Thermoacoustic Tomography in Bounded Domainslipeijun/conference/wip2016/... · 2016-04-27 · TAT:...

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TAT: introduction TAT: models Thermoacoustic Tomography in Bounded Domains Yang Yang Purdue University joint work with: Plamen Stefanov, Purdue University CCAM Workshop, April 23, 2016 1 / 38

Transcript of Thermoacoustic Tomography in Bounded Domainslipeijun/conference/wip2016/... · 2016-04-27 · TAT:...

Page 1: Thermoacoustic Tomography in Bounded Domainslipeijun/conference/wip2016/... · 2016-04-27 · TAT: introduction TAT: models Thermoacoustic Tomography in Bounded Domains Yang Yang

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Thermoacoustic Tomography in Bounded Domains

Yang YangPurdue University

joint work with:Plamen Stefanov, Purdue University

CCAM Workshop, April 23, 2016

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Outline

1 TAT: introduction

2 TAT: modelsIn Rn

In Ω: Full DataIn Ω: Partial Data

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Stand-alone medical imaging modalities

High contrast modalities:Optical Tomography (OT);

Electrical Impedance Tomography(EIT);

Elastographic Imaging (EI).

=⇒ low resolution

High resolution modalities:Computerized Tomography (CT);

Magnetic Resonance Imaging (MRI);

Ultrasound Imaging (UI).

=⇒ sometimeslow contrast.

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Coupled physics medical imaging modalities

Idea: use physical mechanism that couples two modalities toimprove resolution while keeping the high contrast capabilities.

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Photo-acoustic effect

Photo-acoustic effect:

Graham Bell: When rapidpulses of light are inci-dent on a sample of mat-ter they can be absorbedand the resulting energywill then be radiated asheat. This heat causes de-tectable sound waves dueto pressure variation in thesurrounding medium.

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Experimental result

Courtesy UCL (Paul Beard’s Lab).

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In Rn

Outline

1 TAT: introduction

2 TAT: modelsIn Rn

In Ω: Full DataIn Ω: Partial Data

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In Rn

Model in Rn

Let Ω ⊂ Rn be a bounded domain with supp f ⊂ Ω. Assume thespeed c(x) is variable and known in Ω. For T > 0, let u solve theproblem

(∂2t − c2(x)∆)u = 0 in (0,T )× Rn

u|t=0 = f∂tu|t=0 = 0.

Measurement: Λf := u|[0,T ]×∂Ω.

Inverse Problem: recover f from Λf .

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In Rn

Literature in Rn

Previous Results: Agranovsky, Ambartsoumian, Anastasio et. al.,Burcholzer, Cox et. al., Finch, Grun, Haltmeier, Hofer, Hristova,Jin, Kuchment, Nguyen, Paltauff, Patch, Rakesh,Stefanov, Uhlmann, Wang, Xu, ...

For the Riemannian manifold (Ω, c−2dx2), let

T0 := maxΩ

dist(x , ∂Ω).

T1 := length of longest geodesic in Ω.

Theorem (Stefanov and Uhlmann, 2009)

T < T0 ⇒ no uniqueness;

T0 < T < T12 ⇒ uniqueness, no stability;

T12 < T ⇒ stability and explicit reconstruction.

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In Rn

Stefanov and Uhlmann’s Time Reversal

Define A(Λf ) := v(0, ·) (pseudo-inverse of Λ) where v solve thebackward problem

(∂2t − c2(x)∆)v = 0 in (0,T )× Ω

v |t=T = φ∂tv |t=T = 0

v |[0,T ]×∂Ω = Λf

where φ solves

∆φ = 0 in Ω, φ|∂Ω = Λf (T , ·).

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In Rn

Stefanov and Uhlmann’s Time Reversal

Denote the error operator by

Kf := f − A(Λf )

or equivalently(I − K )f = A(Λf ).

Stefanov and Uhlmann showed that ‖K‖ < 1 when T > T12 . This

leads to the Neumann series reconstruction:

f = (I − K )−1A(Λf ) =∞∑

m=0

KmA(Λf ).

Numerical implementation: Qian-Stefanov-Uhlmann-Zhao (2011)11 / 38

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In Ω: Full Data

Outline

1 TAT: introduction

2 TAT: modelsIn Rn

In Ω: Full DataIn Ω: Partial Data

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In Ω: Full Data

Model in Ω

Motivation: placing reflectors around the patient proposed by theUCL photoacoustic group.

Assume the speed c(x) is variable and known in Ω. For T > 0, letu solve the problem

(∂2t − c2(x)∆)u = 0 in (0,T )× Ω

u|t=0 = f∂tu|t=0 = 0

∂νu|(0,T )×∂Ω = 0.

Measurement: Λf := u|[0,T ]×∂Ω.

Inverse Problem: recover f from Λf .

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In Ω: Full Data

Literature in Ω

Previous Results: Kunyansky, Holman, Cox, Acosta, Montalto,Nguyen.

For the Riemannian manifold (Ω, c−2dx2), let

T0 := maxΩ

dist(x , ∂Ω).

T1 := length of longest geodesic in Ω.

Theorem (Stefanov and Y., 2015)

T < T0 ⇒ no uniqueness;

T0 < T < T12 ⇒ uniqueness, no stability;

T12 < T ⇒ stability and explicit reconstruction.

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In Ω: Full Data

Time reversal in Ω fails!

Let v solve the problem(∂2

t − c2(x)∆)v = 0 in (0,T )× Ωv |t=T = φ

∂tu|t=T = 0u|(0,T )×∂Ω = Λf .

Define the error operator Kf := f − v(0), but ‖K‖ = 1!

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In Ω: Full Data

Time reversal in Ω fails!

Figure: Failure of the time reversal to resolve all singularities.T = 0.9× diagonal, c = 1. Increasing T does not help!

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In Ω: Full Data

Propagation of singularities

Figure: Propagation of singularities in [0,T ]× Ω for the positive speedonly with Neumann boundary conditions (left) and time reversal withDirichlet ones (right). In the latter case, the sign changes at eachreflection.

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In Ω: Full Data

Main idea: averaged time reversal

This leads to the following idea:

Average with respect to T !

Then the error will average as well and some of the positive andnegative contributions will cancel out. This will make the erroroperator a microlocal contraction.

Let A(τ) be the time reversal over [0, τ ]. Define the averaged timereversal operator as

A0 :=1

T

∫ T

0A(τ) dτ.

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Averaging works!

Figure: Averaged time reversal. T = 0.9× diagonal, c = 1. This is notour inversion yet!

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In Ω: Full Data

Comparison with non-averaged time reversal

Figure: For comparison: Failure of the time reversal to resolve allsingularities. T = 0.9× diagonal, c = 1.

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In Ω: Full Data

Explicit Inversion

Let T1 be the length of the longest geodesic in (Ω, c−2dx2).

Theorem (Stefanov-Y., 2015)

Let (Ω, c−2e) be non-trapping, strictly convex, and let T > T1.Let Ω0 b Ω. Then A0Λ = Id−K0 on HD(Ω0), where‖K0‖L(HD(Ω0)) < 1. In particular, Id−K0 is invertible on HD(Ω0),and the inverse problem has an explicit solution of the form

f =∞∑

m=0

Km0 A0h, h := Λf .

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In Ω: Full Data

Neumann series inversion

Figure: Full data Neumann series inversion, 10 terms, T = 5, on thesquare [−1, 1]2, variable c = 1 + 0.3 sin(πx1) + 0.2 cos(πx2).The artifacts are mainly due to the presence of corners. The L2 error onthe left is 0.44%; and on the right: 0.34%. The L∞ error on the left isabout 1.2%; and about 3% on the right.

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In Ω: Partial Data

Partial data

Assume the speed c(x) is variable and known in Ω. For T > 0, letu solve the problem

(∂2t − c2(x)∆)u = 0 in (0,T )× Ω

u|t=0 = f∂tu|t=0 = 0

∂νu|(0,T )×∂Ω = 0.

Partial Data Measurement: Λf := u|[0,T ]×Γ where Γ is an opensubset of ∂Ω.

Inverse Problem: recover f from Λf .

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In Ω: Partial Data

Partial data: uniqueness

Uniqueness: follows from unique continuation. Let

T0 := maxΩ

dist(x , Γ).

Theorem (Uniqueness)

Λf = 0 for some f ∈ HD(Ω) implies f (x) = 0 for dist(x , Γ) < T .In particular, if T ≥ T0, then f = 0.

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In Ω: Partial Data

Partial data: stability

Stability: follows from boundary control by Bardos-Lebeau-Rauch.

Theorem (Stability)

If each broken geodesic γ(t) hits Γ for |t| ≤ T =⇒ stability.If some does not hit Γ =⇒ no stability.

Figure: Bardos-Lebeau-Rauch condition: Left: unstable. Right: stable

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In Ω: Partial Data

Partial data: smooth wave speed reconstruction

Figure: Partial data inversion with data on the indicated part of ∂Ω.Neumann series inversion with 10 terms, T = 5, Ω = [−1, 1]2. Left:constant speed c = 1, L2 error = 0.7%. Right: variable speedc = 1 + 0.3 sin(πx1) + 0.2 cos(πx2), L2 error = 2%. Again, the mostvisible artifacts can be explained by the presence of corners.

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In Ω: Partial Data

Partial data: discontinous speed

It works well with the following discontinuous speed AND partialdata.

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

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In Ω: Partial Data

Partial data: discontinuous speed, Iteration = 0

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Partial data: discontinuous speed, Iteration = 1

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In Ω: Partial Data

Partial data: discontinuous speed, Iteration = 2

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In Ω: Partial Data

Partial data: discontinuous speed, Iteration = 3

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In Ω: Partial Data

Partial data: discontinuous speed, Iteration = 4

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In Ω: Partial Data

Partial data: discontinuous speed, Iteration = 5

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Partial data: discontinuous speed, Iteration = 6

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Partial data: discontinuous speed, Iteration = 7

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Partial data: discontinuous speed, Iteration = 8

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Partial data: discontinuous speed, Iteration = 9

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