inverse transport problems and applications guillaume balmueller/graduate...inverse transport theory...

Colorado State University, Ft Collins, August 2nd-3rd, 2007Colorado State University, Ft Collins, August 2nd-3rd, 2007Colorado State University, Ft Collins, August 2nd-3rd, 2007

Inverse Transport Problems

and Applications

Guillaume Bal

Department of Applied Physics & Applied Mathematics

Columbia University

http://www.columbia.edu/∼gb2030

[email protected]


Outline

1. X-ray tomography and Radon transform

2. SPECT and Attenuated Radon transform

3. Optical Tomography and Optical Molecular Imaging

4. Inverse transport theory

5. Diffusion approximation and diffusion-type measurements


Data Acquisition in CT-scan

a(x) is the unknown absorption co-

efficient. For each line in the plane

(full measurements), the measured ra-

tio uout(s, θ)/uin(s, θ) is equal to

exp(−

∫line(s,θ)

a(x)dl). s =line-offset.

The X-ray density u(x, θ) solves the

transport equation

θ · ∇u(x, θ) + a(x)u(x, θ) = 0.

Here, x is position and θ =

(cos θ, sin θ) direction.


Radon transform and transport source problem

We recast the inverse problem for

a(x) as an inverse transport source

problem

θ · ∇u(x, θ) = a(x).

This time, the “measurement”

uout(x, θ) provides∫line

a(x)dl.

The inverse source problem consists

of reconstructing a(x) from all its

line integrals, i.e., from the (unat-

tenuated) Radon transform.


Radon transform. Notation and Inversion

We define the Radon transform

Ra(s, θ) = Rθa(s) =∫Ra(sθ⊥ + tθ)dt

for s ∈ R and θ ∈ S1.

Note the redundancy:

Ra(−s, θ+ π) = Ra(s, θ).

Introduce the adjoint operator and the Hilbert transform

R∗g(x) =∫ 2π

0g(x · θ⊥, θ)dθ, Hf(t) =

1

πp.v.

∫R

f(s)

t− sds.

Then we have (e.g. in the L2-sense) the reconstruction formula

Id =1

4πR∗

∂

∂sHR =

1

4πR∗H

∂

∂sR.


Outline







Data Acquisition in SPECT(Single Photon Emission Computed Tomography)

Here a(x) is a known absorption

coefficient and f(x) an unknown

source term (radioactive particles

are injected into the body and at-

tach differently to different tis-

sues). The γ− ray density solves

θ · ∇u(x, θ) + a(x)u(x, θ) = f(x).

The inverse source problem con-

sists of reconstructing f(x) from

uout(x, θ).


Doppler tomography

So far, the unknown quantities a(x) or f(x) are scalar quantities. There

are applications where vector-valued functions are to be imaged.

For instance in Doppler tomography. Consider a fluid with sound speed

c(x) and velocity v(x). Ultrasounds propagating along the line l(s, θ) have

effective speed c(x) + θ · v(x). Interchanging sources and receivers, we

collect the two travel times

T1 =∫l(s,θ)

ds

c(x) + θ · v(x), T2 =

∫l(s,θ)

ds

c(x)− θ · v(x).

Assuming |v| c, we thus measure∫l(s,θ)

ds

c(x),

∫l(s,θ)

θ · v(x)

c2(x)ds = θ ·

∫l(s,θ)

v(x)

c2(x)ds.

We thus reconstruct c−1(x) from its Radon transform and then want to

reconstruct a field F(x) from its vectorial Radon transform θ ·RF(s, θ).


Some definitions

Let us define

ϕ(x, θ) = (ψ+ − ψ−)(x, θ).

It depends on the measured data and is given by

iϕ(x, θ) = [R∗−a,θHaRa,θf ](x) = [R∗−a,θHag(s, θ)](x),

where

R∗a,θg(x) = eDθa(x)g(x·θ⊥), Ha = (CcHCc + CsHCs)

Ccg(s, θ) = g(s, θ) cos(HRa(s, θ)

2), Csg(s, θ) = g(s, θ) sin(

HRa(s, θ)

2).

Here R∗a,θ is the adjoint operator to Ra,θ. We note that iϕ(x, θ) is real-

valued and that θ ·∇ϕ+ aϕ = 0.


Reconstruction formula

We have:

f(x) =1

4π

∫ 2π

0θ⊥·∇(iϕ)(x, θ)dθ.

This is the Novikov formula. A similar inversion result was obtained at

about the same time by A.L. Bukhgeim.


Application to Doppler tomography.

In Doppler tomography, the source term of interest is of the form

f(x, θ) = F(x) · θ F(x) = (F1(x), F2(x)).

So we define the source term f1(x) = 12(F1(x) − iF2(x)) and fk(x) ≡ 0

for |k| 6= 1. The constraint n = 0 gives

∇× F(x) =∂F2(x)

∂x−∂F1(x)

∂y=

1

2∆(iϕ0)(x).

The constraint n = 1 gives H2f−1(z) = ϕ1(z) so that

1

2

(F1(z) + iF2(z)

)= −

∂

∂z

1

a(z)

∂ϕ1(z)

∂z.

This explicit reconstruction formula is valid on the support of a(x) and

has no equivalent when a ≡ 0.


Outline



2.a. Derivation of the Novikov formula





Mathematical modeling

The transport equation with anisotropic source term is given by

θ · ∇ψ(x, θ) + a(x)ψ(x, θ) = f(x, θ) =N∑

k=−Nfk(x)eikθ, x ∈ R2, θ ∈ S1.

We identify θ = (cos θ, sin θ) ∈ S1 and θ ∈ (0,2π). We assume that

f−k = fk and fk(x) is compactly supported. The boundary conditions

are such that for all x ∈ R2,

lims→+∞

ψ(x− sθ, θ) = 0.

The absorption coefficient a(x) is smooth and decays sufficiently fast at

infinity. The above transport solution admits a unique solution and we

can define the symmetrized beam transform as

Dθa(x) =1

2

∫ ∞

0[a(x− tθ)− a(x + tθ)]dt.


Mathematical modeling (II)

The symmetrized beam transform satisfies θ · ∇Dθa(x) = a(x) so that

the transport solution is given by

eDθa(x)ψ(x, θ) =∫ ∞

0(eDθaf)(x− tθ, θ)dt.

Upon defining θ⊥ = (− sin θ, cos θ) and x = sθ⊥ + tθ, we find that

limt→+∞

eDθa(sθ⊥+tθ)ψ(sθ⊥ + tθ, θ) =

∫R(eDθaf)(sθ⊥ + tθ, θ)dt

Measurements = limt→+∞

ψ(sθ⊥ + tθ, θ) = e−(Rθa)(s)/2(Ra,θf)(s),

where Rθ is the Radon transform and Ra,θ the Attenuated Radon Trans-

form (AtRT) defined by:

Rθf(s) =∫Rf(sθ⊥ + tθ, θ)dt =

∫R2f(x, θ)δ(x · θ⊥ − s)dx

(Ra,θf)(s) = (Rθ(eDθaf))(s) = Data


Inverse Problems

The inverse problem consists then in answering the following questions:

1. Knowing the AtRT Ra,θf(s) for θ ∈ S1 and s ∈ R, and the absorption

a(x), what can we reconstruct in f(x, θ)?

2. Assuming f(x, θ) = f0(x) or f(x, θ) = F(x) · θ, can we obtain explicit

formulas for the source term?

3. Can we reconstruct f(x, θ) = f0(x) from half of the measurements or

do we at least have uniqueness of the reconstruction?

4. Do we have a reliable numerical technique to obtain fast reconstruc-

tions?


Reconstruction as a Riemann Hilbert problem

We recast the inversion as a Riemann Hilbert (RH) problem. Let us define

T = λ ∈ C, |λ| = 1, D+ = λ ∈ C, |λ| < 1, and D− = λ ∈ C, |λ| > 1.Let ϕ(t) be a smooth function defined on T . Then there is a unique

function φ(λ) such that:

(i) φ(λ) is analytic on D+ and D−

(ii) λφ(λ) is bounded at infinity

(iii) ϕ(t) = φ+(t)− φ−(t) ≡ [φ](t)

= lim0<ε→0

(φ((1− ε)t)− φ((1 + ε)t))

Moreover φ(λ) is given by the Cauchy formula

φ(λ) =1

2πi

∫T

ϕ(t)

t− λdt, λ ∈ C\T.


Riemann-Hilbert problem for AtRT, a road map

1. Extend the transport equation to the complex plane (complex-valued

directions of propagation θ → eiθ = λ ∈ C). Replace the transport solu-

tion ψ(x, λ) by φ(x, λ), which is analytic on D+ and D− and O(λ−1) at

infinity by subtracting a finite number of analytic terms on C\0.

2. Verify that the jump of φ(x, λ) at λ ∈ T is a function of the measured

data Ra,θf(s).

3. Read off the constraints imposed on the source terms fk(x) obtained

by Taylor expansion of φ(x, λ) at λ = 0.

4. In simplified settings, reconstruct the fk(x) from the constraints.


Step 1: Complexification of transport equation

Define

λ = eiθ, z = x+ iy with x = (x, y),∂

∂z=

1

2

(∂

∂x− i

∂

∂y

).

The complexified transport equation is then recast as(λ∂

∂z+ λ−1 ∂

∂z+ a(z)

)ψ(z, λ) = f(z, λ).

We consider the above equation for arbitrary complex values of λ. ψ(z, λ)

is analytic on λ ∈ C\(T ∪ 0) and is given by

ψ(z, λ) = e−h(z,λ)∫CG(z − ζ, λ)eh(ζ,λ)f(ζ, λ)dm(ζ),

where h(z, λ) =∫CG(z − ζ, λ)a(ζ)dm(ζ) and

(λ∂

∂z+ λ−1 ∂

∂z

)G(z, λ) = δ(z), so that G(z, λ) =

sign(|λ| − 1)

π(λz − λ−1z).


The source term is given by f(z, λ) =∑Nk=−N fk(z)λ

k. On D+ we have

G(z, λ) =1

πz

∞∑m=0

(z

z

)mλ2m+1, and ψ(·, λ) =

∞∑m=1

(Hmf(·, λ))λm,

where the operators Hm are explicitly computable with

H1 =(∂

∂z

)−1, H2 = −H1aH1,

∂

∂zHk+2 + aHk+1 +

∂

∂zHk = 0.

Using a similar expression on D−, we find that

φ(z, λ) = ψ(z, λ)−−1∑

n=−∞λn

∞∑m=1

(Hmfn−m)(z)−0∑

n=−∞λ−n

∞∑m=1

(Hmfm−n)(z),

satisfies the hypotheses of the RH problem: it is analytic on D+ ∪ D−

and of order O(λ−1) at infinity. Its jump across T is the same as that of

ψ since the difference ψ − φ is analytic in C\0. On D+ it is given by

φ(z, λ) =∞∑n=0

λn∞∑

m=1

(Hmfn−m −Hmfn+m)(z).


Simplifications when f(z, λ) = f0(z)

When f(z, λ) = f0(z), the solution

ψ(z, λ) = e−h(z,λ)∫CG(z − ζ, λ)eh(ζ,λ)f0(ζ)dm(ζ),

of the complexified transport equation(λ∂

∂z+ λ−1 ∂

∂z+ a(z)

)ψ(z, λ) = f0(z),

is analytic on λ ∈ C\T and λψ(z, λ) is bounded at infinity since G(z, λ)

satisfies these properties. So φ(z, λ) = ψ(z, λ) meets the conditions of

the Riemann-Hilbert problem (sectionally analytic and right behavior at

infinity).

Moreover we find using the transport equation and the Cauchy formula:

f0(z) = limλ→0

1

λ

∂

∂zψ(z, λ) = lim

λ→0

1

λ

∂

∂z

1

2πi

∫T

[ψ](t)

t− λdt.


Step 2: jump conditions

Recall that

G(z, λ) =sign(|λ| − 1)

π(λz − λ−1z), h(z, λ) =

∫CG(z − ζ, λ)a(ζ)dm(ζ),

and that the transport solution is given by

ψ(z, λ) = e−h(z,λ)∫CG(z − ζ, λ)eh(ζ,λ)f(ζ, λ)dm(ζ).

Writing λ = reiθ

and sending r − 1 to ±0,

we calculate G±(z, θ) and ψ±(z, θ).


Step 2: jump conditions (ii)

Writing λ = reiθ and sending r − 1 to ±0, we obtain

G±(x, θ) =±1

2πi(θ⊥ · x∓ i0sign(θ · x)),

h±(x, θ) = ±1

2i(HRθa)(x · θ⊥) + (Dθa)(x), Hu(t) =

1

π

∫R

u(s)

t− sds.

Here H is the Hilbert transform. We thus obtain that ψ converges on

both sides of T parameterized by θ ∈ (0,2π) to

ψ±(x, θ) = e−Dθae∓12i (HRθa)(x·θ

⊥)∓1

2iH

(e±12i (HRθa)(s)Rθ(e

Dθaf))(x · θ⊥)

+e−DθaDθ(eDθaf)(x).

Notice that (ψ+ − ψ−) is a function of the measurements Ra,θf(s) =

Rθ(eDθaf)(s) whereas ψ± individually are not.


Jump conditions (ii)

Let us define

ϕ(x, θ) = (ψ+ − ψ−)(x, θ).

It depends on the measured data and is given by

iϕ(x, θ) = [R∗−a,θHaRa,θf ](x) = [R∗−a,θHag(s, θ)](x),

where

R∗a,θg(x) = eDθa(x)g(x·θ⊥), Ha = (CcHCc + CsHCs)

Ccg(s, θ) = g(s, θ) cos(HRa(s, θ)

2), Csg(s, θ) = g(s, θ) sin(

HRa(s, θ)

2).

Here R∗a,θ is the adjoint operator to Ra,θ. We note that iϕ(x, θ) is real-

valued and that θ ·∇ϕ+ aϕ = 0.


Step 3: constraints on source terms

The function φ is sectionally analytic, of order O(λ−1) at infinity and such

that

ϕ(z, θ) = φ+(z, θ)− φ−(z, θ) on T.

So φ is the unique solution to the RH problem given by

φ(z, λ) =1

2πi

∫T

ϕ(z, t)

t− λdt =

∞∑n=0

λn1

2πi

∫T

ϕ(z, t)dt

tn+1

on D+ so that

∞∑m=1

(Hmfn−m −Hmfn+m)(z) =1

2πi

∫T

ϕ(z, t)dt

tn+1≡ ϕn(z), n ≥ 0.

Because ∂∂zϕn+aϕn+1+ ∂

∂zϕn+2 = 0, there are actually only two indepen-

dent constraints for n = 0 and n = 1. This characterizes the redundancy

of order 2 of the AtRT.


Step 4: reconstruction in simplest setting.

Assume that N = 0 so that f(x, λ) = f0(x). Then

0 =1

2πi

∫T

ϕ(z, t)dt

t= ϕ0(z)

H1f0(z) =1

2πi

∫T

ϕ(z, t)dt

t2= ϕ1(z).

Recall that H1 = (∂

∂z)−1. We thus obtain

0 = iϕ0(x),

f0(x) =1

4π

∫ 2π

0θ⊥·∇(iϕ)(x, θ)dθ.

This is the Novikov formula. The first equality is a compatibility condi-

tions.


Step 4 bis: Application to Doppler tomography.

In Doppler tomography, the source term of interest is of the form

f(x, θ) = F(x) · θ F(x) = (F1(x), F2(x)).

So we define the source term f1(x) = 12(F1(x) − iF2(x)) and fk(x) ≡ 0

for |k| 6= 1. The constraint n = 0 gives

∇× F(x) =∂F2(x)

∂x−∂F1(x)

∂y=

1

2∆(iϕ0)(x).

The constraint n = 1 gives H2f−1(z) = ϕ1(z) so that

1

2

(F1(z) + iF2(z)

)= −

∂

∂z

1

a(z)

∂ϕ1(z)

∂z.

This explicit reconstruction formula is valid on the support of a(x) and

has no equivalent when a ≡ 0.


Outline



2.b. Partial Data and Numerical Simulations





Reconstruction from partial measurements

Since we can reconstruct two functions from the AtRT, can we recon-

struct one from half of the measurements? The answer is yes and we

have an explicit reconstruction scheme under a smallness constraint on

the variations of the absorption parameter.

The setting is as follows. We assume that g(s, θ) is available for all

values of s ∈ R and for θ ∈ M ⊂ [0,2π). The assumption on M is that

Mc = [0,2π)\M ⊂M + π; for instance M = [0, π) and Mc = [π,2π).

We also assume that the source term f(x) is compactly supported in the

unit ball B.

The derivation is based on decomposing the explicit reconstruction for-

mula into skew-symmetric and symmetric components in L(L2(B)).


Reconstruction from partial measurements

Using the full-measurement inversion formula, we can recast the recon-

struction problem as

f(x) = d(x) + F af(x) + F sf(x), d(x) = F df(x),

where F a is formally skew-symmetric and F s is formally symmetric.

Theorem 1. The operators F a and F s are bounded in L(L2(B)) and

F s is compact in the same sense with range in H1/2(B).

Theorem 2. Provided that ρ(F s) < 1, we can reconstruct f(x) uniquely

and explicitly from g(s, θ) for θ ∈M . Since F s is compact we can always

reconstruct the singular part of f(x) that is not in the Range of F s.

Theorem 3. [R. Novikov; H. Rullgard] The AtRT g(s, θ) on R×Θ, where

Θ ⊂ S1 has positive measure, uniquely determines f(x). Moreover,

‖f‖L2(R2) ≤ C‖g(s, θ)‖H1/2(R×M), for some C > 0.


Advantage of explicit reconstruction formulas:Fast numerical algorithms such as the slant stack

algorithm

Joint work with Philippe Moireau

Let us represent f(x) by an image with n×n pixels. The objectives are:

• to compute an accurate approximation of g(s, θ) = Ra,θf(s)

• to compute it fast (with a cost of O(n2 logn))

• to invert the AtRT accurately and fast from full or partial measure-

ments.


Implementation of slant stack algorithm (RT)

1. We zero-pad the n× n image F to obtain the n× 2n image F1,

2. We compute a Discrete Fourier Transform (DFT) on the columns,

3. We compute a fractional DFT on the rows,

4. We compute an inverse DFT (IDFT) on the columns.

Each of these operations can be performed in O(n2 logn) operations.

The discrete transform converges to the exact transform with spectral

accuracy. The algorithm is based on a discretization of the Fourier slice

theorem Rf(σ, θ) = f(σθ⊥).


Classical phantom reconstruction: RT data

Image on left; slant-stack (lineogram) data on right.


Classical phantom reconstruction

Left 2 pictures: reconstructions from partial data. Right: full

reconstruction.


Classical phantom reconstruction (ii)

2D reconstruction and 1D cross-section.


Generalization to AtRT data

Left: source. Middle: absorption map. Right: AtRT data.


Example of AtRT reconstruction (i)

Reconstruction from π/2 data.


Example of reconstruction (ii)

Reconstruction from full data and 1D cross section.


Outline



2.c. Hyperbolic geometry and applications to Geophysical Imaging





Ray transforms and inverse problems

Many inverse problems involve integrations along geodesics.

In medical imaging, the geodesics are often lines: CT-scan (X-ray λ =

0.1nm), SPECT (gamma ray 159KeV ), PET (2 gamma rays 511KeV ):

Euclidean geometry is fine.

Earth imaging is mostly based on reconstruction of quantities involving

integration along geodesics. However the geodesics are almost always

curved: non-Euclidean geometry.


Inverse source problem as a linearization

Let l be a geodesic and c(x) = c0(x) + δc(x) a sound speed. Then the

experimentally measured travel time is given by:

T =∫l

ds

c(x)=

∫l

1

c0

(1−

δc

c

)ds + l.o.t.

so that, when c0(x) is assumed to be known, we have access to the

measurements:

δT = −∫l

δc

c20(x) ds.

The reconstruction of the perturbation δc to first order is an inverse

source problem (inverse ray transform) in non-Euclidean geometry.


Crash course in Dutch geometry


Geophysical imaging in hyperbolic geometry

Assuming that speed increases linearly in the Earth C(z) ≈ z ∼ y, energy

propagates along the geodesics of the following Riemannian metric

ds2 =1

y2(dx2 + dy2).

(This is not a ridiculous assumption.)

Let X be the geodesic vector field and a a known absorption parameter.

The forward problem is:

Xu+ au = f, f is the source term.

The Inverse Problem is: reconstruct the source term f from boundary

measurements of u (emission problem).

Below is a recently obtained explicit inversion formula for this problem.


What’s the relationship with Escher?

There are various equivalent ways to look at hyperbolic geometry.

⇐⇒

z(w) =w − i

1− iw

Poincareplane

2; ds2 =

dx2 + dy2

y2⇐⇒ Poincare disc; ds2 =

4dzdz

(1− |z|2)2.


Euclidean geometry, summary

The inversion formula obtained earlier was based on the following com-

plexification. The unit circle is parameterized as

λ = eiθ, θ ∈ (0,2π).

The parameter λ defined on the unit circle T is extended to the whole

complex plane C. The transport equation becomes(λ∂

∂z+ λ−1 ∂

∂z+ a(z)

)u(z, λ) = f(z), z ∈ C, λ ∈ C.

We have used the classical parameterization

∂

∂z=

1

2

(∂

∂x− i

∂

∂y

),

∂

∂z=

1

2

(∂

∂x+ i

∂

∂y

).


Geometry of the extension

λ = eiθ is extended to the complex plane C. Position z ≈ x is a fixed

parameter.


Novikov formula in Euclidean geometry (II)

The reconstruction formula hinges on three ingredients:

(i) We show that u(z, λ) is analytic in D+∪D− = C\T and that λu(z, λ) is

bounded as λ→∞. This comes from the analysis of the fundamental

solution of the ∂ problem.

(ii) We verify that ϕ(x, θ) = u+(x, θ)− u−(x, θ), the jump of u at λ = eiθ

can be written as a function of the measured data Raf(s, θ).

(iii) We solve the Riemann Hilbert problem using the Cauchy formula and

evaluate the complexified transport equation (Xu + au)(λ) = f at

λ = 0 to obtain a reconstruction formula for f(z) = f(x).


How about Hyperbolic Geometry?

θ · ∇ Vector field θi∂

∂xi+ Γkijθ

iθj∂

∂θk

λ∂

∂z+ λ−1 ∂

∂zComplexification ???


Suitable parameterization of geodesics

The vector field X(eiθ) at z ∈ D is parameterized as:

The point at infinity in corresponds to the direction at infinity in .


For the fun of it

On the hyperbolic disc, the geodesic vector field converging to eiθ at

infinity is

X(eiθ) = (1− |z|2)(1− e−iθz

1− eiθzeiθ∂ +

1− eiθz

1− e−iθze−iθ∂

).

It can be complexified for λ ∈ C as

X(λ) = (1− |z|2)(λ− z

1− λz∂ +

1− λz

λ− z∂

).

It generates an elliptic operator for λ ∈ C\T and thus admits a funda-

mental solution (X(λ)G(z;λ, z0) = δg(z − z0)) of the form

G(z;λ, z0) =−P (z0, λ)

2iπ

1

s(z, λ)− s(z0, λ).

We deduce that the solution of the complexified transport equation

X(λ)u(z, λ) + a(z)u(z, λ) = f(z)


is given by

u(z, λ) =∫DG(z;λ, ζ)eh(ζ,λ)−h(z,λ)f(ζ)dmg(ζ).

We have that G, hence u, is sectionally analytic. After an additional

conformal mapping, it is given by the solution of a Riemann Hilbert

problem via the following Cauchy formula

u(z, λ) = u(z, µ) =1

2πi

∫T

ϕ(z, ν)

ν − µdν,

where ϕ depends explicitly only on the measured data.

Once u(z, λ) is reconstructed, we apply the transport operator to it to

obtain the source term f(z).


Reconstruction formula

Let Raf(s, θ) be the attenuated hyperbolic ray transform. Define

Raf(s, θ) =∫ξ(s,θ)

eDθa(z, eiθ)f(z)dmg(z)

X⊥(eiθ) = i(1− |z|2)(−

1− e−iθz

1− eiθzeiθ∂ +

1− eiθz

1− e−iθze−iθ∂

)Hf(t) =

1

πp.v.

∫R

f(s)

t− sds

(R∗a,θg)(z) = P (e−iθz)eDθa(z)g(s(e−iθz)), Ha = CcHCc + CsHCs

Ccg(s, θ) = g(s, θ) cos(Ha(s)

2

), Csg(s, θ) = g(s, θ) sin

(Ha(s)

2

).

Then the source term is given by

f(z) =1

4π

∫ 2π

0X⊥(eiθ)

(R∗−a,θHa[Ra,θf ]

)(z, eiθ)dθ.


Vectorial ray transform

For a vector field F (z), we can consider the vectorial ray transform

RaF (s, θ) ≡ Ra,θF (s) =∫ξ(s,θ)

eDθa(z, eiθ)〈X(eiθ), F 〉dmg(z).

Define F [ = F1dx+ F2dy such that F [X = 〈X(eiθ), F 〉. When a = 0,

curlF ≡ ∗dF [ =∂F2

∂x−∂F1

∂y=

1

2i∆ϕ0,

where ϕ0 depends explicitly on RaF . When a > 0,

F1(z) + iF2(z) = −2∂(1− |z|2

a(z)∂ϕ1(z)

),

where ϕ1 also depends explicitly on RaF . As in Euclidean geometry, we

can reconstruct the full vector field when a > 0 on its support.


Outline







Applications in Optical Tomography

Imaging of human brains.



Brain with clear ventricle in neonate. (A.H.Hielscher, Columbia biomed.)



Brain with blood-filled ventricle in neonate. (A.H. Hielscher)



Detection of Rheumatoid arthritis. (A.H. Hielscher)



Optical imaging of Rheumatoid arthritis. (A.H. Hielscher)


Applications in Optical Molecular Imaging

Optical fluorescence imaging in small animals.


Applications in Waves in random media

Numerical and experimental validations of radiative transfer.


Modeling Tropospheric Ozone layers

Inverse problem: can we reconstruct the ozone concentration profile from

radiation measurements? The answer is yes (there is a theorem!) but

the problem is severely ill-posed, which means that very different profiles

generate very similar measurements.

This implies the following modeling issue: thin layers of ozone in the

troposphere (lower almosphere) can hardly be detected without specific

modeling.


Reconstruction with no specific modeling

Reconstructions from Kevin Bowman et al. (JPL).


Outline







Transport equations in optical tomography

In optical tomography the forward problem is

θ · ∇u(x, θ) + a(x)u(x, θ) =∫S2k(x, θ′, θ)u(x, θ′)dθ′, (x, θ) ∈ Ω× S2

u(x, θ) = g(x, θ), (x, θ) ∈ Γ−(Ω),

where the domains Γ±(Ω) are defined by

Γ±(Ω) = (x, θ) ∈ ∂Ω× S2, such that ± θ · n(x) > 0,

where n(x) is the outward normal to Ω at x ∈ ∂Ω. The Albedo operator

maps the incoming conditions to the outgoing radiation:

A : g 7→ u|Γ+(Ω).

The inverse problem consists of reconstructing a(x) and k(x, θ′, θ) from

knowledge of A.


Transport equations in optical molecular imaging

In optical molecular imaging the forward problem is

θ · ∇u(x, θ) + a(x)u(x, θ)=∫Sd−1k(x, θ′, θ)u(x, θ′)dθ′ + f(x), (x, θ) ∈ Ω×Sd−1

u(x, θ) = 0, (x, θ) ∈ Γ−(Ω),

where a(x) and k(x, θ′, θ) are assumed to be known.

The inverse problem consists of reconstructing f(x) from knowledge of

u|Γ+(Ω).

In both the optical tomography and the optical molecular imaging prob-

lems,∫Sd−1 k(x, θ′, θ)dθ′ ≤ a(x) for the forward problem to be well-posed.


Singular decomposition of the Albedo operator

In the optical tomography framework, define u1(x, θ;x0, θ0) as

θ · ∇u1 + a(x)u1 = 0 (x, θ) ∈ Ω× Sd−1

u1(x, θ;x0, θ0) = δ(x− x0)δ(θ − θ0), (x, θ) ∈ Γ−(Ω),

and u2(x, θ;x0, θ0) as

θ · ∇u2 + a(x)u2 =∫Sd−1

k(x, θ′, θ)u1(x, θ′)dθ′, (x, θ) ∈ Ω× Sd−1

u2(x, θ;x0, θ0) = 0, (x, θ) ∈ Γ−(Ω),

Let u(x, θ;x0, θ0) = u1 + u2 + v be the solution of the full transport

equation (replace u2 and u1 in the above equation by u and the boundary

conditions by δ(x− x0)δ(θ − θ0)).

Knowledge of A is equivalent to that of u(x, θ;x0, θ0). In any dimension,

u1 is more singular than u2 + v. In dimension d ≥ 3, u2 is more singular

that v.


Inverse Transport Problem

Theorem[Choulli-Stefanov].

In any space dimension, knowledge of A implies that of u1(x, θ;x0, θ0) on

Γ+ × Γ−, which uniquely determines a(x) by inverse Radon transform.

In dimension d ≥ 3, knowledge of A implies that of u2(x, θ;x0, θ0) on

Γ+ × Γ−, which uniquely determines k(x, θ, θ′). More precisely, we have

the formula

u2(y + sθ, θ;y − tθ0, θ0)

=exp

(−

∫ s

0a(y + pθ)dp−

∫ t

0a(y − pθ0)dp

)√

1− (θ · θ0)2

k(y, θ0, θ).

Here s and t are chosen so that u2 is evaluated at the domain boundary.

The inverse transport problem for d ≥ 3 is thus solved.


Two-dimensional inverse transport problem

In dimension d = 2, we can reconstruct a(x) from the singularities of

A but not k(x, θ, θ′). We may however uniquely reconstruct k provided

that k(x, θ, θ′) is sufficiently small [Stefanov-Uhlmann]. The method is

perturbative. We now propose an iterative method based on the same

perturbative ideas.

Introduce some notation:

Tu = θ · ∇u+ a(x)u, Ku =∫Sd−1

k(x, θ′, θ)u(x, θ′)dθ′,

L : g|Γ− 7→ u|Ω×V solution of Tu = 0, u = g on Γ−.

Then the transport solution of Tu = Ku, u|Γ− = g is given by

u = T−1Ku+ Lg = (I − T−1K)−1Lg = Lg+ T−1KLg+ (T−1K)2u.

The second term is linear in k while the last term is quadratic in k.


Two-dimensional inverse transport problem (ii)

The transport solution of Tu = Ku, u|Γ− = g is given by

u = T−1Ku+ Lg = (I − T−1K)−1Lg = Lg+ T−1KLg+ (T−1K)2u.

Let g0 = δ(x−x0)δ(θ−θ0) on Γ−. Then knowledge of A is equivalent to

that of its kernel d(x, θ;x0, θ0) and we have

d(x, θ;x0, θ0) = FT−1KLg0(x, θ;x0, θ0)

+F(Lg0 + (T−1K)2(I − T−1K)−1Lg0

)(x, θ;x0, θ0),

where F is restriction on Γ+.

In d ≥ 3, the first term (u2) on the r.h.s. is more singular than the second

term (v) of the r.h.s. for θ 6= θ0. In d = 2, the latter is smaller than the

former when k is sufficiently small.


Two-dimensional inverse transport problem (iii)

Recall that for θ 6= θ0,

d(x, θ;x0, θ0) = FT−1KLg0(x, θ;x0, θ0)+F(T−1K)2(I − T−1K)−1Lg0(x, θ;x0, θ0).

Define k = Bd the solution of d(x, θ;x0, θ0) = FT−1KLg0(x, θ;x0, θ0).

(This is how k is reconstructed from u2 in dimension d ≥ 3.) Then we

can recast the above expression as:

k = Bd− G(k), G = BF( ∞∑n=2

(T−1K)n)Lg0.

The leading term in G is quadratic in K so that morally |G(k1)−G(k2)| .ρ‖k1 − k2‖, where ρ is an a priori bound on k. This shows uniqueness of

the reconstruction when ρ is small; see [Stefanov-Uhlmann].


An iterative reconstruction algorithm

Recall that

k = Bd− G(k), G = BF( ∞∑n=2

(T−1K)n)Lg0.

Let us assume that 0 ≤ k(x, θ, θ′) ≤ 1 (to simplify) and define

Pk = (0 ∨ k) ∧ 1, so that k = P(Bd− G(k)) ≡ H(k).

Define now the iterative algorithm

kn = H(kn−1), k0 = 0.

We can show that H is continuous on L∞(Ω × S1 × S1). Since kn is

bounded in that space and thus converges weakly (*) to k∞, we find

that H(kn−1) converges (weakly *) to H(k∞) so that

k∞ = H(k∞) is a solution.


More general Riemannian geometries

Both results (singular expansion in dimension d ≥ 3 and perturbative

argument in dimension d = 2) have been extended to the case of free

transport along the geodesics of a Riemannian manifold [S. McDowall].

Let (M, g) be a Riemannian manifold with boundary ∂M and let u be the

solution of the transport equation

Xu+ a(x)u =∫ΩxM

k(x, θ, θ′)u(x, θ′)dθ′.

Then knowledge of the corresponding albedo operator uniquely deter-

mines a(x) and k(x, θ, θ′) when the metric is known. In two space dimen-

sions, simple metrics are also uniquely determined by the albedo operator

(which determines the scattering relation).

This has interesting applications in geophysical imaging, as well in optical

tomography when variations in the index of refraction are not neglected.


Inverse source problem in OMI (i)

Recall that the inverse problem in optical molecular imaging consists of

reconstructing the source term f(x) from uΓ+, where

θ · ∇u(x, θ) + a(x)u(x, θ)=∫Sd−1k(x, θ′, θ)u(x, θ′)dθ′ + f(x), (x, θ) ∈ Ω×Sd−1

u(x, θ) = 0, (x, θ) ∈ Γ−(Ω).

No singularity analysis can be used for general f(x). In the absence of

scattering, the problem becomes

θ · ∇u(x, θ) + a(x)u(x, θ) = f(x),

which considered two-dimensional slice by two-dimensional slice, is the

attenuated Radon transform, for which we have an inversion formula.


Inverse source problem in OMI (ii)

Since we can invert the source problem in the absence of scattering, we

also should be able to do it in the presence of little scattering.

Recall the notation of the attenuated Radon transform. We define the

symmetrized beam transform:

Dθa(x) =1

2

∫ ∞

0[a(x− tθ)− a(x + tθ)]dt,

such that θ · ∇Dθa = a and the attenuated Radon transform as

Raf(s, θ) =∫R(eDθaf)(sθ⊥ + tθ, θ)dt

We recall the existence of an operator N (Novikov formula) such that

f(x) = N [Raf(s, θ)](x).


Inverse source problem in OMI (iii)

In the presence of scattering, the attenuated Radon transform is replaced

by the measurements

g(s, θ) = Raf(s, θ) +ReDaKe−DaTf(s, θ).

Above, R is the usual Radon transform. Applying the Novikov inversion

operator, we thus obtain

Ng(x) = f(x) +NReDaKe−DaTf(x) = (I −NK)f(x).

Provided NK is sufficiently small, the following algorithm converges

f(n) = Ng(x) +NKf(n−1), f0 = 0.


Remark on the Inverse source problem in OMI

The above algorithm can be improved upon by remarking that the Novikov

formula can be used in the case of isotropic scattering. Recall the trans-

port equation in that case

θ · ∇u+ a(x)u = σ(x)∫Sn−1

u(x, θ′)dθ′ + f(x) ≡ F (x).

The Novikov formula allows to construct

F (x) = Ng(x).

We then solve the transport equation for u(x, θ) and obtain

f(x) = F (x)− σ(x)∫Sn−1

u(x, θ′)dθ′.

The perturbative algorithm shown earlier can be adapted [Bal-Tamasan]

to solve the inverse transport source problem when anisotropic scattering

is sufficiently small.


Outline







Typical path of a detected photonin a DIFFUSIVE REGION

Source Term

Detector


Same typical path in the presence ofa CLEAR LAYER

Source Term

Detector


Transport Equation and Scaling

The phase-space linear transport equation is given by

1

εv · ∇uε(x, v) +

1

ε2Q(uε)(x, v) + σa(x)uε(x, v) = 0 in Ω× V,

uε(x, v) = g(x, v) on Γ− = (x, v) ∈ ∂Ω× V s.t. v · ν(x) < 0.

uε(x, v) is the particle density at x ∈ Ω ⊂ R3 with direction v ∈ V = S2.

The scattering operator Q is defined by

Q(u)(x, v) = σs(x)(u(x, v)−

∫Vu(x, v′)dµ(v′)

).

The mean free path ε measures the mean distance between successive

interactions of the particles with the background medium.

The diffusion limit occurs when ε→ 0.


Volume Diffusion Equation

Asymptotic Expansion: uε(x, v) = u0(x) + εu1(x, v) + ε2u2(x, v) . . .

Equating like powers of ε in the transport equation yields

Order ε−2 : Q(u0) = 0Order ε−1 : v · ∇u0 +Q(u1) = 0Order ε0 : v · ∇u1 +Q(u2) + σau0 = 0.

Krein-Rutman theory:

Order ε−2 : u0(x, v) = u0(x)

Order ε−1 : u1(x, v) = −1

σs(x)v · ∇u0(x),

Order ε0 : −divD(x) · ∇u0(x) + σa(x)u0(x) = 0 in Ω

where the diffusion coefficient is given by D(x) =1

3σs(x)


Diffusion Equations with BoundaryConditions

The volume asymptotic expansion does not hold in the vicinity of bound-

aries. After boundary layer analysis we obtain

−divD(x) · ∇u0(x) + σa(x)u0(x) = 0 in Ω

u0(x) = Λ(g(x, v)) on ∂Ω.

Λ is a linear form on L∞(V−).

We obtain (in a reasonable sense) that

uε(x, v) = u0(x) +O(ε), d(x, ∂Ω) ≥ ρ > 0

uε(x, v) = H(v · n(x)

)u0(x) +O(ε) x ∈ ∂Ω,

where H(µ) is a (known Chandrasekhar) function.


Diffusion-type measurements

In many practical situations, angularly dependent measurements at the

domain boundary may not be available. In OMI, angular measurements

are necessary as the dimension of the source term is d whereas that of

the measurements is 2(d− 1).

However in optical tomography, the dimension of measurements is 4(d−1), whereas a(x) is typically d-dimensional and k(x, θ, θ′) typically d+2(d−1)-dimensional. Thus there is room for less accurate measurements to

still uniquely define the coefficients a and k.



In practice, the outgoing distribution u|Γ+(x, θ) cannot be measured.

Only the current ∫Sd−1

θ · n(x)u(x, θ)dθ ≈ D(x)∂U

∂n(x)

is measured.

In the diffusive regime, both expressions above agree (for x away from

the boundary) up to a error term O(ε), i.e., proportional to the mean

free path. Recall that for isotropic scattering (k = k(x)) we have

uε(x, θ) = U(x)− εdD(x)θ · ∇U(x) +O(ε2).



For incoming boundary conditions of the form g0(x, θ) = δ(x−x0)δ(θ−θ0),

we have access to measurements of the form J(x;x0, θ0). The latter

kernel is still singular in the x variable. This singularity is sufficient to

reconstruct a(x) by inverse Radon transform. However the scattering ker-

nel k(x) (even assumed independent of angular variables) can no longer

be obtained by the analysis of straightforward singularities of the kernel

J(x;x0, θ0).

We may even further restrict measurements by assuming that the il-

lumination is either isotropic g(x, θ) = g(x) or that it is unidirectional

g(x, θ) = g(x)δ(θ−θ0(x)), where θ0 is normal (θ0 = −n(x)). Both situa-

tions are realistic (the latter more so than the former). There, even a(x)

may no longer be reconstructed from singularities of the measurement

kernel J(x;x0). The latter measurements have the same dimension as in

diffusion theory.


Inverse Transport versus inverse Diffusion

Consider the diffusion equation

iωU −∇ ·D(x)∇U + σa(x)U = 0, Ω

with all possible Cauchy data on ∂Ω known. When ω = 0, theory says

that either D(x) or σa(x) can be reconstructed, and that when ω 6= 0,

both can be reconstructed [e.g. Sylvester-Uhlmann].

In the diffusion approximation, the transport and diffusion coefficients

are related by (in the isotropic scattering case)

D(x) =1

dk(x), σa(x) = a(x)− k(x).

So we expect that diffusion-type measurements allow us to reconstruct

both a(x) and k(x) when ω 6= 0 and one of them when ω = 0. No such

theory exists.


Inverse transport, a summary of results

Knowledge of the full albedo operator A allows us to uniquely reconstruct

a(x) and k(x, θ, θ′) in dimension d ≥ 3. In dimension d = 2, k is uniquely

defined provided that it is sufficiently small. In OMI, the source term is

also uniquely determined by boundary measurements provided that the

the anisotropic part of the scattering coefficient k(x, θ, θ′) is sufficiently

small.

All the results are based on the singular decomposition of the albedo

operator (OT), the Novikov formula (OMI), and perturbative arguments.

Can the theories developed for the diffusion equation be extended to the

transport equations? : partial recent results to follow.


Isotropic Sources and current measurements

Joint work with Ian Langmore and Francois Monard

We consider the inverse transport problem with isotropic sources f = f(x)

on Γ− and current measurements

J(x) =∫Sn−1

u(x,Ω)Ω · ν(x)dΩ.


Objective and sketch of results

For each f(x) incoming source, we measure J(x) outgoing current. These

measurements corresponds to Cauchy data for the inverse elliptic prob-

lem. We cannot expect to reconstruct several coefficients from such

measurements. We assume that σ(x) is known (e.g. has been recon-

structed by Inverse Radon transform with appropriate ballistic-only mea-

surements).

We aim at reconstructing k(x). We obtain a uniqueness result (and a

severe ill-posedness stability estimate) provided that k(x) and σ(x) are

sufficiently small.

The proof is based on using complex exponential functions as in Calderon’s

original problem on inverse elliptic equations.


Derivation

We first write the transport solution as

u(x,Ω) = ballistic part + single scattering + ...

We then introduce the measurement operator

M(f, g) =∫∂X

g(x)J(x)dS(x),

where J(x) is current associated to isotropic source f(x).

Objective: find fξ and gξ such that

M(fξ, gξ) = k(ξ) + small.

This is Calderon’s strategy for the linearized inverse elliptic problem.


Measurement operator

We write

M(f, g) = 〈T0, f ⊗ g〉+ 〈T1k, f ⊗ g〉+∞∑j=2

〈Tj(k), f ⊗ g〉,

where Tj(k) corresponds to measurements with δ−functions f and g for

particles that have scattered exactly j times. Since the ballistic part is

known, then so is 〈T0, f ⊗ g〉. Miraculously,

〈T1k, f ⊗ g〉 = 〈k,Af Ag〉, Af(y) = ωn

∫∂X

f(x)E(x,y)∂N(x,y)

∂nxdS(x),

where

∆yN(x,y) = δ(x− y), Rn, E(x,y) = exp(−

∫ y

xσ(l)dl

).

So Af is almost harmonic (and is harmonic when σ = 0; E ≡ 1).


Constructing adapted measurements

Lemma. Assume that σ = 0 here to simplify. Then there exists a

continuous pseudoinverse A† : H12(X) → L2(∂X) such that

AA†u = u|X , for all harmonic function u.

We use double layer potentials to construct A† explicitly.

It remains to reap the rewards by introducing

fξ = A†eiρ·x, gξ = A†eiρ·x, ρ = ξ + iξ⊥,

so that

〈T1k, fξ ⊗ gξ〉 = k(ξ).


(In)Stability

Recall ρ = ξ + iξ⊥ and f(ξ) ≈ eiρ·x and gξ ≈ eiρ·x

〈T1k, fξ ⊗ gξ〉 = k(ξ).

We can introduce T−11 :

T−11 h(x) =

∫Rneiξ·xT−1

1 h(ξ)dξ =∫Rneiξ·x〈h, fξ ⊗ gξ〉dξ.

Note that f(ξ) and gξ grow like e|ξ|diam|X| at the domain boundary. We

thus have the estimate

|T−11 h(ξ)| . e|ξ|diam|X||h(ξ)|,

which grows exponentially with |ξ|. This shows that reconstruction of

k(x) from angularly averaged measurements is severely ill-posed.


Linear and Nonlinear Inversions

Theorem 1. Let kM = PMk be the orthogonal projection of k onto

wavenumbers |ξ| < M . For k < σ and σ sufficiently small, M(f, g) deter-

mine an approximation k∗ of PMk up to an error

‖PMk − k∗‖∞ . eMdiamX‖k‖2∞.

Proof based on introducing a regularized version TM of T−11 that cuts

frequencies higher than M and to show that TM∑j Tj(k) is bounded.

Theorem 2. For σ and k are sufficiently small, the measurementsM(f, g)

determine an approximation k∗ of PMk up to an error

‖PMk − k∗‖∞ . eMdiamX‖k‖∞‖k − PMk‖∞.

This (iterative) nonlinear inversion shows that we obtain a good recon-

struction when k is sufficiently smooth so that k − PMk is small. Similar

results are obtained for smoother high frequency truncations.


Some References[1] Novikov, R. G., An inversion formula for the attenuated X-ray transformation, Ark.Math., 2002

[2] Bal, G. and Moireau, P., Fast numerical inversion of the attenuated Radon transformwith full and partial measurements, Inverse Problems, 2004

[3] Bal, G., Ray transforms in hyperbolic geometry, J. Math. Pures Appl., 2005

[4] Bal, G. and Ren, K., Atmospheric concentration profile reconstructions from radia-tion measurements, Inverse Problems, 21, pp.151-163, 2005

[5] M.Choulli and P.Stefanov, Reconstruction of the coefficients of the stationary trans-port equation from boundary measurements, Inverse Problems, 12, pp. L19-L23, 1996

[6] P.Stefanov and G.Uhlmann, Optical Tomography in Two Dimensions, MethodsAppl. Anal., 10 pp.1-9, 2003

[7] S.McDowall, An inverse problem for the transport equation in the presence of aRiemannian metric. Pacific J. Math. 216(2), pp.303-326, 2004

[8] G.Bal and A.Tamasan, Inverse source problems in transport equations, SIAM Math-ematical Analysis, 39, pp.57-73, 2007

[9] J.Sylvester and G.Uhlmann, A global Uniqueness theorem for an inverse boundary

value problem, Ann. of Math., 125(1), pp.153-169, 1987

inverse transport problems and applications guillaume balmueller/graduate...inverse transport theory...

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