neighborhood of μ. It is established in [13]that the manifold M0satisfies the strict Stefanov–Uhlmann regularity condition at each point of S∗Mint 0. Notice also that the manifold M0contains conjugate points, so that it is not simple. However, the geodesic X-ray transform on (M0, g0)is injective by [51], and Theorem 1.4 would therefore also follow from [12]. Remark 1.11. We would like to remark that the strict Stefanov-Uhlmann condition is not satisfied for (M0, g0)of Example 1.7 since for any (x 0, ξ 0) ∈S∗Mint 0with ξ 0pointing in the direction of the [0, a]factor, the orthogonal geodesics never reach ∂M0. The proof of Theorem 1.3 depends crucially on the construction of Gaussian beam quasimodes along nontangential geodesics on M0, with exponentially small errors, as stated in the following result. Before stating the result, let us recall from [49, Chapter 1]
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 7 the notion of a classical analytic symbol. Let V⊂Cnbe an open set. We say that a(x; h) =∞ k=0 hkak(x)is a (formal) classical analytic symbol in Vif ak∈Hol(V), k=0, 1, 2, ..., and for every V⊂⊂ V, there exists C=C V>0such that |ak(x)|≤Ck+1kk,x∈ V, (1.3) k=0, 1, 2, .... The classical analytic symbol a(x; h)is said to be elliptic if a0=0. We have the following essentially well known result, see [48]and [49], and see also the work by Babich [1]for a sketch of the proof. Notice that here our quasimode construction is performed along the entire geodesic segment contrary to the standard constructions in a neighborhood of a point, see [9]. We use the notation neigh(p, X) for an open neighborhood of a point p ∈X, and similarly for neighborhoods of general subsets of other topological spaces in place of pand Xabove. Theorem 1.12. Let (X, g)be a compact Riemannian manifold of dimension n ≥2with smooth boundary, contained in a real analytic open manifold ( X, g)of the same dimension with greal analytic in X. Let γ:[−T1, T2] →X, 0 <T 1, T2<∞, be a unit speed non-tangential geodesic in X, and let λ ∈R. For any neighborhood of γ([−T1, T2]), there is a family of C∞functions v(x; h)on X, 0 <h ≤1, supported in the neighborhood, and C>0such that (−h2Δg−(hs)2)vL2(X)=O(e−1 Ch ),vL2(X)1,(1.4) as h →0. Here s =1 h+iλ. The local structure of the family v(x ; h)is as follows: let p ∈γ([−T1, T2]) and let t1<··· <t Npbe the times in (−T1, T2)when γ(tl) =p, l=1, ..., Np. In a sufficiently small neighborhood Vof a point p ∈γ([−T1, T2]), we have v|V=v(1) +···+v(Np), where each v(l)has the form v(l)(x;h)=h−(n−1) 4eisϕ(l)(x)a(l)(x;h). Here ϕ =ϕ(l)is real analytic in Vsatisfying for tnear tl, ϕ(γ(t)) = t, ∇ϕ(γ(t)) = ˙γ(t),Im (∇2ϕ(γ(t))) ≥0,Im (∇2ϕ)|˙γ(t)⊥>0,(1.5) and a(l)is an elliptic classical analytic symbol in a complex neighborhood of p.
8K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 We have chosen to give a fairly complete proof of Theorem 1.12 since we are not aware of a detailed treatment in the literature and since we need to have fairly precise information concerning the quasimodes for our applications. We refer to [33]for a related complex Riccati equation, and to [6]for a geometric interpretation of it. Let us briefly mention how the exponentially small error is achieved in Theorem 1.12. The proof of the theorem is by using the ansatz v(x; h) =eisϕ(x)a(x; h), which, as usual, leads to solving the eikonal equation for the phase function ϕ(x)and a transport equation for the amplitude a(x; h). We first find an exact analytic solution for the eikonal equation near a geodesic segment of γ. Consequently, the transport equation for the amplitude a(x; h) =N k=0 hkak(x)has analytic coefficients and we find a(x; h)as a classical analytic symbol. This involves adapting the nested neighborhood method of [49]. The error for v being a true eigenfunction then is (−h2Δg−(hs)2)eisϕN j=0 hjaj=hN+2T2(aN),(1.6) where T2is a second order operator with analytic coefficients. Cauchy estimates and (1.3) then yield that the error term (1.6)is bounded by hN+2CN+1NN. Letting the order N of the expansions of adepend on has N=N(h) =[ 1 heC ]gives the exponentially small error in the theorem. The above was based on finding first an exact analytic solution to the eikonal equation |dϕ|g=1near a geodesic segment of γ. To find such a solution, we view the eikonal equation as the Hamilton-Jacobi equation, p(x, ϕ x(x)) = 0,(1.7) where p(x, ξ) =|ξ|2 g(x)−1is holomorphically continued to a complex domain. When solving the Hamilton-Jacobi equation (1.7)we proceed by a geometric argument of constructing a complex Lagrangian manifold, Λ⊂p−1(0), in a complex neighborhood of a segment of the graph of ˙γ⊂T∗X, see [48]. The solution ϕis then obtained as a generating function of the Lagrangian Λ, which parametrizes Λ as Λ={(x, ϕ x(x)}. Extending the argument to a neighborhood of the geodesic segment of γrequires some extra work involving positive Lagrangians. Let us proceed to explain the main ideas in the proof of Theorem 1.3. Let α0= (x 0, ξ 0) ∈S∗Mint 0be generated by an admissible pair of geodesics γ1(α0)and γ2(α0)on
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 15 Here σ=n j=1 dξj∧dxjis the complex symplectic form on (the tangent space of) C2n=Cn x×Cn ξ. Indeed, if we fix (x, ξ) ∈Λ, any vector tangent to Λwith base point at (x, ξ)is of the form (0, Vy, Vτ, Vη)with Vη=ψ yy Vyand Vy∈Cn−1. Applying σto two such vectors gives Vy 1·ψ yy Vy 2−Vy 2·ψ yy Vy 1=0, showing (2.10). Note also that Λ∩R2n={(0,0,τ 0,η 0)}.(2.11) Indeed, (0, 0, τ0, η0) ∈Λ∩R2nas (τ0, η0) ∈Rnand ψ y(0) =η0. To see the opposite inclusion, let (0, y, λ(y, η), η=ψ(y)) ∈Λ∩R2nand Taylor expand ψ(y)at y=0, η=ψ(y)=η0+ψ(0)y+O(|y|2),y∈Rn−1. We have Im η=Imψ(0)y+O(|y|2), and therefore, in view of (2.9), Im η= 0 implies that y=0. This shows (2.11). Let Hpbe the complex Hamilton vector field of p, and let us consider the Hpflowout of Λ: Λ=exp t 2Hp(ρ):ρ∈Λ,t∈neigh(I,C)⊂C2n. Here if μ =N j=1 aj(z)∂zjis a holomorphic vector field on an open set V⊂CNin the sense that aj∈Hol(V), j=1, ..., n, we can define the flow exp(tμ)(ρ), ρ ∈V, locally for t ∈neigh(0, C), by solving the system of ODE, ˙zj(t)=aj(z(t)),1≤j≤n, z(0) = ρ, see [16, Section 1] and the references given there. Then Λ⊂Λ, and since the flow of Hppreserves p, we have Λ⊂p−1(0), and Λis a C–Lagrangian submanifold of C2n, see [18, Proposition 5.4] for a proof in the real case. The proof in the present holomorphic setting is similar. Let us also recall from [18, page 60] that the holomorphic Hamilton vector field Hpis tangent to Λat each point of Λ. This is because Λis a Lagrangian contained in p−1(0). The differential of πx|Λis bijective at (0, 0, τ0, η0)since the differential of πxis injective and since any Lagrangian submanifold has dimension dim(X). (The differential of πx|Λis injective since the differential of the exponential map TX →Xis injective.) Consequently, there is a function ϕ ∈Hol(neigh(0, Cn)) such that Λ=Λ ϕ:= {(x, ϕ x(x)) : x∈neigh(0,Cn)},(2.12)
16 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 see [42, Section 5.6, Exercise 4], and also [18, Theorem 5.3] for the real version of this result. We have ϕ x(0) =ξ0and modifying ϕby a constant we get ϕ(0, y) =ψ(y), and such a solution is unique. Step 2. Solving near γ.Let us denote the tangent space of Λat (0, 0, τ0, η0)by Λ0and write Λ0:= T(0,0,τ0,η0)Λ={(δx,δ ξ)∈Cn×Cn:δξ=ϕ xx(0)δx},(2.13) where in the second equality we used (2.12). We claim that Λ0is a positive Lagrangian plane in the sense that 1 iσ(ρ, ρ)≥0,ρ∈Λ0. To this end, letting M0=ϕ xx(0) and using (2.13), we write ρ =(δx, M0δx) ∈Λ0. Then using that M0is symmetric, we get 1 iσ(ρ, ρ)=1 i(M0δx·δx−M0δx·δx)=2Im(M0δx·δx) =2Im(M0)Re δx·Re δx+2Im(M0)Imδx·Imδx, (2.14) and therefore, it suffices to prove that Im M0≥0.(2.15) In doing so, using (2.8), we write M0=ϕ tt(0,0) ϕ ty(0,0) ϕ yt(0,0) ψ yy(0) .(2.16) Using that Hpis tangent to Λϕ, we see that exp( t 2Hp)(0, ξ0) =(x(t), ϕ x(x(t))) is real for t ∈neigh(0, R), so that ϕ t(t, 0), ϕ y(t, 0) are real. Hence, Im M0=00 0Imψ yy(0),(2.17) and therefore, by the condition Im ψ yy(0) >0we imposed on ψin (2.9), (2.15) follows. For future reference, let us remark that Λ0∩R2n=RHp(0,ξ 0),(2.18) where R Hp(0, ξ0) ={s Hp(0, ξ0) :s ∈R}. Indeed, we have Hp(0, ξ0) ∈Λ0∩R2nsince the Hpvector field is tangent to Λ. On the other hand, if (δx, M0δx) ∈Λ0∩R2n, it follows from (2.17)that
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 17 δx=(δt,0) = δt˙x(0) = δtp ξ(0,ξ 0), where δt∈R. Here in the second equality we used that (t, 0) corresponds to the geodesic in Fermi coordinates. We get (δx, M0δx) =δt(p ξ(0, ξ0), M0p ξ(0, ξ0)) =δtHp((0, ξ0)), which shows (2.18). Here in the last equality we used Hp(0, ξ0) ∈Λ0. Let κ(t):=expt 2Hp:Λ→Λ,t∈I⊂R, and therefore, the differential satisfies dκ(t)(0,ξ 0):Λ 0→Tκ(t)(0,ξ0)Λ. As the canonical transformation κ(t)is real for each t ∈I, dκ(t)(0, ξ0) preserves positivity, see [42, Section 5.6, Exercise 8], and therefore, Λt:= Tκ(t)(0,ξ0)Λ⊂C2n is a positive Lagrangian plane, for all t ∈I. We claim that Λtis transversal to the fiber F={(0, η) :η∈Cn} ⊂C2n, for all t ∈I, i.e. Λt+F=C2n. As dim Λt=n, we have to show that Λt∩F={0}. Indeed, let (0, η) ∈Λt∩F. Then (2.13) implies that 0 η=dκ(t)(0,ξ 0)δx M0δx,(2.19) for some δx∈Cn. We have 0=1 iσ0 η,0 η=1 iσdκ(t)(0,ξ 0)δx M0δx,dκ(t)(0,ξ 0)δx M0δx =1 iσ δx M0δx,δx M0δx=2Im(M0δx·δx). As Im M0≥0, we get (Im M0)δx=0, and therefore, (2.17) implies that δx=αp ξ(0, ξ0) for some α∈C. Thus, by (2.19)we obtain that 0 η=dκ(t)(0,ξ 0)(αHp(0,ξ 0)) = αHp(x(t),ξ(t)) = α˙x(t) ˙ ξ(t). Since ˙x(t) =0, we get α=0. Hence, η=0, which establishes the claim.
18 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 As Λtis transversal to the fiber for all t ∈I, by inspection of the proof of Theorem 5.5 in [18], we conclude that there exists ϕ ∈Hol(neigh(I×B, Cn)) such that Λ =Λ ϕand ϕ solves (2.8). The function ϕis a continuation of the one appearing in (2.12). Notice that it is precisely thanks to the fact that the tangent plane Λtdoes not contain any non-zero vector of the form (0, η)for all t ∈Ithat the proof of Theorem 5.5 in [18] applies near each point in I×{0}, see also [26, Section 24.2]. Step 3. Properties of the solution. Next we shall check that the property (2.7), that is Im ϕ(t, y)≥0,Im ϕ(t, 0) = 0,Im ϕ yy(t, 0) >0,t∈I, holds for ϕ. First, ϕ x(x(t)) =ξ(t)is real for t ∈I. Writing d dtϕ(x(t)) = ϕ x(x(t)) ·˙x(t)=ξ(t)·1 2p ξ(x(t),ξ(t)), we have ϕ(t, 0) = ψ(0) + 1 2 t 0 ξ(s)·p ξ(x(s),ξ(s))ds =ψ(0) + t, (2.20) as ξ·p ξ(x, ξ) =2(p(x, ξ) +1). Thus, using that ψ(0) is real, we see that Im ϕ(t, 0) =0 for t ∈I. Furthermore, if ψ(0) =0, we get ϕ(t, 0) =t. Let M(t) =ϕ xx(x(t)). Then M(t)is an n ×ncomplex symmetric matrix depending real analytically on t, such that Im M(t)≥0,(2.21) in view of the positivity of Λt. We claim that Im M(t)|W>0,(2.22) where W⊂Rnis an algebraic supplement to R ˙x(t)so that R ˙x(t) ⊕W=Rn. To that end, let us observe first that Λt∩R2n=dκ(t)(0,ξ 0)(Λ0∩R2n)=dκ(t)(0,ξ 0)(RHp(0,ξ 0)) = RHp(x(t),ξ(t)). Here we have used (2.18)in the second equality. Let v∈Wbe such that Im M(t)v·v=0. Hence, by (2.21), we get Im M(t)v=0.
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 19 Thus, (v, M(t)v) ∈Λt∩R2n=R Hp(x(t), ξ(t)), and therefore, vis proportional to p ξ(x(t), ξ(t)) =˙x(t). This gives that v=0, since v∈W. Hence, (2.22) follows, and we get Im ϕ yy(t, 0) >0for all t ∈I. Finally, we get Im ϕ(t, y) ≥0for all (t, y) ∈Uby Taylor’s formula and by using that ϕ x(x(t)) =ξ(t)is real. We have therefore constructed a real analytic solution ϕof (2.3) such that (2.7)holds. 2.2. Construction of the amplitude We shall follow [49, Theorem 9.3], where the construction of the amplitude as a classical analytic symbol is carried out in a neighborhood of a point, extending the construction to a full neighborhood of a geodesic segment. We look for the amplitude ain the form of a formal power series in h, a(x;h)= ∞ k=0 hkak(x).(2.23) From (2.2), we see that we want to solve the following equation formally in powers of h, e−isϕ(−h2Δg−(hs)2)eisϕa=[−hiL0−ihΔgϕ+h2(−Δg+λL0+λΔgϕ)]a=0,(2.24) in a fixed complex domain U, containing Γ. Here L0=2dϕ, d · g=2G(x)ϕ x·∂x=p ξ(x, ϕ x(x)) ·∂x,(2.25) where pis given in (2.4). The transport equation (2.24)can be written in the following form, (hL0+hf(x)+h2Q(x, Dx))a=0,(2.26) where f(x) =Δ gϕis a holomorphic function on Uand Q(x, Dx) =i(−Δg+λL0+λΔgϕ) is a holomorphic differential operator of order 2. To solve (2.26), we remark first that the holomorphic vector field L0is transversal to each complex hypersurface Ht0={(t, y) ∈ neigh(I, C) ×neigh(0, Cn−1) :t =t0∈I}at (t0, 0). Indeed, p ξ(x(t),ϕ x(x(t))) ·∂x=p τ(x(t),ϕ x(x(t)))∂t+p η(x(t),ϕ x(x(t))) ·∂y, where p τ(x(t), ϕ x(x(t))) =0for all t ∈Isince ∂τp(x(t), ξ(t)) =0for all t ∈Ias noted in (2.6). Thus, substituting (2.23)into (2.26), we get a sequence of transport equations which can all be solved uniquely in a suitable complex domain containing Γ, provided that a|Ht0is prescribed, for some t0∈I. However, the difficulty here is that we would like our solution a(x; h)to be a classical analytic symbol, and following [49, Section 9], we shall establish this fact making use of the method of “nested neighborhoods” introduced
20 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 in [49]. Contrary to [49, Theorem 9.3], where the family of “nested neighborhoods” is considered near a point, here we shall work in such neighborhoods near a piece of the geodesic. For simplicity, let us take t0=0. We look for solution to (2.26)by using convenient coordinates. The coordinates we will use are the usual flowout coordinates (see e.g. [39]), which we show to exist for L0on a neighborhood of a given interval. Lemma 2.1. Let J⊂⊂ Ibe an open interval. There exist local holomorphic coordinates (s, z) ∈neigh(J, C) ×neigh(0, Cn−1)such that the hyperplane H0is given by the equation s =0and L0=∂ ∂s. Proof. We continue to work in the Fermi coordinates x =(t, y)and recall from [28]that G(t, y)=(gjk(t, y)) = 1 + O(|y|2).(2.27) Now (2.3), (2.4), and (2.27)imply that (ϕ t)2(t, 0) + (ϕ y)2(t, 0) = 1,(2.28) and therefore, it follows from (2.20)and (2.28)that ϕ y(t, 0) =0. Hence, Taylor expanding ϕ(t, y)at y=0, we get ϕ(t, y)=ψ(0) + t+O(|y|2).(2.29) It follows from (2.25), (2.27), and (2.29)that L0=2(1+O(|y|2)) 1+O(|y|2) O(|y|)·∂t ∂y=2(1+O(|y|2))∂t+O(|y|)·∂y.(2.30) Consider the initial value problem for the flow exp(sL0)(0, z), ∂s(t, y)(s, z)=L0((t, y)(s, z)), (t, y)(0,z)=(0,z),(2.31) where (s, z) ∈neigh(I, C) ×neigh(0, Cn−1). In particular, y(s, z)|z=0 =0and therefore, y(s, z) =O(|z|). Differentiating the first equation in (2.31)in zjand using (2.30), we get ∂s(∂zjt(s, z)) = O(y(s, z)∂zjy)=O(|z|), ∂zjt(0,z)=0.(2.32) Hence, ∂zjt(s, z)=O(|z|).(2.33)
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 21 Consider the holomorphic map F:neigh(I,C)×neigh(0,Cn−1)(s, z)→ (t, y)(s, z). In view of (2.33), the differential DF(s, 0) is given by DF(s, 0) = ⎛ ⎜ ⎜ ⎝ t s(s, 0) 0 ... 0 y 1s(s, 0) y 1z1(s, 0) ... y 1zn−1(s, 0) . . .. . .. . . y n−1s(s, 0) y n−1z1(s, 0) ... y n−1zn−1(s, 0) ⎞ ⎟ ⎟ ⎠,(2.34) where t s(s, 0) =2(1 +O(|y(s, 0)|2)) =2. By Liouville’s formula, see [24, Theorem 1.2.5], we know that the last n −1 columns in (2.34)are linearly independent, and therefore, det(DF(s, 0)) =0for all s ∈neigh(I, C). Furthermore, F|I×{0}is injective as F(s, 0) = (t(s, 0), 0) =(2s, 0). An application of a holomorphic version of [28, Lemma 7.3] allows us to conclude that Fis a holomorphic diffeomorphism in neigh(J, C) ×neigh(0, Cn−1) where J⊂⊂ Iis an open interval. Now writing x =(t, y), in view of (2.31), we see that ∂ ∂su(x(s, y)) = u x(x(s, y)) ·˙x(s, y)=(L0u)(x(s, y)). Finally, it follows from (2.30)and (2.31)that ∂st(s, z)=2(1+O(|z|2)), t(0,z)=0, and therefore, t(s, z) =2s +O(|z|2)s. Hence, t =0is equivalent to the fact that s =0, showing that the hyperplane H0is given by the equation s =0. Passing to the new holomorphic coordinates provided by Lemma 2.1, and renaming them as x =(t, y), we are led from (2.26)to consider the following initial value problem, h∂ ∂t +hf(x)+h2Q(x, Dx)a=0, a|t=0 =w(y;h),(2.35) where w(y; h)is a classical analytic symbol near 0 ∈Cn−1. We would like to find a classical analytic symbol asolving (2.35). Here fis a holomorphic function, and Qis a holomorphic differential operator of order 2. To that end, it suffices to solve the following problem, h∂ ∂t +hf(x)+h2Q(x, Dx)a=hv, a|t=0 =0,(2.36)
22 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 where v(x; h)is a classical analytic symbol in neigh(J, C) ×neigh(0, Cn−1). This is because a solution ato (2.36)with v=−(∂ ∂t +f(x) +hQ(x, Dx))v0and v0|t=0 =w, implies that a +v0solves (2.35). Using that ∂t+f(t, y)=e−F(t,y)◦∂t◦eF(t,y), where F t(t, y) =f(t, y), we may assume that f(x) =0. We shall first carry out the analysis of (2.36) under the assumption that the interval Jis symmetric about the origin and after a rescaling we may assume that J=[−1, 1]. Let Ω ⊂Cnbe open such that [−1, 1]t×{0}y⊂Ωand Ωis in the domain of definition of various symbols. Then let 0 <ε <1, r>0be small but fixed so that if we set Ω0=(t, y)∈Cn:|y| ε+|Im t| ε+|Re t|<1+r then Ω0⊂Ω. Consider the family of open sets, Ωs=(t, y)∈Cn:|y| ε+|Im t| ε+|Re t|<1+r−s, with 0 ≤s <r. Note that Ωsis a family of “nested neighborhoods” of [−1, 1] ×0in the sense of [49, Theorem 9.3], so that we have (i) if s1>s 2then Ωs1⊂Ωs2, (ii) there exists δ>0such that for all s1>s 2and all x ∈Ωs1we have the inclusion BCn(x, δ(s1−s2)) ⊂Ωs2. Given μ >0, we say that a ∈A μ, if a(x; h) =∞ k=0 ak(x)hk, ais holomorphic in Ω, such that for all s ∈(0, r), sup Ωs |ak|≤f(a, k) skkk,(2.37) where f(a, k)is the best constant for which (2.37)holds, and ∞ k=0 f(a, k)μk:= aμ<∞.(2.38) Now if a ∈A μfor some μ >0then f(a, k) ≤Ck+1, k=0, 1, 2, ..., and therefore, ais a classical analytic symbol on Ω0. Let (∂−1 ta)(t, y)= t 0 a(τ,y)dτ. (2.39) We shall need the following result, see [49, Theorem 9.3] and [46, Lemma 5.5].
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 23 Lemma 2.2. Let a ∈A μbe of the form a= ∞ k=2 hkak, and let b =(h∂t)−1a. Then bμ≤O 1 μaμ.(2.40) Proof. We have b= ∞ k=2 hk−1∂−1 tak= ∞ k=1 hkbk, where bk=∂−1 tak+1. Let us estimate supΩs|bk|. To that end, we write bk(x)=t 1 0 ak+1(σt,y)dσ We claim that for 0 ≤σ≤1, if x =(t, y) ∈Ωsthen (σt,y)∈Ωs+(1−σ)|t|.(2.41) Indeed, using that 0 <ε <1, we get |y| ε+σ|Im t| ε+σ|Re t|<1+r−s−(1 −σ) ε|Im t|−(1 −σ)|Re t| <1+r−(s+(1−σ)|t|), showing (σt, y) ∈Ωs+(1−σ)|t|as claimed. It follows from (2.37), (2.41)that for x ∈Ωs, we have |bk(x)|≤|t|f(a, k +1)(k+1) k+1 1 0 dσ (s+(1−σ)|t|)k+1 =f(a, k +1)(k+1) k+1|t| 1 0 dσ (s+σ|t|)k+1 =f(a, k +1)(k+1) k+1 |t| 0 dσ (s+σ)k+1
24 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 ≤f(a, k +1)(k+1) k+1 ∞ 0 dσ (s+σ)k+1 =f(a, k +1)(k+1) k+1 ∞ s dσ σk+1 =f(a, k +1)(k+1) k+1 ksk. Here we have used that k≥1. Thus, for any 0 <s <r, we get sup Ωs |bk|≤f(a, k +1)(1 + 1/k)kk(1 + 1/k)k sk≤2ef(a, k +1) skkk, and therefore by the definition of f(b, k), see (2.37), we have f(b, k)≤2ef(a, k +1),k=1,2,.... Using (2.38), we obtain that bμ= ∞ k=1 f(b, k)μk≤ ∞ k=1 2ef(a, k +1)μk=2e μaμ, establishing (2.40). Now applying to (h∂t)−1to (2.36), we get a+(h∂t)−1h2Q(x, Dx)a=∂−1 tv. (2.42) Here ∂−1 tvis a classical analytic symbol in Ω0. To proceed, we need the following result. Lemma 2.3. Let a ∈A μ. Then (h∂t)−1h2Q(x, Dx)a ∈A μwith (h∂t)−1h2Q(x, Dx)aμ≤O(μ)aμ(2.43) Proof. Writing a(x) =∞ k=0 ak(x)hk, we get h2Q(x, Dx)a= ∞ k=2 hkQ(x, Dx)ak−2. For s1>s 2, in view of the property (ii) of the “nested neighborhoods” Ωs, and (2.37), we obtain for k=2, 3, ... that sup Ωs1 |Q(x, Dx)ak−2|≤ C (s1−s2)2sup Ωs2 |ak−2|≤ C (s1−s2)2 f(a, k −2) sk−2 2 (k−2)k−2.(2.44) The Cauchy estimate was used here in the first inequality. Taking 0 <s 2=k−2 ks1<s 1 for k=3, 4, ..., we get from (2.44)that
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 31 sufficiently large, we see that γ1(αk)(tk) ∈Xint. As Xis compact, it has a positive injectivity radius Inj(X) >0. Here we have extended Xto a closed manifold to speak about the injectivity radius and the boundary will not cause any problems as γ1(αk)(tk) ∈Xint, for ksufficiently large. Now (3.8) implies that |tk| ≥Inj(X)for all ksufficiently large, which is a contradiction as tk→0. Thus, the claim (3.7) follows. The same is true for the family of geodesics γ2(α)for αin a possibly smaller neighborhood of α0. Finally, we claim there is a neighborhood V⊂ Uof α0such that for all α∈V, we have γ1(α)(t)=γ2(α)(s)=⇒t=s=0.(3.9) Indeed, otherwise, there exists αk→α0as k→∞, and tk∈[−T1(αk), T2(αk)], and sk∈[−S1(αk), S2(αk)] such that γ1(αk)(tk)=γ2(αk)(sk),(3.10) tk=0, and sk=0, for all k. Assuming as we may that tk→t0and sk→s0and passing to the limit in (3.10), we obtain that γ1(α0)(t0)=γ2(α0)(s0), and therefore, as γ1(α0)and γ2(α0)are admissible, we get t0=s0=0. Thus, we get γ1(αk)(tk)=γ2(αk)(sk)→x0, (αk)x=γ1(αk)(0) = γ2(αk)(0) →x0, as k→∞. Note that for ksufficiently large, all the points γ1(αk)(tk), γ1(αk)(0), γ2(αk)(sk), γ2(αk)(0) are in the interior of X. Therefore, |tk| ≥Inj(X)and |sk| ≥Inj(X) for ksufficiently large, as otherwise, the geodesics γ1(αk)and γ2(αk)would intersect at a geodesic ball centered at (αk)x. This contradicts the fact that tk→0and sk→0as k→∞, showing the claim. Hence, the pair of geodesics γ1(α), γ2(α)is admissible, for all α∈V. 4. Analytic families of exponentially accurate Gaussian beam quasimodes When proving Theorem 1.3 below, we shall need the following consequence of Theorem 1.12. Corollary 4.1. Let (X, g)be a compact Riemannian manifold of dimension n ≥2with smooth boundary, contained in an open real analytic manifold ( X, g)of the same dimension with greal analytic in X. Let α0=(x0, ξ0) ∈S∗Xint and let γ0:[−T1, T2] →X, 0 <T 1, T2<∞, be a unit speed nontangential geodesic such that γ0(0) =x0, and γ0does
32 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 not have self-intersections at x0. Let γ(α) :[−T1(α), T2(α)] →X, 0 <T 1(α), T2(α) <∞, α=(αx, αξ) ∈neigh(α0, S∗Xint), be a real analytic family of unit speed nontangential geodesics such that γ(α0) =γ0, and γ(α)(0) =αx. Let λ ∈R. Then there is a real analytic family of C∞functions v(x, α; h)on X, α∈neigh(α0, S∗Xint), 0 <h ≤1, and C>0such that supp (v( · , α; h)) is confined to a small neighborhood of γ(α)([−T1(α), T2(α)]) for each α, and (−h2Δg−(hs)2)v(·,α;h)L2(X)=O(e−1 Ch ),v(·,α;h)L2(X)1,(4.1) as h →0, uniformly in α. Here s =1 h+iλ. The local structure of the family v( · , α; h) in a neighborhood of αxis as follows: v(x, α;h)=h−(n−1) 4eisϕ(x,α)a(x, α;h), where ϕ(x, α)is real analytic in (x, α)for α∈neigh(α0, S∗Xint)and |x −αx| <1 c, c >0, and a(x, α; h)is an elliptic classical analytic symbol near (x0, α0). Furthermore, for tclose to 0and α∈neigh(α0, S∗Xint), we have ϕ(γ(α)(t),α)=t, ∇ϕ(γ(α)(t),α)= ˙γ(α)(t), Im (∇2ϕ(γ(α)(t))) ≥0,Im (∇2ϕ)|˙γ(α)(t)⊥>0. Proof. The functions T1(α)and T2(α) depend continuously on αin a small neighborhood of α0, and shrinking the neighborhood further we may assume that T1(α), T2(α)are bounded. Let ε >0be such that γ(α)(t) ∈ X\Xand γ(α)(t)has no self-intersection for t ∈[−T1(α) −2ε, −T1(α)) ∪(T2(α), T2(α) +2ε]for all α∈neigh(α0, S∗Xint). This choice of εis possible since γ(α)are non-tangential and depend smoothly on α. It follows from [28, Lemma 7.2] that γ(α)|[−T1(α)−ε,T2(α)+ε]self-intersects only at finitely many times tj(α), 1 ≤j≤N(α), with −T1(α)<t 1(α)<···<t N(α)(α)<T 2(α). First we claim that there is N0such that N(α) ≤N0<∞for all αin a small neighborhood of α0. This follows by inspection of the arguments in the proof of [28, Lemma 7.2]. Indeed, as explained in [28, Lemma 7.2], if γ(α)(t) =γ(α)(s)for some t =s then ˙γ(α)(t) =±˙γ(α)(s). Furthermore, if ris smaller than the injectivity radius of some closed manifold containing a fixed neighborhood of X∪γ(α)([−T1(α) −2ε, T2(α) +2ε] for α∈neigh(α0, S∗Xint), then any two distinct geodesic segments of length ≤rcan intersect in at most one point. Partitioning [−T1(α) −2ε, T2(α) +2ε]in disjoint intervals {Jk}K(α) k=1 of length ≤r, we get an injective map {(t, s)∈[−T1(α)−2ε, T2(α)+2ε]2:s<tand γ(α)(t)=γ(α)(s)} −→ { (k,l)∈{1,...,K(α)}2}, (t, s)→ (k,l) such that t∈Jk,s∈Jl. (4.2)
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 33 Since T1(α)and T2(α)are bounded for αin a small neighborhood of α0, we may assume that K(α)is bounded. Consequently, the cardinality of the set {(k, l) ∈{1, ..., K(α)}2} is bounded uniformly in α. The claim follows. We also set t0(α) := −T1(α) −εand tN+1(α) := T2(α) +ε. An inspection of the proof of [12, Lemma 3.5] allows us to conclude that there exists an open cover {(Uj(α), κj(α))}N(α)+1 j=0 of γ(α)([−T1(α) −ε, T2(α) +ε]) consisting of coordinate neighborhoods Uj(α)and real analytic diffeomorphisms κj(α), depending real analytically on α, such that the following properties hold, (i) κj(α)(Uj(α)) =Ij×B, where Ijare fixed open intervals and B=B(0, δ)is an open ball in Rn−1. Here δ>0can be taken arbitrarily small and the same for each Uj(α), uniformly for αclose to α0, (ii) κj(α)(γ(α)(t)) =(t, 0) for each t ∈Ij, (iii) tjonly belongs to Ijand Ij∩Ik=∅unless |j−k| ≤1, (iv) κj(α) =κk(α)on κ−1 j((Ij∩Ik) ×B). In particular, the open sets Uj(α)are bounded uniformly in αand contain a fixed open set. Following the proof of Theorem 1.12, and making use of the fact that the geodesics γ(α)do not have self-intersections at αx, for αclose to α0, we obtain the statement of Corollary 4.1. Remark 4.2. Let us also note that in general the number of self-intersecting times N(α) need not depend continuously on α. To this end, assume that the dimension of the manifold Xis >2and that the geodesic γ0in Xhas a self-intersection at some point x1∈γ0((−T1, T2)) so that x1=γ0(t1) =γ0(t2), t1<t 2. Then one can show that by means of a small perturbation, that one can unwind the loop in the direction orthogonal to the plane spanned by the velocity vectors ˙γ(t1)and ˙γ(t2). 5. Construction of families of harmonic functions based on Gaussian beam quasimodes Let (M, g)be a transversally anisotropic manifold of dimension n ≥3with transversal manifold (M0, g0), and assume that Mint 0and g0|Mint 0are real analytic. First assume, as we may, that (M, g)is embedded in a compact smooth manifold (N, g) without boundary of the same dimension, and let Ube open in Nsuch that M⊂U. Our starting point is the following Carleman estimate for −h2Δ, established in [10]. Proposition 5.1. Let φbe a limiting Carleman weight for −h2Δon U. Then for all 0 <h 1, we have huL2(N)≤Ceφ h(−h2Δ)e−φ huL2(N),C>0,(5.1)
34 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 for all u ∈C∞ 0(Mint). Using a standard argument, see [10], we convert the Carleman estimate (5.1)into the following solvability result. Proposition 5.2. Let φbe a limiting Carleman weight for −h2Δon U. If h >0is small enough, then for any v∈L2(M), there is a solution u ∈L2(M)of the equation eφ h(−h2Δ)e−φ hu=vin Mint, which satisfies uL2(M)≤C hvL2(M). Now as M⊂⊂ R ×Mint 0, there is a compact Riemannian manifold M0of dimension n − 1with smooth boundary such that M⊂⊂ R × M0⊂⊂ R ×Mint 0. Note that (Mint 0, g0|Mint 0) is an open real analytic manifold with real analytic metric, and we can use Corollary 4.1 to construct a real analytic family of Gaussian beam quasimodes along nontangential geodesics on M0. Let us write x =(x1, x)for local coordinates in R × M0. Let s=1 h+iλ, λ ∈R,λfixed. Note by [10, Lemma 2.9] that φ(x) =±x1is a limiting Carleman weight for −h2Δon U. We are interested in finding harmonic functions, −Δgu=0 in Mint,(5.2) having the form u=u(x, α;h)=e−sx1(v(x,α;h)+r(x, α;h)), where v=v(x, α; h)is the Gaussian beam quasimode constructed in Corollary 4.1 on the transversal manifold M0, associated to a nontangential unit speed geodesic γ(α)on M0depending analytically on α∈neigh(α0, S∗ Mint 0), and ris a remainder term. Thus, uis a solution of (5.2) provided that rsolves ex1 h(−h2Δg)e−x1 h(e−iλx1r)=−e−iλx1(−h2Δg0−(hs)2)v(x,α;h).(5.3) Proposition 5.2 and Corollary 4.1 imply that there is r=r( · ; α; h) ∈L2(M)solving (5.3)such that rL2(M)=O(e−1 Ch ),C>0,
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 35 as h →0, uniformly in α∈neigh(α0, S∗ Mint 0). To summarize, we have the following result. Proposition 5.3. Let s =1 h+iλ with λ ∈Rbeing fixed. For all h >0small enough, there are families of harmonic functions u1, u2∈L2(M), i.e. −Δguj=0in Mint, having the form u1(x, α;h)=e−sx1(v(x,α;h)+r1(x, α;h)), u2(x, α;h)=esx1(v(x,α;h)+r2(x, α;h)), where v=v( · , α; h)is the family of Gaussian beam quasimodes constructed in Corollary 4.1 on M0, and r∈L2(M)is such that rjL2(M)=O(e−1 Ch ), C>0, as h →0, uniformly in α∈neigh(α0, S∗ Mint 0), j=1, 2. 6. Proofs of Theorem 1.3 and Theorem 1.4 6.1. Some facts about analytic wave front sets We shall rely on the following characterization of the analytic wave front set, which we recall from [49, Definition 6.1] for the convenience of the reader. In our applications, we have m =n −1. Definition 6.1. Let α0=(x0, ξ0) ∈T∗Rm\{0}, and let ϕ(x, α), x ∈Rm, α=(αx, αξ) ∈ T∗Rm\{0}, be analytic defined in a neighborhood of (x0, α0)such that ϕ(x, α)|x=αx=0,ϕ x(x, α)|x=αx=αξ,(6.1) and Im ϕ(x, α)≥C0|x−αx|2,x,αreal,(6.2) for some C0>0. Let a(x, α; h)be an elliptic classical analytic symbol defined in a neighborhood of (x0, α0), and let u ∈D (X), where X⊂Rmis an open set containing x0. We have α0/∈WFa(u)if and only if there is a real neighborhood Uof α0and C>0 such that sup α∈U |Tu(α;h)|≤Ce−1 Ch ,(6.3) for 0 <h ≤1, where Tu(α;h)=eiϕ(x,α) ha(x, α;h)χ(x)u(x)dx, and χ ∈C∞ 0(X)is supported in a small neighborhood of x0and χ =1near x0.
36 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 Remark 6.2. It is established in [49, Proposition 6.2] that the condition (6.3)is independent of the choice of χ, a, and ϕ. Remark 6.3. Assume that ϕ, a, and usatisfy the same conditions as in Definition 6.1, and let ψ∈C∞ 0(Rn)be supported in a small neighborhood of 0and ψ=1near 0. We have α0/∈WFa(u)if and only if there is a real neighborhood Uof α0and C>0such that sup α∈ U | Tu(α;h)|≤ Ce−1 Ch ,(6.4) for 0 <h ≤1, where Tu(α;h)=eiϕ(x,α) ha(x, α;h)ψ(x−αx)u(x)dx. The condition (6.4)is independent of the choice of ψ. Remark 6.4. The condition (6.1)in Definition 6.1 and Remark 6.3 can be replaced by the following ϕ(x, α)|x=αx=f(α)realvalued,ϕ x(x, α)|x=αx=t0αξ,(6.5) for some fixed t0>0. Indeed, we apply Definition 6.1 and Remark 6.3 with ϕ(x, α) replaced by 1 t0(ϕ(x, α) −f(α)) and with hreplaced by h/t0. Remark 6.5. Since the wave front set WFa(u)is conic, we may restrict the attention in Definition 6.1 to ξ0∈Rmsuch that |ξ0| =1. 6.2. Proof of Theorem 1.3 Let α0=(x 0, ξ 0) ∈S∗Mint 0be generated by an admissible pair of geodesics γ1(α0) : [−T1(α0), T2(α0)] →M0and γ2(α0) :[−S1(α0), S2(α0)] →M0. As M⊂⊂ R ×Mint 0, there is a compact Riemannian manifold M0of dimension n −1with smooth boundary such that M⊂⊂ R × M0⊂⊂ R ×Mint 0, and x 0∈ Mint 0. Furthermore, we can choose M0 so that the geodesics γ1(α0)and γ2(α0)are nontangential on M0, and hence, γ1(α0)and γ2(α0)are admissible on M0. Then by Lemma 3.2, there exists a neighborhood Vof α0in S∗ Mint 0such that every point α=(αx, αξ) ∈Vis generated by an admissible pair of geodesics γ1(α) :[−T1(α), T2(α)] → M0and γ2(α) :[−S1(α), S2(α)] → M0, which depend realanalytically on α. Thus, for all α∈V, we have γ1(α)(0) = γ2(α)(0) = αx,(6.6) ˙γ1(α)(0) + ˙γ2(α)(0) = t0αξ,(6.7)
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 37 for some 0 <t 0<2fixed, γ1(α), γ2(α)do not have self-intersections at αx, and αxis the only point where γ1(α)and γ2(α) intersect, for all α∈V. Let s1=1 h+iλ and s2=1 h, where λ ∈R. Applying Corollary 4.1, we get vj(x, α; h), j=1, 2, Gaussian beam quasimodes on M0, associated to γj(α)on M0, depending real analytically on α∈Vsuch that vj(·,α;h)L2(M)1,(−h2Δg0−(hs)2)vj(·,α;h)L2(M)=O(e−1 Ch ),(6.8) as h →0, for some C>0, uniformly in α∈V. An application of Proposition 5.3 gives harmonic functions on Mhaving the form u1(x, α;h)=e−s1x1(v1(x,α;h)+r1(x, α;h)), u2(x, α;h)=es2x1(v2(x,α;h)+r2(x, α;h)),(6.9) where rjL2(M)=O(e−1 Ch ),C>0,(6.10) as h →0, uniformly in α∈V. Substituting the harmonic functions u1and u2given by (6.9)into (1.2), we get M fe−iλx1(v1(x,α;h)+r1)(v2(x,α;h)+r2)dVg=0.(6.11) Using (6.10)and (6.8), we see that M fe−iλx1v1(x,α;h)v2(x,α;h)dVg=O(e−1 Ch ),C>0,(6.12) uniformly in α∈V. Let us extend f∈L∞(M)by zero to (R ×M0) \Mand set f(λ, x)= ∞ −∞ e−iλx1f(x1,x )dx1 for the Fourier transform with respect to x1. Using the fact that dVg=dx1dVg0, we obtain from (6.12)that M0 f(λ, x)v1(x,α;h)v2(x,α;h)dVg0=O(e−1 Ch ),C>0,(6.13) uniformly in α∈V. Recalling that the geodesics γ1(α)and γ2(α) intersect at αxonly and that
38 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 supp (vj(·,α;h)) ⊂small neigh(γj(α)),j=1,2, we conclude from (6.13)that neigh(αx,M0) f(λ, x)v1(x,α;h)v2(x,α;h)g0(x)dx=O(e−1 Ch ),(6.14) uniformly in α∈V. Recalling that the geodesics γ1(α), γ2(α)do not have selfintersections at αx, by Corollary 4.1, we have in a small neighborhood of αx, v1(x,α;h)=h−(n−2) 4eis1ϕ1(x,α)a1(x,α;h), v2(x,α;h)=h−(n−2) 4eis2ϕ2(x,α)a2(x,α;h). (6.15) Here ϕj(x, α)are real analytic in (x, α)in a region of the form α∈Vand |x−αx| < 1/c, which is an open neighborhood of (x 0, α0). Furthermore, aj(x, α; h)are elliptic classical analytic symbols in a neighborhood of (x 0, α0), j=1, 2. It follows that the neighborhood of αxoccurring as the domain of integration in (6.14)can be taken to be fixed and independent of α. We also have for the geodesic parameters tand snear 0that (ϕ1) x(γ1(α)(t),α)=˙γ1(α)(t),Im ((ϕ1) xx(γ1(α)(t),α)) ≥0, Im ((ϕ1) xx(γ1(α)(t),α)|[˙γ1(α)(t)]⊥)>0,(6.16) and (ϕ2) x(γ2(α)(s),α)=˙γ2(α)(s),Im ((ϕ2) xx(γ2(α)(s),α)) ≥0, Im ((ϕ2) xx(γ2(α)(s),α)|[˙γ2(α)(s)]⊥)>0.(6.17) Now substituting (6.15)into (6.14), we see that neigh(αx,M0) eiϕ(x,α) h f(λ, x)a(x,α;h)dx=O(e−1 Ch ),h→0,(6.18) uniformly in α∈V. Here ϕ(x,α)=ϕ1(x,α)+ϕ2(x,α) (6.19) is analytic in a neighborhood of (x 0, α0), and a(x,α;h)=e−λϕ1(x,α)a1(x,α;h)a2(x,α;h)g0(x) is an elliptic classical analytic symbol in a neighborhood of (x 0, α0), since the product of two classical analytic symbols is a classical analytic symbol.
K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 39 We now claim that the phase function ϕ(x, α)in (6.19)satisfies the conditions (6.5) and (6.2). First, in view of (6.6)and (1.5), we have ϕ(x,α)|x=αx=ϕ1(γ1(α)(0),α)+ϕ2(γ2(α)(0),α)=0.(6.20) Using (6.16), (6.17), and (6.7), we get ϕ x(x,α)|x=αx=(ϕ1) x(γ1(α)(0),α)+(ϕ2) x(γ2(α)(0),α) =˙γ1(α)(0) + ˙γ2(α)(0) = t0αξ.(6.21) It follows from (6.20)and (6.21)that the condition (6.5)holds. Let us now check the condition (6.2). To this end, Taylor expanding ϕ(x, α)at x=αx, we get ϕ(x,α)=ϕ(αx,α)+ϕ x(αx,α)·(x−αx) +1 2ϕ xx(αx,α)(x−αx)·(x−αx)+O(|x−αx|3), and therefore, in view of (6.20)and (6.21), when xand αare real, we see that Im ϕ(x,α)=1 2Im ϕ xx(αx,α)(x−αx)·(x−αx)+O(|x−αx|3). Hence, the condition (6.2)is equivalent the following condition, Im ϕ xx(αx,α)>0.(6.22) Using (6.16), (6.17), and the fact that the vectors ˙γ1(α)(0) and ˙γ2(α)(0) are not parallel, we have Im ϕ xx(αx,α)=Im(ϕ1) xx(γ1(α)(0),α)+Im(ϕ2) xx(γ2(α)(0),α)>0, showing (6.22). Thus, by Remarks 6.3 and 6.4, in view of (6.18), we get α0/∈ WFa( f(λ, ·)) =WFa( f(−λ, ·)) for all λ ∈R. Noting that if (1.2)holds for f, it also holds for fand λ ∈Ris arbitrary, we get α0/∈WFa( f(λ, ·)) for all λ ∈R. This completes the proof of Theorem 1.3. 6.3. Proof of Theorem 1.4 Now since every point (x 0, ξ 0) ∈S∗Mint 0is generated by an admissible pair of geodesics, by Theorem 1.3, we get f(λ, ·)is real-analytic in Mint 0for all λ ∈R. The fact that f(λ, ·)has a compact support in Mint 0and that M0is connected implies that f(λ, ·) =0for all λ ∈R, and therefore, f=0. This completes the proof of Theorem 1.4.
40 K. Krupchyk et al. / Advances in Mathematics 403 (2022) 108362 Appendix A. Discussion related to Example 1.7 Let M0=S1×[0, a], with a >0, be a cylinder with its usual flat metric g0. The purpose of this Appendix is to show that every point (x0, ξ0) ∈S∗Mint 0is generated by an admissible pair of geodesics. We have T∗(S1×(0,a)) ≃T∗S1×T∗(0,a)≃(S1×R)×((0,a)×R)≃(S1×(0,a)) ×R2, and therefore, we may identify S∗ x0Mint 0with the unit circle S1in R2≃C. Given S1ξ0=(ξ01, ξ02) ≃ξ01 +iξ02, we set ξ1=eiα(ξ01 +iξ02)∈S1,ξ 2=e−iα(ξ01 +iξ02)∈S1,(A.1) with α∈(0, 2π)to be chosen. The geodesics γ1and γ2on M0such that γj(0) =x0and ˙γj(0) =ξj, j=1, 2, are given by γ1(t)=(x01 +ξ11t, x02 +ξ12t)∈R/2πZ×[0,a], γ2(s)=(x01 +ξ21s, x02 +ξ22s)∈R/2πZ×[0,a]. The geodesics γ1and γ2are nontangential provided that ξ12 =Im(eiα(ξ01 +iξ02)) = ξ02 cos α+ξ01 sin α=0, ξ22 =Im(e−iα(ξ01 +iξ02)) = ξ02 cos α−ξ01 sin α=0.(A.2) Note that if γ1and γ2are nontangential then they do not have self-intersections. We have in view of (A.1), ξ1+ξ2=(2cosα)ξ0,(A.3) and therefore, the property (ii) of Definition 1.2 follows with t0=2 cos α, provided that 0<cos α<1.(A.4) Note that γ1and γ2intersect each other if there exist tand ssuch that ξ11t−ξ21s∈2πZ,ξ 12t=ξ22s. (A.5) Now if we choose αso that |ξ11t−ξ21s|<2π, ξ12t=ξ22s, (A.6) then (A.5) implies that ξ1t =ξ2s, and therefore, |t| =|s|. In view of (A.3)and (A.4), we get t =s =0, and hence, x0is the only point of intersections of γ1and γ2.