Broken ray tomography

Broken ray tomography Applied Inverse Problems Helsinki Joonas Ilmavirta University of Jyväskylä 28 May 2015 Joonas Ilmavirta (Jyväskylä) Broken ray...
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Broken ray tomography Applied Inverse Problems Helsinki Joonas Ilmavirta University of Jyväskylä

28 May 2015 Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Outline

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Broken ray tomography Broken rays Broken ray transform Applications Counterexamples

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Results

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More details

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Broken rays

Let M be a compact Riemannian manifold with boundary (or the closure of a bounded smooth Euclidean domain). Split its boundary in two disjoint parts: ∂M = E ∪ R.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Broken rays

Let M be a compact Riemannian manifold with boundary (or the closure of a bounded smooth Euclidean domain). Split its boundary in two disjoint parts: ∂M = E ∪ R. Broken rays have endpoints on (or exits through) E and reflect finitely many times on R (like light reflecting from a perfect mirror).

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Broken rays

Let M be a compact Riemannian manifold with boundary (or the closure of a bounded smooth Euclidean domain). Split its boundary in two disjoint parts: ∂M = E ∪ R. Broken rays have endpoints on (or exits through) E and reflect finitely many times on R (like light reflecting from a perfect mirror). If E = ∂M and R = ∅, then broken rays are just maximal geodesics in M.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Broken rays

A domain. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Broken rays

A line through the domain. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Broken rays

A broken ray. E is blue and R is red. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Broken rays

Large E. Joonas Ilmavirta (Jyväskylä)

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Broken rays

A broken ray. Joonas Ilmavirta (Jyväskylä)

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Broken rays

Another broken ray. Joonas Ilmavirta (Jyväskylä)

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Broken rays

Small E. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Broken rays

A broken ray. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Broken rays

An annulus. Joonas Ilmavirta (Jyväskylä)

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Broken rays

A broken ray. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Broken rays

Another broken ray. Joonas Ilmavirta (Jyväskylä)

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Broken ray transform Broken ray tomography problem: Can we recover a function from its integrals over all broken rays?

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Broken ray transform Broken ray tomography problem: Can we recover a function from its integrals over all broken rays? Let Γ be the set broken rays in M . (This set depends on E and R!) The broken ray transform of a continuous function f : M → R is Gf : Γ → R, Z Gf (γ) = f ds. γ

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Broken ray transform Broken ray tomography problem: Can we recover a function from its integrals over all broken rays? Let Γ be the set broken rays in M . (This set depends on E and R!) The broken ray transform of a continuous function f : M → R is Gf : Γ → R, Z Gf (γ) = f ds. γ

If E = ∂M , then G is the X-ray transform.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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. Broken ray transform Broken ray tomography problem: Can we recover a function from its integrals over all broken rays? Let Γ be the set broken rays in M . (This set depends on E and R!) The broken ray transform of a continuous function f : M → R is Gf : Γ → R, Z Gf (γ) = f ds. γ

If E = ∂M , then G is the X-ray transform.

Broken ray tomography problem: Is G injective? How does this depend on M , E, and regularity assumptions on f ? Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Applications

Calderón’s problem with partial data: Can we find the conductivity at every point inside an object by making electrical measurements on a part of the boundary? (Some of the boundary is completely inaccessible.)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Applications

Calderón’s problem with partial data: Can we find the conductivity at every point inside an object by making electrical measurements on a part of the boundary? (Some of the boundary is completely inaccessible.) Kenig and Salo (2013) showed that this partial data problem on a tube-shaped domain can be reduced to the injectivity of the broken ray transform on the cross section of the tube.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Applications

Calderón’s problem with partial data: Can we find the conductivity at every point inside an object by making electrical measurements on a part of the boundary? (Some of the boundary is completely inaccessible.) Kenig and Salo (2013) showed that this partial data problem on a tube-shaped domain can be reduced to the injectivity of the broken ray transform on the cross section of the tube. Eskin (2004) related the broken ray transform to an inverse boundary value problem for the magnetic Schrödinger equation.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Applications

Many inverse problems with full data have been previously reduced to the X-ray transform. It seems that in the case of partial data, the X-ray transform is often replaced by the broken ray transform.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Applications

Many inverse problems with full data have been previously reduced to the X-ray transform. It seems that in the case of partial data, the X-ray transform is often replaced by the broken ray transform. Example: Linearization (w.r.t. the metric) of broken ray scattering relation leads to broken ray transform of 2-tensor fields.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Counterexamples

The broken ray transform can only be injective if the X-ray transform is injective. The integral over a broken ray is a sum of integrals over geodesics. (The broken ray transform can be seen as a partial data version of the X-ray transform.)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Counterexamples

The broken ray transform can only be injective if the X-ray transform is injective. The integral over a broken ray is a sum of integrals over geodesics. (The broken ray transform can be seen as a partial data version of the X-ray transform.) There are counterexample to the injectivity of the broken ray transform in Euclidean domains. (Pictures coming.)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Counterexamples

Some functions supported in the gray area are invisible. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Counterexamples

Some functions supported in the gray area are invisible. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Outline

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Broken ray tomography

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Results Four angles of attack Reflection Energy estimate for a PDE Direct calculation Boundary reconstruction

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More details

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Four angles of attack

As far as I know, there are four known ways to prove anything about the broken ray transform:

Joonas Ilmavirta (Jyväskylä)

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Four angles of attack

As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments (I., Hubenthal)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Four angles of attack

As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments (I., Hubenthal) energy estimates (Eskin, I.–Salo)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Four angles of attack

As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments (I., Hubenthal) energy estimates (Eskin, I.–Salo) explicit calculation of the transform (I.)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Four angles of attack

As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments (I., Hubenthal) energy estimates (Eskin, I.–Salo) explicit calculation of the transform (I.) boundary reconstruction via boundary geodesics (I.)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Four angles of attack

As far as I know, there are four known ways to prove anything about the broken ray transform: reflection arguments (I., Hubenthal) energy estimates (Eskin, I.–Salo) explicit calculation of the transform (I.) boundary reconstruction via boundary geodesics (I.) The main goal is to reduce the problem to a more tractable one, like the X-ray transform, manipulation of Fourier series, or unique solvability of a PDE.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

Reflections at the boundary are what makes the broken ray transform difficult. We want to get rid of them.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

Reflections at the boundary are what makes the broken ray transform difficult. We want to get rid of them. Can we make the domain reflect but keep the rays straight? This would turn broken rays in our domain to straight lines on a reflected domain.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Reflection

Reflections at the boundary are what makes the broken ray transform difficult. We want to get rid of them. Can we make the domain reflect but keep the rays straight? This would turn broken rays in our domain to straight lines on a reflected domain. Yes, at least if R is flat. This works in great generality, but we will only discuss an Euclidean example.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Reflection

Theorem (I. 2013) Let C ⊂ R2 be a cone. (In polar coordinates, r > 0 and 0 < θ < θ0 for some θ0 .) Let Ω ⊂ C be a bounded domain. Then the broken ray transform in Ω with R = ∂C ∩ ∂Ω is injective. An example of a domain on the next slide.

Joonas Ilmavirta (Jyväskylä)

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Reflection

A planar domain contained in a cone. The conic part of the boundary is R (narrow line) and the rest is E (strong line). Joonas Ilmavirta (Jyväskylä)

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Reflection

Application for Calderón’s problem in tubular domains with partial data:

Theorem (Kenig–Salo 2012) Let I ⊂ R be an interval and Ω0 ⊂ Rn , n ≥ 2, a sufficiently nice domain. Let ∂Ω0 = E ∪ R be a disjoint union. Let Ω ⊂ R × Ω0 be a sufficiently nice domain such that I × Ω0 ⊂ Ω. Suppose I × R ⊂ ∂Ω is inaccessible. If the broken ray transform on Ω0 is injective, then the partial data for Calderón’s problem determines the conductivity uniquely.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

Example of a tubular domain Ω. The transversal domain Ω0 is a planar cone. Joonas Ilmavirta (Jyväskylä)

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Energy estimate for a PDE

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Energy estimate for a PDE

Theorem (I.–Salo 2014) ˆ be a nonpositively curved strictly convex Riemannian surface. Let Let M ˆ be a strictly convex open subset. Let M = M ˆ \ O with E = ∂M O⊂M and R = ∂O. Then the broken ray transform on M is injective on C 2 (M ).

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Energy estimate for a PDE

Theorem (I.–Salo 2014) ˆ be a nonpositively curved strictly convex Riemannian surface. Let Let M ˆ be a strictly convex open subset. Let M = M ˆ \ O with E = ∂M O⊂M and R = ∂O. Then the broken ray transform on M is injective on C 2 (M ). The proof is based on an energy identity (Pestov identity) and establishing sufficient regularity to use the identity. For regularity, we introduce and study Jacobi fields along broken rays.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Energy estimate for a PDE Let SM be the sphere bundle of M . (In the Euclidean case, SM = M × S 1 .)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Energy estimate for a PDE Let SM be the sphere bundle of M . (In the Euclidean case, SM = M × S 1 .) For a C 2 function f : M → R, let u : SM → R be the integral of f along this broken ray: Z f ds. u(x, v) = γx,v

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Energy estimate for a PDE Let SM be the sphere bundle of M . (In the Euclidean case, SM = M × S 1 .) For a C 2 function f : M → R, let u : SM → R be the integral of f along this broken ray: Z f ds. u(x, v) = γx,v

We need two differential operators: X is the geodesic vector field (X = v · ∇x ) and V is the vertical vector field (∂v ). Now Xu = −f and V f = 0, so V Xu = 0. Assuming Gf = 0 gives some boundary conditions for u.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Energy estimate for a PDE Let SM be the sphere bundle of M . (In the Euclidean case, SM = M × S 1 .) For a C 2 function f : M → R, let u : SM → R be the integral of f along this broken ray: Z f ds. u(x, v) = γx,v

We need two differential operators: X is the geodesic vector field (X = v · ∇x ) and V is the vertical vector field (∂v ). Now Xu = −f and V f = 0, so V Xu = 0. Assuming Gf = 0 gives some boundary conditions for u. We want to show that V Xu = 0 with suitable boundary conditions on u implies u = 0. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Energy estimate for a PDE

Lemma (Pestov identity for the broken ray transform) If u : SM → R has sufficient regularity and boundary conditions on ∂(SM ), then kV Xuk2L2 (SM ) = kXV uk2L2 (SM ) + kXuk2L2 (SM ) Z Z 2 − K(x) |V u(x, v)| − SM

κ(x) |V u(x, v)|2 ,

∂(SM )

where K is the curvature of M and κ is the curvature of ∂M .

Joonas Ilmavirta (Jyväskylä)

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Direct calculation

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Direct calculation

If M is a disc in the plane, we can try to calculate the broken ray transform explicitly.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Direct calculation

If M is a disc in the plane, we can try to calculate the broken ray transform explicitly. In polar coordinates (r, θ), we can write a function as a Fourier series: X f (r, θ) = eikθ ak (r). k∈Z

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Direct calculation

If M is a disc in the plane, we can try to calculate the broken ray transform explicitly. In polar coordinates (r, θ), we can write a function as a Fourier series: X f (r, θ) = eikθ ak (r). k∈Z

A line (a chord) in the disc can be parametrized by the polar coordinates of the closest point to the center.

Joonas Ilmavirta (Jyväskylä)

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Direct calculation The integral of f over a line given by (r, θ) is X If (r, θ) = eikθ A|k| ak (r) k∈Z

where Ak is a generalized Abel transform. (A0 is the usual Abel transform.)

Joonas Ilmavirta (Jyväskylä)

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Direct calculation The integral of f over a line given by (r, θ) is X If (r, θ) = eikθ A|k| ak (r) k∈Z

where Ak is a generalized Abel transform. (A0 is the usual Abel transform.) These integral transforms Ak are injective, and this procedure can be used to invert the X-ray transform: If = 0 ⇐⇒ all Fourier components of If are zero ⇐⇒ A|k| ak = 0 for all k ⇐⇒ ak = 0 for all k ⇐⇒ f = 0. (Cormack used this method for inversion.)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Direct calculation The integral of f over a line given by (r, θ) is X If (r, θ) = eikθ A|k| ak (r) k∈Z

where Ak is a generalized Abel transform. (A0 is the usual Abel transform.) These integral transforms Ak are injective, and this procedure can be used to invert the X-ray transform: If = 0 ⇐⇒ all Fourier components of If are zero ⇐⇒ A|k| ak = 0 for all k ⇐⇒ ak = 0 for all k ⇐⇒ f = 0. (Cormack used this method for inversion.) A broken ray consists of finitely many line segments which are rotations of each other. The integral of f over a broken ray can be expressed explicitly, but it is more complicated than the above sum for the X-ray transform. Careful analysis of this expression shows that the broken ray transform is injective with some regularity assumptions. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Direct calculation

Theorem (I. 2013) Suppose f : D → C is given by f (r, θ) =

X

eikθ ak (r)

|k|≤K

for some integer K and suppose the functions ak are Hölder continuous. If E ⊂ ∂D is open and Gf = 0, then f = 0. The function can be much more general, but this version is easy to state.

Joonas Ilmavirta (Jyväskylä)

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Direct calculation

Theorem (I. 2013) Suppose f : D → C is given by f (r, θ) =

X

eikθ ak (r)

|k|≤K

for some integer K and suppose the functions ak are Hölder continuous. If E ⊂ ∂D is open and Gf = 0, then f = 0. The function can be much more general, but this version is easy to state. It is an open problem whether the theorem is true for smooth functions.

Joonas Ilmavirta (Jyväskylä)

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Boundary reconstruction

Suppose ∂M is strictly convex. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Boundary reconstruction

Suppose ∂M is strictly convex. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time. We can write a geodesic on R ⊂ ∂M as a limit of broken rays. Boundary geodesics as limits of broken rays (= billiard trajectories) are known by many names: ‘glancing billiards’, ‘creeping rays’, ‘whispering gallery trajectories’, ‘gliding rays’.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Boundary reconstruction

Suppose ∂M is strictly convex. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time. We can write a geodesic on R ⊂ ∂M as a limit of broken rays. Boundary geodesics as limits of broken rays (= billiard trajectories) are known by many names: ‘glancing billiards’, ‘creeping rays’, ‘whispering gallery trajectories’, ‘gliding rays’. If we know the broken ray transform of a function f : M → R, we can recover the X-ray transform of f |R . If this X-ray transform (on the boundary manifold!) is injective, we can recover f on ∂M . In a similar way one can also recover normal derivatives of f at ∂M .

Joonas Ilmavirta (Jyväskylä)

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End

Thank you. Slides and papers available at http://users.jyu.fi/~jojapeil.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Outline

1

Broken ray tomography

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Results

3

More details Reflection Reflection Boundary reconstruction

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Reflection

Proposition The broken ray transform is injective for the domain Ω = {x ∈ B(0, 1) ⊂ R2 ; x1 > 0} with R = {x ∈ ∂Ω; x1 = 0}.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Reflection

Proposition The broken ray transform is injective for the domain Ω = {x ∈ B(0, 1) ⊂ R2 ; x1 > 0} with R = {x ∈ ∂Ω; x1 = 0}.

Proof ¯ integrates to zero over all broken rays. Let Ω ˜ = B(0, 1) Suppose f ∈ C(Ω) ¯ ˜ →Ω ¯ be the folding map π(x1 , x2 ) = (|x1 | , x2 ). Define and let π : Ω ¯ ˜ ˜ ˜ f : Ω → R by f (x) = f (π(x)). Now f˜ integrates to zero over all lines in ˜ Because the X-ray transform is injective, f˜ = 0. Thus also f = 0. Ω.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Reflection

Proposition The broken ray transform is injective for the domain Ω = {x ∈ B(0, 1) ⊂ R2 ; x1 > 0} with R = {x ∈ ∂Ω; x1 = 0}.

Proof ¯ integrates to zero over all broken rays. Let Ω ˜ = B(0, 1) Suppose f ∈ C(Ω) ¯ ˜ →Ω ¯ be the folding map π(x1 , x2 ) = (|x1 | , x2 ). Define and let π : Ω ¯ ˜ ˜ ˜ f : Ω → R by f (x) = f (π(x)). Now f˜ integrates to zero over all lines in ˜ Because the X-ray transform is injective, f˜ = 0. Thus also f = 0. Ω. The idea of reducing the broken ray transform on Ω to the X-ray transform ˜ works in great generality. of a reflected domain Ω

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

Lemma (Helgason’s support theorem) Let f : Rn → R be a compactly supported continuous function and K ⊂ Rn a compact, convex set. If the integral of f is zero over every line that does not meet K, then f is zero outside K.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Reflection

Lemma (Helgason’s support theorem) Let f : Rn → R be a compactly supported continuous function and K ⊂ Rn a compact, convex set. If the integral of f is zero over every line that does not meet K, then f is zero outside K.

Theorem (I. 2013) Let C ⊂ R2 be a cone. (In polar coordinates, r > 0 and 0 < θ < θ0 for some θ0 .) Let Ω ⊂ C be a bounded domain. Then the broken ray transform in Ω with R = ∂C ∩ ∂Ω is injective.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Reflection

Lemma (Helgason’s support theorem) Let f : Rn → R be a compactly supported continuous function and K ⊂ Rn a compact, convex set. If the integral of f is zero over every line that does not meet K, then f is zero outside K.

Theorem (I. 2013) Let C ⊂ R2 be a cone. (In polar coordinates, r > 0 and 0 < θ < θ0 for some θ0 .) Let Ω ⊂ C be a bounded domain. Then the broken ray transform in Ω with R = ∂C ∩ ∂Ω is injective. Proof will be given as pictures.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

A domain with opening angle θ0 = π/m for an integer m. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

Glue together m copies of Ω (every other one reflected). The broken ray folds out to a straight line. Use injectivity of X-ray transform. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

A domain with general opening angle θ0 . Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

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Reflection

Glue together enough copies of Ω so that the shaded sector becomes convex. Apply Helgason’s support theorem. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Boundary reconstruction Suppose ∂M is strictly convex and E ⊂ ∂M is open. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Boundary reconstruction Suppose ∂M is strictly convex and E ⊂ ∂M is open. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time. We can control how well it stays near the boundary. This is similar how an ideally bouncing ball stays near the surface in slowly varying gravitation. (Now gravitation is replaced by the second fundamental form of the boundary.)

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Boundary reconstruction Suppose ∂M is strictly convex and E ⊂ ∂M is open. If a broken ray starts very close to the boundary and almost tangential to it, it will remain this way for some time. We can control how well it stays near the boundary. This is similar how an ideally bouncing ball stays near the surface in slowly varying gravitation. (Now gravitation is replaced by the second fundamental form of the boundary.) If we take a sequence of broken rays that stay closer and closer to the boundary, the limit curve is a curve on the boundary ∂M . This curve is actually a geodesic on ∂M . Boundary geodesics as limits of broken rays (= billiard trajectories) are known by many names: ‘glancing billiards’, ‘creeping rays’, ‘whispering gallery trajectories’, ‘gliding rays’. These boundary geodesics can be assumed to have endpoints on ∂E. Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

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Boundary reconstruction

If we know the broken ray transform of f , we can find its integrals over boundary geodesics. If the X-ray transform on ∂M (or more properly R) is injective, we can recover f at ∂M from its broken ray transform.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

∞/∞

Boundary reconstruction

If we know the broken ray transform of f , we can find its integrals over boundary geodesics. If the X-ray transform on ∂M (or more properly R) is injective, we can recover f at ∂M from its broken ray transform. If certain weighted X-ray transforms on R are injective, then we can also recover the normal derivatives (of all orders) of f at the boundary. The weight depends on curvature. Intuitively, this is because a broken ray is on average further away from the boundary where the curvature is largest.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

∞/∞

.

Boundary reconstruction

If we know the broken ray transform of f , we can find its integrals over boundary geodesics. If the X-ray transform on ∂M (or more properly R) is injective, we can recover f at ∂M from its broken ray transform. If certain weighted X-ray transforms on R are injective, then we can also recover the normal derivatives (of all orders) of f at the boundary. The weight depends on curvature. Intuitively, this is because a broken ray is on average further away from the boundary where the curvature is largest. When recovering derivatives of order k, the weight is the second fundamental form to the power −k/3.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

∞/∞

.

Boundary reconstruction

A broken ray in an ellipse.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

∞/∞

.

End

Thank you. Slides and papers available at http://users.jyu.fi/~jojapeil.

Joonas Ilmavirta (Jyväskylä)

Broken ray tomography

28 May 2015

∞/∞

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