Determining the first order perturbation of a polyharmonic operator on admissible manifolds
aa r X i v : . [ m a t h . A P ] A ug DETERMINING THE FIRST ORDER PERTURBATION OF APOLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS
YERNAT M. ASSYLBEKOV AND YANG YANG
Abstract.
We consider the inverse boundary value problem for the first or-der perturbation of the polyharmonic operator L g,X,q , with X being a W , ∞ vector field and q being an L ∞ function on compact Riemannian manifoldswith boundary which are conformally embedded in a product of the Euclideanline and a simple manifold. We show that the knowledge of the Dirichlet-to-Neumann determines X and q uniquely. The method is based on the construc-tion of complex geometrical optics solutions using the Carleman estimate forthe Laplace-Beltrami operator due to Dos Santos Ferreira, Kenig, Salo andUhlmann. Notice that the corresponding uniqueness result does not hold forthe first order perturbation of the Laplace-Beltrami operator. Introduction
Let (
M, g ) be a compact oriented Riemannian smooth manifold with boundary.Throughout this paper, the word “smooth” will be used as the synonym of “ C ∞ ”.Let ∆ g be the Laplace-Beltrami operator associated to the metric g which is givenin local coordinates by ∆ g u = | g | − / ∂∂x j (cid:18) | g | / g jk ∂u∂x k (cid:19) , where as usual ( g jk ) is the matrix inverse of ( g jk ), and | g | = det( g jk ). If F denotesa function or distribution space ( C k , L p , H k , D ′ , etc.), then we will denote by F ( M, T M ) the corresponding space of vector fields on M .Let X ∈ W , ∞ ( M, T M ) and q ∈ L ∞ ( M ). Consider the polyharmonic operator( − ∆ g ) m , m ≥
1, with the first order perturbation induced by X and q L g,X,q = ( − ∆ g ) m + X + q The operator L g,X,q equipped with the domain D ( L g,X,q ) = { u ∈ H m ( M ) : γu = 0 } = H m ( M ) ∩ H m ( M )is an unbounded closed operator on L ( M ) with purely discrete spectrum; see [8].Here and in what follows, γu := ( u | ∂M , ∆ g u | ∂M , . . . , ∆ m − g u | ∂M )is the Dirichlet trace of u , and H s ( M ) is the standard Sobolev space on M , s ∈ R . We make the assumption that 0 is not a Dirichlet eigenvalue of L g,X,q in M . Underthis assumption, for any f = ( f , . . . , f m − ) ∈ H m ( ∂M ) := Q m − j =0 H m − j − / ( ∂M ),the Dirichlet problem L g,X,q u = 0 in M,γu = f in ∂M, (1)has a unique solution u ∈ H m ( M ). Let ν be an outer unit normal to ∂M . Intro-ducing the Neumann trace operator e γ by e γ : H m ( M ) → m − Y j =0 H m − j − / ( ∂M ) , e γu = ( ∂ ν u | ∂M , ∂ ν ∆ g u | ∂M , . . . , ∂ ν ∆ m − g u | ∂M ) , we define the Dirichlet-to-Neumann map N g,X,q by N g,X,q : H m ( ∂M ) → m − Y j =0 H m − j − / ( ∂M ) , N g,X,q ( f ) = e γu, where u ∈ H m ( M ) is the unique solution to the boundary value problem (1). Letus also introduce the set of the Cauchy data for the operator L g,X,q C g,X,q = { ( γu, e γu ) : u ∈ H m ( M ) , L g,X,q u = 0 } . When 0 is not a Dirichlet eigenvalue of L g,X,q in M , the set C g,X,q is the graph ofthe Dirichlet-to-Neumann map N g,X,q .The inverse problem we are concerned in this paper is to recover the vector field X and the function q from the knowledge of the Dirichlet-to-Neumann map N g,X,q onthe boundary ∂M .When m = 1, the Dirichlet-to-Neumann map N g,X,q is invariant under gauge trans-formations in the following sense. Let ψ be a C ( M ) such that ψ | ∂M = 0 and ∂ ν ψ | ∂M = 0. Then e − iψ L g,X,q e iψ = L g, e X, e q , N g, e X, e q = N g,X,q , where e X = X + 2 ∇ ψ, e q = q + h X, ∇ ψ i g + |∇ ψ | g − i ∆ g ψ. Therefore, we may hope to recover X and q from boundary measurements onlymodulo the above gauge transformations.In the Euclidean setting, this inverse boundary value problem has been extensivelystudied, usually in the context of magnetic Schr¨odinger operators [15, 18, 19, 22, 24].In the case of Riemannian manifolds, this was proved in [3] for the special class ofso-called admissible manifolds.Let us now introduce admissible manifolds. For this we need the notion of sim-ple manifolds [21]. The notion of simplicity arises naturally in the context of theboundary rigidity problem [17]. OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 3
Definition 1.1.
A compact Riemannian manifold (
M, g ) with boundary is said tobe simple if the boundary ∂M is strictly convex, and for any point x ∈ M theexponential map exp x is a diffeomorphism from its maximal domain in T x M onto M . Definition 1.2.
A compact Riemannian manifold (
M, g ) with boundary of dimen-sion n ≥
3, is said to be admissible if it is conformal to a submanifold with boundaryof R × ( M , g ) where ( M , g ) is a simple ( n − M in R n , endowed with a metric which in some coordinateshas the form g ( x , x ′ ) = c ( x ) (cid:18) g ( x ′ ) (cid:19) , with c > g simple, is admissible.4. The class of admissible metrics is stable under C -small perturbations of g .It was shown in [12] that, in the Euclidean case, the obstruction to uniquenesscoming from the gauge equivalence when m = 1 can be eliminated by consideringoperators of higher order. The purpose of this paper is to extend this result for thecase of admissible manifolds.Our main result is as follows. Theorem 1.3.
Let ( M, g ) be admissible, and let m ≥ be an integer. Suppose that X , X ∈ W , ∞ ( R × M , T ( R × M )) ∩E ′ ( M, T M ) and q , q ∈ L ∞ ( M ) are such that is not a Dirichlet eigenvalue of L g,X ,q and L g,X ,q in M . If N g,X ,q = N g,X ,q ,then X = X and q = q . The key ingredient in the proof of Theorem 1.3 is the construction of complex geo-metric optics solutions for the operator L g,X,q with X being a W , ∞ vector fieldand q an L ∞ ( M ) function. For this, we use the method of Carleman estimateswhich is based on the corresponding Carleman estimate for the Laplacian due toDos Santos Ferreira, Kenig, Salo and Uhlmann [3].In Theorem 1.3, the condition that X = X = 0 on ∂M is needed to extend thevector fields X and X to a slightly larger simple manifold than M while preservingthe W , ∞ regularities. When more regularities on X j and q j ( j = 1 ,
2) are available,we can show a boundary determination result for the vector fields and thus dropsuch an assumption. This is the following theorem.
YERNAT M. ASSYLBEKOV AND YANG YANG
Theorem 1.4.
Let ( M, g ) be admissible, and let m ≥ be an integer. Supposethat X , X ∈ C ∞ ( M, T M ) and q , q ∈ C ∞ ( M ) are such that is not a Dirichleteigenvalue of L g,X ,q and L g,X ,q in M . If N g,X ,q = N g,X ,q , then X = X and q = q . Let π : R × M → M be the canonical projection π ( x , x ′ ) = x ′ . It is interestingto notice that the boundary determination becomes unnecessary if ( π ( M ) , g ) is asimple ( n − ∂M is connected. Theorem 1.5.
Let ( M, g ) be admissible, and let m ≥ be an integer. Suppose that X , X ∈ W , ∞ ( M, T M ) and q , q ∈ L ∞ ( M ) are such that is not a Dirichleteigenvalue of L g,X ,q and L g,X ,q in M . Suppose further that ( π ( M ) , g ) is asimple ( n − -dimensional manifold and ∂M is connected. If N g,X ,q = N g,X ,q ,then X = X and q = q . In the case of Euclidean space, the recovery of a zeroth order perturbation of thebiharmonic operator, that is when m = 2, has been studied by Isakov [11], wherea uniqueness result was obtained, similarly to the case of the Schr¨odinger operator.The recovery of a first order perturbation of the biharmonic operator from partialdata was studied in [13] in a bounded domain, and in [25] in an infinite slab. Higherorder operators occur in the areas of physics and geometry such as the study of theKirchhoff plate equation in the theory of elasticity, and the study of the Paneitz-Branson operator in conformal geometry; for more details see [7].Finally, we would like to remark that the problem considered in this paper can beviewed as generalization of the Calder´on’s inverse conductivity problem [1], knownalso as electrical impedance tomography. In the fundamental paper by Sylvester andUhlmann [23] it was shown that C conductivities can be uniquely determined fromboundary measurements. A corresponding result was proved by Dos Santos Ferreira,Kenig, Salo and Uhlmann [3] in the setting of admissible geometries.The structure of the paper is as follows. In Section 2 a Carleman estimate isderived for polyharmonic operators based on a similar estimate for the Laplace-Beltrami operator. Section 3 is devoted to the construction of complex geometricoptics solutions for the perturbed polyharmonic operator L g,X,q with X being a W , ∞ vector field and q ∈ L ∞ ( M ). Then the proof of Theorem 1.3 is given inSection 4. Attenuated ray transform is the subject of Section 5. In Section 6, weshow that the Dirichlet-to-Neumann map determines X on the boundary, this leadsto the proof of Theorem 1.4. Finally, the proof of Theorem 1.5 is given in Section7. 2. Carleman estimates for polyharmonic operators
Let (
M, g ) be a Riemannian manifold with boundary. In this section, following [3,14], we shall use the method of Carleman estimates to construct complex geometricoptics solutions for the equation L g,X,q u = 0 in M , with X being a W , ∞ vectorfield on M and q ∈ L ∞ ( M ).We start by recalling the definition of the Carleman weight for the semiclassicalLaplace-Beltrami operator − h ∆ g . Let U be an open manifold without boundart OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 5 such that M ⊂⊂ U and let ϕ ∈ C ∞ ( U, R ). Consider the conjugated operator P ϕ = e ϕ/h ( − h ∆ g ) e − ϕ/h . Following [3, 14], we say that ϕ is a limiting Carleman weight for − h ∆ g in U , if ithas non-vanishing differential, and if it satisfies the Poisson bracket condition { p ϕ , p ϕ } ( x, ξ ) = 0 when p ϕ ( x, ξ ) = 0 , ( x, ξ ) ∈ T ∗ M, where p ϕ is the semiclassical principal symbol of P ϕ .First we shall derive a Carleman estimate for the semiclassical polyharmonic oper-ator ( − h ∆ g ) m , where h > − h ∆ g , whichwe now proceed to recall the following [3, 14].We use the notation d Vol g for the volume form of ( M, g ). For any two functions u, v on M , define an inner product( u | v ) := Z M u ( x ) v ( x ) d Vol g ( x ) , and the corresponding norm will be denoted by k · k L ( M ) . We also write for short k∇ u k L ( M ) = k|∇ u |k L ( M ) = (cid:18)Z M |∇ u ( x ) | g d Vol g ( x ) (cid:19) / . We assume that (
M, g ) is embedded in a compact manifold (
N, g ) without boundary,and ϕ is a limiting Carleman weight on ( U, g ), where U is an open submanifold of N such that M ⊂⊂ U . By semiclassical spectral theorem one can define for s ∈ R the semiclassical Bessel potentials J s = (1 − h ∆ g ) s/ . One has J s J t = J s + t ,and Bessel potentials commute with any function of − ∆ g . Define for s ∈ R thesemiclassical Sobolev space associated to the norm k u k H s scl ( N ) = k J s u k L ( N ) . Our starting point is the following Carleman estimate for the semiclassical Laplace-Beltrami operator − h ∆ g which is due to Dos Santos Ferreira, Kenig, Salo andUhlmann [3, Lemma 4.3]. In what follows, A . B means that A ≤ CB where C > h and A, B . Proposition 2.1.
Let ( U, g ) be an open Riemannian manifold and ( M, g ) be asmooth compact Riemannian submanifold with boundary such that M ⊂⊂ U . Let ϕ be a limiting Carleman weight on ( U, g ) . Then for all h > small enough and s ∈ R , we have h k u k H s +1scl ( N ) . k e ϕ/h ( − h ∆ g ) e − ϕ/h u k H s scl ( N ) for all u ∈ C ∞ ( M ) . Next we shall derive a Carleman estimate for the operator L g,X,q with X being a W , ∞ vector field on M and q ∈ L ∞ ( M ). To that end we shall use Proposition 2.1with s = −
1. We have the following result.
YERNAT M. ASSYLBEKOV AND YANG YANG
Proposition 2.2.
Let ( U, g ) be an open Riemannian manifold and ( M, g ) be asmooth compact Riemannian submanifold with boundary such that M ⊂⊂ U . Let ϕ be a limiting Carleman weight on ( U, g ) . Suppose that X is a W , ∞ vector field on M and q ∈ L ∞ ( M ) . Then for all h > small enough, we have k u k L ( N ) . h m k e ϕ/h ( h m L g,X,q ) e − ϕ/h u k H − ( N ) , (2) for all u ∈ C ∞ ( M ) .Proof. Iterating the Carleman estimate in Proposition 2.2 m times, m ≥
2, we getthe following Carleman estimate for the polyharmonic operator, h m k u k H s + m scl ( N ) . k e ϕ/h ( − h ∆ g ) m e − ϕ/h u k H s scl ( N ) , for all u ∈ C ∞ (Ω), s ∈ R and h > s = − h m k u k H m − ( N ) . k e ϕ/h ( − h ∆ g ) m e − ϕ/h u k H − ( N ) , (3)for all u ∈ C ∞ ( N ) and h > m ≥
2, the following weakenedversion of (3) will be sufficient for our purposes h m k u k L ( N ) . k e ϕ/h ( − h ∆ g ) m e − ϕ/h u k H − ( N ) , (4)for all u ∈ C ∞ ( M ) and h > k e ϕ/h h m qe − ϕ/h u k L ( N ) . h m k q k L ∞ ( M ) k u k H ( N ) . (5)Note that e ϕ/h h m X ( e − ϕ/h u ) = − h m − h X, ∇ ϕ i g u + h m h X, ∇ u i g . Therefore,since m ≥ k h m − h X, ∇ ϕ i g u k L ( N ) ≤ h m − kh X, ∇ ϕ i g k L ∞ ( M ) k u k L ( N ) ≤ h m kh X, ∇ ϕ i g k L ∞ ( M ) k u k H ( N ) and k h m h X, h ∇ u i g k L ( N ) ≤ h m − k X k L ∞ ( M ) k u k H ( N ) ≤ h m k X k L ∞ ( M ) k u k H ( N ) . imply k e ϕ/h h m X ( e − ϕ/h u ) k L ( N ) . h m k u k H ( N ) . Combining this together with estimates (4) and (5), we get the result. (cid:3)
Set L ϕ := e ϕ/h ( h m L g,X,q ) e − ϕ/h . Then we have hL ϕ u, v i Ω = h u, L ∗ ϕ v i Ω , u, v ∈ C ∞ (Ω) , where L ∗ ϕ = e − ϕ/h ( h L g, − X, − div g X + q ) e ϕ/h is the formal adjoint of L ϕ , and h· , ·i M is the distribution duality on M . The estimate in Proposition 2.2 holds for L ∗ ϕ ,since − ϕ is a limiting Carleman weight as well.To construct the complex geometric optics solutions for the operator L g,X,q , we needto convert the Carleman estimate (2) for L ∗ ϕ into the following solvability result. OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 7
The proof is essentially well-known, and we include it here for the convenience ofthe reader. We shall use the following notation for the semiclassical Sobolev normon M k u k H ( M ) = k u k L ( M ) + k h ∇ u k L ( M ) . Proposition 2.3.
Let X be a W , ∞ vector field on M and q ∈ L ∞ ( M ) and assumethat m ≥ . If h > is small enough, then for any v ∈ L ( M ) there is a solution u ∈ H ( M ) of the equation e ϕ/h h m L g,X,q e − ϕ/h u = v satisfying k u k H ( M ) ≤ Ch m k v k L ( M ) . Proof.
Let v ∈ H − ( M ) and let us consider the following complex linear functional, L : L ∗ ϕ C ∞ ( M ) → C , L ∗ ϕ w
7→ h w, v i M . By the Carleman estimate (2) for L ∗ ϕ , the map L is well-defined. Let w ∈ C ∞ ( M ).Then we have | L ( L ∗ ϕ w ) | = |h w, v i M | ≤ k w k L ( N ) k v k L ( M ) . h m k v k L ( M ) kL ∗ ϕ w k H − ( N ) . By the Hahn-Banach theorem, we may extend L to a linear continuous functional ˜ L on L ( N ), without increasing its norm. By the Riesz representation theorem, thereexists u ∈ H ( R n ) such that for all ψ ∈ H − ( R n ),˜ L ( ψ ) = h ψ, u i R n , and k u k H ( R n ) . h m k v k L (Ω) . Let us now show that L ϕ u = v in Ω. To that end, let w ∈ C ∞ (Ω). Then hL ϕ u, w i Ω = h u, L ∗ ϕ w i R n = ˜ L ( L ∗ ϕ w ) = h w, v i Ω = h v, w i Ω . The proof is complete. (cid:3) Complex geometric optics solutions
Let ϕ be a limiting Carleman weight in an admissible manifold ( M, g ). We willconstruct solutions to L g,X,q u = 0 in M of the form u = e − ( ϕ + iψ ) /h ( a + r ) , (6)where a is an amplitude, r is a correction term which is small when h > ψ is a real valued phase.Set ρ = ϕ + iψ for the complex valued phase. Consider the conjugated operator P ρ = e ρ/h h m L g,X,q e − ρ/h , which has the following expression P ρ = ( − h ∆ g − |∇ ρ | g + h ∆ g ρ + 2 h ∇ ρ ) m + h m X − h m − h−∇ ρ, X i g + h m q. YERNAT M. ASSYLBEKOV AND YANG YANG
Here and in what follows, the norm | · | g and the inner product h· , ·i g are extendedto complex valued tangent vectors by h ζ, η i g = h Re ζ, Re η i g −h Im ζ, Im η i g + i ( h Re ζ, Im η i g + h Im ζ, Re η i g ) , | ζ | g = h ζ, ζ i g . Since m ≥
2, in order to get e ϕ/h h m L g,X,q ( e − ϕ/h a ) = O ( h m +1 ) , in L ( M ), we should choose ρ satisfying the following eikonal equation |∇ ρ | g = 0 in M, (7)and choose a ∈ C ∞ ( M ) satisfying the following transport equation(2 ∇ ρ + ∆ g ρ ) m a = 0 in M. (8)Recall that ( M, g ) is conformally embedded in R × ( M , g ), where ( M , g ) is somesimple ( n − M with a slightlylarger simple manifold. Therefore, we can and shall assume that for some simple( D, g ) ⊂⊂ ( M int0 , g ) one has( M, g ) ⊂⊂ ( R × D int , g ) ⊂ ( R × M int0 , g ) . (9)Note that R × M has global coordinate chart in which the metric g has the followingform g ( x ) = c ( x ) (cid:18) g ( x ′ ) (cid:19) , (10)where c > g is simple. A natural choice of the limiting Carleman weight is ϕ ( x ) = x . Then the equation (7) for the complex valued phase ρ becomes |∇ ψ | = 1 c , ∂ x ψ = 0 . This equation will be solved using special coordinates on (
M, g ). This is based on theso-called polar coordinates on the transversal simple manifold ( M , g ). Let ω ∈ D be such that ( x , ω ) / ∈ M for all x . Points of M have the form x = ( x , r, θ ) where( r, θ ) are polar normal coordinates in ( D, g ) with center ω . That is, x ′ = exp Dω ( rθ )where r > θ ∈ S n − . In terms of these coordinates the metric g has the form g ( x , r, θ ) = c ( x , r, θ ) m ( r, θ ) , where m is a smooth positive definite matrix.We solve (7) by simply taking ψ ( x ) = ψ ω ( x ) = r . Thus, the complex valued phasehas the form ρ = x + ir and its gradient is ∇ ρ = c ∂ , where ∂ = 12 (cid:18) ∂∂x + i ∂∂r (cid:19) . Next, we solve transport equation (8). In the coordinates ( x , r, θ ) equation (8)becomes (cid:18) c ∂ + 1 c log | g | c (cid:19) m a = 0 . OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 9
Consider a as the function having the following form a = | g | − / c / a ( x , r, θ ) b ( θ )where b is smooth and a is such that ∂a = ca for some a satisfying ∂a = 0.Note that (6) will be a solution for L g,X,q u = 0 if P ρ ( a + hr ) = 0. Then, with thechoice of ϕ and ψ made above, this equation is equivalent to the following e ϕ/h h m L g,X,q e − ϕ/h ( e − iψ/h hr ) = − e − iψ/h ( h m L g,X,q a + h m − h∇ ρ, X i g a ) . This will be solved by using Proposition 2.3. We find r ∈ H ( M ) satisying k r k H ( M ) = O (1) . The discussion of this section can be summarized in the following proposition.
Proposition 3.1.
Assume that ( M, g ) satisfies (9) and (10) , and let m ≥ be aninteger. Suppose that X is a W , ∞ vector field on M and q ∈ L ∞ ( M ) . Let ω ∈ D such that ( x , ω ) / ∈ M for all x . If ( r, θ ) are polar normal coordinates in ( D, g ) with center ω , then the equation L g,X,q u = 0 in M has a solution of the form u = e − h ( ϕ + iψ ) ( | g | − / c / a ( x , r, θ ) b ( θ ) + hr ) , where ∂a = ca for some a depending on ( x , r ) and satisfying ∂a = 0 , b issmooth and the remainder term r ∈ H ( M ) such that k r k H ( M ) = O (1) . Remark 3.2.
In fact, we need complex geometric optics solutions belonging to H m ( M ). Such solutions can be obtained in the following way. Extend X and q smoothly to R × M . By elliptic regularity, the complex geometric optics solutionsconstructed as above in M will belong to H m ( M ). Remark 3.3.
It is easy to check that if a depends only on ( x , r ) and satisfies ∂a = 0, then the equation L g,X,q u = 0 in M has a solution as in Proposition 3.1.4. Proof of Theorem 1.3
Let (
M, g ) be an admissible manifold and let m ≥ Proposition 4.1.
Let
M, M be compact manifolds with boundary such that M ⊂⊂ M , and let m ≥ be an integer. Assume that X , X are W , ∞ vector fields on M and q , q ∈ L ∞ ( M ) . Suppose that X = X , q = q in M \ M. If C Mg,X ,q = C Mg,X ,q , then C M g,X ,q = C M g,X ,q , where C M g,X j ,q j denotes the set ofthe Cauchy data for L g,X j ,q j in M , j = 1 , . Proof.
Let u ∈ H m ( M ) be a solution of L g,X ,q u = 0 in M . Since C Mg,X ,q = C Mg,X ,q , there exists v ∈ H m ( M ), solving L g,X ,q v = 0 in M , and satisfying γv = γu in ∂M and ˜ γv = ˜ γu in ∂M . Setting v = ( v in M,u in M \ M, we get v ∈ H m ( M ) and L g,X ,q v = 0 in M . Thus, C M g,X ,q ⊂ C M g,X ,q . Exactlythe same way but in the other direction finishes the proof. (cid:3) The second ingredient is the derivation of the following integral identity based onthe assumption that C Mg,X ,q = C Mg,X ,q . Proposition 4.2.
Let ( M, g ) be a compact Riemannian manifold with boundary,and let m ≥ be an integer. Assume that X , X are W , ∞ vector fields on M and q , q ∈ L ∞ ( M ) . If C g,X ,q = C g,X ,q , then Z M [ h X − X , v ∇ u i g + ( q − q ) uv ] d Vol g ( x ) = 0 , for any u, v ∈ H m ( M ) satisfying L g, − X , − div g X + q v = 0 and L g,X ,q u = 0 in M .Proof. We will use the following consequence of the Green’s formula, see [8],( L g,X ,q u, v ) L ( M ) = ( u, L ∗ g,X ,q v ) L ( M ) (11)for all u, v ∈ H m ( M ) such that γu = γv = 0, where L ∗ g,X ,q = L g, − X , − div g X + q .Now, let u, v ∈ H m ( M ) be such that L g, − X , − div g X + q v = 0 and L g,X ,q u = 0 in M . The hypothesis that N g,X ,q = N g,X ,q implies the existence of ˜ u ∈ H m ( M )such that L g,X ,q ˜ u = 0 and γ ˜ u = γu , e γ ˜ u = e γu . We have L g,X ,q ( u − ˜ u ) = ( X − X )˜ u + ( q − q )˜ u. Using (11), this implies the result. (cid:3)
According to hypothesis, that X = X in ( R × M ) \ M int . We also extend q and q to R × M by zero outside M int . Let, as in Section 3, ( D, g ) ⊂⊂ ( M int0 , g ) be simplesuch that ( M, g ) ⊂⊂ ( R × D int , g ) ⊂ ( R × M int0 , g ). Let ( M , g ) be also admissibleand simply connected such that ( M, g ) ⊂⊂ ( M int1 , g ) and ( M , g ) ⊂⊂ ( R × D int , g ).According to Proposition 4.1, we know that C M g,X ,q = C M g,X ,q is true.According to Proposition 4.2 the following integral identity holds for all u, v ∈ H m ( M ) satisfying L g,X ,q u = 0 and L g, − X , − div g X + q v = 0 in M , respectively: Z M [ h X − X , v ∇ u i g + ( q − q ) uv ] d Vol g ( x ) = 0 . (12)The main idea of the proof of Theorem 1.3 is to use the integral identity (12)with u, v ∈ H m ( M ) being complex geometric optics solutions for the equations OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 11 L g,X ,q u = 0 and L g, − X , − div g X + q v = 0 in M , respectively. We use Proposi-tion 3.1, Remark 3.2 and Remark 3.3 to choose solutions of the form u = e − h ( x + ir ) ( | g | − / c / e iλ ( x + ir ) b ( θ ) + hr ) ,v = e h ( x + ir ) ( | g | − / c / + hr ) , where λ ∈ R and k r j k H ( M ) = O (1), j = 1 ,
2. Substituting these solutions in (12),multiplying the resulting equality by h and letting h →
0, we getlim h → Z M h X − X , ∇ ρ i g uv d Vol g ( x ) = 0 , where ρ = x + ir . Let us rewrite the integral in ( x , r, θ ) coordinates. Write X = X − X , and let X ♭ be a 1-form dual to X . Let X ♭x and X ♭r denote thecomponents of X ♭ in the x and r coordinates. Then Z R Z M ,x ( X ♭x + iX ♭r ) e iλ ( x + ir ) b ( θ ) dr dθ dx = 0 , (13)where M ,x = { ( r, θ ) : ( x , r, θ ) ∈ M } . Since X = X in ( R × M ) \ M int , we mayassume that the integral is over R × D . Taking x -integral inside gives Z S n − Z e − λr (cid:18)Z R e iλx ( X ♭x + iX ♭r )( x , r, θ ) dx (cid:19) dr dθ = 0 . Define f ( x ′ ) = Z R e iλx X ♭x ( x , x ′ ) dx , α ( x ′ ) = n X j =2 (cid:18)Z R e iλx X ♭j ( x , x ′ ) dx (cid:19) dx j . Then f ∈ W , ∞ ( D ) and α is a 1-form which is W , ∞ on D , and the integral identityabove can be rewritten as Z S n − Z e − λr [ f ( γ w,θ ( r )) + iα ( ˙ γ w,θ ( r ))] dr dθ = 0 , where γ w,θ is a geodesic in ( D, g ) issued from the point ω in the direction θ . For ω ∈ ∂D , the integral above is related to the attenuated ray transform of function f and 1-form iα in D with constant attenuation − λ . Therefore, by varying thepoint ω in Proposition 3.1 on ∂D and using Proposition 5.1 in Section 5, for smallenough λ , we have f = − λp and α = − idp where p ∈ W , ∞ ( D ) and p | ∂D = 0. Thedefinition of α and analyticity of the Fourier transform imply that ∂ k X ♭j − ∂ j X ♭k = 0 , j, k = 2 , . . . , n. Also Z e iλx ( ∂ j X ♭ − ∂ X ♭j )( x , x ′ ) dx = ∂ j f + iλα j = 0 , showing that dX ♭ = 0 in M . Since M is simply connected, there is φ ∈ W , ∞ ( M )such that φ | ∂M = 0 and X = ∇ φ .Since X = X − X in the neighborhood of the boundary ∂M , we conclude that φ is a constant, say c ∈ C , on ∂M . Therefore, considering φ − c , we may and will assume that φ = 0 on ∂M . Since X = X in ( R × M ) \ M int , we also may andshall assume that φ is zero outside M . In particular, φ is compactly supported.Next, we show that X = X . For this, using Proposition 3.1 and Remark 3.2,consider u = e − h ( x + ir ) ( | g | − / c / e iλ ( x + ir ) b ( θ ) + hr ) ,v = e h ( x + ir ) ( | g | − / c / a + hr ) , where a satisfies ∂a = c . Such a can be constructed using Cauchy’s integralformula in [6] as a ( x , r, θ ) = a ( ρ, θ ) = 12 π Z B c ( z, θ ) z − ρ dz ∧ dz, for all θ ∈ S n − , where ρ = x + ir , B is a bounded domain in the upper half plane H ⊂ C suchthat the map B × S n − → R × M , ( x , r, θ ) ( x , exp Dω ( rθ )) covers M and theboundary ∂ B is piecewise smooth. Here and in what follows, ω ∈ D such that ω ∈ M in Proposition 3.1.Substituting these solutions and X − X = ∇ φ in (12), multiplying the resultingequality by h and letting h →
0, we getlim h → Z M h∇ φ, ∇ ρ i g uv d Vol g ( x ) = 0 , where ρ = x + ir . Rewriting the integral in ( x , r, θ ) coordinates and taking x -integral inside, we obtain2 Z S n − (cid:18)Z ∞ Z R ∂φ a e iλ ( x + ir ) b ( θ ) dx dr (cid:19) dθ = 0 . Since φ is compactly supported, integrating by parts, in ( x , r ), gives0 = − Z S n − (cid:18)Z ∞ Z R ∂φ a e iλ ( x + ir ) b ( θ ) dx dr (cid:19) dθ = Z S n − (cid:18)Z ∞ Z R φ ∂a e iλ ( x + ir ) b ( θ ) dx dr (cid:19) dθ = Z S n − (cid:18)Z ∞ Z R φc e iλ ( x + ir ) b ( θ ) dx dr (cid:19) dθ. (14)Set Φ λ ( r, θ ) = Z R φc e iλx dx , i.e. Φ λ is the Fourier transform of φc in x -variable. Then (14) can be written as Z S n − Z e − λr Φ λ ( γ ω,θ ( r )) b ( θ ) dr dθ = 0 . By varying the point ω in Proposition 3.1 on ∂D and using [5, Lemma 5.1], for smallenough λ , we have Φ λ = 0. Since φc is compactly supported, its Fourier transformΦ λ is analytic. Therefore we obtain, φ = 0 which shows that X = X . OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 13
To show that q = q , consider (12) with X = X which becomes Z M ( q − q ) uv d Vol g ( x ) = 0 (15)holds for all u, v ∈ H m ( M ) satisfying L g,X ,q u = 0 and L g, − X , − div g X + q v = 0in M , respectively. Use Proposition 3.1, Remark 3.2 and Remark 3.3 to choosesolutions of the form u = e − h ( x + ir ) ( | g | − / c / e iλ ( x + ir ) b ( θ ) + hr ) ,v = e h ( x + ir ) ( | g | − / c / + hr ) , where λ ∈ R and k r j k H ( M ) = O (1), j = 1 ,
2. Substituting these solutions in (15)and letting h →
0, we get Z R Z M ,x e iλ ( x + ir ) ( q − q ) c ( x , r, θ ) b ( θ ) dr dθ dx = 0 . Taking x -integral inside and varying b gives Z S n − Z ∞ Z R e iλ ( x + ir ) ( q − q ) c ( x , r, θ ) dx dr dθ = 0 . Set Q λ ( r, θ ) = Z R ( q − q ) c e iλx dx , i.e. Q λ is the Fourier transform of ( q − q ) c in x -variable. Then, as in the case ofΦ λ , one can show that Q λ = 0 for all λ small enough. We have extended q and q to R × M by zero outside M , which implies that q − q is compactly supported.Hence, Q λ is analytic. This together with Q λ = 0 for all λ small enough, allows usto conclude that q = q .5. Attenuated ray transform
The aim of this section is to prove the following proposition which was used in theproof of Theorem 1.3. We will closely follow the arguments in [5].
Proposition 5.1.
Let ( D, g ) be an ( n − -dimensional simple manifold. Let f ∈ L ∞ ( D ) and α be a -form which is L ∞ on D . Consider the integrals Z S n − Z τ ( ω,θ )0 [ f ( γ ω,θ ( r )) + α k ( γ ω,θ ( r )) ˙ γ kω,θ ( r )] e − λr b ( θ ) dr dθ, where ( r, θ ) are polar normal coordinates in ( D, g ) centered at some ω ∈ ∂D , and τ ( ω, θ ) is the time when the geodesic r ( r, θ ) exits D . If | λ | is sufficiently small,and if these integrals vanish for all ω ∈ ∂D and all b ∈ C ∞ ( S n − ) , then there is p ∈ W , ∞ ( D ) with p | ∂D = 0 such that f = − λp and α = dp . This is related to the injectivity of attenuated ray transform acting on function and1-form on D . Let us introduce some notions and facts; see [21] for more details. By SD we will denote its unit sphere bundle SD := { ( x, v ) ∈ T D : | v | g ( x ) = 1 } . On the boundary of D , we consider the set of inward and outward unit vectors definedas ∂ + SD = { ( x, v ) ∈ SD : x ∈ ∂D, h v, ν ( x ) i g ( x ) ≤ } ,∂ − SD = { ( x, v ) ∈ SD : x ∈ ∂D, h v, ν ( x ) i g ( x ) ≥ } , where ν is the unit outer normal to ∂D . The geodesics entering D can be param-eterized by ∂ + SD . For any ( x, v ) ∈ SD the first non-negative exit time of thegeodesic γ x,v , with x = γ x,v (0), v = ˙ γ x,v (0), will be denoted as τ ( x, v ). Simplic-ity assumption guarantees that τ ( x, v ) is finite for all ( x, v ) ∈ SD . We also write φ t ( x, v ) = ( γ x,v ( t ) , ˙ γ x,v ( t )) for the geodesic flow.We endow the unit sphere bundle SD with its usual Liouville (local product) mea-sure d Σ n − , and endow the bundle ∂ + SD with its standard measure d Σ n − . By dσ x we denote the measure on S x D .Let f be a function and α be a 1-form on D . The geodesic ray transform of f and α , with constant attenuation − λ , is defined as T λ [ f, α ]( x, v ) = Z τ ( x,v )0 [ f ( γ x,v ( t )) + α k ( γ x,v ( t )) ˙ γ kx,v ( t )] e − λt dt, ( x, v ) ∈ ∂ + SD.
In Propostion 5.1, if f and α were a continuous function and 1-form, respectively,one could choose b ( θ ) to approximate a delta function at fixed angles θ and obtainthat Z τ ( ω,θ )0 [ f ( γ ω,θ ( r )) + α k ( γ ω,θ ( r )) ˙ γ kω,θ ( r )] e − λr dr = 0for all ω ∈ ∂D and all θ ∈ S n − . This would imply that T λ [ f, α ]( x, v ) = 0 for all ( x, v ) ∈ ∂ + SD.
We will use the following result from [3, Theorem 7.1].
Proposition 5.2.
Let ( D, g ) be a compact simple manifold with smooth boundary.There exists ε > such that the following assertion holds for a real number λ with | λ | < ε : If f ∈ C ∞ ( D ) and α be a smooth -form on D , then T λ [ f, α ]( x, v ) = 0 for all ( x, v ) ∈ ∂ + SD implies the existence of p ∈ C ∞ ( D ) with p | ∂D = 0 such that f = − λp and α = dp . The previous argument together with the above theorem proves Proposition 5.1 forsmooth f and α . However, this requires f and α to be C ∞ -smooth in D and it isnot obvious how to do this when f and α are L ∞ on D . We resolve this problemby using duality and the ellipticity of the normal operator T ∗ λ T λ .In the space of functions on ∂ + SD define the inner product( h, h ′ ) L µ ( ∂ + SD ) := Z D h h ′ dµ where dµ ( x, v ) = h v, ν i g ( x ) d Σ n − . Denote the corresponding Hilbert space andthe norm by L µ ( ∂ + SD ) and k · k L µ ( ∂ + SD ) , respectively. We will also write h ψ ( x, v ) = h ( φ − τ ( x, − v ) ( x, v )) , ( x, v ) ∈ SD, for h ∈ C ∞ ( ∂ + SD ). OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 15 If F is a notation for a function space ( C k , L p , H k , etc.), then we will denote by F ( D ) the corresponding space of pairs [ f, α ] with f a function and α a 1-form on D . In particular, L ( D ) is the space of square integrable pairs [ f, α ], and we endowthis space with the inner product([ f, α ] , [ f ′ , α ′ ]) L ( D ) = Z D ( f f ′ + h α, α ′ i g ) d Vol g . Lemma 5.3. If f ∈ C ∞ ( D ) , α is a smooth -form on D and h ∈ C ∞ (( ∂ + SD ) int ) ,then ( T λ [ f, α ] , h ) L µ ( ∂ + SD ) = ([ f, α ] , T ∗ λ h ) L ( D ) , where T ∗ λ h is a pair defined as T ∗ λ h ( x ) = (cid:20)Z S x D h ψ ( x, v ) e − λτ ( x, − v ) dσ x ( v ) , Z S x D v k h ψ ( x, v ) e − λτ ( x, − v ) dσ x ( v ) (cid:21) . Proof.
By Santal´o formula (see [21] or [2])( T λ [ f, α ] ,h ) L µ ( ∂ + SD ) = Z ∂ + SD Z τ ( x,v )0 [ f ( γ x,v ( t )) + α k ( γ x,v ( t )) ˙ γ kx,v ( t )] e − λt dt h ( x, v ) dµ ( x, v )= Z SD [ f ( x ) + α k ( x ) v k ] h ψ ( x, v ) e − λτ ( x, − v ) d Σ n − ( x, v )= Z D f ( x ) (cid:18)Z S x D h ψ ( x, v ) e − λτ ( x, − v ) dσ x ( v ) (cid:19) d Vol g ( x )+ Z D α k ( x ) (cid:18)Z S x D v k h ψ ( x, v ) e − λτ ( x, − v ) dσ x ( v ) (cid:19) d Vol g ( x ) . This proves the statement. (cid:3)
Proof of Proposition 5.1.
First, we extend (
D, g ) to a slightly larger simple man-ifold and to extend both f and α by zero. Then f and α are still in L ∞ , and inparticular in L p for all 1 < p < ∞ . In this way we can assume that both f and α are compactly supported in D int .We let b also depend on ω and change notations to write the assumption in theform Z S x D Z τ ( x,v )0 e − λt [ f ( γ x,v ( t )) + α k ( γ x,v ( t )) ˙ γ kx,v ( t )] b ( x, v ) dt dσ x ( v ) = 0for all x ∈ ∂D and b ∈ C ∞ (( ∂ + SD ) int ). Let e D be a compact submanifold of D with boundary such that ( e D, g ) is also simple, e D ⊂⊂ D int and supports of f and α are compact subsets of e D int . Note that α is L ∞ on D (and in particularbeing in L on D ) implies that in particular δα ∈ H − ( e D ). Then we obtain thesolenoidal decomposition α = α s + dp on e D , where δ g α s = 0 and p ∈ H ( e D )with − ∆ g p = δα . Here δ g α = ∇ ig α i , where ∇ g is the covariant derivative corresponding to the metric g . Extend p to D by zero, so that p ∈ H ( D ) with p = 0 in D \ e D . An integration by parts shows that we have Z S x D Z τ ( x,v )0 e − λt [ f ( γ x,v ( t )) + λp ( γ x,v ( t )) + α sk ( γ x,v ( t )) ˙ γ kx,v ( t )] b ( x, v ) dt dσ x ( v ) = 0(16)for all x ∈ ∂D and b ∈ C ∞ (( ∂ + SD ) int ). Next, we make the choice b ( x, v ) = h ( x, v ) µ ( x, v ) for h ∈ C ∞ (( ∂ + SD ) int ) and integrate (16) over ∂D and get( T λ [ f + λp, α s ] , h ) L µ ( ∂ + SD ) = 0 . We are now in the same situation as in the proof of Lemma 5.3, and using theSantal´o formula implies ([ f + λp, α s ] , T ∗ λ h ) L ( D ) = 0for all h ∈ C ∞ (( ∂ + SD ) int ). Note that the last integral is absolutely convergentbecause f ∈ L ∞ ( D ) and α is 1-form which is L ∞ on D , and also the previous stepsare justified by Fubini’s theorem.It remains to choose h = T λ [ ϕ, β ] for ϕ ∈ C ∞ ( D int ) and β being C ∞ -smooth 1-formin D int , to obtain that ([ f + λp, α s ] , T ∗ λ T λ [ ϕ, β ]) L ( D ) = 0 . Since T ∗ λ T λ is self-adjoint, we have( T ∗ λ T λ [ f + λp, α s ] , [ ϕ, β ]) L ( D ) = 0for all ϕ ∈ C ∞ ( D int ) and for all C ∞ -smooth 1-form β in D int . Therefore, T ∗ λ T λ [ f + λp, α s ] = 0. By [10, Proposition 1], T ∗ λ T λ is an elliptic pseudodifferential operatorof order − D int . Here, ellipticity of T ∗ λ T λ is in the sense that whenever f ′ , α ′ are in L ( D int ) and T ∗ λ T λ [ f ′ , α ′ ] = 0 and δ g α ′ = 0, then f ′ , α ′ are smooth. Since f + λp and α s were compactly supported in D int , this implies that f + λp and α s are smooth and compactly supported in D int . Hence f + λp and α s are smooth in D and compactly supported in D int . Now we can use the argument for smooth f and α given above, together with Proposition 5.2 to conclude that f = − λp − λψ and α = α s + dp = dψ + dp for some ψ ∈ C ∞ ( D ) with ψ | ∂D = 0. To finish theproof, it remains to show that p ∈ W , ∞ ( D ). But this is clear from dp = α − α s and from α is L ∞ on D and α s is C ∞ on D . (cid:3) Boundary Determination and proof of Theorem 1.4
In this part we show boundary determination of the vector field X . For the gener-ality of the statement we will assume the knowledge of the Cauchy data set C g,X,q .It is easy to see that when 0 is not a Dirichlet eigenvalue of L g,X,q , knowledge ofthe Cauchy data set is equivalent to knowledge of the Dirichlet-to-Neumann map N g,X,q . Moreover, we can determine not only the boundary values of X , but alsothe boundary values of q . This is the following proposition. Proposition 6.1.
Let ( M, g ) be admissible, and let m ≥ be an integer. Supposethat X is a C ∞ vector field on M and q ∈ C ∞ ( M ) . Then the knowledge of the OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 17
Cauchy data set C g,X,q determines the boundary values of X and the boundaryvalues of q . To prove this proposition, it suffices to show that for any p ∈ ∂M , C Mg,X,q determines X ( p ) and q ( p ). In the following we will consider this local problem.Fix a point p ∈ ∂M and let ( x ′ , x n ) be the boundary normal coordinates near p ,where x ′ = ( x , . . . , x n − ). In these coordinates ∂M corresponds to { x n = 0 } , thevector field X becomes the differential operator X = X j ∂∂x j , and the metric tensorcan be written as g = g αβ dx α ⊗ dx β + dx n ⊗ dx n . Here the in the following we use the convention that Greek indices run from 1 to n − n . Denote D j = i ∂∂x j , then the Laplace-Beltramioperator in the boundary normal coordinates takes the form − ∆ g = D n + iE ( x ) D n + Q ( x, D x ′ ) + Q ( x, D x ′ ) (17)with E, Q , Q given by E ( x ) = 12 g αβ ∂ n g αβ , (18) Q ( x, D x ′ ) = g αβ D α D β , (19) Q ( x, D x ′ ) = − i ( 12 g αβ ∂ α (log | g | ) + ∂ α g αβ ) D β . (20)Next we would like to write the 2 m order equation L g,X,q u = ( − ∆ g ) m u + Xu + qu = 0 in M, m ≥ u = u, u = ( − ∆ g ) u, . . . , u m = ( − ∆ g ) m − u and let U = ( u , . . . , u m ) T . By a standard reduction, (21) can be written as asystem of equations in U : L A ,A ,A U := ( − ∆ g ⊗ I + iA ( x, D x ′ )+ iA ( x ) D n + A ( x )) U = 0 in M, (22)where I is the m × m identity matrix, A ( x, D x ′ ), A ( x ) and A ( x ) are definedby A ( x, D x ′ ) := . . . . . . ...0 0 . . . X α ( x ) D α . . . , A ( x ) := . . . . . . ...0 0 . . . X n ( x ) 0 . . . , A ( x ) := − . . .
00 0 − . . . . . . ...0 0 0 . . . − q ( x ) 0 0 . . . . The associated Cauchy data set to the system (22) is C A ,A ,A := { ( U | ∂M , ∂ ν U | ∂M ) : L A ,A ,A U = 0 in M, U ∈ ( H ( M )) m } . It is easy to see that C A ,A ,A and C g,X,q are mutually determined, hence it suf-fices to show C A ,A ,A determines A , A and A at p ∈ ∂M .The following result gives a factorization of the operator L A ,A ,A . Similar tech-niques are employed in [3, 12, 16, 18]. Proposition 6.2.
There is a matrix-valued pseudodifferential operator B ( x, D x ′ ) of order in x ′ , depending smoothly on x n , such that L A ,A ,A = ( D n ⊗ I + iE ( x ) ⊗ I + iA ( x ) − iB ( x, D x ′ ))( D n ⊗ I + iB ( x, D x ′ )) (23) modulo a smoothing operator. Moreover, the principle symbol of the operator B ( x, D x ′ ) is − p Q ( x, ξ ′ ) I . Here E ( x ) and Q ( x, D x ′ ) are given by (18) and (19) respectively.Proof. Plug (17) into (22) we have L A ,A ,A =( D n + iE ( x ) D n + Q ( x, D x ′ ) + Q ( x, D x ′ )) ⊗ I + iA ( x, D x ′ ) + iA ( x ) D n + A ( x ) . Comparing this expression with (23) gives the following constrains on B ( x, D x ′ )modulo a smoothing operator: B ( x, D x ′ ) + i [ D n ⊗ I, B ( x, D x ′ )] − E ( x ) B ( x, D x ′ ) − A ( x ) B ( x, D x ′ )= Q ( x, D x ′ ) ⊗ I + Q ( x, D x ′ ) ⊗ I + iA ( x, D x ′ ) + A ( x ) . (24)Let b ( x, ξ ′ ) be the full symbol of B ( x, D x ′ ), then (24) implies on the level of symbolsthat X | α |≥ α ! ∂ αx ′ bD αξ ′ b + ∂ n b − E ( x ) b − A ( x ) b = Q ( x, ξ ′ ) I + Q ( x, ξ ′ ) I + iA ( x, ξ ′ )+ A ( x ) . (25)Let b ∼ P j ≤ b j where b j ( x, ξ ′ ) is an m × m matrix with entries homogeneous ofdegree j in ξ ′ = ( ξ , . . . , ξ n − ). Collecting the terms homogeneous of degree 2 in(25) yields b ( x, ξ ′ ) = Q ( x, ξ ′ ) I, from which we can choose b ( x, ξ ′ ) = − p Q ( x, ξ ′ ) I. (26) OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 19
Collecting the terms homogeneous of degree 1 in (25) yields b b + b b + X | α | =1 ∂ αx ′ b ∂ αξ ′ b + ∂ n b − E ( x ) b − A ( x ) b = Q ( x, ξ ′ ) I + iA ( x, ξ ′ ) . (27)Since b ( x, ξ ′ ) has been determined above, E ( x ) and Q ( x, ξ ′ ) are known from (18)(20), by some elementary linear algebra there exists a unique b ( x, ξ ′ ) satisfying thisidentity. Next collecting the terms of homogeneous of degree 0 in (25) implies b + b b − + b − b + X | α | =1 ∂ αx ′ b ∂ αξ ′ b + X | α | =1 ∂ αx ′ b ∂ αξ ′ b + X | α | =2 ∂ αx ′ b ∂ αξ ′ b + ∂ n b − E ( x ) b − A ( x ) b = A ( x ) , (28)From which we can solve for b − ( x, ξ ′ ). In general, the term b j ( x, ξ ′ ) can be deter-mined by considering the terms homogeneous of degree j +1 in (25). This completesthe proof. (cid:3) Proof of Proposition 6.1.
Using a similar argument as in [16, Proposition 1.2], weconclude that the Cauchy data set C A ,A ,A determines the operator B ( x ′ , , D x ′ )modulo a smoothing operator. Consequently each b j | x n =0 is determined, j ≤
1. Itfollows from (27) that the following expression is determined by the Cauchy dataset C A ,A ,A : − X n | x n =0 p Q ( x ′ , , ξ ′ ) + iX α | x n =0 ξ α , ξ ′ ∈ R n − . Varying ξ ′ determines X n | x n =0 and X α | x n =0 , α = 1 , . . . , n −
1. Evaluating (28) on { x n = 0 } shows that the Cauchy data set C A ,A ,A determines A | x n =0 , hence q | x n =0 . (cid:3) Proof of Theorem 1.4.
If 0 is not a Dirichlet eigenvalue of L g,X ,q and L g,X ,q ,then N g,X ,q = N g,X ,q implies C g,X ,q = C g,X ,q . By Proposition 6.1 we con-clude that X = X on ∂M . The result then follows from Theorem 1.3. (cid:3) Proof of Theorem 1.5
We will follow the argument of [12, Theorem 1.3] and [3, Theorem 4]. Proceedingas in the proof of Theorem 1.3,We can derive the following integral identity (see (13)) Z M ( X ♭x + iX ♭r ) e iλ ( x + ir ) b ( θ ) dr dθ dx = 0 . (29)This is similar to (13) but this time the integral is over M instead of R × M ,x since we cannot extend the vector fields X and X any more. Varying the smoothfunction b ( θ ) leads to Z M θ ( X bx + iX br ) e iλ ( x + ir ) d ¯ ρ ∧ dρ = 0 for all θ ∈ S n −
20 YERNAT M. ASSYLBEKOV AND YANG YANG where M θ := { ( x , r ) ∈ R : ( x , r, θ ) ∈ M } and ρ = x + ir . Define f ( x ′ ) = Z R e iλx X ♭x ( x , x ′ ) dx , α ( x ′ ) = n X j =2 (cid:18)Z R e iλx X ♭j ( x , x ′ ) dx (cid:19) dx j . Here we extend X as zero outside of M so that the integral in x can be over R .The above argument shows that Z e − λr [ f ( γ ω,θ ( r )) + iα ( ˙ γ ω,θ ( r ))] dr = 0 for all θ ∈ S n − , where the r -integrals are integrals over geodesics γ ω,θ in π ( M ) ⊂ M . Observethat α ( x ′ ) is W , ∞ on π ( M ) and f ( x ′ ) is L ∞ on π ( M ). Under the assumption that( π ( M ) , g ) is a simple ( n − D := π ( M ) and conclude that for small enough λ , we have f = − λp and α = − idp where p ∈ W , ∞ ( π ( M )) and p | ∂π ( M ) = 0. The definition of α and analyticity of theFourier transform imply that ∂ k X ♭j − ∂ j X ♭k = 0 , j, k = 2 , . . . , n in M int . Also Z e iλx ( ∂ j X ♭ − ∂ X ♭j )( x , x ′ ) dx = ∂ j f + iλα j = 0 in M int , showing that dX ♭ = 0 in M . Since M is simply connected, there exists a function φ such that ∇ φ = X ∈ W , ∞ ( M ). By [9, Theorem 4.5.12 and Theorem 3.1.7] wehave φ ∈ C , ( M ).Next we need to show that φ is constant on ∂M . In the case where we can extend X to be a compactly supported W , ∞ vector field on a larger manifold, that φ isconstant on ∂M simply follows from the construction, but here we have to prove it.This is the content of the next proposition. Proposition 7.1.
The function φ is constant on the connnected boundary ∂M .Proof. Let us start by constructing more complex geometric optics solutions. FromProposition 3.1, Remark 3.2 and Remark 3.3 we can choose complex geometricoptics solutions of the form u = e − h ( x + ir ) ( | g | − / c / a ( x , r, θ ) b ( θ ) + hr ) ,v = e h ( x + ir ) ( | g | − / c / + hr ) , where ¯ ∂a = 0, λ ∈ R and k r j k H ( M ) = O (1), j = 1 ,
2. Note that in the previousconstruction we choose a = e iλ ( x + ir ) but this time we need more a ’s. Substitutingthese solutions in (12), multiplying the resulting equality by h and letting h → h → Z M h X, ∇ ρ i g uv d Vol g ( x ) = 0 , OLYHARMONIC OPERATOR ON ADMISSIBLE MANIFOLDS 21 where ρ = x + ir and X = X − X . Recall that X := ∇ φ . Insert the abovecomplex geometric optics solutions yields Z M ¯ ∂φa ( x , r, θ ) b ( θ ) dr dθ dx = 0 . Varying the smooth function b ( θ ) leads to Z M θ ¯ ∂φa d ¯ ρ ∧ dρ = 0 for all θ ∈ S n − . Integrating by parts and using that ¯ ∂a = 0 gives Z ∂M θ φa dρ = 0 for all θ ∈ S n − (30)and for every a with ¯ ∂a = 0.On the other hand, noticing that in solving the eikonal equation (7), we may choose ϕ = x but ψ = − r . Then we can construct complex geometric optics solutions ofthe form u = e − h ( x − ir ) ( | g | − / c / a ( x , r, θ ) b ( θ ) + hr ) ,v = e h ( x − ir ) ( | g | − / c / + hr ) , where ∂a = 0, λ ∈ R and k r j k H ( M ) = O (1), j = 1 ,
2. Here ∂ = 12 (cid:18) ∂∂x − i ∂∂r (cid:19) . Using a similar argument as in the preceding paragraph we can derive Z ∂M θ φ ˜ a d ¯ ρ = 0 for all θ ∈ S n − and for every ˜ a with ∂ ˜ a = 0. In particular we can choose ˜ a = ¯ a where a solves¯ ∂a = 0. Then taking complex conjugate gives Z ∂M θ ¯ φa dρ = 0 for all θ ∈ S n − . (31)Combining (30) and (31) we see Z ∂M θ Re φ a dρ = 0 , Z ∂M θ Im φ a dρ = 0for all a with ¯ ∂a = 0. Using the argument in [4, Section 5] implies that Re φ | ∂M θ = F | ∂M θ for some non-vanishing holomorphic function F on M θ . Observing that Im F is a harmonic function in M θ and Im F | ∂M θ = 0, we conclude that F is real-valuedand hence is constant on each connected component of ∂M θ . Varying θ showsthat Re φ is constant along ∂M . Likewise we can show Im φ is also constant along ∂M . (cid:3) Proof of Theorem 1.5.
Since φ = c for some constant c along ∂M , replacing φ by φ − c if necessary, we may assume φ = 0 on ∂M . The rest part of the proof is thesame as that of Theorem 1.3. (cid:3) Acknowledgement:
The authors would like to thank Prof. Katya Krupchykfor her suggestions on an earlier version of this paper. The authors are also deeplygrateful to Prof. Gunther Uhlmann for his generous support relating to this project.
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