Inverse Obstacle Scattering: Local uniqueness
and iterative methods in reconstruction
William Rundell
Department of Mathematics
Texas A&M University
Joint work with:
Frank Hettlich (Karlsruhe)
Rainer Kress (Göttingen)
Program Script managed in C: The Kernighan-Ritchie version.
Set in plain TEX: No pretentious, aftermarket “improvements”.
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
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..............
Far Field
u1
ui = ei x:d
ui = i Ho kjx yj
us
D
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
...................
....
...
.
.
...
...
.
..
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4 ...
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Far Field
u1
ui = ei x:d
us
D
ui = i Ho kjx yj
The total wave u = ui + us is modeled by an exterior boundary value problem
for the Helmholtz equation:
4u + 2u = 0
with positive wave number
in IRn n D
and a boundary condition on @D :
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
...................
....
...
.
.
...
...
.
..
.
..
..
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)
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. (
4 ...
..
..
..
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..
...
...
...
...
.
.
.....
.
..............
Far Field
u1
ui = ei x:d
us
D
ui = i Ho kjx yj
The total wave u = ui + us is modeled by an exterior boundary value problem
for the Helmholtz equation:
4u + 2u = 0
with positive wave number
u=0
in IRn n D
and a boundary condition on @D :
“sound soft”
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
...................
....
...
.
.
...
...
.
..
.
..
..
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.....
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..
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.. ............. .
..
.
.... ........... ....
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..............................
..
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.. ...
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..
..
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..
.
)
.
. (
4 ...
..
..
..
.
.
..
...
...
...
...
.
.
.....
.
..............
Far Field
u1
ui = ei x:d
us
D
ui = i Ho kjx yj
The total wave u = ui + us is modeled by an exterior boundary value problem
for the Helmholtz equation:
4u + 2u = 0
with positive wave number
@u
@
=0
in IRn n D
and a boundary condition on @D :
“sound hard”
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
...................
....
...
.
.
...
...
.
..
.
..
..
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........ ...................
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..... .. .....
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.......
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.....
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..............................
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..
........... .... ......... .... ... .................
.. ............. .
..
.
.... ........... ....
.
..............................
..
.
.. ...
....
..
..
..
..
.
)
.
. (
4 ...
..
..
..
.
.
..
...
...
...
...
.
.
.....
.
..............
Far Field
u1
ui = ei x:d
us
D
ui = i Ho kjx yj
The total wave u = ui + us is modeled by an exterior boundary value problem
for the Helmholtz equation:
4u + 2u = 0
with positive wave number
@u
@
in IRn n D
and a boundary condition on @D :
+ iu = 0
impedance condition
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
...................
....
...
.
.
...
...
.
..
.
..
..
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........ ...................
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..... .. .....
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.......
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.....
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..............................
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..
........... .... ......... .... ... .................
.. ............. .
..
.
.... ........... ....
.
..............................
..
.
.. ...
....
..
..
..
..
.
)
.
. (
4 ...
..
..
..
.
.
..
...
...
...
...
.
.
.....
.
..............
Far Field
u1
ui = ei x:d
us
D
ui = i Ho kjx yj
The total wave u = ui + us is modeled by an exterior boundary value problem
for the Helmholtz equation:
4u + 2u = 0
with positive wave number
in IRn n D
and a boundary condition on @D :
buc@D = b @u
@ c@D = 0
transmission condition
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
...................
....
...
.
.
...
...
.
..
.
..
..
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.......
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.....
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..............................
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..
........... .... ......... .... ... .................
.. ............. .
..
.
.... ........... ....
.
..............................
..
.
.. ...
....
..
..
..
..
.
)
.
. (
4 ...
..
..
..
.
.
..
...
...
...
...
.
.
.....
.
..............
Far Field
u1
ui = ei x:d
us
D
ui = i Ho kjx yj
The scattered wave us is required to satisfy the Sommerfeld radiation condition
uniformly in all directions x
^ = x=jxj
@us
@r
ius = o
p1r ; r = jxj ! 1; :
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
...................
....
...
.
.
...
...
.
..
.
..
..
.
..
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........ ...................
.......
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..... .. .....
........
.......
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.....
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..............................
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..
........... .... ......... .... ... .................
.. ............. .
..
.
.... ........... ....
.
..............................
..
.
.. ...
....
..
..
..
..
.
)
.
. (
4 ...
..
..
..
.
.
..
...
...
...
...
.
.
.....
.
..............
Far Field
u1
D
ui = i Ho kjx yj
This implies the asymptotic behaviour
eix
us
ui = ei x:d
1
u1 (^x; d) + O
; jxj ! 1; :
(
n
1)
=
2
jxj
jxj
The Amplitude factor u1 is the far field pattern of the scattered wave.
us (x) =
Introduction
Inverse Obstacle Scattering
A time-harmonic acoustic or
electromagnetic plane wave
ui = ei x:d or point source
i H (kjx yj) is fired at a
4 o
cylindrical obstacle D of
unknown shape and location.
The wave us scattered from this
object is measured at “infinity”
– the far field pattern u1 .
...................
....
...
.
.
...
...
.
..
.
..
..
.
..
.
.
..
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..
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......................
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........ ...................
.......
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..... .. .....
........
.......
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.....
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..............................
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..
........... .... ......... .... ... .................
.. ............. .
..
.
.... ........... ....
.
..............................
..
.
.. ...
....
..
..
..
..
.
)
.
. (
4 ...
..
..
..
.
.
..
...
...
...
...
.
.
.....
.
..............
Far Field
u1
D
ui = i Ho kjx yj
This implies the asymptotic behaviour
eix
us
ui = ei x:d
1
u1 (^x; d) + O
; jxj ! 1; :
(
n
1)
=
2
jxj
jxj
The Amplitude factor u1 is the far field pattern of the scattered wave.
F : X ! L2 (S 1 ) maps a set X of admissible boundaries onto the far field
pattern, F (@D) = u1 . F is nonlinear and ill-posed.
us (x) =
Mathematical Issues
The Issues
Mathematical Issues
The Issues
We need uniqueness results.
Mathematical Issues
The Issues
We need uniqueness results.
Æ
Global would be ideal, but local ( := injectivity of the derivative of the
obstacle to data map) would be of practical value.
Mathematical Issues
The Issues
We need uniqueness results.
Æ
Æ
Global would be ideal, but local ( := injectivity of the derivative of the
obstacle to data map) would be of practical value.
We would like to know the minimum amount of data that allows unique
determination.
Mathematical Issues
The Issues
We need uniqueness results.
Æ
Æ
Global would be ideal, but local ( := injectivity of the derivative of the
obstacle to data map) would be of practical value.
We would like to know the minimum amount of data that allows unique
determination.
We know the problems are (highly) ill-posed. Can we measure the degree of
ill-posedness in a useful way?
Mathematical Issues
The Issues
We need uniqueness results.
Æ
Æ
Global would be ideal, but local ( := injectivity of the derivative of the
obstacle to data map) would be of practical value.
We would like to know the minimum amount of data that allows unique
determination.
We know the problems are (highly) ill-posed. Can we measure the degree of
ill-posedness in a useful way?
We would like a robust computational algorithm for computing
@D .
Mathematical Issues
The Issues
We need uniqueness results.
Æ
Æ
Global would be ideal, but local ( := injectivity of the derivative of the
obstacle to data map) would be of practical value.
We would like to know the minimum amount of data that allows unique
determination.
We know the problems are (highly) ill-posed. Can we measure the degree of
ill-posedness in a useful way?
We would like a robust computational algorithm for computing
Æ
Ideally, this algorithm should be maximally efficient.
@D .
Mathematical Issues
The Issues
We need uniqueness results.
Æ
Æ
Global would be ideal, but local ( := injectivity of the derivative of the
obstacle to data map) would be of practical value.
We would like to know the minimum amount of data that allows unique
determination.
We know the problems are (highly) ill-posed. Can we measure the degree of
ill-posedness in a useful way?
We would like a robust computational algorithm for computing
Æ
Æ
Ideally, this algorithm should be maximally efficient.
Good performance under data error is essential.
@D .
Mathematical Issues
The Issues
We need uniqueness results.
Æ
Æ
Global would be ideal, but local ( := injectivity of the derivative of the
obstacle to data map) would be of practical value.
We would like to know the minimum amount of data that allows unique
determination.
We know the problems are (highly) ill-posed. Can we measure the degree of
ill-posedness in a useful way?
We would like a robust computational algorithm for computing
Æ
Æ
Æ
Ideally, this algorithm should be maximally efficient.
Good performance under data error is essential.
The regularisation question must be addressed.
@D .
Mathematical Issues
The Issues
We need uniqueness results.
Æ
Æ
Global would be ideal, but local ( := injectivity of the derivative of the
obstacle to data map) would be of practical value.
We would like to know the minimum amount of data that allows unique
determination.
We know the problems are (highly) ill-posed. Can we measure the degree of
ill-posedness in a useful way?
We would like a robust computational algorithm for computing
Æ
Æ
Æ
Ideally, this algorithm should be maximally efficient.
Good performance under data error is essential.
The regularisation question must be addressed.
Clearly, some compromises will have to be made
@D .
Global Uniqueness
Global Uniqueness Results
Global Uniqueness
Global Uniqueness Results
Lemma: (Rellich) If
u1 = 0 then us (x) = 0 in IRn n D .
Global Uniqueness
Global Uniqueness Results
u1 = 0 then us (x) = 0 in IRn n D .
Lemma: (Holmgren) If u is a solution of the Helmoltz equation and both u and
@u vanish along an arc, then u is identically zero.
@
Lemma: (Rellich) If
Global Uniqueness
Global Uniqueness Results
u1 = 0 then us (x) = 0 in IRn n D .
Lemma: (Holmgren) If u is a solution of the Helmoltz equation and both u and
@u vanish along an arc, then u is identically zero.
@
Lemma: (Rellich) If
Theorem: (Schiffer) If D 1 and D 2 are two sound soft scatterers such that
the far field patterns from an infinite number of distinct plane waves with fixed
wavenumber are identical, then D 1 = D 2 .
Global Uniqueness
Global Uniqueness Results
u1 = 0 then us (x) = 0 in IRn n D .
Lemma: (Holmgren) If u is a solution of the Helmoltz equation and both u and
@u vanish along an arc, then u is identically zero.
@
Lemma: (Rellich) If
Theorem: (Schiffer) If D 1 and D 2 are two sound soft scatterers such that
the far field patterns from an infinite number of distinct plane waves with fixed
wavenumber are identical, then D 1 = D 2 .
us
Proof:
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D
D
Define D := ( D 1 [ D 2 )n D 2 .
For each incident field, ui :
us is common to both D 1 and
D 2 and is defined on D .
u := ui + us is zero on @D .
Global Uniqueness
Global Uniqueness Results
u1 = 0 then us (x) = 0 in IRn n D .
Lemma: (Holmgren) If u is a solution of the Helmoltz equation and both u and
@u vanish along an arc, then u is identically zero.
@
Lemma: (Rellich) If
Theorem: (Schiffer) If D 1 and D 2 are two sound soft scatterers such that
the far field patterns from an infinite number of distinct plane waves with fixed
wavenumber are identical, then D 1 = D 2 .
Proof:
.......................................
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2 ......
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............................................
D
4u = u
u=0
Define D := ( D 1 [ D 2 )n D 2 .
For each incident field, ui :
us is common to both D 1 and
D 2 and is defined on D .
u := ui + us is zero on @D .
Global Uniqueness
Global Uniqueness Results
u1 = 0 then us (x) = 0 in IRn n D .
Lemma: (Holmgren) If u is a solution of the Helmoltz equation and both u and
@u vanish along an arc, then u is identically zero.
@
Lemma: (Rellich) If
Theorem: (Schiffer) If D 1 and D 2 are two sound soft scatterers such that
the far field patterns from an infinite number of distinct plane waves with fixed
wavenumber are identical, then D 1 = D 2 .
Proof:
.......................................
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D
4u = u
u=0
2 , u are Dirichlet eigenvalue and eigenfunctions for 4 on D ; – for all incident ui . But,
distinct plane waves are linearly independent.
A contradiction; there is only a finite number
of linearly independent eigenfunctions corresponding to 2 for 4 in H01 (D) .
Global Uniqueness
Global Uniqueness Results
u1 = 0 then us (x) = 0 in IRn n D .
Lemma: (Holmgren) If u is a solution of the Helmoltz equation and both u and
@u vanish along an arc, then u is identically zero.
@
Lemma: (Rellich) If
Theorem: (Schiffer) If D 1 and D 2 are two sound soft scatterers such that
the far field patterns from an infinite number of distinct plane waves with fixed
wavenumber are identical, then D 1 = D 2 .
Proof:
.......................................
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............................................
D
4u = u
u=0
2 , u are Dirichlet eigenvalue and eigenfunctions for 4 on D ; – for all incident ui . But,
distinct plane waves are linearly independent.
A contradiction; there is only a finite number
of linearly independent eigenfunctions corresponding to 2 for 4 in H01 (D) .
(A similar result holds if there is an infinite number of plane waves with
constant direction d , but varying over an finite interval.)
Global Uniqueness
Global Uniqueness Results
u1 = 0 then us (x) = 0 in IRn n D .
Lemma: (Holmgren) If u is a solution of the Helmoltz equation and both u and
@u vanish along an arc, then u is identically zero.
@
Lemma: (Rellich) If
Theorem: (Schiffer) If D 1 and D 2 are two sound soft scatterers such that
the far field patterns from an infinite number of distinct plane waves with fixed
wavenumber are identical, then D 1 = D 2 .
Proof:
.......................................
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D
4u = u
u=0
What about other boundary conditions?
Define D := ( D 1 [ D 2 )n D 2 .
For each incident field, ui :
us is common to both D 1 and
D 2 and is defined on D .
u := ui + us is zero on @D .
Global Uniqueness
Global Uniqueness Results
u1 = 0 then us (x) = 0 in IRn n D .
Lemma: (Holmgren) If u is a solution of the Helmoltz equation and both u and
@u vanish along an arc, then u is identically zero.
@
Lemma: (Rellich) If
Theorem: (Schiffer) If D 1 and D 2 are two sound soft scatterers such that
the far field patterns from an infinite number of distinct plane waves with fixed
wavenumber are identical, then D 1 = D 2 .
Proof:
.......................................
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D
4u = u
u=0
Define D := ( D 1 [ D 2 )n D 2 .
For each incident field, ui :
us is common to both D 1 and
D 2 and is defined on D .
u := ui + us is zero on @D .
Unfortunately, Schiffer’s proof fails for non sound soft conditions due to the
potential cusps in the domain D above. However . . . .
Global Uniqueness
Global Uniqueness Results
Results for other boundary conditions:
Global Uniqueness
Global Uniqueness Results
Results for other boundary conditions:
Neumann and impedance conditions;
Kress and Kirsch (ff. Isakov).
Global Uniqueness
Global Uniqueness Results
Results for other boundary conditions:
Neumann and impedance conditions;
For the transmission problem;
Kress and Kirsch (ff. Isakov).
Kress and Kirsch (ff. Potthast).
Global Uniqueness
Global Uniqueness Results
Results for other boundary conditions:
Neumann and impedance conditions;
For the transmission problem;
Kress and Kirsch (ff. Isakov).
Kress and Kirsch (ff. Potthast).
Unique determination of the support of a scatterer; Sun and Uhlmann.
Global Uniqueness
Global Uniqueness Results
Results for other boundary conditions:
Neumann and impedance conditions;
For the transmission problem;
Kress and Kirsch (ff. Isakov).
Kress and Kirsch (ff. Potthast).
Unique determination of the support of a scatterer; Sun and Uhlmann.
Determination of both
D and the impedance ;
Kress and Rundell.
Global Uniqueness
Global Uniqueness Results
Results for other boundary conditions:
Neumann and impedance conditions;
For the transmission problem;
Kress and Kirsch (ff. Isakov).
Kress and Kirsch (ff. Potthast).
Unique determination of the support of a scatterer; Sun and Uhlmann.
Determination of both
D and the impedance ;
Kress and Rundell.
But the information count is wrong! We are giving an infinite number of
complex valued functions on S just to obtain one real curve @D .
Global Uniqueness
Global Uniqueness Results
Results for other boundary conditions:
Neumann and impedance conditions;
For the transmission problem;
Kress and Kirsch (ff. Isakov).
Kress and Kirsch (ff. Potthast).
Unique determination of the support of a scatterer; Sun and Uhlmann.
Determination of both
D and the impedance ;
Kress and Rundell.
But the information count is wrong! We are giving an infinite number of
complex valued functions on S just to obtain one real curve @D .
It has been a long-standing conjecture that the far field pattern from a single
incident plane wave suffices to determine an obstacle D .
Uniqueness, Single Incident Wave
Theorem: (Colton, Sleeman) If D 1 and D 2 are two sound soft scatterers contained in a ball of radius < = and if the far field patterns coincide for a single
incident plane wave with wavenumber , then D 1 = D 2 .
Uniqueness, Single Incident Wave
Theorem: (Colton, Sleeman) If D 1 and D 2 are two sound soft scatterers contained in a ball of radius < = and if the far field patterns coincide for a single
incident plane wave with wavenumber , then D 1 = D 2 .
Theorem: (Liu, Nachman) If D 1 and D 2 are convex polyhedra with the same
far field pattern from a single incident plane wave, then D 1 = D 2 .
Uniqueness, Single Incident Wave
Theorem: (Colton, Sleeman) If D 1 and D 2 are two sound soft scatterers contained in a ball of radius < = and if the far field patterns coincide for a single
incident plane wave with wavenumber , then D 1 = D 2 .
Theorem: (Liu, Nachman) If D 1 and D 2 are convex polyhedra with the same
far field pattern from a single incident plane wave, then D 1 = D 2 .
Theorem: (Potthast) For all > 0 , there exists M () and N () such that
if the far field patterns coincide for M incident directions and N observation
directions, then Æ ( D 1 ; D 2 ) < .
Uniqueness, Single Incident Wave
Theorem: (Colton, Sleeman) If D 1 and D 2 are two sound soft scatterers contained in a ball of radius < = and if the far field patterns coincide for a single
incident plane wave with wavenumber , then D 1 = D 2 .
Theorem: (Liu, Nachman) If D 1 and D 2 are convex polyhedra with the same
far field pattern from a single incident plane wave, then D 1 = D 2 .
Theorem: (Potthast) For all > 0 , there exists M () and N () such that
if the far field patterns coincide for M incident directions and N observation
directions, then Æ ( D 1 ; D 2 ) < .
Warning!
Let the incident wave be a superposition of plane waves:
Z
sin
j
x
j
i x:d
i
u (x) =
Let
D
jxj = 4
S
ds(d)
e
be a ball of radius R , center the origin. Then us
Thus the total field u(x) = sin (jxj
radius R + m= for all integers m .
R)e
sin R eijxj
(x) = eiR jxj
jxj vanishes on the sphere with
iR =
Uniqueness, Single Incident Wave
Theorem: (Colton, Sleeman) If D 1 and D 2 are two sound soft scatterers contained in a ball of radius < = and if the far field patterns coincide for a single
incident plane wave with wavenumber , then D 1 = D 2 .
Theorem: (Liu, Nachman) If D 1 and D 2 are convex polyhedra with the same
far field pattern from a single incident plane wave, then D 1 = D 2 .
Theorem: (Potthast) For all > 0 , there exists M () and N () such that
if the far field patterns coincide for M incident directions and N observation
directions, then Æ ( D 1 ; D 2 ) < .
Warning!
Let the incident wave be a superposition of plane waves:
Z
sin
j
x
j
i x:d
i
u (x) =
Let
D
jxj = 4
S
ds(d)
e
be a ball of radius R , center the origin. Then us
Thus the total field u(x) = sin (jxj
radius R + m= for all integers m .
R)e
sin R eijxj
(x) = eiR jxj
jxj vanishes on the sphere with
iR =
We must use special features of the incident wave.
There are several methods that can be used to prove uniquness and constructibilty
results. There is simply no such thing as the “best method.’
In this lecture we will concentrate on local uniqueness results; that is, we will
attempt to prove injectivity of the derivative map F 0 . Since we will be using
iteration schemes, such as Newton’s method, this will be a necessary step.
There are several methods that can be used to prove uniquness and constructibilty
results. There is simply no such thing as the “best method.’
In this lecture we will concentrate on local uniqueness results; that is, we will
attempt to prove injectivity of the derivative map F 0 . Since we will be using
iteration schemes, such as Newton’s method, this will be a necessary step.
Typically such methods can give information under minimal amounts of data, but
they have drawbacks; an initial ansatz (such as a single obstacle) and “reasonable”
initial approximation to the solution is required.
There are several methods that can be used to prove uniquness and constructibilty
results. There is simply no such thing as the “best method.’
In this lecture we will concentrate on local uniqueness results; that is, we will
attempt to prove injectivity of the derivative map F 0 . Since we will be using
iteration schemes, such as Newton’s method, this will be a necessary step.
Typically such methods can give information under minimal amounts of data, but
they have drawbacks; an initial ansatz (such as a single obstacle) and “reasonable”
initial approximation to the solution is required.
There are some methods that can give good reconstructions of an obstacle even
when there are several of these and we do not know in advance what the boundary
conditions on the obstacle(s) are.
Of course, such luxuries have to be paid for; often in the amount of “extra data”
required.
There are several methods that can be used to prove uniquness and constructibilty
results. There is simply no such thing as the “best method.’
In this lecture we will concentrate on local uniqueness results; that is, we will
attempt to prove injectivity of the derivative map F 0 . Since we will be using
iteration schemes, such as Newton’s method, this will be a necessary step.
Typically such methods can give information under minimal amounts of data, but
they have drawbacks; an initial ansatz (such as a single obstacle) and “reasonable”
initial approximation to the solution is required.
There are some methods that can give good reconstructions of an obstacle even
when there are several of these and we do not know in advance what the boundary
conditions on the obstacle(s) are.
Of course, such luxuries have to be paid for; often in the amount of “extra data”
required.
We mention one such approach: the sampling method due originally to Colton
and Kirsch - with a critical mathematical refinement due to Kirsch.
Sampling Method
Define the far field operator
G[f ](x) =
G : L2 (S ) ! L2 (S ) by
Z
S
u1 (x; d)f (d)ds(d)
x 2 S:
This is an integral operator with kernel u1 (x; d) , which is the far-field pattern
at a point x from an incident plane wave from angle d .
Further, let 1 (x; z ) be the far field pattern of the fundamental solution:
1 (x; z ) =
1 ix:z
e
4
x 2 S; z 2 IR2;3
Sampling Method
Define the far field operator
G[f ](x) =
G : L2 (S ) ! L2 (S ) by
Z
S
u1 (x; d)f (d)ds(d)
x 2 S:
This is an integral operator with kernel u1 (x; d) , which is the far-field pattern
at a point x from an incident plane wave from angle d .
Further, let 1 (x; z ) be the far field pattern of the fundamental solution:
1 (x; z ) =
1 ix:z
e
4
x 2 S; z 2 IR2;3
Colton and Kirsch gave mathematical reasons why one should expect that the
solution of the above equation should have significantly larger norm for z 62 D
than for z 2 D .
Led to a practical reconstruction method (and an industry!).
Sampling Method
Define the far field operator
G[f ](x) =
G : L2 (S ) ! L2 (S ) by
Z
S
u1 (x; d)f (d)ds(d)
x 2 S:
This is an integral operator with kernel u1 (x; d) , which is the far-field pattern
at a point x from an incident plane wave from angle d .
Further, let 1 (x; z ) be the far field pattern of the fundamental solution:
1 (x; z ) =
1 ix:z
e
4
x 2 S; z 2 IR2;3
Theorem: (Kirsch) Assume that is not an eigenvalue for
D subject to an impedance or transmission condition. Then
if the ill-posed equation
has a solution
f
2 L2(S ) .
(G G)1=4 [f ]( ; z ) = 1 ( ; z )
4 on the domain
z 2 D if and only
Sampling Method
Define the far field operator
G[f ](x) =
G : L2 (S ) ! L2 (S ) by
Z
S
u1 (x; d)f (d)ds(d)
x 2 S:
This is an integral operator with kernel u1 (x; d) , which is the far-field pattern
at a point x from an incident plane wave from angle d .
Further, let 1 (x; z ) be the far field pattern of the fundamental solution:
1 (x; z ) =
1 ix:z
e
4
x 2 S; z 2 IR2;3
Theorem: (Kirsch) Assume that is not an eigenvalue for
D subject to an impedance or transmission condition. Then
if the ill-posed equation
has a solution
f
2 L2(S ) .
4 on the domain
z 2 D if and only
(G G)1=4 [f ]( ; z ) = 1 ( ; z )
Advantages:
Requires (virtually) no apriori knowledge of the domain
Is easy and fast to implement.
D.
Sampling Method
Define the far field operator
G[f ](x) =
G : L2 (S ) ! L2 (S ) by
Z
S
u1 (x; d)f (d)ds(d)
x 2 S:
This is an integral operator with kernel u1 (x; d) , which is the far-field pattern
at a point x from an incident plane wave from angle d .
Further, let 1 (x; z ) be the far field pattern of the fundamental solution:
1 (x; z ) =
1 ix:z
e
4
x 2 S; z 2 IR2;3
Theorem: (Kirsch) Assume that is not an eigenvalue for
D subject to an impedance or transmission condition. Then
if the ill-posed equation
has a solution
f
2 L2(S ) .
4 on the domain
z 2 D if and only
(G G)1=4 [f ]( ; z ) = 1 ( ; z )
Disadvantages:
Requires enormous amounts of data; full far field pattern from all incident
directions.
Must solve a highly ill-conditioned integral equation whose kernel contains
the data - very susceptible to noise.
Local Uniqueness
Local Uniqueness Results
Local Uniqueness
Local Uniqueness Results
D and a function h 2 C 2 (@D) ! IRn , we
@Dh := fx + h(x) : x 2 @Dg, Br := fh 2 C 2 (@D) : khkC (@D) < rg
then the boundary to far field map F , is defined by F : Br ! L2 (S ) .
Define h = : h , the normal component of the vector field h .
For a reference domain
2
define
Local Uniqueness
Local Uniqueness Results
D and a function h 2 C 2 (@D) ! IRn , we
@Dh := fx + h(x) : x 2 @Dg, Br := fh 2 C 2 (@D) : khkC (@D) < rg
then the boundary to far field map F , is defined by F : Br ! L2 (S ) .
Define h = : h , the normal component of the vector field h .
For a reference domain
define
2
Theorem: For D sound soft, F : @Dh ! u1 is Fréchet differentiable with
derivative given by the far field pattern v1 where v satisfies
4v + 2v = 0
in IRn n D;
the Sommerfeld condition and the boundary condition
v = h
@u .
@
Local Uniqueness
Local Uniqueness Results
D and a function h 2 C 2 (@D) ! IRn , we
@Dh := fx + h(x) : x 2 @Dg, Br := fh 2 C 2 (@D) : khkC (@D) < rg
then the boundary to far field map F , is defined by F : Br ! L2 (S ) .
Define h = : h , the normal component of the vector field h .
For a reference domain
define
2
Theorem: For D sound soft, F : @Dh ! u1 is Fréchet differentiable with
derivative given by the far field pattern v1 where v satisfies
4v + 2v = 0
in IRn n D;
the Sommerfeld condition and the boundary condition
v = h
@u .
@
Since v satisfies the same boundary value problem as the function u (with
different Dirichlet values) it can be computed very quickly from u . The largest
part of the computation of F and F 0 is the common inversion of the matrix
representing 4v + k2 v .
Local Uniqueness
Local Uniqueness Results
D and a function h 2 C 2 (@D) ! IRn , we
@Dh := fx + h(x) : x 2 @Dg, Br := fh 2 C 2 (@D) : khkC (@D) < rg
then the boundary to far field map F , is defined by F : Br ! L2 (S ) .
Define h = : h , the normal component of the vector field h .
For a reference domain
define
2
Theorem: For D sound soft, F : @Dh ! u1 is Fréchet differentiable with
derivative given by the far field pattern v1 where v satisfies
4v + 2v = 0
in IRn n D;
the Sommerfeld condition and the boundary condition
Theorem: For sound soft scatterering obstacles
N (F 0 (@D)) = fh 2 C 2 (@D) : h = 0g .
v = h
@u .
@
D , the nullspace of F 0 (@D) is
Local Uniqueness
Local Uniqueness Results
D and a function h 2 C 2 (@D) ! IRn , we
@Dh := fx + h(x) : x 2 @Dg, Br := fh 2 C 2 (@D) : khkC (@D) < rg
then the boundary to far field map F , is defined by F : Br ! L2 (S ) .
Define h = : h , the normal component of the vector field h .
For a reference domain
define
2
Theorem: For D sound soft, F : @Dh ! u1 is Fréchet differentiable with
derivative given by the far field pattern v1 where v satisfies
4v + 2v = 0
in IRn n D;
the Sommerfeld condition and the boundary condition
Theorem: For sound soft scatterering obstacles
N (F 0 (@D)) = fh 2 C 2 (@D) : h = 0g .
v = h
@u .
@
D , the nullspace of F 0 (@D) is
Proof: Let F 0 (@D) = 0 . Then v1 = 0 and by Rellich’s lemma, v = 0
and so v = 0 on @D . Thus h @u
in IRn n D
@ = 0 on @D . However, by
Holmgren’s theorem, @u
vanish on any arc of @D since this would mean
@ cannot
u is identically zero in IRn n D . Hence h = 0 .
Frozen Newton Schemes
We derive an explicit representation of
F0
when
@D is the unit circle.
Frozen Newton Schemes
We derive an explicit representation of
The Jacobi–Anger expansion gives
ui (x) = ei xd =
1
X
n=
1
F0
u(x) =
in
(1) ()
H
n= 1 n
@D is the unit circle.
in Jn ()ein(
From this it can be seen that the total field
1
X
when
0 ) ;
u = ui + us
x 2 IRn ;
has the form
fJn ()Hn(1)() Jn ()Hn(1)()gein(
0 )
Frozen Newton Schemes
We derive an explicit representation of
The Jacobi–Anger expansion gives
ui (x) = ei xd =
1
X
n=
1
F0
u(x) =
in
(1) ()
H
n= 1 n
@D is the unit circle.
in Jn ()ein(
From this it can be seen that the total field
1
X
when
0 ) ;
u = ui + us
x 2 IRn ;
has the form
fJn ()Hn(1)() Jn ()Hn(1)()gein(
0 )
From the asymptotics of the Hankel functions we obtain
F () = u1 = e
i
4
r
2
1
X
Jn () in(
e
(1)
n= 1 Hn ()
0 ) ;
0 2:
Frozen Newton Schemes
We derive an explicit representation of
The Jacobi–Anger expansion gives
ui (x) = ei xd =
1
X
n=
1
F0
u(x) =
in
(1) ()
H
n= 1 n
@D is the unit circle.
in Jn ()ein(
From this it can be seen that the total field
1
X
when
0 ) ;
u = ui + us
x 2 IRn ;
has the form
fJn ()Hn(1)() Jn ()Hn(1)()gein(
0 )
From the asymptotics of the Hankel functions we obtain
r
4
2
1
X
Jn () in( )
e
; 0 2:
(1)
n= 1 Hn ()
We compute @u
@ (x) and then v in the above theorem to obtain
r
1 X
1
m ei(m n) ein
X
i
8
F 0 [1] h = 3
hm
(1) ()
(1) ()
H
H
n
n= 1 m= 1
n m
in terms of the Fourier coefficients h ; h ; : : : , of the perturbation h .
F () = u1 = e
i
0
0
0
1
Frozen Newton Schemes
For fixed N , the quasi-Newton scheme to recover the leading Fourier coefficients h0 ; h1 ; : : : ; hN of the perturbation h from the corresponding Fourier
coefficients of the far field pattern un1 has the form
Anm hm
= Bn (un un1 )
where A is a (complex) 2N +1 2N +1 matrix, B a 2N +1 vector
Anm () =
N
X
m= N
eim(0 2 )
H (1)
n
m ()
Bn () =
r
3 i(n0 4 ) (1)
e
H ()
8
n
Frozen Newton Schemes
For fixed N , the quasi-Newton scheme to recover the leading Fourier coefficients h0 ; h1 ; : : : ; hN of the perturbation h from the corresponding Fourier
coefficients of the far field pattern un1 has the form
Anm hm
= Bn (un un1 )
where A is a (complex) 2N +1 2N +1 matrix, B a 2N +1 vector
Anm () =
N
X
m= N
eim(0 2 )
H (1)
n
m ()
Theorem. For suÆciently small
Bn () =
r
3 i(n0 4 ) (1)
e
H ()
8
n
k , the matrix A of the nite linear system
is regular and its condition number (with respect to the maximum norm)
is uniformly bounded with respect to N .
Frozen Newton Schemes
For fixed N , the quasi-Newton scheme to recover the leading Fourier coefficients h0 ; h1 ; : : : ; hN of the perturbation h from the corresponding Fourier
coefficients of the far field pattern un1 has the form
Anm hm
= Bn (un un1 )
where A is a (complex) 2N +1 2N +1 matrix, B a 2N +1 vector
Anm () =
N
X
m= N
eim(0 2 )
H (1)
n
m ()
Theorem. For suÆciently small
Bn () =
r
3 i(n0 4 ) (1)
e
H ()
8
n
k , the matrix A of the nite linear system
is regular and its condition number (with respect to the maximum norm)
is uniformly bounded with respect to N .
Doesn’t this mean that the inversion is well-posed?
Frozen Newton Schemes
For fixed N , the quasi-Newton scheme to recover the leading Fourier coefficients h0 ; h1 ; : : : ; hN of the perturbation h from the corresponding Fourier
coefficients of the far field pattern un1 has the form
Anm hm
= Bn (un un1 )
where A is a (complex) 2N +1 2N +1 matrix, B a 2N +1 vector
Anm () =
N
X
m= N
eim(0 2 )
H (1)
n
Bn () =
m ()
Theorem. For suÆciently small
r
3 i(n0 4 ) (1)
e
H ()
8
n
k , the matrix A of the nite linear system
is regular and its condition number (with respect to the maximum norm)
is uniformly bounded with respect to N .
Doesn’t this mean that the inversion is well-posed?
N n
5
10
15
j
Right hand side BN ()
j
The bad news,
0.5
1.0
1.5
2.0
2.5
3.0
7:910113 2:61028 3:71016
9:9 5
3:8 4 1:9 3
1:21019 1:21014 2:21012 1:31010 1:5109 2:6107
3:010 9:310 2:210 3:010 1:110 7:410
Frozen Newton Schemes
What can we conclude from this?
Frozen Newton Schemes
What can we conclude from this?
If the data measurement gives an error in the Fourier coefficients of the far
field then the approximate error in the updated perturbation of the curve @D
is cond (A)jBN j Frozen Newton Schemes
What can we conclude from this?
If the data measurement gives an error in the Fourier coefficients of the far
field then the approximate error in the updated perturbation of the curve @D
is cond (A)jBN j Giving further incident waves does not allow any more Fourier coefficients
to be computed.
Frozen Newton Schemes
What can we conclude from this?
If the data measurement gives an error in the Fourier coefficients of the far
field then the approximate error in the updated perturbation of the curve @D
is cond (A)jBN j Giving further incident waves does not allow any more Fourier coefficients
to be computed.
Using additional frequencies does not allow more Fourier coefficients to
be computed (should use largest ).
Frozen Newton Schemes
What can we conclude from this?
If the data measurement gives an error in the Fourier coefficients of the far
field then the approximate error in the updated perturbation of the curve @D
is cond (A)jBN j Giving further incident waves does not allow any more Fourier coefficients
to be computed.
Using additional frequencies does not allow more Fourier coefficients to
be computed (should use largest ).
Our invertibility theorem shows the linear system is uniquely invertible at
each step; no such guarantee can be given for full-Newton.
Frozen Newton Schemes
What can we conclude from this?
If the data measurement gives an error in the Fourier coefficients of the far
field then the approximate error in the updated perturbation of the curve @D
is cond (A)jBN j Giving further incident waves does not allow any more Fourier coefficients
to be computed.
Using additional frequencies does not allow more Fourier coefficients to
be computed (should use largest ).
Our invertibility theorem shows the linear system is uniquely invertible at
each step; no such guarantee can be given for full-Newton.
We actually find that this frozen-Newton scheme gives reconstructions that
are virtually identical to a full-Newton implementation.
Frozen Newton Schemes
Single Frequency, Multiple Direction Incident Waves
For a fixed offset angle and
each direction d , 0 d 2 ,
measure the far field pattern
u1 (d) at the single point given
by the direction d + .
The frequency is fixed.
= is backscattering case,
= 0 is forward scattering.
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u1
ui
D
Frozen Newton Schemes
Single Frequency, Multiple Direction Incident Waves
For a fixed offset angle and
each direction d , 0 d 2 ,
measure the far field pattern
u1 (d) at the single point given
by the direction d + .
The frequency is fixed.
= is backscattering case,
= 0 is forward scattering.
..............
...
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...................
u1
ui
D
Frozen Newton Schemes
Single Frequency, Multiple Direction Incident Waves
For a fixed offset angle and
each direction d , 0 d 2 ,
measure the far field pattern
u1 (d) at the single point given
by the direction d + .
The frequency is fixed.
= is backscattering case,
= 0 is forward scattering.
..............
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...
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.....
..............
ui
D
u1
Frozen Newton Schemes
Single Frequency, Multiple Direction Incident Waves
For a fixed offset angle and
each direction d , 0 d 2 ,
measure the far field pattern
u1 (d) at the single point given
by the direction d + .
The frequency is fixed.
= is backscattering case,
= 0 is forward scattering.
..............
.....
...
...
...
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ui
D
u1
Theorem. Let 0 < < 2 . Then if the wavenumber is sufficiently small the
derivative map F 0 is injective. With M incident waves from distinct directions
and a finite trigonometric basis the resulting Jacobian matrix has trivial nullspace
provided M > 2 N + 1 .
In the forward scattering case, = 0 , the odd cosine and sine coefficients
of q (as measured from the origin) cannot be recovered.
Frozen Newton Schemes
Single Direction, Multiple Frequency Incident Waves
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............
For a fixed offset angle and
fixed direction d0 , measure the
far field pattern u1 () at the i i x:d
u =e
single point given by the direc
tion d0 + .
The frequency is varied over
the range [min ; max ] .
u1 ()
D
Frozen Newton Schemes
Single Direction, Multiple Frequency Incident Waves
............
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..
..
.
...
.
...
...
.....
...
............
For a fixed offset angle and
fixed direction d0 , measure the
far field pattern u1 () at the i i x:d
u =e
single point given by the direc
tion d0 + .
The frequency is varied over
the range [min ; max ] .
u1 ()
D
Theorem. For any with 0 < < 2 , the dimension of the nullspace of
F 0 is exactly N . The nullspace consists of the odd cosine and the even sine
coefficients (with the origin at t = 2 ).
Frozen Newton Schemes
Single Direction, Multiple Frequency Incident Waves
............
.....
...
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...
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..
.
..
.
..
..
..
..
.
...
.
...
...
.....
...
............
For a fixed offset angle and
fixed direction d0 , measure the
far field pattern u1 () at the i i x:d
u =e
single point given by the direc
tion d0 + .
The frequency is varied over
the range [min ; max ] .
u1 ()
D
Theorem. For any with 0 < < 2 , the dimension of the nullspace of
F 0 is exactly N . The nullspace consists of the odd cosine and the even sine
coefficients (with the origin at t = 2 ).
Is there complete loss of information if
= 0?
No, we can show that the nullspace of F 0 consists of all the sine
coefficients in addition to all the odd cosine coefficients.
Frozen Newton Schemes
Single Direction, Multiple Frequency Incident Waves
............
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0
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For a fixed offset angle and
fixed direction d0 , measure the
far field pattern u1 () at the i i x:d
u =e
single point given by the direc
tion d0 + .
The frequency is varied over
the range [min ; max ] .
u1 ()
D
Theorem. For any with 0 < < 2 , the dimension of the nullspace of
F 0 is exactly N . The nullspace consists of the odd cosine and the even sine
coefficients (with the origin at t = 2 ).
The obvious question is, does measurements at a scan of frequencies
at two distinct points recover full information?
Frozen Newton Schemes
Single Direction, Multiple Frequency Incident Waves
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0
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...
............
For a fixed offset angle and
fixed direction d0 , measure the
far field pattern u1 () at the i i x:d
u =e
single point given by the direc
tion d0 + .
The frequency is varied over
the range [min ; max ] .
u1 ()
D
Theorem. For any with 0 < < 2 , the dimension of the nullspace of
F 0 is exactly N . The nullspace consists of the odd cosine and the even sine
coefficients (with the origin at t = 2 ).
Corollary. From a measurement of the far field pattern at two angles 1 and
2 we can recover all Fourier coefficients of q provided 1 6= 0 and 2 6= 0 and
sin m (1 2 2 ) 6= 0 , for m = 1; : : : ; N .
Impedance Problem
Impedance Boundary Conditions
@u
+
iku = 0
@
On physical grounds
on @D:
() is real and non-negative.
Impedance Problem
Impedance Boundary Conditions
@u
+
iku = 0
@
On physical grounds
on @D:
() is real and non-negative.
Theorem. If D1 and D2 have impedances 1 and 2 , respectively, such that
the far field patterns for both scatterers coincide for an infinite number of distinct
plane waves or for an infinite number of point sources on the boundary @B of a
1 [ D 2 B , then D1 = D2 and 1 = 2 .
domain B such that D
Impedance Problem
Impedance Boundary Conditions
@u
+
iku = 0
@
On physical grounds
on @D:
() is real and non-negative.
Theorem. If D1 and D2 have impedances 1 and 2 , respectively, such that
the far field patterns for both scatterers coincide for an infinite number of distinct
plane waves or for an infinite number of point sources on the boundary @B of a
1 [ D 2 B , then D1 = D2 and 1 = 2 .
domain B such that D
What about local uniqueness for a single incident plane wave?
Impedance Problem
Impedance Boundary Conditions
@u
+
iku = 0
@
On physical grounds
on @D:
() is real and non-negative.
Theorem. If D1 and D2 have impedances 1 and 2 , respectively, such that
the far field patterns for both scatterers coincide for an infinite number of distinct
plane waves or for an infinite number of point sources on the boundary @B of a
1 [ D 2 B , then D1 = D2 and 1 = 2 .
domain B such that D
Theorem. The map F (@D; ) ! u1 is Fréchet differentiable with respect to
the boundary and the derivative Fr0 q in the direction h is given by the far field
pattern vr;1 of the solution vr of the impedance boundary value problem with
boundary condition
@vr
d
du
@u
2
+ ikvr = k h u +
h
ik h
Hu
()
@
ds
ds
@
H
is the mean curvature on
@D ,
h =
hq
p
q2
.
0
2
+ (q )
Impedance Problem
Theorem. Assume that (x) > 0 and H (x) > 0 for all x 2 @D . Then the
derivative of the map F with respect to the boundary is injective, i.e., Fq0 h = 0
implies h = 0 .
Impedance Problem
Theorem. Assume that (x) > 0 and H (x) > 0 for all x 2 @D . Then the
derivative of the map F with respect to the boundary is injective, i.e., Fq0 h = 0
implies h = 0 .
Proof: If Fq0 h = 0 then the solution to the scattering problem has vanishing far
field pattern. Rellich’s lemma and (*) shows
k2 (1
d
du
2
)h u +
h
ds
ds
+ ikHh u = 0
on @D;
u and take the imaginary part to obtain
d
(
hU ) = kHh juj2
on @D;
ds
i u
du . Assume that h is not identically zero.
where U := 2i u ddsu
2 ds
the set := fx 2 @D : h (x) > 0g is nonempty and
multiply this by
Z
Thus
Hh juj2 ds = 0:
u = 0 on . Holmgren’s theorem gives a contradiction.
Then,
Impedance Problem
Theorem. Assume that (x) > 0 and H (x) > 0 for all x 2 @D . Then the
derivative of the map F with respect to the boundary is injective, i.e., Fq0 h = 0
implies h = 0 .
Proof: If Fq0 h = 0 then the solution to the scattering problem has vanishing far
field pattern. Rellich’s lemma and (*) shows
k2 (1
d
du
2
)h u +
h
ds
ds
+ ikHh u = 0
on @D;
u and take the imaginary part to obtain
d
(
hU ) = kHh juj2
on @D;
ds
i u
du . Assume that h is not identically zero.
where U := 2i u ddsu
2 ds
the set := fx 2 @D : h (x) > 0g is nonempty and
multiply this by
Z
Thus
Then,
Hh juj2 ds = 0:
u = 0 on . Holmgren’s theorem gives a contradiction.
The farfield u1 is a complex-valued function and we are only obtaining a real
valued curve @D in return.
Impedance Problem
Theorem. Assume that (x) > 0 and H (x) > 0 for all x 2 @D . Then the
derivative of the map F with respect to the boundary is injective, i.e., Fq0 h = 0
implies h = 0 .
Proof: If Fq0 h = 0 then the solution to the scattering problem has vanishing far
field pattern. Rellich’s lemma and (*) shows
k2 (1
d
du
2
)h u +
h
ds
ds
+ ikHh u = 0
on @D;
u and take the imaginary part to obtain
d
(
hU ) = kHh juj2
on @D;
ds
i u
du . Assume that h is not identically zero.
where U := 2i u ddsu
2 ds
the set := fx 2 @D : h (x) > 0g is nonempty and
multiply this by
Z
Thus
Then,
Hh juj2 ds = 0:
u = 0 on . Holmgren’s theorem gives a contradiction.
Is there any chance of recovering both
u1 ?
@D and from a single measurement of
Impedance Problem
Theorem. Assume that (x) > 1 for all x 2 @D . Then the total derivative of
the map F is injective, i.e., Fq0 h + F0 = 0 implies h = 0 and = 0 .
Impedance Problem
Theorem. Assume that (x) > 1 for all x 2 @D . Then the total derivative of
the map F is injective, i.e., Fq0 h + F0 = 0 implies h = 0 and = 0 .
Proof: Assume that Fr0 q + F0 = 0 . Then again using by Rellich’s lemma we
and therefore
have v = 0 in IRn n D
k2 (1
d
2
)h u +
du
h
+ ik(Hh
ds ds
)u = 0
on @D;
u and taking the real part we obtain
2
2
du
1
d
d
j
u
j
k2 (1 2 )h juj2 h +
ds
2 ds h ds = 0 on @D:
Assume that h is not identically zero. Then, without loss of generality, the set
:= fx 2 @D : h (x) > 0g is nonempty and therefore
2 )
Z (
du 2
2
2
k (1 )juj h ds = 0:
ds
Since (x) > 1 for all x 2 @D we obtain u = 0 on . By the boundary
condition for u this implies @u=@ = 0 on and employing Holmgren’s
theorem we arrive at the contradiction that u = 0 in IRn n D .
Multiplying this by
Impedance Problem
Theorem. Assume that (x) > 1 for all x 2 @D . Then the total derivative of
the map F is injective, i.e., Fq0 h + F0 = 0 implies h = 0 and = 0 .
Is the assumption
> 1 necessary?
Impedance Problem
Theorem. Assume that (x) > 1 for all x 2 @D . Then the total derivative of
the map F is injective, i.e., Fq0 h + F0 = 0 implies h = 0 and = 0 .
Is the assumption
> 1 necessary?
For a plane wave incident field we suspect not. However, for the cylindrical wave
ui (x) = Jn (kjxj)ein arg x we can show that the resulting total wave satsifies the
radiating condition, the Helmholtz equation and if 0 < 1 , for h = 1 the
critical equation is also satisfied
k2 (1 2 )h juj2
2
du h +
ds
1 d h djuj2 = 0
2 ds ds
on @D:
Thus studying this equation without regard to the incident wave is not enough.
Impedance Problem
0.75
y
0.0
-0.75-1.5
1.0
0.5
0.0
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r ( )
0.0
x
1.5
( )
......
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.. .....
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.. .......
..... .
.. ............................
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2
Transmission Problem
Transmission boundary conditions
@u @ = 0...............................................................
....
...
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....
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...
u = 0......
...
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D
.
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.. ... .... ....
..
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.. .... ....
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......
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................................................
D
On @D the total field
satisfies the conditions
u
+=
u
@u @u =
@ +
@ Transmission Problem
Transmission boundary conditions
@u @ = 0...............................................................
....
...
....
...
....
...
.
....
.
.
.
...
u = 0......
...
...
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...
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...
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..
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D
.
..
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.
..
. ..
.. ... .... ....
..
. ...
... ...
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.. .... ....
... .
...
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......
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.....
..
.......
D
On @D the total field
satisfies the conditions
u
+=
u
@u @u =
@ +
@ The map F : @D ! u1 is Fréchet differentiable and
field pattern u01 of the radiating solution to
4u0 + kD2 u0 = 0
in D
F 0 :h
4u0 + k2u0 = 0
is given by the far
in Rn =D
satisfying the Sommerfeld condition and the boundary conditions
[u0 ] = 1 @u
h
@
Div u and
0
@u
=
(
k2
@ kD2 )uh + (1 )Div(h r u)
r u are the surface divergence and gradients respectively.
Transmission Problem
Transmission boundary conditions
@u @ = 0...............................................................
....
...
....
...
....
...
.
....
.
.
.
...
u = 0......
...
...
.
...
.
.
...
.
.
.
..
.
D
.
..
.
.
.
.
.
........
....
.....
..
.......
..
. ..
.. ... .... ....
..
. ...
... ...
.
.
..
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.
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.
.. .... ....
... .
...
...
....
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..
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..
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..
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.....
...
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.....
.
.
.
.
...
...
......
...
.......
.
....
.
.
.
.
.
.
......
...
................................................
D
On @D the total field
satisfies the conditions
u
+=
u
@u @u =
@ +
@ The map F : @D ! u1 is Fréchet differentiable and
field pattern u01 of the radiating solution to
4u0 + kD2 u0 = 0
in D
F 0 :h
4u0 + k2u0 = 0
is given by the far
in Rn =D
satisfying the Sommerfeld condition and the boundary conditions
[u0 ] = 1 @u
h
@
0
@u
=
(
k2
@ Uniqueness questions here are largely open.
kD2 )uh + (1 )Div(h r u)
Iteration Schemes
Iteration Schemes
We attempt to solve the nonlinear equation
the form
F (x) = g
by an iterative method of
xn+1 = xn + An (F (xn ) g)
Some possible choices for
A are:
Iteration Schemes
Iteration Schemes
We attempt to solve the nonlinear equation
the form
F (x) = g
by an iterative method of
xn+1 = xn + An (F (xn ) g)
Some possible choices for
8
<
A=:
A are:
Iteration Schemes
Iteration Schemes
We attempt to solve the nonlinear equation
the form
F (x) = g
by an iterative method of
xn+1 = xn + An (F (xn ) g)
Some possible choices for
8
<
A=:
A are:
(F 0 )
(Landweber)
Iteration Schemes
Iteration Schemes
We attempt to solve the nonlinear equation
the form
F (x) = g
by an iterative method of
xn+1 = xn + An (F (xn ) g)
Some possible choices for
8
<
A=:
A are:
(F 0 )
(F 0 ) 1
(Landweber)
(Newton)
Iteration Schemes
Iteration Schemes
We attempt to solve the nonlinear equation
the form
F (x) = g
by an iterative method of
xn+1 = xn + An (F (xn ) g)
Some possible choices for
8
<
A are:
(F 0 )
A = : (F 0) 1
f (F 0 ; F 00 )
(Landweber)
(Newton)
(involving a second derivative)
Iteration Schemes
Iteration Schemes
We attempt to solve the nonlinear equation
the form
F (x) = g
by an iterative method of
xn+1 = xn + An (F (xn ) g)
Some possible choices for
8
<
A are:
(F 0 )
A = : (F 0) 1
f (F 0 ; F 00 )
(Landweber)
(Newton)
(involving a second derivative)
Let’s look at one possibility for the latter case.
Suppose we have the map F (x) = g and a starting approximation
~ be computed by a Newton step,
compute the step n + 1 , let h
F 0 [xn ] h~ = g F (xn ):
x0 .
To
Iteration Schemes
Iteration Schemes
We attempt to solve the nonlinear equation
the form
F (x) = g
by an iterative method of
xn+1 = xn + An (F (xn ) g)
Some possible choices for
8
<
A are:
(F 0 )
A = : (F 0) 1
f (F 0 ; F 00 )
(Landweber)
(Newton)
(involving a second derivative)
Let’s look at one possibility for the latter case.
Suppose we have the map F (x) = g and a starting approximation
~ be computed by a Newton step,
compute the step n + 1 , let h
x0 .
To
F 0 [xn ] h~ = g F (xn ):
A linear equation. The next iteration xn+1 = xn + h is defined from the 2nd
degree Taylor remainder by the solution of the linear equation
F 0 [xn ]h + 12 F 00 [xn ](h~ ; h) = g F (xn ):
Iteration Schemes
Iteration Schemes
We thus obtain a predictor-corrector scheme:
h~ = (F 0 [xn ]) 1 (g F (xn ))
1
h = F 0 [xn ] + 12 F 00 [xn ](h~ ; ) (g F (xn ))
xn+1 = xn + h
Predictor
Corrector
Update
Iteration Schemes
We thus obtain a predictor-corrector scheme:
h~ = (F 0 [xn ]) 1 (g F (xn ))
1
h = F 0 [xn ] + 12 F 00 [xn ](h~ ; ) (g F (xn ))
xn+1 = xn + h
In one dimension this gives Halley’s method.
Predictor
Corrector
Update
Iteration Schemes
We thus obtain a predictor-corrector scheme:
h~ = (F 0 [xn ]) 1 (g F (xn ))
1
h = F 0 [xn ] + 12 F 00 [xn ](h~ ; ) (g F (xn ))
xn+1 = xn + h
Predictor
Corrector
Update
In one dimension this gives Halley’s method.
denote a solution of (1). Assume F 0 [^
x] admits
a bounded inverse and F 0 and F 00 are uniformly bounded in U . Then
there exists Æ > 0 such that the above iteration with starting guess x0 2
B (^x; Æ ) = fx 2 X : kx^ xk < Æ g converges quadraticly to x^ . Additionally,
if the second derivative is Lipschitz continuous, i.e.
Theorem. Let
x^ 2 U
X
kF 00[x](h; h~ ) F 00[y](h; h~ )k Lkx yk khk kh~ k
for all x; y 2 U with h; h~ 2 X and a constant L > 0 , then
kxn+1 x^k c kxn x^k3
holds for n = 0; 1; 2; : : : with a constant c > 0 .
Halley Applied to Scattering
Back to Scattering
F is twice differentiable. F 0 is represented by the far field pattern
F 0 [@D] h = u01 of the solution u0 of the exterior Dirichlet problem,
on @D
4u0 + k2 u0 = 0 in IRn n D u0 = h @u
@
Theorem.
F 00 [@D](h1 ; h2 ) = u001 where u001 is the far field pattern of the radiating solution
1 (IRn n D
) of the exterior Dirichlet problem
u00 2 Hloc
4 u00 + k2 u00 = 0
@u02
00
u = h1;
n
in R
I n D̄;
@u01
h2;
+ h1; h2;
@
@u
h1; h2; H
@
@
@u
+ h1; ( r(h2; )) + h2; ( r(h1; ))
on @D
@
u is the solution of the scattering problem, u0j ( j = 1; 2 ) is the solution of the
boundary value problem with respect to the variation hj .
Halley Applied to Scattering
Back to Scattering
F is twice differentiable. F 0 is represented by the far field pattern
F 0 [@D] h = u01 of the solution u0 of the exterior Dirichlet problem,
on @D
4u0 + k2 u0 = 0 in IRn n D u0 = h @u
@
Theorem.
F 00 [@D](h1 ; h2 ) = u001 where u001 is the far field pattern of the radiating solution
1 (IRn n D
) of the exterior Dirichlet problem
u00 2 Hloc
4 u00 + k2 u00 = 0
@u02
00
u = h1;
n
in R
I n D̄;
@u01
h2;
+ h1; h2;
@
@u
h1; h2; H
@
@
@u
+ h1; ( r(h2; )) + h2; ( r(h1; ))
on @D
@
u is the solution of the scattering problem, u0j ( j = 1; 2 ) is the solution of the
boundary value problem with respect to the variation hj .
Note: The largest part of the computation of F , F 0 and F 00 is the common
inversion of the matrix representing 4v + k2 v .
Halley Applied to Scattering
....
..
...
..
%
...
The size of the obstacle is 1:2 1
units and the value of is one.
....
....
Exact Obstacle
of
% Direction
Incident Wave
Reconstruction of a sound soft
rectangular object from a single
incident field using accurate data
( 0:1% noise).
..
Halley Applied to Scattering
....
..
...
..
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The graphic below plots the relative
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Tichonov regularisation is used.
Halley Applied to Scattering
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Halley Applied to Scattering
Reconstructions from noise-free data
Newton method: 1
Newton method: 20
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Halley Applied to Scattering
Reconstructions from data with 10% noise
Newton method: 1
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Halley Applied to Scattering
Some Open Issues
~ = g F (x) where F 0 [x] is
We are solving the Newton equation F 0 [x] h
singular. Let us suppose that it turns out that F 0 is a positive semidefinite matrix
(this may come from the maximum principle in the underlying pde) and is also
symmetric (just to simplify the idea and avoid transposes). Then we might
regularise by
(F 0 [x] + R) h~ = g F (x)
where R is a positive (semi-definite) matrix and > 0 .
Open Issues
Some Open Issues
~ = g F (x) where F 0 [x] is
We are solving the Newton equation F 0 [x] h
singular. Let us suppose that it turns out that F 0 is a positive semidefinite matrix
(this may come from the maximum principle in the underlying pde) and is also
symmetric (just to simplify the idea and avoid transposes). Then we might
regularise by
(F 0 [x] + R) h~ = g F (x)
where R is a positive (semi-definite) matrix and > 0 .
What if (again from the underlying pde) it could be shown that the Hessian
F 00 [x]( ; h) was a positive semidefinite matrix? Would then the choice
R = 12 F 00 [x]( ; h) be a regularisation?
Even if it only partly (we cannot expect full) did so, it might allow a reduction
in the level of regularisation required by another scheme.
Open Issues
Some Open Issues
~ = g F (x) where F 0 [x] is
We are solving the Newton equation F 0 [x] h
singular. Let us suppose that it turns out that F 0 is a positive semidefinite matrix
(this may come from the maximum principle in the underlying pde) and is also
symmetric (just to simplify the idea and avoid transposes). Then we might
regularise by
(F 0 [x] + R) h~ = g F (x)
where R is a positive (semi-definite) matrix and > 0 .
What if (again from the underlying pde) it could be shown that the Hessian
F 00 [x]( ; h) was a positive semidefinite matrix? Would then the choice
R = 12 F 00 [x]( ; h) be a regularisation?
Even if it only partly (we cannot expect full) did so, it might allow a reduction
in the level of regularisation required by another scheme.
Can we effectively use a higher (than two) order MacLaurin expansion in an
attempt to better model the nonlinear map? For most problems we think the
answer is no. However, we were once convinced that a single derivative was
the sensible limit.
Open Issues
Is Halley’s method a Panacea?
We have painted a very rosy picture of the power of Halley’s method. To be fair
we should point out difficulties.
Open Issues
Is Halley’s method a Panacea?
We have painted a very rosy picture of the power of Halley’s method. To be fair
we should point out difficulties.
There are several important inverse problems where we are unable (or more
precisely, we don’t know how) to compute derivatives of the map F using the
construction from F itself.
Open Issues
Is Halley’s method a Panacea?
We have painted a very rosy picture of the power of Halley’s method. To be fair
we should point out difficulties.
There are several important inverse problems where we are unable (or more
precisely, we don’t know how) to compute derivatives of the map F using the
construction from F itself.
Regularisation issues are more complex. With any regularisation method there
is always the tricky problem of choosing the optimal value of the regularisation
parameter With the second degree method we have to select two regularisation
parameters, one each for the predictor and the corrector; experience shows that
best results are obtained when these are not the same.
Open Issues
Is Halley’s method a Panacea?
We have painted a very rosy picture of the power of Halley’s method. To be fair
we should point out difficulties.
There are several important inverse problems where we are unable (or more
precisely, we don’t know how) to compute derivatives of the map F using the
construction from F itself.
Regularisation issues are more complex. With any regularisation method there
is always the tricky problem of choosing the optimal value of the regularisation
parameter With the second degree method we have to select two regularisation
parameters, one each for the predictor and the corrector; experience shows that
best results are obtained when these are not the same.
While the scattering example shows that Halley’s method can provide superior
as well as faster reconstructions, there are problems (even those involving the
detection of obstacles) for which this seems not to be the case. This may be
due to difficulties in selecting an optimal level of regularisation or it may be
inherent in the problem.
References
Colton, D., and Kress, R: Inverse Acoustic and Electromagnetic Scattering Theory.
Springer-Verlag, Berlin Heidelberg New York 1992.
Colton, D., and Kirsch, A.: A simple method for solving inverse scattering problems in
the resonance region, Inverse Problems, 12, 383–393, (1996).
Colton, D., and Sleeman, B.D: Uniqueness theorems for the inverse problem of acoustic
scattering. IMA J. Appl. Math. 31, 253–259 (1983).
Hettlich, F.: Fréchet derivatives in inverse obstacle scattering. Inverse Problems 11,
371–382 (1995), Erratum: Inverse Problems 14, 209–210 (1998).
Hettlich, F. and Rundell, W., A Second Degree Method for Nonlinear Inverse Problems,
SIAM J. Numer. Anal, 37, No. 2, pp 587–620.
Kirsch, A: The domain derivative and two applications in inverse scattering theory. Inverse
Problems 9, 81–96 (1993).
Kress, R., and Rundell, W: Inverse obstacle scattering with modulus of the far field pattern
as data. In: Inverse Problems in Medical Imaging and Nondestructive Testing,
(Engl, Louis, Rundell eds.) Springer-Verlag, Wien, New York 1997.
Kress, R., and Rundell, W: A quasi-Newton method in inverse obstacle scattering, Inverse
Problems 10, 1145–1157 (1994).
Kress, R., and Rundell, W.: Inverse obstacle scattering using reduced data. SIAM J. Appl.
Math. 59, 442–454 (1999).
Kress, R., and Rundell, W: Inverse scattering for shape and impedance, Inverse Problems
16, (2000).