Microlocal analysis and inverse problems Lecture 4 : Uniqueness results in admissible geometries

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1 Microlocal analysis and inverse problems Lecture 4 : Uniqueness results in admissible geometries David Dos Santos Ferreira LAGA Université de Paris 13 Wednesday May 18 Instituto de Ciencias Matemáticas, Madrid David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 1 / 24

2 Outline Introduction 1 Introduction 2 Complex Geometrical Optics 3 Attenuated X-ray transform 4 Uniqueness of unbounded potentials David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 2 / 24

3 Introduction Introduction It is time now to state some precise uniqueness results. So far we have seen that: 1 In order to have Carleman estimates with opposite weights, one has to work with limiting Carleman weights. This is in order to comply with Hörmander s (necessary) criterium of solvability for non-selfadjoint operators. 2 On Riemannian manifolds, the existence of LCW is a limiting condition. It implies that manifolds have to be locally conformal to a product. 3 For reasons related to the global solvability of the transport equation, we will ask that the manifolds under scope be globally conformal to a product. 4 In fact, we need more conditions: for reasons related to the global solvability of the eikonal eqution, we ask the cutlocus of the manifold to be empty. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 3 / 24

4 Introduction Introduction It is time now to state some precise uniqueness results. So far we have seen that: 1 In order to have Carleman estimates with opposite weights, one has to work with limiting Carleman weights. This is in order to comply with Hörmander s (necessary) criterium of solvability for non-selfadjoint operators. 2 On Riemannian manifolds, the existence of LCW is a limiting condition. It implies that manifolds have to be locally conformal to a product. 3 For reasons related to the global solvability of the transport equation, we will ask that the manifolds under scope be globally conformal to a product. 4 In fact, we need more conditions: for reasons related to the global solvability of the eikonal eqution, we ask the cutlocus of the manifold to be empty. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 3 / 24

5 Introduction Introduction It is time now to state some precise uniqueness results. So far we have seen that: 1 In order to have Carleman estimates with opposite weights, one has to work with limiting Carleman weights. This is in order to comply with Hörmander s (necessary) criterium of solvability for non-selfadjoint operators. 2 On Riemannian manifolds, the existence of LCW is a limiting condition. It implies that manifolds have to be locally conformal to a product. 3 For reasons related to the global solvability of the transport equation, we will ask that the manifolds under scope be globally conformal to a product. 4 In fact, we need more conditions: for reasons related to the global solvability of the eikonal eqution, we ask the cutlocus of the manifold to be empty. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 3 / 24

6 Introduction Introduction It is time now to state some precise uniqueness results. So far we have seen that: 1 In order to have Carleman estimates with opposite weights, one has to work with limiting Carleman weights. This is in order to comply with Hörmander s (necessary) criterium of solvability for non-selfadjoint operators. 2 On Riemannian manifolds, the existence of LCW is a limiting condition. It implies that manifolds have to be locally conformal to a product. 3 For reasons related to the global solvability of the transport equation, we will ask that the manifolds under scope be globally conformal to a product. 4 In fact, we need more conditions: for reasons related to the global solvability of the eikonal eqution, we ask the cutlocus of the manifold to be empty. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 3 / 24

7 Introduction Admissible geometries The former remarks justify the introduction of: Definition A compact Riemannian manifold (M, g), with dimension n 3 and with boundary M, is called admissible if M R M 0 for some (n 1)-dimensional simple manifold (M 0, g 0 ), and if g = c(e g 0 ) where e is the Euclidean metric on R and c is a smooth positive function on M. Definition Here, a compact manifold (M 0, g 0 ) with boundary is simple if for any p M 0 the exponential map exp p with its maximal domain of definition is a diffeomorphism onto M 0, and if M 0 is strictly convex (that is, the second fundamental form of M 0 M 0 is positive definite). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 4 / 24

8 Introduction Admissible geometries The former remarks justify the introduction of: Definition A compact Riemannian manifold (M, g), with dimension n 3 and with boundary M, is called admissible if M R M 0 for some (n 1)-dimensional simple manifold (M 0, g 0 ), and if g = c(e g 0 ) where e is the Euclidean metric on R and c is a smooth positive function on M. Definition Here, a compact manifold (M 0, g 0 ) with boundary is simple if for any p M 0 the exponential map exp p with its maximal domain of definition is a diffeomorphism onto M 0, and if M 0 is strictly convex (that is, the second fundamental form of M 0 M 0 is positive definite). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 4 / 24

9 Introduction Admissible geometries The former remarks justify the introduction of: Definition A compact Riemannian manifold (M, g), with dimension n 3 and with boundary M, is called admissible if M R M 0 for some (n 1)-dimensional simple manifold (M 0, g 0 ), and if g = c(e g 0 ) where e is the Euclidean metric on R and c is a smooth positive function on M. Definition Here, a compact manifold (M 0, g 0 ) with boundary is simple if for any p M 0 the exponential map exp p with its maximal domain of definition is a diffeomorphism onto M 0, and if M 0 is strictly convex (that is, the second fundamental form of M 0 M 0 is positive definite). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 4 / 24

10 Introduction Main results (smooth potentials) Theorem Let (M, g) be admissible, and let q 1 and q 2 be two smooth functions on M. If Λ g,q1 = Λ g,q2, then q 1 = q 2. (In fact, we have results for anisotropic magnetic Schrödinger operators). Theorem Let (M, g 1 ) and (M, g 2 ) be two admissible Riemannian manifolds in the same conformal class. If Λ g1 = Λ g2, then g 1 = g 2. The above theorems were proved in a joint paper with Carlos Kenig, Mikko Salo and Gunther Uhlmann. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 5 / 24

11 Introduction Main results (smooth potentials) Theorem Let (M, g) be admissible, and let q 1 and q 2 be two smooth functions on M. If Λ g,q1 = Λ g,q2, then q 1 = q 2. (In fact, we have results for anisotropic magnetic Schrödinger operators). Theorem Let (M, g 1 ) and (M, g 2 ) be two admissible Riemannian manifolds in the same conformal class. If Λ g1 = Λ g2, then g 1 = g 2. The above theorems were proved in a joint paper with Carlos Kenig, Mikko Salo and Gunther Uhlmann. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 5 / 24

12 Introduction Main results (unbounded potentials) Theorem Let (M, g) be admissible and let q 1, q 2 be complex functions in L n/2 (M). If Λ g,q1 = Λ g,q2, then q 1 = q 2. The above theorem was proved in a joint paper with Carlos Kenig and Mikko Salo. From the point of view of unique continuation, the regularity L n/2 is optimal. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 6 / 24

13 Introduction Main results (unbounded potentials) Theorem Let (M, g) be admissible and let q 1, q 2 be complex functions in L n/2 (M). If Λ g,q1 = Λ g,q2, then q 1 = q 2. The above theorem was proved in a joint paper with Carlos Kenig and Mikko Salo. From the point of view of unique continuation, the regularity L n/2 is optimal. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 6 / 24

14 Outline Complex Geometrical Optics 1 Introduction 2 Complex Geometrical Optics 3 Attenuated X-ray transform 4 Uniqueness of unbounded potentials David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 7 / 24

15 Complex Geometrical Optics Reminder of the formal WKB expansion Conjugated operator : P ϕ = e ϕ/h h 2 g e ϕ/h P ϕ = P ϕ Principal symbol : p ϕ = ξ 2 dϕ 2 + 2i ξ, dϕ We have g (e 1 h (ϕ+iψ) a) = e 1 h ϕ P ϕ (e i h ψ a) = e 1 h (h (ϕ+iψ) 0 p ϕ (x, dψ) [ + 2h (grad g ϕ + igrad g ψ)a + 1 ] 2 g(ϕ + iψ)a ) + h 2 g a. Eikonal equation: p ϕ (x, dψ) = 0 Transport equation: (grad g ϕ + igrad g ψ)a g(ϕ + iψ)a = 0 David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 8 / 24

16 Complex Geometrical Optics Reminder of the formal WKB expansion Conjugated operator : P ϕ = e ϕ/h h 2 g e ϕ/h P ϕ = P ϕ Principal symbol : p ϕ = ξ 2 dϕ 2 + 2i ξ, dϕ We have g (e 1 h (ϕ+iψ) a) = e 1 h ϕ P ϕ (e i h ψ a) = e 1 h (h (ϕ+iψ) 0 p ϕ (x, dψ) [ + 2h (grad g ϕ + igrad g ψ)a + 1 ] 2 g(ϕ + iψ)a ) + h 2 g a. Eikonal equation: p ϕ (x, dψ) = 0 Transport equation: (grad g ϕ + igrad g ψ)a g(ϕ + iψ)a = 0 David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 8 / 24

17 Complex Geometrical Optics Reminder of the formal WKB expansion Conjugated operator : P ϕ = e ϕ/h h 2 g e ϕ/h P ϕ = P ϕ Principal symbol : p ϕ = ξ 2 dϕ 2 + 2i ξ, dϕ We have g (e 1 h (ϕ+iψ) a) = e 1 h ϕ P ϕ (e i h ψ a) = e 1 h (h (ϕ+iψ) 0 p ϕ (x, dψ) [ + 2h (grad g ϕ + igrad g ψ)a + 1 ] 2 g(ϕ + iψ)a ) + h 2 g a. Eikonal equation: p ϕ (x, dψ) = 0 Transport equation: (grad g ϕ + igrad g ψ)a g(ϕ + iψ)a = 0 David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 8 / 24

18 Complex Geometrical Optics Reminder of the formal WKB expansion Conjugated operator : P ϕ = e ϕ/h h 2 g e ϕ/h P ϕ = P ϕ Principal symbol : p ϕ = ξ 2 dϕ 2 + 2i ξ, dϕ We have g (e 1 h (ϕ+iψ) a) = e 1 h ϕ P ϕ (e i h ψ a) = e 1 h (h (ϕ+iψ) 0 p ϕ (x, dψ) [ + 2h (grad g ϕ + igrad g ψ)a + 1 ] 2 g(ϕ + iψ)a ) + h 2 g a. Eikonal equation: p ϕ (x, dψ) = 0 Transport equation: (grad g ϕ + igrad g ψ)a g(ϕ + iψ)a = 0 David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 8 / 24

19 Complex Geometrical Optics Complex geometrical optics (eikonal equation) We suppose that the metric has the form ( ) 1 0 g(x) = c(x) 0 g 0 (x ) and we choose ϕ(x) = x 1. We have dϕ = dx 1 and grad g ϕ = c 1 x1. Eikonal equation: dψ 2 g = dϕ 2 g = c 1 dϕ, dψ g = c 1 x1 ψ We choose ψ to be independent of x 1, and solution to dψ g0 = 1. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 9 / 24

20 Complex Geometrical Optics Complex geometrical optics (eikonal equation) We suppose that the metric has the form ( ) 1 0 g(x) = c(x) 0 g 0 (x ) and we choose ϕ(x) = x 1. We have dϕ = dx 1 and grad g ϕ = c 1 x1. Eikonal equation: dψ 2 g = dϕ 2 g = c 1 dϕ, dψ g = c 1 x1 ψ We choose ψ to be independent of x 1, and solution to dψ g0 = 1. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 9 / 24

21 Complex Geometrical Optics Complex geometrical optics (eikonal equation) Eikonal equation: dψ 2 g 0 = 1 In simple manifolds, it is easy to give explicit solutions of this eikonal equation ψ(x) = d g0 (x, ω 0 ), ω 0 M \ M where d g0 is the geodesical distance ( M is a simple extension of M). We have grad g0 ψ = (dψ) = exp 1 ω 0 (x ) ψ(x ) In fact, one can use geodesical polar coordinates x = exp ω0 (rθ), r = d g0 (x, ω 0 ) > 0, θ S ω0 M. In those coordinates, the metric reads g 0 = dr 2 + h 0 (r) David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 10 / 24

22 Complex Geometrical Optics Complex geometrical optics (eikonal equation) Eikonal equation: dψ 2 g 0 = 1 In simple manifolds, it is easy to give explicit solutions of this eikonal equation ψ(x) = d g0 (x, ω 0 ), ω 0 M \ M where d g0 is the geodesical distance ( M is a simple extension of M). We have grad g0 ψ = (dψ) = exp 1 ω 0 (x ) ψ(x ) In fact, one can use geodesical polar coordinates x = exp ω0 (rθ), r = d g0 (x, ω 0 ) > 0, θ S ω0 M. In those coordinates, the metric reads g 0 = dr 2 + h 0 (r) David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 10 / 24

23 Complex Geometrical Optics Complex geometrical optics (eikonal equation) Eikonal equation: dψ 2 g 0 = 1 In simple manifolds, it is easy to give explicit solutions of this eikonal equation ψ(x) = d g0 (x, ω 0 ), ω 0 M \ M where d g0 is the geodesical distance ( M is a simple extension of M). We have grad g0 ψ = (dψ) = exp 1 ω 0 (x ) ψ(x ) In fact, one can use geodesical polar coordinates x = exp ω0 (rθ), r = d g0 (x, ω 0 ) > 0, θ S ω0 M. In those coordinates, the metric reads g 0 = dr 2 + h 0 (r) David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 10 / 24

24 Complex Geometrical Optics Complex geometrical optics (transport equation) Transport equation: c 1 x1 a + igrad g ψa g(ϕ + iψ)a = 0 In polar coordinates grad g0 ψ = r, L gradg0 ψ = r, g0 ψ = 1 ( 1/2 ψ ) g 0 1/2 g 0 = 1 r r 2 r log g 0 hence the transport equation reads and can easily be solved where h is a holomorphic function. x1 a + r a r log g 0 a = 0 a = h(x 1 + ir) g 0 1/4 b(θ) David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 11 / 24

25 Complex Geometrical Optics Complex geometrical optics (transport equation) Transport equation: c 1 x1 a + igrad g ψa g(ϕ + iψ)a = 0 In polar coordinates grad g0 ψ = r, L gradg0 ψ = r, g0 ψ = 1 ( 1/2 ψ ) g 0 1/2 g 0 = 1 r r 2 r log g 0 hence the transport equation reads and can easily be solved where h is a holomorphic function. x1 a + r a r log g 0 a = 0 a = h(x 1 + ir) g 0 1/4 b(θ) David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 11 / 24

26 Complex Geometrical Optics Complex geometrical optics (L p case) Proposition Assume that q L n/2 (M). Let ω M 0 \ M 0 be a fixed point, let λ R be fixed, and let b C (S n 2 ) be a function. Write x = (x 1, r, θ) where (r, θ) are polar normal coordinates with center ω in ( M 0, g 0 ). For τ sufficiently large outside a countable set, there exists u 0 H 1 (M) satisfying where r 0 satisfies ( g + q)u = 0 in M, u = e τx 1 (e iτr g 1/4 e iλ(x 1+ir) b(θ) + r) τ r L 2 (M) + r H 1 (M) + r 2n 1. L n 2 (M) David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 12 / 24

27 Outline Attenuated X-ray transform 1 Introduction 2 Complex Geometrical Optics 3 Attenuated X-ray transform 4 Uniqueness of unbounded potentials David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 13 / 24

28 Attenuated X-ray transform Simple manifolds Definition A compact manifold (M 0, g 0 ) with boundary is simple if for any p M 0 the exponential map exp p with its maximal domain of definition is a diffeomorphism onto M 0, and if M 0 is strictly convex (that is, the second fundamental form of M 0 M 0 is positive definite). 1 Simple manifolds are non-trapping. 2 Simple manifolds are diffeomorphic to a ball. 3 A hemisphere is not simple. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 14 / 24

29 Attenuated X-ray transform Simple manifolds Definition A compact manifold (M 0, g 0 ) with boundary is simple if for any p M 0 the exponential map exp p with its maximal domain of definition is a diffeomorphism onto M 0, and if M 0 is strictly convex (that is, the second fundamental form of M 0 M 0 is positive definite). 1 Simple manifolds are non-trapping. 2 Simple manifolds are diffeomorphic to a ball. 3 A hemisphere is not simple. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 14 / 24

30 Attenuated X-ray transform Simple manifolds Definition A compact manifold (M 0, g 0 ) with boundary is simple if for any p M 0 the exponential map exp p with its maximal domain of definition is a diffeomorphism onto M 0, and if M 0 is strictly convex (that is, the second fundamental form of M 0 M 0 is positive definite). 1 Simple manifolds are non-trapping. 2 Simple manifolds are diffeomorphic to a ball. 3 A hemisphere is not simple. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 14 / 24

31 Attenuated X-ray transform Simple manifolds Definition A compact manifold (M 0, g 0 ) with boundary is simple if for any p M 0 the exponential map exp p with its maximal domain of definition is a diffeomorphism onto M 0, and if M 0 is strictly convex (that is, the second fundamental form of M 0 M 0 is positive definite). 1 Simple manifolds are non-trapping. 2 Simple manifolds are diffeomorphic to a ball. 3 A hemisphere is not simple. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 14 / 24

32 Attenuated X-ray transform Attenuated X-ray transform The unit sphere bundle : SM 0 = { S x, S x = (x, ξ) Tx M 0 ; ξ g = 1 }. x M 0 Boundary: (SM 0 ) = {(x, ξ) SM 0 ; x M 0 } union of inward and outward pointing vectors: ± (SM 0 ) = { (x, ξ) SM 0 ; ± ξ, ν 0 }. Denote by t γ(t, x, ξ) the unit speed geodesic starting at x in direction ξ, and let τ(x, ξ) be the time when this geodesic exits M 0. Godesic ray transform with constant attenuation λ: T λ f(x, ξ) = τ(x,ξ) 0 f(γ(t, x, ξ))e λt dt, (x, ξ) + (SM 0 ). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 15 / 24

33 Attenuated X-ray transform Attenuated X-ray transform The unit sphere bundle : SM 0 = { S x, S x = (x, ξ) Tx M 0 ; ξ g = 1 }. x M 0 Boundary: (SM 0 ) = {(x, ξ) SM 0 ; x M 0 } union of inward and outward pointing vectors: ± (SM 0 ) = { (x, ξ) SM 0 ; ± ξ, ν 0 }. Denote by t γ(t, x, ξ) the unit speed geodesic starting at x in direction ξ, and let τ(x, ξ) be the time when this geodesic exits M 0. Godesic ray transform with constant attenuation λ: T λ f(x, ξ) = τ(x,ξ) 0 f(γ(t, x, ξ))e λt dt, (x, ξ) + (SM 0 ). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 15 / 24

34 Attenuated X-ray transform Attenuated X-ray transform The unit sphere bundle : SM 0 = { S x, S x = (x, ξ) Tx M 0 ; ξ g = 1 }. x M 0 Boundary: (SM 0 ) = {(x, ξ) SM 0 ; x M 0 } union of inward and outward pointing vectors: ± (SM 0 ) = { (x, ξ) SM 0 ; ± ξ, ν 0 }. Denote by t γ(t, x, ξ) the unit speed geodesic starting at x in direction ξ, and let τ(x, ξ) be the time when this geodesic exits M 0. Godesic ray transform with constant attenuation λ: T λ f(x, ξ) = τ(x,ξ) 0 f(γ(t, x, ξ))e λt dt, (x, ξ) + (SM 0 ). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 15 / 24

35 Attenuated X-ray transform Injectivity of the attenuated X-ray transform Proposition Let (M 0, g 0 ) be a simple manifold. There exists ε > 0 such that if λ is a real number with λ < ε and if f C (M), then the condition T λ f(x, ξ) = 0 for all (x, ξ) + (SM 0 ) implies that f = 0. This was known when λ = 0. The proof in the attenuated case uses perturbation arguments. What about nonsmooth functions? David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 16 / 24

36 Attenuated X-ray transform Injectivity of the attenuated X-ray transform Proposition Let (M 0, g 0 ) be a simple manifold. There exists ε > 0 such that if λ is a real number with λ < ε and if f C (M), then the condition T λ f(x, ξ) = 0 for all (x, ξ) + (SM 0 ) implies that f = 0. This was known when λ = 0. The proof in the attenuated case uses perturbation arguments. What about nonsmooth functions? David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 16 / 24

37 Attenuated X-ray transform Injectivity of the attenuated X-ray transform Proposition Let (M 0, g 0 ) be a simple manifold. There exists ε > 0 such that if λ is a real number with λ < ε and if f C (M), then the condition T λ f(x, ξ) = 0 for all (x, ξ) + (SM 0 ) implies that f = 0. This was known when λ = 0. The proof in the attenuated case uses perturbation arguments. What about nonsmooth functions? David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 16 / 24

38 Normal operator Attenuated X-ray transform Notations: µ(x, ξ) = ξ, ν(x) and dn is the volume form on N. Scalar product: (h, h) L 2 µ ( + (SM 0 )) = h hµ d( (SM 0 )) Adjoint of the ray transform: + (SM 0 ) T λ h(x) = S x e λτ(x, ξ) h(ϕ τ(x, ξ) (x, ξ)), ds x (ξ), x M 0. where ϕ t (x, ξ) = (γ(t, x, ξ), γ(t, x, ξ)) is the geodesic flow. Lemma T λ T λ is a self-adjoint elliptic pseudodifferential operator of order 1 in M int 0. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 17 / 24

39 Normal operator Attenuated X-ray transform Notations: µ(x, ξ) = ξ, ν(x) and dn is the volume form on N. Scalar product: (h, h) L 2 µ ( + (SM 0 )) = h hµ d( (SM 0 )) Adjoint of the ray transform: + (SM 0 ) T λ h(x) = S x e λτ(x, ξ) h(ϕ τ(x, ξ) (x, ξ)), ds x (ξ), x M 0. where ϕ t (x, ξ) = (γ(t, x, ξ), γ(t, x, ξ)) is the geodesic flow. Lemma T λ T λ is a self-adjoint elliptic pseudodifferential operator of order 1 in M int 0. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 17 / 24

40 Normal operator Attenuated X-ray transform Notations: µ(x, ξ) = ξ, ν(x) and dn is the volume form on N. Scalar product: (h, h) L 2 µ ( + (SM 0 )) = h hµ d( (SM 0 )) Adjoint of the ray transform: + (SM 0 ) T λ h(x) = S x e λτ(x, ξ) h(ϕ τ(x, ξ) (x, ξ)), ds x (ξ), x M 0. where ϕ t (x, ξ) = (γ(t, x, ξ), γ(t, x, ξ)) is the geodesic flow. Lemma T λ T λ is a self-adjoint elliptic pseudodifferential operator of order 1 in M int 0. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 17 / 24

41 Normal operator Attenuated X-ray transform Notations: µ(x, ξ) = ξ, ν(x) and dn is the volume form on N. Scalar product: (h, h) L 2 µ ( + (SM 0 )) = h hµ d( (SM 0 )) Adjoint of the ray transform: + (SM 0 ) T λ h(x) = S x e λτ(x, ξ) h(ϕ τ(x, ξ) (x, ξ)), ds x (ξ), x M 0. where ϕ t (x, ξ) = (γ(t, x, ξ), γ(t, x, ξ)) is the geodesic flow. Lemma T λ T λ is a self-adjoint elliptic pseudodifferential operator of order 1 in M int 0. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 17 / 24

42 Attenuated X-ray transform Injectivity of the attenuated X-ray transform (non-smooth case) Lemma Let (M 0, g 0 ) be an (n 1)-dimensional simple manifold, and let f L 1 (M 0 ). Consider the integrals S n 2 τ(ω,θ) 0 f(r, θ)e λr b(θ) dr dθ where (r, θ) are polar normal coordinates in (M 0, g 0 ) centered at some ω M 0, and τ(ω, θ) is the time when the geodesic r (r, θ) exits M 0. If λ is sufficiently small, and if these integrals vanish for all ω M 0 and all b C (S n 2 ), then f = 0. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 18 / 24

43 Attenuated X-ray transform Using Elliptic regularity Preliminary step: extend (M 0, g 0 ) to a slightly larger simple manifold and f by zero. In this way we can assume that f is compactly supported in M int 0. Let b also depend on ω and change notations to write τ(x,ξ) e λt f(γ(t, x, ξ))b(x, ξ) dt ds x (ξ) = 0. S x 0 Next we make the choice b(x, ξ) = h(x, ξ)µ(x, ξ) and integrate the last identity over M 0 to obtain τ(x,ξ) e λt f(γ(t, x, ξ))h(x, ξ)µ dt d( (SM 0 )) = 0. + (SM 0 ) 0 By adjunction, we get f(x)tλ h(x) dv (x) = 0 M 0 for all h C 0 (( +(SM 0 )) int ). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 19 / 24

44 Attenuated X-ray transform Using Elliptic regularity Preliminary step: extend (M 0, g 0 ) to a slightly larger simple manifold and f by zero. In this way we can assume that f is compactly supported in M int 0. Let b also depend on ω and change notations to write τ(x,ξ) e λt f(γ(t, x, ξ))b(x, ξ) dt ds x (ξ) = 0. S x 0 Next we make the choice b(x, ξ) = h(x, ξ)µ(x, ξ) and integrate the last identity over M 0 to obtain τ(x,ξ) e λt f(γ(t, x, ξ))h(x, ξ)µ dt d( (SM 0 )) = 0. + (SM 0 ) 0 By adjunction, we get f(x)tλ h(x) dv (x) = 0 M 0 for all h C 0 (( +(SM 0 )) int ). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 19 / 24

45 Attenuated X-ray transform Using Elliptic regularity Preliminary step: extend (M 0, g 0 ) to a slightly larger simple manifold and f by zero. In this way we can assume that f is compactly supported in M int 0. Let b also depend on ω and change notations to write τ(x,ξ) e λt f(γ(t, x, ξ))b(x, ξ) dt ds x (ξ) = 0. S x 0 Next we make the choice b(x, ξ) = h(x, ξ)µ(x, ξ) and integrate the last identity over M 0 to obtain τ(x,ξ) e λt f(γ(t, x, ξ))h(x, ξ)µ dt d( (SM 0 )) = 0. + (SM 0 ) 0 By adjunction, we get f(x)tλ h(x) dv (x) = 0 M 0 for all h C 0 (( +(SM 0 )) int ). David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 19 / 24

46 Attenuated X-ray transform Using Elliptic regularity Choose h = T λ ϕ for ϕ C0 Since T λ T λ is self-adjoint, we have (M int 0 ) so that M 0 f(x)t λ T λϕ(x) dv (x) = 0. M 0 (T λ T λf(x))ϕ(x) dv (x) = 0 for all test functions ϕ, so T λ T λf = 0. By ellipticity, since f was compactly supported in M0 int, it follows that f C0 (M int 0 ). One can now use the injectivity result for f smooth to conclude that f = 0. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 20 / 24

47 Attenuated X-ray transform Using Elliptic regularity Choose h = T λ ϕ for ϕ C0 Since T λ T λ is self-adjoint, we have (M int 0 ) so that M 0 f(x)t λ T λϕ(x) dv (x) = 0. M 0 (T λ T λf(x))ϕ(x) dv (x) = 0 for all test functions ϕ, so T λ T λf = 0. By ellipticity, since f was compactly supported in M0 int, it follows that f C0 (M int 0 ). One can now use the injectivity result for f smooth to conclude that f = 0. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 20 / 24

48 Outline Uniqueness of unbounded potentials 1 Introduction 2 Complex Geometrical Optics 3 Attenuated X-ray transform 4 Uniqueness of unbounded potentials David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 21 / 24

49 Using CGOs Uniqueness of unbounded potentials Starting point: If q = q 1 q 2 where M qu 1 u 2 dv g = 0 with b C (S n 2 ) and u 1 = e τ(x 1+ir) ( g 1/4 e iλ(x 1+ir) b(θ) + r 1 ), u 2 = e τ(x 1+ir) ( g 1/4 + r 2 ). r j L 2n n 2 (M) = O(1), r j L 2 (M) = o(1) as τ. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 22 / 24

50 Using CGOs Uniqueness of unbounded potentials Noting that dv g = g 1/2 dx 1 dr dθ, we obtain that qe iλ(x1+ir) b(θ) dx 1 dr dθ = q(a 1 r 2 + a 2 r 1 + r 1 r 2 ) dv M The RHS converges to 0 as τ. Taking the limit as τ, we obtain that 0 M S n 2 q(x 1, r, θ)e iλ(x 1+ir) b(θ) dx 1 dr dθ = 0. This statement is true for all choices of ω M 0 \ M 0, for all real numbers λ, and for all functions b C (S n 2 ). Hence f λ (r, θ)e λr b(θ) dr dθ = 0 S n 2 where f λ L 1 (M 0 ) is the function given by 0 f λ (r, θ) = e iλx 1 q(x 1, r, θ) dx 1. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 23 / 24

51 Using CGOs Uniqueness of unbounded potentials Noting that dv g = g 1/2 dx 1 dr dθ, we obtain that qe iλ(x1+ir) b(θ) dx 1 dr dθ = q(a 1 r 2 + a 2 r 1 + r 1 r 2 ) dv M The RHS converges to 0 as τ. Taking the limit as τ, we obtain that 0 M S n 2 q(x 1, r, θ)e iλ(x 1+ir) b(θ) dx 1 dr dθ = 0. This statement is true for all choices of ω M 0 \ M 0, for all real numbers λ, and for all functions b C (S n 2 ). Hence f λ (r, θ)e λr b(θ) dr dθ = 0 S n 2 where f λ L 1 (M 0 ) is the function given by 0 f λ (r, θ) = e iλx 1 q(x 1, r, θ) dx 1. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 23 / 24

52 Using CGOs Uniqueness of unbounded potentials Noting that dv g = g 1/2 dx 1 dr dθ, we obtain that qe iλ(x1+ir) b(θ) dx 1 dr dθ = q(a 1 r 2 + a 2 r 1 + r 1 r 2 ) dv M The RHS converges to 0 as τ. Taking the limit as τ, we obtain that 0 M S n 2 q(x 1, r, θ)e iλ(x 1+ir) b(θ) dx 1 dr dθ = 0. This statement is true for all choices of ω M 0 \ M 0, for all real numbers λ, and for all functions b C (S n 2 ). Hence f λ (r, θ)e λr b(θ) dr dθ = 0 S n 2 where f λ L 1 (M 0 ) is the function given by 0 f λ (r, θ) = e iλx 1 q(x 1, r, θ) dx 1. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 23 / 24

53 End of the proof Uniqueness of unbounded potentials If λ is sufficiently small, it follows that f λ = 0. Since q(, r, θ) is a compactly supported function in L 1 (R) for a.e. (r, θ), the Paley-Wiener theorem shows that q(, r, θ) = 0 for such (r, θ). Consequently q 1 = q 2. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 24 / 24

54 End of the proof Uniqueness of unbounded potentials If λ is sufficiently small, it follows that f λ = 0. Since q(, r, θ) is a compactly supported function in L 1 (R) for a.e. (r, θ), the Paley-Wiener theorem shows that q(, r, θ) = 0 for such (r, θ). Consequently q 1 = q 2. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 24 / 24

55 End of the proof Uniqueness of unbounded potentials If λ is sufficiently small, it follows that f λ = 0. Since q(, r, θ) is a compactly supported function in L 1 (R) for a.e. (r, θ), the Paley-Wiener theorem shows that q(, r, θ) = 0 for such (r, θ). Consequently q 1 = q 2. David Dos Santos Ferreira (LAGA) Inverse Problems 4 ICMAT 24 / 24