Topological derivatives for shape and
parameter reconstruction
Ana Carpio
1
María–Luisa Rapún
2
1
Matemática Aplicada, Universidad Complutense de Madrid, Spain
2Fundamentos Matemáticos, Universidad Politécnica de Madrid, Spain
1
Inverse scattering problems
2
Topological derivative methods
TD for shape reconstruction
TD for shapes and parameters
3
Conclusions
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)
Scattering problem
An incident wave
u
inc
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 "
×
"
Scattering problem
An incident wave
u
inc
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 "
×
"
Scattering problem
An incident wave
u
inc
interacts with a medium
R
containing objects
Ω
.
Inverse problem
Measurements
umeas
are taken at the receptors
Find the scatters
Ω
and the interior parameters s.t.
u
=
u
meas
on
Γ
meas
,
u= sol. forward problem
Model problem
We assume that
We generate
acustic waves
Incident waves are
time–harmonic
U
inc
(x
,
t
) =
Re
[e
−
i
ω
t
u
inc
(x)]
,
u
inc
is a
planar wave
in the direction
d
,
u
inc
(x) =
e
ik
x
·
d
The solution to the direct problem is time–harmonic
We assume that
We generate
acustic waves
Incident waves are
time–harmonic
U
inc
(
x
,
t
) =
Re
[
e
−
i
ωt
u
inc
(
x
)]
,
u
inc
is a
planar wave
in the direction
d
,
u
inc
(x) =
e
ik
x
·
d
The solution to the direct problem is time–harmonic
Model problem
We assume that
We generate
acustic waves
Incident waves are
time–harmonic
U
inc
(
x
,
t
) =
Re
[
e
−
i
ωt
u
inc
(
x
)]
,
u
inc
is a
planar wave
in the direction
d
,
u
inc
(
x
) =
e
ik
x
·
d
The solution to the direct problem is time–harmonic
We assume that
We generate
acustic waves
Incident waves are
time–harmonic
U
inc
(
x
,
t
) =
Re
[
e
−
i
ωt
u
inc
(
x
)]
,
u
inc
is a
planar wave
in the direction
d
,
u
inc
(
x
) =
e
ik
x
·
d
The solution to the direct problem is time–harmonic
A simple forward problem
Ω
is a penetrable known obstacle. The incident field generates
a scattered wave
usc
in
R
n
\
Ω
and a transmitted wave
utr
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
i
2
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
1
Inverse scattering problems
2
Topological derivative methods
TD for shape reconstruction
TD for shapes and parameters
3
Conclusions
Constrained optimization
Original problem
(we assume 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
Γ
meas|u
−
u
meas
|
2
for
u
solving the forward problem with objects
Ω
The domain
Ω
is the variable
Original problem
(we assume 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
Γ
meas|u
−
u
meas
|
2
for
u
solving the forward problem with objects
Ω
The domain
Ω
is the variable
Constrained optimization
Original problem
(we assume 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
Γ
meas|u
−
u
meas
|
2
for
u
solving the forward problem with objects
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
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
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
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
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
x
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
x
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
x
D
T
(
x
,
R
)
<<
0
=
⇒
high probability of finding an object
Equivalently, for
x
∈ R
and
h
(
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
Γ
meas|
u
−
u
meas
|
2
is
D
T
(x
,
R
n
) =
Re
(k
i
2
−
k
e
2
)
u
(x)w
(x)
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
Γ
meas|u
−
u
meas
|
2
is
D
T
(
x
,
R
n
) =
Re
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
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
ike
x
·
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
=
Γ
meas
G
k
e(x
−
y)(
u
meas
−
u
)(y)
dl
y
The true obstacles enter in 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
ike
x
·
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
=
Γ
meas
G
ke
(
x
−
y
)(
umeas
−
u
)(
y
)
dly
The true obstacles enter in 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
ike
x
·
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
=
Γ
meas
G
ke
(
x
−
y
)(
umeas
−
u
)(
y
)
dly
The true obstacles enter in 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
ike
x
·
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
=
Γ
meas
G
ke
(
x
−
y
)(
umeas
−
u
)(
y
)
dly
The true obstacles enter in the TD through the measured
data at the adjoint field
"
×
"= observation points, 24 incident directions in
[
0
,
2
π
)
,
ke
=
2 and
k
i
=
1
/
2. Level of noise=1%
Some examples
"
×
"= observation points, 24 incident directions in
[
0
,
2
π
)
,
ke
=
2 and
k
i
=
1
/
2. Level of noise=1%
Results depend on the wave length (1 w.l.=2
π/k
):
1
st
row:
k
e
=
2 and
k
i
=
1
/
2
TD with an initial guess
Case II:
Ω
ap
first guess,
R
=
R
n
\
Ω
ap
,
Ω = Ω
ap
Theorem.
For any
x
∈
R
n
\
Ω
ap
the topological derivative of
J
(
R
n
\
Ω
ap
) =
1
2
Γ
meas|
u
−
u
meas
|
2
is
D
T
(x
,
R
n
\
Ω
ap
) =
Re
(k
2
i
−
k
e
2
)
u
(x)w
(x)
Case II:
Ω
ap
first guess,
R
=
R
n
\
Ω
ap
,
Ω = Ω
ap
Theorem.
For any
x
∈
R
n
\
Ω
ap
the topological derivative of
J
(
R
n
\
Ω
ap
) =
1
2
Γ
meas|u
−
umeas
|
2
is
D
T
(
x
,
R
n
\
Ω
ap
) =
Re
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
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
∂
Ω
ap
lim
r
→∞
r
(
n
−
1
)
/
2
(
∂r
(
u
−
u
inc
)
−
ike
(
u
−
u
inc
)) =
0
Adjoint problem with
Ω = Ω
ap
:
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
∆
w
+
k
e
2
w
= (
umeas
−
u
)
δ
Γ
measin
R
n
\
Ω
ap
∆
w
+
k
i
2
w
=
0
in
Ω
ap
w
−
=
w
+
, ∂
nw
−
=
∂
n
w
+
on
∂
Ω
ap
lim
r
→∞
r
(
n
−
1
)
/
2
(
∂r
w
−
ike
w
) =
0
Same examples as before with
Ω =
∅
An iterative method
Algorithm
1
Compute the TD when
Ω =
∅
2
Take
Ω1
=
{x
,
D
T
(
x
,
R
n
)
<
−C
1
},
C
1
>
0
3
For j=1:jmax
Compute the TD in
R
n
\
Ω
j
Select
Ω
j
+
1
⊃
Ω
j
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
<
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?
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
<
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?
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
<
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?
1
Inverse scattering problems
2
Topological derivative methods
TD for shape reconstruction
TD for shapes and parameters
3
Conclusions
Direct problem
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
∆
u
+
k
e
2
u
=
0
in
R
n
\
Ω
∆
u
+
k
i
2
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
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
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
where
u
=
u
inc
and
w
=
Γ
meas
G
ke
(
x
−
y
)(
umeas
−
u
)(
y
)
dly
We compute the TD taking
k
i
0
≈
k
e
to get an initial guess
Ω1
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
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
where
u
=
u
inc
and
w
=
Γ
meas
G
ke
(
x
−
y
)(
umeas
−
u
)(
y
)
dly
We compute the TD taking
k
i
0
≈
k
e
to get an initial guess
Ω
1
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
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
where
u
=
u
inc
and
w
=
Γ
meas
G
ke
(
x
−
y
)(
umeas
−
u
)(
y
)
dly
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
(
k
i
2
−
k
e
2
)
u
(
x
)
w
(
x
)
where
u
=
u
inc
and
w
=
Γ
meas