Variational methods for inverse scattering
problems
Ana Carpio
1
María–Luisa Rapún
2
1
Matemática Aplicada, Universidad Complutense de Madrid, Spain
2
Fundamentos Matemáticos, Universidad Politécnica de Madrid, Spain
Outline
1
Inverse scattering problem
2
Topological derivative methods
TD for shape reconstruction
Iterative methods
TD for shapes and parameters
Description of the problem
Medium
R
with obstacles
Ω
:
How many? how big? where?
physical properties in
Ω
?
Some applications
Medicine (tumors, fracture)
Geophysics (oil, gas)
Materials (damage, cracks)
Scattering problem
An incident acoustic radiation
u
inc
=
e
ikx·d
interacts with a
medium
R
containing objects
Ω
.
Forward (direct) problem
The shape, size, location and physical properties of the
objects are known
Compute the response of the system at the detectors "×"
A well–posed problem: it has a unique solution that
depends continuously on the data
Scattering problem
An incident acoustic radiation
u
inc
=
e
ikx·d
interacts with a
medium
R
containing objects
Ω
.
Forward (direct) problem
The shape, size, location and physical properties of the
objects are known
Compute the response of the system at the detectors "×"
A well–posed problem: it has a unique solution that
depends continuously on the data
Scattering problem
An incident acoustic radiation
u
inc
=
e
ikx·d
interacts with a
medium
R
containing objects
Ω
.
Inverse problem
Measurements
u
meas
are taken at the receptors
Find the scatterers
Ω
and the interior parameters s.t.
u
=
u
meas
on
Γ
meas
,
u= sol. forward problem
An ill–posed problem: it may not have a solution and if it
has one, it may not depend continuously on the data
A simple forward problem
Ω
is a penetrable known obstacle. The incident field generates
a scattered wave
u
sc
in
R
n
\
Ω
and a transmitted wave
u
tr
in
Ω
.
The total field
u
=
u
inc
+
u
sc
in
R
n
\
Ω
and
u
=
u
tr
in
Ω
solves
∆u
+
k
e
2
u
=
0
in
R
n
\
Ω
∆u
+
k
2
i
u
=
0
in
Ω
u
−
=
u
+
,
∂
nu
−
=
∂
nu
+
on
∂
Ω
lim
r
→∞
r
(
n
−
1
)
/
2
(
∂
r
(u
−
u
inc)
−
ik
e(u
−
u
inc)) =
0
Other boundary conditions can be handled in a similar way
:
∆
u
+
k
e
2
u
=
0
in
R
n
\
Ω
∂
nu
+
=
0
on
∂
Ω
lim
r→∞
r
(n−
1
)
/
2
(
∂r
(
u
−
u
inc
)
−
ik
e
(
u
−
u
inc
)) =
0
Non-constant parameters
∇ ·
(
α
e
(
x)
∇
u) +
k
e
2
(
x
)
u
=
0
in
R
n
\
Ω
∇ ·
(
α
i
(
x)
∇
u) +
k
i
2
(
x
)
u
=
0
in
Ω
u
−
=
u
+
,
α
e
(
x)∂
n
u
−
=
α
i
(
x)∂
n
u
+
on
∂Ω
lim
r
→∞
r
(
n
−
1
)
/
2
(∂
r
(u
−
u
inc
)
−
i
κ
e(u
−
u
inc
)) =
0
where
k
s
(x)
≥
k
s
,
0
>
0
, α
s
(x)
≥
a
s
,
0
>
0
,
s
=
e,
i
and
k
e
(x)/
p
Other boundary conditions can be handled in a similar way
:
∆
u
+
k
e
2
u
=
0
in
R
n
\
Ω
∂
nu
+
=
0
on
∂
Ω
lim
r→∞
r
(n−
1
)
/
2
(
∂r
(
u
−
u
inc
)
−
ik
e
(
u
−
u
inc
)) =
0
Non-constant parameters
∇ ·
(
α
e
(
x
)
∇
u
) +
k
e
2
(
x
)
u
=
0
in
R
n
\
Ω
∇ ·
(
α
i(
x
)
∇
u
) +
k
2
i
(
x
)
u
=
0
in
Ω
u
−
=
u
+
,
α
e(
x
)
∂n
u
−
=
α
i
(
x
)
∂n
u
+
on
∂
Ω
lim
r→∞
r
(n−
1
)
/
2
(
∂r
(
u
−
u
inc
)
−
i
κ
e
(
u
−
u
inc
)) =
0
where
k
s
(
x
)
≥
k
s
,
0
>
0
, α
s
(
x
)
≥
a
s
,
0
>
0
,
s
=
e
,
i
and
k
e
(
x
)
/
p
Outline
1
Inverse scattering problem
2
Topological derivative methods
TD for shape reconstruction
Iterative methods
TD for shapes and parameters
Constrained optimization
Original problem
(we assume first that
k
i
is known)
Find
Ω
such that
u
=
u
meas
on
Γ
meas
A weaker formulation
Find
Ω
minimizing
J
(Ω) =
1
2
Z
Γ
meas|
u
−
u
meas
|
2
for
u
solving the forward problem in
R
n
\
Ω
,
Ω
The domain
Ω
is the variable
Constrained optimization
Original problem
(we assume first that
k
i
is known)
Find
Ω
such that
u
=
u
meas
on
Γ
meas
A weaker formulation
Find
Ω
minimizing
J
(Ω) =
1
2
Z
Γmeas
|
u
−
u
meas
|
2
for
u
solving the forward problem in
R
n
\
Ω
,
Ω
The domain
Ω
is the variable
Constrained optimization
Original problem
(we assume first that
k
i
is known)
Find
Ω
such that
u
=
u
meas
on
Γ
meas
A weaker formulation
Find
Ω
minimizing
J
(Ω) =
1
2
Z
Γmeas
|
u
−
u
meas
|
2
for
u
solving the forward problem in
R
n
\
Ω
,
Ω
The domain
Ω
is the variable
Some alternatives
Modified gradient methods:
differ on how an initial guess is
deformed from one iteration to the next in such a way that the
cost functional decreases
Classical deformations
following a vector field
Problem: The number of scatterers has to be known from
the beginning
Kirsch 1993, Hettlich 1995, Potthast 1996
Level set based deformations
allow changes in topology
Problem: Slow evolution. Initial guess?
Santosa 1996, Dorn 2005
Topological derivatives
Provide good initial guesses
Fast and allow topological changes
Some alternatives
Modified gradient methods:
differ on how an initial guess is
deformed from one iteration to the next in such a way that the
cost functional decreases
Classical deformations
following a vector field
Problem: The number of scatterers has to be known from
the beginning
Kirsch 1993, Hettlich 1995, Potthast 1996
Level set based deformations
allow changes in topology
Problem: Slow evolution. Initial guess?
Santosa 1996, Dorn 2005
Topological derivatives
Provide good initial guesses
Fast and allow topological changes
Some alternatives
Modified gradient methods:
differ on how an initial guess is
deformed from one iteration to the next in such a way that the
cost functional decreases
Classical deformations
following a vector field
Problem: The number of scatterers has to be known from
the beginning
Kirsch 1993, Hettlich 1995, Potthast 1996
Level set based deformations
allow changes in topology
Problem: Slow evolution. Initial guess?
Santosa 1996, Dorn 2005
Topological derivatives
Provide good initial guesses
Fast and allow topological changes
Some alternatives
Modified gradient methods:
differ on how an initial guess is
deformed from one iteration to the next in such a way that the
cost functional decreases
Classical deformations
following a vector field
Problem: The number of scatterers has to be known from
the beginning
Kirsch 1993, Hettlich 1995, Potthast 1996
Level set based deformations
allow changes in topology
Problem: Slow evolution. Initial guess?
Santosa 1996, Dorn 2005
Topological derivatives
Provide good initial guesses
Fast and allow topological changes
Definition of Topological Derivative (Sokowloski–Zochowski ’99)
The TD of a shape functional
J
(
R
)
at a point
x
∈
R
is
D
T
(
x
,
R
) =
lim
ε
→
0
J
(
R
\
B
ε
(
x
)
)
−
J
(
R
)
Vol(
B
ε
(
x
)
)
It is a scalar function of
x
It measures sensitivity to removing balls around a point
D
T
(
x
,
R)
<<
0
=
⇒
high probability of finding an object
Equivalently, for
x
∈ R
and
h
(ε) =
Vol
(
B
ε
(
x
))
Definition of Topological Derivative (Sokowloski–Zochowski ’99)
The TD of a shape functional
J
(
R
)
at a point
x
∈
R
is
D
T
(
x
,
R
) =
lim
ε
→
0
J
(
R
\
B
ε
(
x
)
)
−
J
(
R
)
Vol(
B
ε
(
x
)
)
It is a scalar function of
x
It measures sensitivity to removing balls around a point
D
T
(
x
,
R)
<<
0
=
⇒
high probability of finding an object
Equivalently, for
x
∈ R
and
h
(ε) =
Vol
(
B
ε
(
x
))
Definition of Topological Derivative (Sokowloski–Zochowski ’99)
The TD of a shape functional
J
(
R
)
at a point
x
∈
R
is
D
T
(
x
,
R
) =
lim
ε
→
0
J
(
R
\
B
ε
(
x
)
)
−
J
(
R
)
Vol(
B
ε
(
x
)
)
It is a scalar function of
x
It measures sensitivity to removing balls around a point
D
T
(
x
,
R)
<<
0
=
⇒
high probability of finding an object
Equivalently, for
x
∈ R
and
h
(
ε
) =
Vol
(
B
ε
(
x
))
How to obtain
D
T
(
x
,
R
)
(Feijoo ’04)
1
Given
x
∈ R, take the ball
B
ε
(x)
. Choose the vector field
V(z) =
−
n(z),
z
∈
∂Bε(x)
and extend
V
to
R
n
s.t.
V
=
0
far from
∂B
ε(x)
2Consider the domain
R
τ
:=
{
z
+
τ
V(z)
|
z
∈ R \
B
ε
(x)
}.
Then,
J(
R
τ
)
is a scalar function of
τ
3
Compute the
shape derivative
(Lagrangian formulation)
D
S
:=
d
d
τ
J
(
R
τ
)
τ
=
0
4Use the relation (
asymptotic expansions
)
D
T
(
x
,
R
) =
lim
ε
→
0
−
1
V
0
(ε)
D
S
,
V
(ε) =
Vol(
B
ε(x)
)
How to obtain
D
T
(
x
,
R
)
(Feijoo ’04)
1
Given
x
∈ R
, take the ball
B
ε
(
x
)
. Choose the vector field
V
(
z
) =
−n
(
z
)
,
z
∈
∂
Bε
(
x
)
and extend
V
to
R
n
s.t.
V
=
0
far from
∂
B
ε
(
x
)
2Consider the domain
R
τ
:=
{
z
+
τ
V(z)
|
z
∈ R \
B
ε
(x)
}.
Then,
J(
R
τ
)
is a scalar function of
τ
3
Compute the
shape derivative
(Lagrangian formulation)
D
S
:=
d
d
τ
J
(
R
τ
)
τ
=
0
4Use the relation (
asymptotic expansions
)
D
T
(
x
,
R
) =
lim
ε
→
0
−
1
V
0
(ε)
D
S
,
V
(ε) =
Vol(
B
ε(x)
)
How to obtain
D
T
(
x
,
R
)
(Feijoo ’04)
1
Given
x
∈ R
, take the ball
B
ε
(
x
)
. Choose the vector field
V
(
z
) =
−n
(
z
)
,
z
∈
∂
Bε
(
x
)
and extend
V
to
R
n
s.t.
V
=
0
far from
∂
B
ε
(
x
)
2
Consider the domain
R
τ
:=
{z
+
τ
V
(
z
)
|
z
∈ R \
B
ε
(
x
)
}
.
Then,
J
(
R
τ
)
is a scalar function of
τ
3
Compute the
shape derivative
(Lagrangian formulation)
D
S
:=
d
d
τ
J
(
R
τ
)
τ
=
0
4Use the relation (
asymptotic expansions
)
D
T
(
x
,
R
) =
lim
ε
→
0
−
1
V
0
(ε)
D
S
,
V
(ε) =
Vol(
B
ε(x)
)
How to obtain
D
T
(
x
,
R
)
(Feijoo ’04)
1
Given
x
∈ R
, take the ball
B
ε
(
x
)
. Choose the vector field
V
(
z
) =
−n
(
z
)
,
z
∈
∂
Bε
(
x
)
and extend
V
to
R
n
s.t.
V
=
0
far from
∂
B
ε
(
x
)
2
Consider the domain
R
τ
:=
{z
+
τ
V
(
z
)
|
z
∈ R \
B
ε
(
x
)
}
.
Then,
J
(
R
τ
)
is a scalar function of
τ
3
Compute the
shape derivative
(Lagrangian formulation)
D
S
:=
d
d
τ
J
(R
τ
)
τ
=
0
4
Use the relation (
asymptotic expansions
)
D
T
(
x
,
R
) =
lim
ε
→
0
−
1
V
0
(ε)
D
S
,
V
(ε) =
Vol(
B
ε(x)
)
How to obtain
D
T
(
x
,
R
)
(Feijoo ’04)
1
Given
x
∈ R
, take the ball
B
ε
(
x
)
. Choose the vector field
V
(
z
) =
−n
(
z
)
,
z
∈
∂
Bε
(
x
)
and extend
V
to
R
n
s.t.
V
=
0
far from
∂
B
ε
(
x
)
2
Consider the domain
R
τ
:=
{z
+
τ
V
(
z
)
|
z
∈ R \
B
ε
(
x
)
}
.
Then,
J
(
R
τ
)
is a scalar function of
τ
3
Compute the
shape derivative
(Lagrangian formulation)
D
S
:=
d
d
τ
J
(R
τ
)
τ
=
0
4
Use the relation (
asymptotic expansions
)
D
T
(
x
,
R) =
lim
ε
→
0
−
1
V
0
(ε)
D
S
,
V
(
ε
) =
Vol(
B
ε
(
x
)
)
Transmission problem:
u
−
=
u
+
,
∂
n
u
−
=
∂
n
u
+
Case I: No a priori information on the obstacles,
R
=
R
n
,
Ω =
∅
Theorem.
For any
x
∈
R
n
the topological derivative of
J
(
R
n
) =
1
2
R
Γ
meas|u
−
u
meas
|
2
is
D
T
(
x
,
R
n
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
Transmission problem:
u
−
=
u
+
,
∂
n
u
−
=
∂
n
u
+
Case I: No a priori information on the obstacles,
R
=
R
n
,
Ω =
∅
Theorem.
For any
x
∈
R
n
the topological derivative of
J
(
R
n
) =
1
2
R
Γmeas
|
u
−
u
meas
|
2
is
D
T
(
x,
R
n
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
Forward problem with
Ω =
∅:
(
∆
u
+
k
e
2
u
=
0
in
R
n
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
(
u
−
u
inc
)
−
ik
e
(
u
−
u
inc
)) =
0
Therefore,
u
=
u
inc
(
x
) =
e
ikex·d
Adjoint problem with
Ω =
∅
:
(
∆
w
+
k
e
2
w
= (u
meas
−
u)δ
Γ
measin
R
n
lim
r
→∞
r
(
n
−
1
)
/
2
(∂
r
w
−
ik
e
w
) =
0
Therefore,
w
=
R
Γ
measG
k
e(x
−
y)(u
meas
−
u)(y)dl
y
The true obstacle enters the TD through the measured
data at the adjoint field
Forward problem with
Ω =
∅:
(
∆
u
+
k
e
2
u
=
0
in
R
n
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
(
u
−
u
inc
)
−
ik
e
(
u
−
u
inc
)) =
0
Therefore,
u
=
u
inc
(
x
) =
e
ikex·d
Adjoint problem with
Ω =
∅
:
(
∆
w
+
k
e
2
w
= (
u
meas
−
u
)
δΓmeas
in
R
n
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
w
−
ik
e
w
) =
0
Therefore,
w
=
R
Γmeas
G
ke
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
The true obstacle enters the TD through the measured
data at the adjoint field
Forward problem with
Ω =
∅:
(
∆
u
+
k
e
2
u
=
0
in
R
n
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
(
u
−
u
inc
)
−
ik
e
(
u
−
u
inc
)) =
0
Therefore,
u
=
u
inc
(
x
) =
e
ikex·d
Adjoint problem with
Ω =
∅
:
(
∆
w
+
k
e
2
w
= (
u
meas
−
u
)
δΓmeas
in
R
n
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
w
−
ik
e
w
) =
0
Therefore,
w
=
R
Γmeas
G
ke
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
The true obstacle enters the TD through the measured
data at the adjoint field
Forward problem with
Ω =
∅:
(
∆
u
+
k
e
2
u
=
0
in
R
n
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
(
u
−
u
inc
)
−
ik
e
(
u
−
u
inc
)) =
0
Therefore,
u
=
u
inc
(
x
) =
e
ikex·d
Adjoint problem with
Ω =
∅
:
(
∆
w
+
k
e
2
w
= (
u
meas
−
u
)
δΓmeas
in
R
n
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
w
−
ik
e
w
) =
0
Therefore,
w
=
R
Γmeas
G
ke
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
The true obstacle enters the TD through the measured
data at the adjoint field
Other problems
For Helmholtz transmission problem with
u
−
=
u
+
and
∂
n
u
−
=
∂
n
u
+
D
T
(
x,
R
2
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
where
u
=
u
inc
and
w
=
R
Γmeas
G
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
For rigid obstacles
D
T
(x,
R
2
) =
Re
h
2
∇u(x)∇
w
(x)
−
k
e
2
u(x)
w
(x)
i
where
u
=
u
inc
and
w
=
R
Other problems
For Helmholtz transmission problem with
u
−
=
u
+
and
∂
n
u
−
=
∂
n
u
+
D
T
(
x,
R
2
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
where
u
=
u
inc
and
w
=
R
Γmeas
G
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
For rigid obstacles
D
T
(
x,
R
2
) =
Re
h
2
∇u
(
x
)
∇w
(
x
)
−
k
e
2
u
(
x
)
w
(
x
)
i
where
u
=
u
inc
and
w
=
R
Other problems
For the HTP with non–constant parameters
D
T
(
x,
R
2
)
=
Re
2
αe
(
x
)(
αe
(
x
)
−
αi
(
x
))
αe
(
x
) +
α
i
(
x
)
∇u
(
x
)
∇w
(
x
)
+(
k
i
2
(
x
)
−
k
e
2
(
x
))
u
(
x
)
w
(
x
)
where
u
and
w
solve forward and adjoint problems.
Since
k
e
is non–constant, both have to be computed
numerically
Some examples
"
×
"= observation points, 24 incident directions in
[
0
,
2
π
)
,
Similar results when
Observation points are further
+
observation points
+
incident directions
Results depend on the wave length (1 w.l.=2
π/
k
e
):
1
st
row:
k
e
=
2 and
k
i
=
1
/
2
2
nd
row:
k
TD with an initial guess
Case II:
Ωap
first guess,
R
=
R
n
\
Ωap,
Ω = Ωap
Theorem.
For any
x
∈
R
n
\
Ω
ap
the topological derivative of
J
(R
n
\
Ω
ap
) =
1
2
R
Γ
meas|u
−
u
meas
|
2
is
D
T
(
x
,
R
n
\
Ω
ap
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
TD with an initial guess
Case II:
Ωap
first guess,
R
=
R
n
\
Ωap,
Ω = Ωap
Theorem.
For any
x
∈
R
n
\
Ω
ap
the topological derivative of
J
(
R
n
\
Ω
ap
) =
1
2
R
Γmeas
|
u
−
u
meas
|
2
is
D
T
(
x,
R
n
\
Ω
ap
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
Forward problem with
Ω = Ω
ap
:
∆
u
+
k
e
2
u
=
0
in
R
n
\
Ω
ap
∆
u
+
k
i
2
u
=
0
in
Ω
ap
u
−
=
u
+
,
∂n
u
−
=
∂n
u
+
on
∂Ωap
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
(
u
−
u
inc
)
−
ik
e
(
u
−
u
inc
)) =
0
Adjoint problem with
Ω = Ω
ap
:
∆
w
+
k
e
2
w
= (
u
meas
−
u
)
δ
Γmeas
in
R
n
\
Ω
ap
∆
w
+
k
i
2
w
=
0
in
Ω
ap
w
−
=
w
+
,
∂
n
w
−
=
∂
n
w
+
on
∂Ω
ap
lim
r→∞
r
(n−
1
)
/
2
(
∂
r
w
−
ik
e
w
) =
0
Remarks
1
If
Ω = Ω
ap
,
u
and
w
are computed numerically. For our
model problem we use BEM
2
For Neumann or general transmission problems the
Remarks
1
If
Ω = Ω
ap
,
u
and
w
are computed numerically. For our
model problem we use BEM
2
For Neumann or general transmission problems the
Same examples as before with
Ω =
∅
Outline
1
Inverse scattering problem
2
Topological derivative methods
TD for shape reconstruction
Iterative methods
TD for shapes and parameters
A monotone iterative method
Idea
By removing from the initial domain regions where the TD is
negative and large, we obtain a new domain in which the
magnitude of the topological derivative is smaller
Algorithm
1
Compute the TD when
Ω =
∅
2
Take
Ω
1
=
{
x
,
D
T
(
x
,
R
n
)
<
−C
1
},
C
1
>
0
3For j=1:jmax
Compute the TD in
R
n
\
Ω
j
Select
Ω
j
+
1
⊃
Ω
j
Ω
j
+
1
= Ω
j
∪ {
x,
D
T
(
x,
R
n
\
Ω
j
)
<
−
C
j
+
1
}
How to choose
C
j
?
First step
:
Ω
1
=
{x,
D
T
(
x,
R
2
)
<
−
C
1
}
C
1
=
3
5
|
min
D
T
|
Accept
C
1
if
J
1
<
J
0
Otherwise
C
1
0
<
C
1
Iterations:
Ω
j
+1
= Ω
j
∪ {x
,
D
T
(
x
,
R
2
\
Ω
j
)
<
−C
j
+1
}
C
j
+1
=
9
10
|min
D
T
|
Stopping criteria?
J
≈
0
or
Ω
j
≈
Ω
j
+
1
or
|
u
δ
−
u
meas
|
<
1
.
2
δ
How to choose
C
j
?
First step
:
Ω
1
=
{x,
D
T
(
x,
R
2
)
<
−
C
1
}
C
1
=
3
5
|
min
D
T
|
Accept
C
1
if
J
1
<
J
0
Otherwise
C
1
0
<
C
1
Iterations:
Ω
j+
1
= Ω
j
∪ {x,
D
T
(
x,
R
2
\
Ω
j
)
<
−
C
j+
1
}
C
j+
1
=
9
10
|
min
D
T
|
Stopping criteria?
J
≈
0
or
Ω
j
≈
Ω
j
+
1
or
|
u
δ
−
u
meas
|
<
1
.
2
δ
How to choose
C
j
?
First step
:
Ω
1
=
{x,
D
T
(
x,
R
2
)
<
−
C
1
}
C
1
=
3
5
|
min
D
T
|
Accept
C
1
if
J
1
<
J
0
Otherwise
C
1
0
<
C
1
Iterations:
Ω
j+
1
= Ω
j
∪ {x,
D
T
(
x,
R
2
\
Ω
j
)
<
−
C
j+
1
}
C
j+
1
=
9
10
|
min
D
T
|
Stopping criteria?
J
≈
0
or
Ω
j
≈
Ω
j+
1
or
|
u
δ
−
u
meas
|
<
1
.
2
δ
Implementation in 2D:
TD
=
Re
(k
i
2
−
k
e
2
)
u(
x
)w(
x
)
Computation of
TD
when
Ω =
∅
:
u
=
u
inc
and
w
=
R
Γ
measG
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
Computation of
TD
when
Ω = Ω
j
:
u
and
w
solve HTP with
Ω = Ω
j
=
∪
d
i
=
1
Ω
i
j
To apply BEM,
we assume that
Ω
i
j
is star–shaped
:
x
i
(
t
) = (
c
x
i
,
c
y
i
) +
r
i
(
t
)(
cos
(
t
)
,
sin
(
t
))
We approximate
r
i
(
t
)
≈
a
i
0
+
K
X
k=
1
(
a
i
k
cos
(
kt
) +
b
i
k
sin
(
kt
))
Solve a least squares problem
to obtain
a
0
s
,
b
0
s
such that
A non–monotone iterative method
The previous scheme was monotone: it generates a
sequence of increasing approximations.
If at some step a
spurious region is included, it cannot be removed
Non–monotone schemes need a
generalized definition of
topological derivative
:
x
∈
R
n
\
Ω
ap
,
D
T
(
x,
R
n
\
Ω
ap
) =
lim
ε
→
0
J
(
R
n
\
Ω
ap
)
−
J
(
R
n
\
(Ω
ap
∪
B
ε
(
x
))
)
Vol
(
B
ε
(
x
))
x
∈
Ω
ap
,
D
T
(
x,
R
n
\
Ω
ap
) =
lim
ε
→
0
J
(
R
n
\
Ω
ap
)
−
J
(
R
n
\
(Ω
ap
\
B
ε
(
x
))
)
Vol
(
B
ε
(
x
))
A non–monotone iterative method
The previous scheme was monotone: it generates a
sequence of increasing approximations.
If at some step a
spurious region is included, it cannot be removed
Non–monotone schemes need a
generalized definition of
topological derivative
:
x
∈
R
n
\
Ω
ap
,
D
T
(
x,
R
n
\
Ω
ap
) =
lim
ε
→
0
J
(
R
n
\
Ω
ap
)
−
J
(
R
n
\
(Ω
ap
∪
B
ε
(
x
))
)
Vol
(
B
ε
(
x
))
x
∈
Ω
ap
,
D
T
(
x,
R
n
\
Ω
ap
) =
lim
ε
→
0
J
(
R
n
\
Ω
ap
)
−
J
(
R
n
\
(Ω
ap
\
B
ε
(
x
))
)
Vol
(
B
ε
(
x
))
Algorithm
1
Compute the TD when
Ω =
∅
(as before)
2
Take
Ω
1
=
{x,
D
T
(
x,
R
n
)
<
−
C
1
},
C
1
>
0 (as before)
3
For j=1:jmax
Compute the TD in
R
n
when
Ω = Ω
j
Select
Ω
j
+
1
if
x
∈
R
n
\
Ω
j
and
TD
<
−C
j
=
⇒
x
∈
Ω
j
+
1
Reconstruction of an annular region. Points are added or
removed at each step. The hole is recovered after 6 steps
Outline
1
Inverse scattering problem
2
Topological derivative methods
TD for shape reconstruction
Iterative methods
TD for shapes and parameters
Direct and inverse problems
Direct problem
∆
u
+
k
e
2
u
=
0
in
R
n
\
Ω
∆
u
+
k
2
i
u
=
0
in
Ω
u
−
=
u
+
,
∂n
u
−
=
∂n
u
+
on
∂Ω
lim
r→∞
r
(n−
1
)
/
2
(
∂r
(
u
−
u
inc
)
−
ik
e
(
u
−
u
inc
)) =
0
Inverse problem
Idea
In the first computation of the TD, i.e. when
Ω =
∅
, we do
not need to know
k
i
:
D
T
(
x,
R
2
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
where
u
=
u
inc
and
w
=
R
Γmeas
G
ke
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
We compute the TD taking
k
i
0
≈
k
e
to get an initial guess
Ω1
In the next step, we update
k
i
by a gradient method
Idea
In the first computation of the TD, i.e. when
Ω =
∅
, we do
not need to know
k
i
:
D
T
(
x,
R
2
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
where
u
=
u
inc
and
w
=
R
Γmeas
G
ke
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
We compute the TD taking
k
i
0
≈
k
e
to get an initial guess
Ω
1
In the next step, we update
k
i
by a gradient method
Idea
In the first computation of the TD, i.e. when
Ω =
∅
, we do
not need to know
k
i
:
D
T
(
x,
R
2
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
where
u
=
u
inc
and
w
=
R
Γmeas
G
ke
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
We compute the TD taking
k
i
0
≈
k
e
to get an initial guess
Ω
1
In the next step, we update
k
i
by a gradient method
Idea
In the first computation of the TD, i.e. when
Ω =
∅
, we do
not need to know
k
i
:
D
T
(
x,
R
2
) =
Re
h
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
i
where
u
=
u
inc
and
w
=
R
Γmeas
G
ke
(
x
−
y
)(
u
meas
−
u
)(
y
)
dl
y
We compute the TD taking
k
i
0
≈
k
e
to get an initial guess
Ω
1
In the next step, we update
k
i
by a gradient method
Materiales heterogéneos
Outline
1
Inverse scattering problem
2
Topological derivative methods
TD for shape reconstruction
Iterative methods
TD for shapes and parameters
Conclusions
We have computed topological derivatives for inverse
scattering problems involving Helmholtz equations with
constant and non–constant parameters
The TD gives a
good approximation
of the number, size
and location of objects buried in the medium
An iterative procedure improves
their shape, and catches
smaller objects, if missed in the first trial
An extended notion of topological derivative
allows to
design non–monotone schemes, useful to capture holes
The algorithm for shape reconstruction can be combined
with a gradient method to recover both
shapes and
parameters
Conclusions
We have computed topological derivatives for inverse
scattering problems involving Helmholtz equations with
constant and non–constant parameters
The TD gives a
good approximation
of the number, size
and location of objects buried in the medium
An iterative procedure improves
their shape, and catches
smaller objects, if missed in the first trial
An extended notion of topological derivative
allows to
design non–monotone schemes, useful to capture holes
The algorithm for shape reconstruction can be combined
with a gradient method to recover both
shapes and
parameters
Conclusions
We have computed topological derivatives for inverse
scattering problems involving Helmholtz equations with
constant and non–constant parameters
The TD gives a
good approximation
of the number, size
and location of objects buried in the medium
An iterative procedure improves
their shape, and catches
smaller objects, if missed in the first trial
An extended notion of topological derivative
allows to
design non–monotone schemes, useful to capture holes
The algorithm for shape reconstruction can be combined
with a gradient method to recover both
shapes and
parameters
Conclusions
We have computed topological derivatives for inverse
scattering problems involving Helmholtz equations with
constant and non–constant parameters
The TD gives a
good approximation
of the number, size
and location of objects buried in the medium
An iterative procedure improves
their shape, and catches
smaller objects, if missed in the first trial
An extended notion of topological derivative
allows to
design non–monotone schemes, useful to capture holes
The algorithm for shape reconstruction can be combined
with a gradient method to recover both
shapes and
parameters
Conclusions
We have computed topological derivatives for inverse
scattering problems involving Helmholtz equations with
constant and non–constant parameters
The TD gives a
good approximation
of the number, size
and location of objects buried in the medium
An iterative procedure improves
their shape, and catches
smaller objects, if missed in the first trial
An extended notion of topological derivative
allows to
design non–monotone schemes, useful to capture holes
The algorithm for shape reconstruction can be combined
with a gradient method to recover both
shapes and
More information
A Carpio, ML Rapún
.
Solving inhomogeneous inverse
problems by topological derivative methods
. Inverse
Problems 24 (2008) Art. 045014
A Carpio, ML Rapún
.
An iterative method for parameter
identification and shape reconstruction
. Inv Probl Sci Eng
18 (2010) 35–50
Ana Carpio:
