LocalInversionsinUltrasound ModulatedOp5calTomography GuillaumeBal - PowerPoint PPT Presentation

Local
Inversions
in
Ultrasound
 Modulated
Op5cal
Tomography
 Guillaume
Bal
 Shari
Moskow


Ultrasound
Modulated
Op5cal
 Tomography
(Acousto‐Op5cs)
 • Acous5c
waves
are
emiDed
which
perturb
the
 op5cal
proper5es
of
the
medium
 • Light
propaga5ng
through
the
medium
is
used
 to

recover
the
original
op5cal
parameters
 • Model
by
G.
Bal
and
J.
C.
Schotland
Phys.
Rev.
 LeDers,
2010.



Op5cal
proper5es
perturbed
by
acous5c
waves
 σ ǫ = σ + ǫ (2 β + 1) cos( k · x + φ ) γ ǫ = γ + ǫ (2 β − 1) cos( k · x + φ ) Lineariza5on
wrt
epsilon
and
some
manipula5on
yields
 boundary
data
 � (2 β − 1) γ ( ∇ φ ) 2 + (2 β + 1) σφ 2 � � Σ ( k, φ ) = cos( k · x + φ ) Ω Which
is
the
Fourier
transform
of
some
internal
data


Mathema5cal
Problem
 Given
internal
data
of
the
form
 H ij ( x ) = γ ∇ u i · ∇ u j + ησ u i u j , where







is
a
known
fixed
constant
and








 η −∇ · γ ∇ u j + σ u j 0 in Ω = f j on ∂ Ω u j = γ Find










and





 σ

Previous
work
 • Recovery
of








only,

 σ = 0 γ Capdeboscq,
Fehrenback,
De
Gournay,
Kavian
(n=2)
 Bal,
Bonne5er,
Monard,
Triki
(n=3)
 Bal,
Monard
(n>=4)
 Kuchment,
Kunyansky
 Kuchment,
Steinhauer‐

pseudo‐differen5al
calculus
 Ammari,
Capdeboscq,
Triki
2012‐
separa5on
of
terms


Assume
we
have
some
known
 background






and







.
 γ 0 σ 0 γ γ 0 + δγ = σ σ 0 + δσ = u 0 j + δ u j = u j Where
the
background
solu5ons
sa5sfy
 −∇ · γ 0 ∇ u 0 j + σ 0 u 0 0 in Ω = j u 0 f j on ∂ Ω = j

L 0 := −∇ · γ 0 ∇ + σ 0 δ u j = L − 1 0 ( ∇ · δγ ∇ u 0 j − δσ u 0 j )

linearized
problem
 δγ ∇ u 0 i · ∇ u 0 j + γ 0 ∇ δ u i · ∇ u 0 j + γ 0 ∇ u 0 i · ∇ δ u 0 dH ij = j + ηδσ u 0 i u 0 j + ησ 0 δ u i u 0 j + ησ 0 u 0 i δ u j δ u j , δγ , δσ 
Really
have
3
unknowns
here









 L 0 δ u j = ∇ · δγ ∇ u 0 j − δσ u 0 j 
But
they
are
coupled

 δ u j One
approach:
solve
for










and
subs5tute
back
in


Take
Laplacian
of
data

 ∆ dH ij ( δγ , δσ ) = G ij ( δγ , δσ ) + ∆ T ij ( δγ , δσ ) , where T ij is compact, and j ) s : D 2 δγ + η u 0 G ij ( δγ , δσ ) = ∇ u 0 i · ∇ u 0 j ∆ δγ − 2( ∇ u 0 i ⊗ ∇ u 0 i u 0 j ∆ δσ

Simplest
case

 • Case
where
n=2
and


 γ 0 = 1 , σ 0 = 0 • Take



















to
get

 u 0 = 1 dH 00 ( δγ , δσ ) = ηδσ • Eliminate









 δσ • Take

 u 0 i = x i

then
 ∆ ˜ dH 11 = ( ∂ 2 x 2 − ∂ 2 x 1 ) ∆ ˜ dH 12 = − 2 ∂ x 1 x 2 ∆ ˜ dH 22 = ( ∂ 2 x 1 − ∂ 2 x 2 ) Separately
get
hyperbolic,
not
ellip5c
 Together
ellip5c
as
a
redundant
system
 Hard
to
invert
because
redundant



but
consider
 Γ T Γ = ( ∆ d ˜ � H ij ) 2 ij Γ T Γ = 2 ∂ 4 x 1 + 2 ∂ 4 x 2 Which
is
ellip5c.



And
for


 n ≥ 3 n ∆ d ˜ � H ii = ( n − 2) ∆ i =1 Which
we
can
invert


For
general




this
doesn’t
work
 σ 0 So
let
us
consider
for
general


 γ 0 , σ 0 The
highest
order
part
 j ) s : D 2 δγ + η u 0 G ij ( δγ , δσ ) = ∇ u 0 i · ∇ u 0 j ∆ δγ − 2( ∇ u 0 i ⊗ ∇ u 0 i u 0 j ∆ δσ Define
 ∇ u 0 θ i = i | ∇ u 0 i |

Then
we
are
interested
in
the
system
 A ij δγ + B ij δσ = F ij , where
 A ij = θ i · θ j ∆ − 2( θ i ⊗ θ j ) s : ∇ ⊗ ∇ , and
 u i B ij = η d i d j ∆ , d i = | ∇ u i | .

Define
the
operator

 � A ij � B ij Γ = With
a
row
for
each
pair
(i,j)


 We
want
to
show
this
operator
is
ellip5c
so
that
we
can
get
a
parametrix,
or

 Inver5bility
of
the
highest
order
part.
 Construc5on
of
parametrices
for
similar
problems
in
 Kuchment
and
Steinhauer
for
one
coefficient.


Consider
the
2x2
system
 � δγ � Γ T Γ = Γ T F. δσ ij A T ij A T �� � � ij A ij ij B ij Γ T Γ = . ij B T ij B T � � ij A ij ij B ij

Using
symbols,
the
system
is
inver5ble
 when
we
always
have
at
least
one
of

 the
sub‐determinants
not
vanishing

 � A ij � B ij Det A kl B kl

These
determinants
are
zero
when

 ( θ i · θ j − 2 θ i · ˆ ξθ j · ˆ ξ ) d p d q = ( θ p · θ q − 2 θ p · ˆ ξθ q · ˆ ξ ) d i d j ∀ ( i, j, p, q ) . ∇ u 0 u 0 θ i = i d i = i | ∇ u 0 i | | ∇ u 0 i |

Thanks
to
Gunther
Uhlmann
and
CGOs
 e ρ · x = u ρ e ρ r · x (cos ρ I · x + i sin ρ I · x ) = e ρ r · x [( ρ r cos ρ I · x − ρ I sin ρ I · x ) + i ( ρ r sin ρ I · x + ρ I cos ρ I · x )] = ∇ u ρ Which
gives,
by
taking
real
and
imaginary
parts
 � � � � cos ρ I · x sin ρ I · x θ 1 = θ 2 = − sin ρ I · x cos ρ I · x d 1 = cos ρ I · x d 2 = sin ρ I · x | ρ | | ρ |

(1 − 2( θ 1 · ξ ) 2 ) d 2 (1 − 2( θ 2 · ξ ) 2 ) d 2 = 2 1 − 2 θ 1 · ξθ 2 · ξ d 2 (1 − 2( θ 1 · ξ ) 2 ) d 1 d 2 = 1 − 2 θ 1 · ξθ 2 · ξ d 2 (1 − 2( θ 2 · ξ ) 2 ) d 1 d 2 = 2 Which
is
 ( s 2 − c 2 ) d 2 ( c 2 − s 2 ) d 2 = 2 1 − 2 csd 2 ( s 2 − c 2 ) d 1 d 2 = 1 − 2 csd 2 ( c 2 − s 2 ) d 1 d 2 = 2 (1) ⇒ s 2 − c 2 = 0 (2) or (3) ⇒ sc = 0

• But
ellip5city
doesn’t
guarantee
injec5vity
 • Need
injec5vity
for
extensions
to
nonlinear
 problem


One
approach:
view
as
a
differen5al
 operator
with
its
natural
square
 bilinear
form
 �� v � v � v � v � �� � � := � B Ω Γ · Γ , w w w w 
on

 H 2 0 ( Ω ) × H 2 0 ( Ω )

Varia5onal
fomula5on:

find

 ( δγ , δσ ) ∈ H 2 0 ( Ω ) × H 2 0 ( Ω ) Such
that

 �� δγ �� δγ � v � v � v � �� � �� � � Ω F · Γ T B , + L , = δσ δσ w w w ∀ ( v, w ) ∈ H 2 0 ( Ω ) × H 2 0 ( Ω ) Where
L
is
a
lower
order
operator
 (generally
nonlocal)



• B
is
clearly
bounded
above
on
 H 2 0 ( Ω ) × H 2 0 ( Ω ) • Know
ellip5c,
can
get
coercivity
bounds
 explicitly
in
some
cases


Case
n=2,
constant




 σ 0 , γ 0 • Have
the
two
background
solu5ons
 q σ 0 γ 0 x 1 u 0 1 = e q σ 0 γ 0 x 2 , u 0 2 = e • Which
give
 � γ 0 θ i = e i and d i = σ 0 .

Γ =( A ij B ij ) Corresponding
to
(i,j)=(1,1),(1,2),(2,2)
where

 A 11 = ∂ yy − ∂ xx A 12 = − 2 ∂ xy A 22 = ∂ xx − ∂ yy B := B 11 = B 12 = B 22 = η γ 0 ∆ . σ 0

�� v � v 2( v xx ) 2 + 2( v yy ) 2 + 3 η 2 γ 2 � �� � ( ∆ w ) 2 − 2 η γ 0 0 = v xy ∆ w B , σ 2 w w σ 0 Ω 0 Use
Cauchy’s
inequality
 xy + ( ∆ w ) 2 | v xy ∆ w | ≤ ǫ v 2 4 ǫ 

and
integra5on
by
parts
 � Ω v 2 � Ω v xx v yy xy =

�� v � v � �� B , w w � 2 − | η | γ 0 � � � � � σ 0 ǫ � 3 η 2 γ 2 � v xx v xx � ( ∆ w ) 2 · + 0 − | η | γ 0 ≥ 0 σ 2 − | η | γ 0 2 Ω 2 ǫσ 0 σ 0 ǫ v yy v yy σ 0 choose
 To
get
 ǫ = γ 0 | η | . γ 2 L 2 + 3 ≥ � v xx � 2 L 2 + � v yy � 2 0 η 2 � ∆ w � 2 0 L 2 . σ 2 2

• If
we
have
injec5vity,
this
means
that
the
 linearized
solu5ons
 � ˆ 0 ( Ω ) , � ˆ δγ � H 2 δσ � H 2 0 ( Ω ) ≤ C � F � L 2 ( Ω ) • 
and
we
have
explicit
knowledge
of
C


• System
is
ellip5c‐
but
don’t
yet
know
if
 injec5ve.

 • But
since
problem
is
square:
 � 0 � v � v � v � � � � � Ω Γ · Γ = 0 ⇒ Γ = 0 w w w

Case
where
domain
is
small
 • 
If
the
domain
is
small,












and








are
 u 0 ∇ u 0 i i close
to
constants
 δγ ∇ u 0 i · ∇ u 0 j + γ 0 ∇ δ u i · ∇ u 0 j + γ 0 ∇ u 0 dH ij ( δγ , δσ ) = i · ∇ δ u j + ηδσ u 0 i u 0 j + ησ 0 δ u i u 0 j + ησ 0 u 0 i δ u j So
when
we
take
L_0
of
data,
lower
order
terms
 are
differen5al
operators.


 L 0 = −∇ · γ 0 ∇ + σ

• 
if







is
a
differen5al
operator
and
 Γ � v � � 0 � Γ = . 0 w v = ∂ v ∂ν = w = ∂ w • 

since




















































 ∂ν = 0 on ∂ Ω We
can
get
that

























from
Holmgren’s
 v = w = 0 theorem