Modeling and Simulation in Photoacousticsmath.mit.edu/~vjugnon/pdf/inha_slides.pdf · 2012. 9. 7. · NIMS 2010. Introduction Boundary conditions Limited view Attenuation Quantitative

IntroductionBoundary conditions

Limited viewAttenuation

Quantitative PhotoAcousticsConclusion

Modeling and Simulation in Photoacoustics

Vincent Jugnon

6 aout 2010

NIMS 2010




Summary

1 Introduction

2 Imposed boundary conditions

3 Limited view

4 AttenuationFree spaceBounded Domain

5 Quantitative PhotoAcousticsRTEDiffusion

6 Conclusion

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The photoacoustic effect

phyisical phenomenon observed when a medium isexposed to an electromagnetic wave.the EM wave drops energy in the medium, which heats up,dilates and emits an (ultra-sonic) acoustic wave.

Mary Evans Picture Library

Photoacoustic imaging

The sound of lightJun 4th 2009From The Economist print edition

Biomedical technology: A novel scanning technique that combines optics with ultrasound couldprovide detailed images at greater depths

IF LIGHT passed through objects, rather than bouncing off them, people might now talk to each other on“photophones”. Alexander Graham Bell demonstrated such a device in 1880, transmitting a conversation on abeam of light. Bell’s invention stemmed from his discovery that exposing certain materials to focused,flickering beams of light caused them to emit sound—a phenomenon now known as the photoacoustic effect.

It was the world’s first wireless audio transmission, and Bell regarded the photophone as his most importantinvention. Sadly its use was impractical before the development of optical fibres, so Bell concentrated insteadon his more successful idea, the telephone. But more than a century later the photoacoustic effect is making acomeback, this time transforming the field of biomedical imaging.

A new technique called photoacoustic (or optoacoustic) tomography, which marries optics with ultrasonicimaging, should in theory be able to provide detailed scans comparable to those produced by magnetic-resonance imaging (MRI) or X-ray computerised tomography (CT), but with the cost and convenience of ahand-held scanner. Since the technology can operate at depths of several centimetres, its champions hopethat within a few years it will be able to help guide biopsy needles deep within tissue, assist withgastrointestinal endoscopies and measure oxygen levels in vascular and lymph nodes, thereby helping todetermine whether tumours are malignant or not. There is even scope to use photoacoustic imaging tomonitor brain activity and gene expression within cells.

To create a photoacoustic image, pulses of laser light are shone onto the tissue being scanned. This heats thetissue by a tiny amount—just a few thousandths of a degree—that is perfectly safe, but is enough to cause thecells to expand and contract in response. As they do so, they emit sound waves in the ultrasonic range. Anarray of sensors placed on the skin picks up these waves, and a computer then uses a process of triangulationto turn the ultrasonic signals into a two- or three-dimensional image of what lies beneath.

The technique works at far greater depths (up to seven centimetres) than other optical-imaging techniquessuch as confocal microscopy or optical-coherence tomography, which penetrate to depths of only about amillimetre. And because the degree to which a particular wavelength of light is absorbed depends on the type

Economist.com http://www.economist.com/science/tq/PrinterFriendly.cfm?story_id=1...

1 of 3 15/06/2009 09:46

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Interest for medical imaging

advantages of both pure optical imaging and ultrasonicimaging without their drawbacks.possible functional imaging.

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The physical equations - direct problem

ρ(x , t)Cp

∂T∂t

(x , t)− κ∆T (x , t) = E(x , t)

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ρ(x , t)Cp

∂T∂t

(x , t)− κ∆T (x , t) = E(x , t)∂ρ

∂t(x , t) +∇.(ρv)(x , t) = ρ(x , t)β

∂T∂t

(x , t)

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ρ(x , t)Cp∂T∂t

(x , t)− κ∆T (x , t) = E(x , t)∂ρ

∂t(x , t) +∇.(ρv)(x , t) = ρ(x , t)β

∂T∂t

(x , t)

ρ(x , t)[∂v∂t

(x , t)(v.∇)v(x , t)]

= −∇p(x , t) + µ∆v(x , t) +(ζ + 1

3η)∇(∇.v(x , t))

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Some reasonable approximations

The problem is simplified using the following assumptions :

non-visquous and incompressible medium.thermal and stress confinement (very different time scales)short EM pulse :

E(x , t) = A(x)δ(t)

small variations (linearization)adiabatic reversible process

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The problem is simplified using the following assumptions :non-visquous and incompressible medium.

thermal and stress confinement (very different time scales)short EM pulse :

E(x , t) = A(x)δ(t)


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The problem is simplified using the following assumptions :non-visquous and incompressible medium.thermal and stress confinement (very different time scales)

short EM pulse :

E(x , t) = A(x)δ(t)


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The problem is simplified using the following assumptions :non-visquous and incompressible medium.thermal and stress confinement (very different time scales)short EM pulse :

E(x , t) = A(x)δ(t)


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E(x , t) = A(x)δ(t)

small variations (linearization)

adiabatic reversible process

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E(x , t) = A(x)δ(t)


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The considered model

∂2p∂t2 (x , t)− c2∆p(x , t) = 0

p(x ,0) =β

CpA(x)

∂p∂t

(x ,0) = 0

where A(x) is the ”instantaneous” energy deposition of the EMwave in the medium.

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The inverse problem

Reconstruct :p(x ,0) = p0(x) =

β

CpA(x)

from boundary measurements∂p∂ν

(y , t) or p(y , t), y ∈ ∂Ω

NIMS 2010




The free-space case

In the free space, we can express the pressure p on theboundary of the medium Ω in terms of spherical means of p0.We can then use a wide range of inversion formulas for thespherical Radon transform to reconstruct p0.

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NIMS 2010




1 Introduction


3 Limited view



6 Conclusion

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The free-space hypothesis implies that ultrasonic waves ”do notsee” the boundary of the observed medium.In some imaging devices, we have to account for acousticboundary conditions :

1c2

0

∂2p∂t2 (x , t)−∆p(x , t) =

1c2

0δ′(t)p0(x)

p(x ,0) =∂p∂t

(x ,0) = 0

p(y , t) = 0 y ∈ ∂Ω

with boundary measurements∂p∂ν

(y , t) on some time interval(0,T ).

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We cannot express the measurements in terms ofspherical means anymore.However, a simple duality approach gives access to the(line) Radon transform of p0.

NIMS 2010




Consider the solutions of the wave equation in the free-space :

1c2

0

∂2v∂t2 (x , t)−∆v(x , t) = 0

with final conditions :

v(x ,T ) =∂v∂t

(x ,T ) = 0

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Testing the measures against those ”probe functions”, we get :∫ T

0

∫∂Ω

∂p∂ν

(y , t)v(y , t)dσ(y) =

∫Ω

p0(x)∂v∂t

(x ,0)dx

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We first considered plane test functions of the form :

v(x , t ; θ, s) = δ

(x .θc0

+ s − t)

which enables to compute :

F (s, θ) =

∫Ω

p0(x)δ′(

x .θc0

+ s)

dx

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In the frame of small inclusions :

p0(x) =

Ninc∑i=1

aiχDi (x)

we see that F (s, θ) will be non-zero if and only if the linex .θc0

+ s = 0 intersects an inclusion Di .Combining the information for several (s, θ) (=intersectingstripes), we can determine the position of the smallinclusions.using asymptotic development, we can get |ai |ε where ε isthe radius of the inclusion.

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Now, choosing plane test functions of the form :

v(x , t ; θ, s) = 1− H(

x .θc0

+ s − t)

where H is the Heaviside function. we will have :∫ T

0

∫∂Ω

∂p∂ν

(y , t)v(y , t)dσ(y) = R[p0](s, θ)

where R denotes the Radon transform. It is then straightforwardto obtain p0 using classical back-projection algorithms.

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Note that using the previous probe functions :

v(x , t ; θ, s) = δ

(x .θc0

+ s − t)

we have indeed :

F (s, θ) =∂

∂sR[p0](s, θ)

and we can also reconstruct p0 in a ”cheaper” way.

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1 Introduction


3 Limited view



6 Conclusion

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In some experimental settings, measurements are acquiredonly on a part of the boundary Γc ⊂ ∂Ω.

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We still consider the wave equation :∂2p∂t2 (x , t)− c2∆p(x , t) = 0

p(x ,0) = p0(x)∂p∂t

(x ,0) = p1(x)

p(y , t) = 0 y ∈ ∂Ω

and we want to reconstruct p0,p1 from the measurements :∂p∂ν

(x , t) x ∈ Γc .

NIMS 2010




The previous duality approach is quite robust with respect toview limitation...

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But in some case, it fails :

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Once again we are interested in obtaining information of thekind :

M(v0, v1) =

∫Ω

p0(x)v1(x)− p1(x)v0(x)dx

for example we would have :

M(

0, δ(

x .θc0

+ s))

= R[p0](s, θ)

andM(

0,eik .x)

= F [p0](k)

NIMS 2010




Once again we are interested in obtaining information of thekind :

M(v0, v1) =

∫Ω

p0(x)v1(x)− p1(x)v0(x)dx

for example we would have :

M(

0, δ(

x .θc0

+ s))

= R[p0](s, θ)

andM(

0,eik .x)

= F [p0](k)

NIMS 2010




If we can find gv0,v1 such that v solution of :

1c2

s

∂2v∂t2 (x , t)−∆v(x , t) = 0

v(x ,0) = v0(x),∂v∂t

(x ,0) = v1(x)

v(y , t) = 0 y ∈ ∂Ω\Γcv(y , t) = gv0,v1 (y , t) y ∈ Γc

(1)

vansihes at time T :

v(x ,T ) = 0,∂v∂t

(x ,T ) = 0

Then we get what we want :∫ T

0

∫Γc

gv0,v1 (y , t)∂p∂ν

(y , t)dσ(y)dt = M(v0, v1)

NIMS 2010




If we can find gv0,v1 such that v solution of :

1c2

s

∂2v∂t2 (x , t)−∆v(x , t) = 0

v(x ,0) = v0(x),∂v∂t

(x ,0) = v1(x)

v(y , t) = 0 y ∈ ∂Ω\Γcv(y , t) = gv0,v1 (y , t) y ∈ Γc

(1)

vansihes at time T :

v(x ,T ) = 0,∂v∂t

(x ,T ) = 0

Then we get what we want :∫ T

0

∫Γc

gv0,v1 (y , t)∂p∂ν

(y , t)dσ(y)dt = M(v0, v1)

NIMS 2010




Such a control exists :

Theorem J.L. LionsIf T and Γc are ”large enough”, for all (v0, v1) inL2(Ω) ∈ H−1(Ω), there exists a control gv0,v1 solution of thiscontrol problem.

+ constructive proof.

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Large enough ?

Theorem Bardos, Lebeau, RauchEvery ray of geometrical optics, starting at any point x ∈ Ω, at timet = 0, hits Γc before time T at a nondiffractive point.

⇒ No “glancing” nor “trapped” rays.

Uncontrollable and controllable geometries

NIMS 2010




Let (e0,e1) ∈ H10 (Ω)× L2(Ω). Solve :

1c2

s

∂2φ

∂t2 (x , t)−∆φ(x , t) = 0

φ(x ,0) = e0(x),∂φ

∂t(x ,0) = e1(x)

φ(y , t) = 0 y ∈ ∂Ω

and then backwards :

1c2

s

∂2ψ

∂t2 (x , t)−∆ψ(x , t) = 0

ψ(x ,T ) = 0,∂ψ

∂t(x ,T ) = 0

ψ(y , t) =

0 y ∈ ∂Ω\Γc∂φ∂ν y ∈ Γc

NIMS 2010




Let (e0,e1) ∈ H10 (Ω)× L2(Ω). Solve :

1c2

s

∂2φ

∂t2 (x , t)−∆φ(x , t) = 0

φ(x ,0) = e0(x),∂φ

∂t(x ,0) = e1(x)

φ(y , t) = 0 y ∈ ∂Ω

and then backwards :

1c2

s

∂2ψ

∂t2 (x , t)−∆ψ(x , t) = 0

ψ(x ,T ) = 0,∂ψ

∂t(x ,T ) = 0

ψ(y , t) =

0 y ∈ ∂Ω\Γc∂φ∂ν y ∈ Γc

NIMS 2010




Define :

Λ(e0,e1) =

(ψ(x ,0),

∂ψ

∂t(x ,0)

)

The operator Λ is linear.Under the controllability conditions, it is an isomorphismfrom

(H1

0 (Ω)× L2(Ω))

onto(L2(Ω)× H−1(Ω)

).

Finding gv0,v1 is equivalent to solving :

Λ(e0,e1) = (v0, v1)

Then we get ψ ≡ v and gv0,v1 = ψ|Γc=∂φ

∂ν

∣∣∣∣Γc

NIMS 2010




Define :

Λ(e0,e1) =

(ψ(x ,0),

∂ψ

∂t(x ,0)

)


(H1

0 (Ω)× L2(Ω))


).


Λ(e0,e1) = (v0, v1)


∂ν

∣∣∣∣Γc

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Define :

Λ(e0,e1) =

(ψ(x ,0),

∂ψ

∂t(x ,0)

)


(H1

0 (Ω)× L2(Ω))


).


Λ(e0,e1) = (v0, v1)


∂ν

∣∣∣∣Γc

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So where’s the catch ?

Numerically, the HUM produces an ill-posed problem...

Ill-posedness

stability + consistency V/ convergence

Various solutions :Tykhonov regularization.Mixed finite elements.Bi-grid filtering.Uniformly controllable schemes.Spectral approach.

NIMS 2010




So where’s the catch ?

Numerically, the HUM produces an ill-posed problem...

Ill-posedness

stability + consistency V/ convergence

Various solutions :Tykhonov regularization.Mixed finite elements.Bi-grid filtering.Uniformly controllable schemes.Spectral approach.

NIMS 2010




We solve the wave equation (in 2D) using a P1 finiteelements scheme (space) - explicit finite differences (time)+ mass lumpingTo solve Λ(e0,e1) = (v0, v1), we use a conjugate gradientmethod on the operator Λ.To deal with ill-posedness, we use a bigrid approach : wesolve the wave equation on a fine mesh and carry out theupdate step of the CG on a coarse mesh.Convergence results are available on homogeneousmeshes.

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We now have the liberty to choose the parametrized testfunctions v1.

for general p0, we can use

v1(x) = eik .x or δ(x .θ/c0 − s)

to get and inverse integral transforms of p0.

for point-like initial conditions : p0(x) =∑

i

δ(zi), we can

use less expensive adapted approaches (arrival time,backpropagation, MUSIC, Kirchhoff migration).

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Although this approach is time-expensive : O(n2d+1x ), it is highly

parallelizable and we applied it successfully.

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Consider point-like initial condition :

p0(x) = δz(x)

We are interested in getting z.Choose v1, carry out the control and compute

M(0, v1) := N(v1) =

∫Ω

v1(x)δz(x)dx = v1(z)

for some ”smartly” parametrized v1 .

NIMS 2010




Arrival time

Simplest choice :

v1(x ; xr ) = |xr − x |

We get :N(v1(.; xr )) = |z − xr |

So that z ∈ C(xr ,N(v1(.; xr ))).Intersecting several circles, we can uniquely determine z.No practical interest since it requires to know the intensityof the (unique), source.

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virtual receiversreal acoustic receiverstrue position of the sourceintersections of the circles

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Time harmonic based algorithms

Considering :v1(x ; θ) = eiωθ·x

We getN(v1(.; θ)) = eiωθ.z

Now compute the imaging function :

Fθ(y) = <(

N(v1(.; θ, ω))e−iωθ·y)

= cos (ωθ · (z − y))

Summing for a discretization of the parameter∑θ

Fθ(y)

reaches its maximum at point z.

NIMS 2010




Time harmonic based algorithms

Considering :v1(x ; θ) = eiωθ·x

We getN(v1(.; θ)) = eiωθ.z

Now compute the imaging function :

Fθ(y) = <(

N(v1(.; θ, ω))e−iωθ·y)

= cos (ωθ · (z − y))

Summing for a discretization of the parameter∑θ

Fθ(y)

reaches its maximum at point z.

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Back Propagation imaging function

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Back Propagation imaging function

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MUSIC imaging function

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Kirchhoff imaging function

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This geometric control approach is expensive forreconstructing general distributions (O(n5

x )).in the case of point-like sources, we have faster algorithmswhich make this approach interesting.

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Free spaceBounded Domain

1 Introduction


3 Limited view



6 Conclusion

NIMS 2010





So far, we have worked in (acoustically) non-attenuatingmedia.In real experiments, ultra-sonic waves interact with themedium and lose energy.If not accounted for, this phenomenon generates a blurringin the reconstruction.

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We now consider the attenuated wave equation :

1c2

0

∂2pa

∂t2 (x , t)−∆pa(x , t)− L(t) ∗ pa(x , t) =1c2

0δ′(t)p0(x)

where L takes the form :

L(t) =1√2π

∫R

(K 2(ω)− ω2

c20

)eiωtdω

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We propose algorithms to deal with power-law attenuationmodels of the form :

K (ω) =ω

c0+ iCa|ω|ν

with special emphasis on the case ν = 2, which is close to thethermo-visquous model :

K 2(ω) =ω2

c20

11− aiω

at low frequencies.

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In free-space, we link pa(x , t) to p(x , t) (solution with noattenuation) via an integral operator. Inverting this operator,we get p and using usual algorithms, we reconstruct p0.In a bounded medium, testing the attenuated measuresagainst an appropriate set of probe functions, we can get a(similar) operator of p0 that we have to invert.

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In both case, the inversion of the operator is achieved using anasymptotic expansion (w/r to a) given by a stationnary phasetheorem. We will write :

Lφ = Lkφ+ o(ak+1)

and :L−1ψ = L−1

k ψ + o(ak+1)

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Going in the Fourier domain, we can write :(∆ + K 2(ω)

)pa(x , ω) =

iωc2

0p0(x)

(∆ +

ω2

c20

)p(x , ω) =

iωc2

0p0(x)

Green’s functions are known for both equations, so we get :

pa(x , t) =1

2π

∫R

ω

c0K (ω)e−iωt

∫ ∞0

p(x , s)eic0K (ω)sdsdω

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We want to invert the operator :

Lφ(t) =1

2π

∫R

ω

c0K (ω)e−iωt

∫ ∞0

φ(s)eic0K (ω)sdsdω

NIMS 2010





In the case ν = 2, we have K (ω) = ωc0

+ iaω2

2 and :

ω

c0K (ω)= 1− i

ac0

2ω +

∞∑k=2

ik (−1)k(ac0ω

2

)k

Lφ(t) =1

2π

∫ ∞0

φ(x , s)

∫R

(1− i

ac0

2ω +

∞∑k=2

(−iω)k(a

2

)k)

eiω(s−t)e−12 c0aω2sdωds

=

(1 +

ac0

2∂

∂t+∞∑

k=2

(−

ac0

2

)k ∂k

∂tk

)(1√

2π

∫ ∞0

φ(x , s)1

√c0as

e− 1

2(s−t)2

c0as ds

)

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We have to investigate the operator :

Mφ =1√2π

∫ ∞0

φ(x , s)1√

c0ase−

12

(s−t)2

c0as ds

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Stationnary phase theoremFor f and ψ sufficienty smooth, if :

Im(f (t0)) = 0 f ′(t0) = 0 f ′′(t0) 6= 0 f ′ 6= 0 in K\t0

then for ε > 0, we have the following approximation :∣∣∣∣∣∣∫

Kψ(t)ei f (t)

ε dt − eif (t0)/ε

(ε−1 f ′′(t0)

2iπ

)− 12 ∑

j<k

εjLjψ

∣∣∣∣∣∣ ≤ Cεk∑α≤2k

sup |ψ(α)(x)|

where the Lj can be explicitely expressed as differentialoperators.

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In our case, we obtain for k = 1 :∣∣∣Mφ(t)−(φ(t) +

ac0

2(tφ(t))′′

)∣∣∣ ≤ Ca32∑α≤4

sup |φ(α)(t)|

At the end we have a third order ODE to solve :

pa(x , t) = Lp(x , t) =

(1 +

ac0

2∂

∂t

)(p(x , t) +

ac0

2

(t∂2p∂t2 (x , t) +

∂p∂t

(x , t)))

+o(a)

for higher order asymptotics, we use :

Mφ(t) =k∑

i=0

(c0a)i

2i i!(t iφ(t))(2i)(t) + o(ak )

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In a bounded domain, we consider :1c2

0

∂2pa

∂t2 (x , t)−∆pa(x , t)− L(t) ∗ pa(x , t) =1c2

0δ′(t)p0(x)

pa(y , t) = 0 y ∈ ∂Ω

Which reads in the Fourier domain : (∆ + K 2(ω)

)pa(x , ω) = iω

c20p0(x)

pa(y , t) = 0 y ∈ ∂Ω

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We will consider duality with solutions of the backwardsattenuated wave equation in the free space.

We define K (ω) =

√K (ω)2 and :(

∆ + K 2(ω))

va(x , ω) = 0

to get :∫R

∫∂Ω

∂pa

∂ν(x , ω)va(x , ω)dσ(x)dω =

i√2π

∫Ω

p0(x)

(∫Rωva(x , ω)dω

)dx

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More precisely we look at plane solutions :

va(x , ω) = g(ω)e−K (ω)(x .θ−c0τ)

where we have to choose g carefully.

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In the thermo-visquous case we have :

K (ω) ≈ ω

c0− iaω2

2

Setting :

g(ω) =1iω

e−12 ac0Tω2

where T is such that(

T + x .θc0− τ)

stays positive for all x ∈ Ω,and for all parameters θ, τ .

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We have :

va(x , t) ≈ erf

(x .θ − c0τ + t)2

2ac0

(T + x .θ

c0− τ)

and in the end :∫ T

0

∫∂Ω

∂pa

∂ν(x , t) va(x , t)dσ(x)dt ≈∫ smax

smin

R[p0](θ, s)1√

ac0

(T + s

c0− τ)e

− (s−c0τ)2

2ac0

(T + s

c0−τ

)ds

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We have to study the operator :

M2φ(τ) =

∫ smax

smin

φ(s)1√

ac0

(T + s

c0− τ)e− (s−c0τ)2

2ac0

(T + s

c0−τ

)ds

which looks a lot likeM.

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Using the same stationnary phase approach, we get at order 1 :

M2φ(τ) = φ(c0τ) +ac0T

2

(∂2φ

∂t2 (c0τ) +2

c0T∂φ

∂t(c0τ)

)+ o(a)

and for higher orders :

M2φ(τ) =k∑

i=0

(ac0)i

2i i!Diφ(c0τ) + o(ak )

where :

Diφ(t) =

((T +

sc0− τ)i

φ(s)

)(2i)∣∣∣∣∣∣s=t

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RTEDiffusion

1 Introduction


3 Limited view



6 Conclusion

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RTEDiffusion

A second inverse problem

good reconstruction of the absorbed energy density p0(x)inside the medium is not the true (relevant) piece ofinformation.It is not intrinsic and depends on the illumination (lightfluence or photon density) and on optical coefficients.Need of light propagation model, in terms of intrinsicoptical coefficients (µa, µs, ...) that we will aim atreconstructing from the knowledge of p0(x) everywhereinside the medium.

NIMS 2010




RTEDiffusion

We address 2 main ways of modeling light in tissues.First the radiative tranfer equation (RTE) :

s.∇xϕ(x , s) + µt (x , s)ϕ(x , s) =

∫Sn−1

k(x , s′, s)ϕ(x , s′)dσ(s′)

with µt = µa + µs, µs(x , s) =

∫Sn−1

k(x , s, s′)dσ(s′).

In this case we have : p0(x) ∝∫

Sn−1µa(x , s′)ϕ(x , s′)dσ(s′).

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RTEDiffusion

A theoretical result for the RTE

We define an Albedo operator A :

L1(∂Ω−) → L1(Ω)

ϕ(x , s) 7→ Aϕ(x) =

∫Sn−1

µa(x , s′)ϕ(x , s′)dσ(s′)

Under the following assumptions :µt and k are continuous with bounded support.µa(x , s) ≥ µ0 > 0µt is known in a vincinity of ∂Ω

µt (x , s) = µt (x ,−s) and µa(x , s) = µa(x ,−s)

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RTEDiffusion

We have the following stability estimates :

Theorem

If (µt , µa, k) generates the Albedo A and (µt , µa, k) generates Athen :

‖µt − µt‖L∞(Sn−1,W−1,1(X)) +‖µa − µa‖L∞(Sn−1,L1(X))

≤ C‖A− A‖L(L1(∂Ω−),L1(Ω))

Assuming a model for k , we can even get stability estimates onanisotropy.

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RTEDiffusion

The diffusion equation reads :(µa(x)− 1

3∇. 1µt (x)

∇)

Φ(x) = 0

In this case, p0(x) ∝ µa(x)Φ(x).

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RTEDiffusion

In the case of small inclusions, we proposed 2 approaches :assuming µs is known, asymptotic approach givesestimates of µa and ε.assuming that the dependence of µa, µs with respect tolight wavelength is known, we can also estimate µa and ε.

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Photoacoustic imaging is a promising tool with immediateapplicability.

Most approaches rely on simplifying (often valid)assumptions that we tried to ”relax”

free-space→ duality.full-view measurements→ geometric control.non-attenuating medium→ asymptotic correction.

PA yields a second inverse problem (light propagation) thathas been less addressed.

NIMS 2010




Photoacoustic imaging is a promising tool with immediateapplicability.Most approaches rely on simplifying (often valid)assumptions that we tried to ”relax”



NIMS 2010




Photoacoustic imaging is a promising tool with immediateapplicability.Most approaches rely on simplifying (often valid)assumptions that we tried to ”relax”



NIMS 2010




Future work involves :taking into acount inhomogeneous (random) sound speed.dealing with real experimental datas.

0 500 1000 1500 2000 2500 3000 3500−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

0 1000 2000 3000 4000−1.5

−1

−0.5

0

0.5

1

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Documents

Modeling and Simulation in Photoacousticsmath.mit.edu/~vjugnon/pdf/inha_slides.pdf · 2012. 9. 7. · NIMS 2010. Introduction Boundary conditions Limited view Attenuation Quantitative