AIP Kress - Texas A&M University

Inverse Obstacle Scattering

Rainer Kress, Göttingen

AIP 2011, Pre-Conference WorkshopTexas A&M University, May 2011

Rainer Kress Inverse Obstacle Scattering

Scattering theory

Scattering theory is concerned with the effects that obstaclesand inhomogenities have on the propagation of waves

Restrict to time-harmonic waves and obstacles


Scattering theory

Scattering theory is concerned with the effects that obstaclesand inhomogenities have on the propagation of waves

Restrict to time-harmonic waves and obstacles


Incident plane wave


Scattered wave


Total wave


Direct and inverse Scattering

Direct scattering problemGiven: Incident field

and scattererFind: Scattered field

Inverse scattering problemGiven: Incident and

scattered fieldFind: Position and shape

of scatterer


Direct and inverse Scattering

Direct scattering problemGiven: Incident field

and scattererFind: Scattered field

Inverse scattering problemGiven: Incident and

scattered fieldFind: Position and shape

of scatterer


Outline

1 The Helmholtz equation2 The direct scattering problem

b) Uniquenessb) Existencec) Numerical solution

3 The inverse scattering problemb) Uniquenessb) Iterative solution methodsc) Decomposition methodsd) Sampling and probe methods


Time-harmonic waves

Wave equation: ∆U =1c2

∂2U∂t2

U = velocity potential, electric fieldc = speed of sound, speed of light

U(x , t) = <{

u(x)e−iωt}Helmholtz equation: ∆u + k2u = 0

ω = frequency, k = ω/c = wave number


Time-harmonic waves

Wave equation: ∆U =1c2

∂2U∂t2

U = velocity potential, electric fieldc = speed of sound, speed of light

U(x , t) = <{

u(x)e−iωt}Helmholtz equation: ∆u + k2u = 0

ω = frequency, k = ω/c = wave number


Helmholtz equation

Two main tools:Fundamental solution and Green’s integral theorems

Φ(x , y) :=1

4πeik |x−y |

|x − y |, x 6= y , in IR3

Φ(x , y) :=i4

H(1)0 (k |x − y |), x 6= y , in IR2

H(1)0 = J0 + iY0 Hankel function

(∆x + k2)Φ(x , y) = 0, x 6= y

(∆ + k2)Φ(·, y) = −δy


Helmholtz equation


Φ(x , y) :=1

4πeik |x−y |

|x − y |, x 6= y , in IR3

Φ(x , y) :=i4

H(1)0 (k |x − y |), x 6= y , in IR2


(∆x + k2)Φ(x , y) = 0, x 6= y

(∆ + k2)Φ(·, y) = −δy


Helmholtz equation


Φ(x , y) :=1

4πeik |x−y |

|x − y |, x 6= y , in IR3

Φ(x , y) :=i4

H(1)0 (k |x − y |), x 6= y , in IR2


(∆x + k2)Φ(x , y) = 0, x 6= y

(∆ + k2)Φ(·, y) = −δy


Fundamental solution in three dimensions

<eir

r = cos rr =eir

r = sin rr


Fundamental solution in two dimensions

<iH(1)0 (r) = −Y0(r) =iH(1)

0 (r) = J0(r)


Green’s integral formula

��ν

∂D

D

∂D ∈ C2

u ∈ C2(D) ∩ C1(D)

∆u + k2u = 0 in D

u(x) =

∫∂D

{∂u∂ν

(y) Φ(x , y)− u(y)∂Φ(x , y)

∂ν(y)

}ds(y), x ∈ D

Solutions to Helmholtz equation inherit properties offundamental solution.Solutions to Helmholtz equation are analytic

Theorem (Holmgren)

u|Γ = ∂νu|Γ = 0 for Γ ⊂ ∂D open ⇒ u = 0 in D



��ν

∂D

D

∂D ∈ C2

u ∈ C2(D) ∩ C1(D)

∆u + k2u = 0 in D

u(x) =

∫∂D

{∂u∂ν

(y) Φ(x , y)− u(y)∂Φ(x , y)

∂ν(y)

}ds(y), x ∈ D


Theorem (Holmgren)




��ν

∂D

D

∂D ∈ C2

u ∈ C2(D) ∩ C1(D)

∆u + k2u = 0 in D

u(x) =

∫∂D

{∂u∂ν

(y) Φ(x , y)− u(y)∂Φ(x , y)

∂ν(y)

}ds(y), x ∈ D


Theorem (Holmgren)



Laplace versus Helmholtz equation

Helmholtz equation shares many properties with Laplaceequation.

14π

eik |x−y |

|x − y |=

14π

1|x − y |

+ik4π

+ O(|x − y |)

i4

H(1)0 (k |x−y |) =

12π

ln1

|x − y |+

i4− 1

2πln

k2− C

2π+O(|x−y |)

However no maximum-minimum principle and no coercitivity




14π

eik |x−y |

|x − y |=

14π

1|x − y |

+ik4π

+ O(|x − y |)

i4

H(1)0 (k |x−y |) =

12π

ln1

|x − y |+

i4− 1

2πln

k2− C

2π+O(|x−y |)

However no maximum-minimum principle and no coercitivity∫∂D

u∂u∂ν

ds =

∫D

{| grad u|2 + u∆u

}dx




14π

eik |x−y |

|x − y |=

14π

1|x − y |

+ik4π

+ O(|x − y |)

i4

H(1)0 (k |x−y |) =

12π

ln1

|x − y |+

i4− 1

2πln

k2− C

2π+O(|x−y |)


u∂u∂ν

ds =

∫D

{| grad u|2 − k2|u|2

}dx




14π

eik |x−y |

|x − y |=

14π

1|x − y |

+ik4π

+ O(|x − y |)

i4

H(1)0 (k |x−y |) =

12π

ln1

|x − y |+

i4− 1

2πln

k2− C

2π+O(|x−y |)


u∂u∂ν

ds =

∫D

{| grad u|2 − k2|u|2

}dx

In particular, there exist Dirichlet and Neumann eigenvalues,that is, wave numbers k and solutions u 6= 0 to ∆u + k2u = 0in D with homogeneous boundary data u = 0 or ∂νu = on ∂D,respectively.


Sommerfeld radiation condition

Consider Helmholtz equation in IR3 \ D.Sommerfeld radiation condition requires

∂u∂r− iku = o

(1r

), r = |x | → ∞

uniformly for all directions. Radiating solutions

Equivalent to

u(x) =eik |x |

|x |

{u∞

(x|x |

)+ O

(1|x |

)}, |x | → ∞

u∞ is called the far field patternand defined on the unit sphere S2.

Lemma (Rellich)

u∞(x) = 0 for all x ∈ S2 (or all x ∈ Γ for an open Γ ⊂ S2)⇒ u(x) = 0 for all x ∈ IR3 \ D




∂u∂r− iku = o

(1r

), r = |x | → ∞

uniformly for all directions. Radiating solutionsEquivalent to

u(x) =eik |x |

|x |

{u∞

(x|x |

)+ O

(1|x |

)}, |x | → ∞


Lemma (Rellich)





∂u∂r− iku = o

(1r

), r = |x | → ∞

uniformly for all directions. Radiating solutionsEquivalent to

u(x) =eik |x |

|x |

{u∞

(x|x |

)+ O

(1|x |

)}, |x | → ∞


Lemma (Rellich)



Direct obstacle scattering

U(x , t) = <{

u(x)e−iωt}

XXXz -��

uius

D

ui : incident field, plane wave, ui(x ,d) = eik x ·d , |d | = 1

us : scattered field

u = ui + us : total field



��ν

XXXz -��

uius, u = ui + us

D

∆u + k2u = 0 in IR3 \ D

Bu = 0 on ∂D

∂us

∂r− ikus = o

(1r

), r = |x | → ∞

Boundary condition:

Bu = u sound-soft

Bu =∂u∂ν

+ ikλu impedance, λ ≥ 0



��ν

XXXz -��

uius, u = ui + us

D

∆u + k2u = 0 in IR3 \ D

Bu = 0 on ∂D

∂us

∂r− ikus = o

(1r

), r = |x | → ∞

Boundary condition:

Bu = u sound-soft

Bu =∂u∂ν




��ν

XXXz -��

uius, u = ui + us

D

∆u + k2u = 0 in IR3 \ D

Bu = 0 on ∂D

∂us

∂r− ikus = o

(1r

), r = |x | → ∞

Boundary condition:

Bu = u sound-soft

Bu =∂u∂ν




��ν

XXXz -��

uius, u = ui + us

D

∆u + k2u = 0 in IR3 \ D

Bu = 0 on ∂D

∂us

∂r− ikus = o

(1r

), r = |x | → ∞

Boundary condition:

Bu = u sound-soft

Bu =∂u∂ν



Existence and uniqueness

TheoremThe direct scattering problem for a sound-soft obstacle has aunique solution u ∈ H1

loc(IR3 \ D).

Uniqueness: DR :={

x ∈ IR3 \ D : |x | ≤ R}

∫|x |=R

us ∂us

∂νds−

∫∂D

us ∂us

∂νds =

∫DR

[| grad us|2 + us∆us

]dx

⇒ ik∫

S2|u∞|2 ds + O

(1R

)=

∫DR

[| grad us|2 − k2|us|2

]dx

⇒ u∞ = 0 on S2 ⇒ us = 0 in IR3 \ D




loc(IR3 \ D).

Uniqueness: DR :={

x ∈ IR3 \ D : |x | ≤ R}

∫|x |=R

us ∂us

∂νds−

∫∂D

us ∂us

∂νds =

∫DR


]dx

⇒ ik∫

S2|u∞|2 ds + O

(1R

)=

∫DR


]dx

⇒ u∞ = 0 on S2 ⇒ us = 0 in IR3 \ D




loc(IR3 \ D).

Uniqueness: DR :={

x ∈ IR3 \ D : |x | ≤ R}

∫|x |=R

us ∂us

∂νds−

∫∂D

us ∂us

∂νds =

∫DR


]dx

⇒ ik∫

S2|u∞|2 ds + O

(1R

)=

∫DR


]dx

⇒ u∞ = 0 on S2 ⇒ us = 0 in IR3 \ D




loc(IR3 \ D).

Uniqueness: DR :={

x ∈ IR3 \ D : |x | ≤ R}

∫|x |=R

us ∂us

∂νds−

∫∂D

us ∂us

∂νds =

∫DR


]dx

⇒ ik∫

S2|u∞|2 ds + O

(1R

)=

∫DR


]dx

⇒ u∞ = 0 on S2 ⇒ us = 0 in IR3 \ D



Combined double- and single-layer potential

us(x) =

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y), x ∈ IR3 \ D

solves sound-soft scattering problem if ϕ ∈ H1/2(∂D) satisfies

12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D

Let ϕ solve the homogeneous equation and set u := us

⇒ u+ = 0 on ∂D ⇒ u = 0 in IR3 \ D⇒ −u− = ϕ, −∂νu− = iϕ on ∂D

⇒ i∫∂D|ϕ|2ds =

∫∂D

u−∂u−∂ν

ds =

∫D

{| grad u|2 − k2|u|2

}dx

⇒ ϕ = 0 ⇒ Apply Riesz theory for compact operators




us(x) =

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y), x ∈ IR3 \ D


12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D


⇒ u+ = 0 on ∂D ⇒ u = 0 in IR3 \ D

⇒ −u− = ϕ, −∂νu− = iϕ on ∂D

⇒ i∫∂D|ϕ|2ds =

∫∂D

u−∂u−∂ν

ds =

∫D

{| grad u|2 − k2|u|2

}dx





us(x) =

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y), x ∈ IR3 \ D


12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D



⇒ i∫∂D|ϕ|2ds =

∫∂D

u−∂u−∂ν

ds =

∫D

{| grad u|2 − k2|u|2

}dx





us(x) =

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y), x ∈ IR3 \ D


12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D



⇒ i∫∂D|ϕ|2ds =

∫∂D

u−∂u−∂ν

ds =

∫D

{| grad u|2 − k2|u|2

}dx





us(x) =

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y), x ∈ IR3 \ D


12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D



⇒ i∫∂D|ϕ|2ds =

∫∂D

u−∂u−∂ν

ds =

∫D

{| grad u|2 − k2|u|2

}dx



Numerical solution

12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D


Numerical solution

12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D

Advantages of boundary integral equation method versus varia-tional approach

1. Radiation condition automatically satisfied2. Reduce dimension by one


Numerical solution

12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D

Solve numerically via collocation method plus quadrature.1. Approximate ϕ by low order spline functions2. Map boundary ∂D on circle or sphere and approximate ϕ bytrigonometric polynomials or spherical harmonics, respectively


Numerical solution

12ϕ(x)+

∫∂D

{∂Φ(x , y)

∂ν(y)− iΦ(x , y)

}ϕ(y) ds(y) = −ui(x), x ∈ ∂D

Solve numerically via collocation method plus quadrature.1. Approximate ϕ by low order spline functions2. Map boundary ∂D on circle or sphere and approximate ϕ bytrigonometric polynomials or spherical harmonics, respectively

K.E. Atkinson 1997The most efficient numerical methods for solving boundary in-tegral equations on smooth planar boundaries are those basedon trigonometric polynomial approximations. When calculationsusing piecewise polynomial approximations are compared withthose using trigonometric polynomial approximations, the latterare almost always the more efficient.


Inverse obstacle scattering

��ν

XXXz -��

uius, u = ui + us

D

∆u + k2u = 0 in IR3 \ D

Bu = 0 on ∂D

∂us

∂r− ikus = o

(1r

), r = |x | → ∞

us(x) =eik |x |

|x |

{u∞

(x|x |

)+ O

(1|x |

)}, |x | → ∞

Given: Far field u∞ for one incident plane waveFind: Shape and location of scatterer DNonlinear and ill-posedModel problem: nondestructive evaluation, radar, sonar etc.



��ν

XXXz -��

uius, u = ui + us

D

∆u + k2u = 0 in IR3 \ D

Bu = 0 on ∂D

∂us

∂r− ikus = o

(1r

), r = |x | → ∞

us(x) =eik |x |

|x |

{u∞

(x|x |

)+ O

(1|x |

)}, |x | → ∞

Given: Far field u∞ for one incident plane waveFind: Shape and location of scatterer D

Nonlinear and ill-posedModel problem: nondestructive evaluation, radar, sonar etc.



��ν

XXXz -��

uius, u = ui + us

D

∆u + k2u = 0 in IR3 \ D

Bu = 0 on ∂D

∂us

∂r− ikus = o

(1r

), r = |x | → ∞

us(x) =eik |x |

|x |

{u∞

(x|x |

)+ O

(1|x |

)}, |x | → ∞

Given: Far field u∞ for one incident plane waveFind: Shape and location of scatterer DNonlinear and ill-posedModel problem: nondestructive evaluation, radar, sonar etc.


Uniqueness, i.e., identifiability

Recall Rellich’s Lemma:u∞(x) = 0 for all directions x (or all x from a limited aperture)⇒ us(x) = 0 for all x ∈ IR3 \ D

Far field u∞ uniquely determines total field u = ui + us

DBu = 0

uius

XXXz -��

Question of uniqueness: ⇔Existence of additional closed surfaces with Bu = 0





DBu = 0

uius

XXXz -��






DBu = 0

uius

XXXz -��




Question of uniqueness: ⇔Existence of additional closed surfaces on which Bu = 0

DBu = 0

uius

XXXz -��

Bu = 0

No!!



Question of uniqueness: ⇔Existence of additional closed surfaces on which Bu = 0

DBu = 0

uius

XXXz -��

Bu = 0

Do not know???


Schiffer’s theorem

Theorem (Schiffer ≈ 1960)For a sound-soft scatterer, assume that

u∞,1(x ,d) = u∞,2(x ,d)

for all observation directions x and all incident directions d.Then D1 = D2.

D1 D2


Idea of Schiffer’s proof

D1 D2

D∗ = unbounded componentof IR3 \ (D1 ∪ D2)

us1(· ,d) = us

2(· ,d) in D∗

u1(x ,d) = eik x ·d + us1(x ,d)

In shaded domain: 4u1 + k2u1 = 0

On boundary: u1 = 0

{u1(· ,d) : d ∈ S2} linearly independent



D1 D2


us1(· ,d) = us

2(· ,d) in D∗

u1(x ,d) = eik x ·d + us1(x ,d)


On boundary: u1 = 0




D1 D2


us1(· ,d) = us

2(· ,d) in D∗

u1(x ,d) = eik x ·d + us1(x ,d)


On boundary: u1 = 0




D1 D2


us1(· ,d) = us

2(· ,d) in D∗

u1(x ,d) = eik x ·d + us1(x ,d)


On boundary: u1 = 0




D1 D2


us1(· ,d) = us

2(· ,d) in D∗

u1(x ,d) = eik x ·d + us1(x ,d)


On boundary: u1 = 0




D1 D2


us1(· ,d) = us

2(· ,d) in D∗

u1(x ,d) = eik x ·d + us1(x ,d)


On boundary: u1 = 0




Correct shaded domain: (IR3 \ D∗) \ D1

D1 D2

D∗

Incorrect shaded domain: D2 \ (D1 ∩ D2)

D1 D2

D∗

Lax and Philipps 1967This proof does not work for other boundary conditions!




D1 D2

D∗


D1 D2

D∗





D1 D2

D∗


D1 D2

D∗

Lax and Philipps 1967

This proof does not work for other boundary conditions!




D1 D2

D∗


D1 D2

D∗



Uniqueness for one incident wave

Quasi counter example for D = ball of radius R

ui(x) =sin k |x ||x |

us(x) = −sin kReikR

eik |x |

|x |

u(x) =sin k(|x | − R)

eikR |x |

u = 0 on spheres |x | = R +nπk, n = 0,1,2 . . .






eik |x |

|x |

u(x) =sin k(|x | − R)

eikR |x |







eik |x |

|x |

u(x) =sin k(|x | − R)

eikR |x |







eik |x |

|x |

u(x) =sin k(|x | − R)

eikR |x |




Theorem (Colton, Sleeman 1983)Under the a priori assumption k diamD < 2π a sound-softobstacle D is uniquely determined by the far field for oneincident plane wave.

Schiffer’s proof plus monotonicity of eigenvalues with respect tothe domain.

Gintides 2005 k diamD < 8.99 . . .



Theorem (Colton, Sleeman 1983)Under the a priori assumption k diamD < 2π a sound-softobstacle D is uniquely determined by the far field for oneincident plane wave.

Schiffer’s proof plus monotonicity of eigenvalues with respect tothe domain.

Gintides 2005 k diamD < 8.99 . . .



Theorem (Liu 1997)A sound-soft ball is uniquely determined by the far field patternfor one incident plane wave.

Theorem ( Cheng, Yamamoto 2003, Alessandrini, Rondi 2005,Elschner,Yamamoto 2006, Liu, Zou 2006)A polyhydral sound-soft obstacle is uniquely determined by thefar field pattern for one incident plane wave.

D��

��

��

PPPPP��

XXXXX

��D0

��

��

XXXXXX

��

��u = 0

General case: Use reflection principle to find a path to infinity





D��

��

��

PPPPP��

XXXXX

��

D0

��

��

XXXXXX

��

��u = 0






D��

��

��

PPPPP��

XXXXX

��D0

��

��

XXXXXX

��

��u = 0






D��

��

��

PPPPP��

XXXXX

��D0

��

��

XXXXXX

��

��u = 0






D��

��

��

PPPPP��

XXXXX

��D0

��

��

XXXXXX

��

��u = 0



Uniqueness of obstacle plus boundary condition

Theorem (Kirsch, K. 1992)Assume that

u∞,1(x ,d) = u∞,2(x ,d)

for all observation directions x and all incident directions d.Then D1 = D2 and B1 = B2.

D1

B1

D2B2


Idea of proof

z∗

x∗

D

w i(x , z) =eik |x−z|

|x − z|= incident field, point source

ws(x , z) = scattered field, w∞(x , z) = far field

reciprocity: u∞(x ,d) = u∞(−d ,−x), ws(x , z) = ws(z, x)

mixed reciprocity: us(z,d) = w∞(−d , z)


Idea of proof

z∗

x∗

D







Idea of proof

z∗

x∗

D







Idea of proof


D1D2

D∗

x??z

B1

B2

u∞,1(x ,d) = u∞,2(x ,d) for |x | = |d | = 1

us1(z,d) = us

2(z,d) for z ∈ D∗, |d | = 1

w∞,1(d , z) = w∞,2(d , z) for z ∈ D∗, |d | = 1

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗


Idea of proof


D1D2

D∗

x??z

B1

B2

u∞,1(x ,d) = u∞,2(x ,d) for |x | = |d | = 1

us1(z,d) = us

2(z,d) for z ∈ D∗, |d | = 1

w∞,1(d , z) = w∞,2(d , z) for z ∈ D∗, |d | = 1

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗


Idea of proof


D1D2

D∗

x??z

B1

B2

u∞,1(x ,d) = u∞,2(x ,d) for |x | = |d | = 1

us1(z,d) = us

2(z,d) for z ∈ D∗, |d | = 1

w∞,1(d , z) = w∞,2(d , z) for z ∈ D∗, |d | = 1

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗


Idea of proof


D1D2

D∗

x??z

B1

B2

u∞,1(x ,d) = u∞,2(x ,d) for |x | = |d | = 1

us1(z,d) = us

2(z,d) for z ∈ D∗, |d | = 1

w∞,1(d , z) = w∞,2(d , z) for z ∈ D∗, |d | = 1

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗


Idea of proof


D1D2

D∗

x??z

B1

B2

u∞,1(x ,d) = u∞,2(x ,d) for |x | = |d | = 1

us1(z,d) = us

2(z,d) for z ∈ D∗, |d | = 1

w∞,1(d , z) = w∞,2(d , z) for z ∈ D∗, |d | = 1

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗


Idea of proof

D1D2

D∗

x∗??z

B1

B2

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗

limz→x∗

B1ws1 (x∗, z) =∞, lim

z→x∗B1ws

2 (x∗, z) = finite

⇒ D1 = D2

Holmgren’s theorem yields B1 = B2Use of mixed reciprocity: Potthast 1999Has been extended to a variety of other scattering problems.


Idea of proof

D1D2

D∗

x∗??z

B1

B2

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗

limz→x∗

B1ws1 (x∗, z) =∞,

limz→x∗

B1ws2 (x∗, z) = finite

⇒ D1 = D2



Idea of proof

D1D2

D∗

x∗??z

B1

B2

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗

limz→x∗

B1ws1 (x∗, z) =∞, lim

z→x∗B1ws


⇒ D1 = D2



Idea of proof

D1D2

D∗

x∗??z

B1

B2

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗

limz→x∗

B1ws1 (x∗, z) =∞, lim

z→x∗B1ws


⇒ D1 = D2



Idea of proof

D1D2

D∗

x∗??z

B1

B2

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗

limz→x∗

B1ws1 (x∗, z) =∞, lim

z→x∗B1ws


⇒ D1 = D2

Holmgren’s theorem yields B1 = B2

Use of mixed reciprocity: Potthast 1999Has been extended to a variety of other scattering problems.


Idea of proof

D1D2

D∗

x∗??z

B1

B2

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗

limz→x∗

B1ws1 (x∗, z) =∞, lim

z→x∗B1ws


⇒ D1 = D2

Holmgren’s theorem yields B1 = B2Use of mixed reciprocity: Potthast 1999

Has been extended to a variety of other scattering problems.


Idea of proof

D1D2

D∗

x∗??z

B1

B2

ws1 (x , z) = ws

2 (x , z) for x , z ∈ D∗

limz→x∗

B1ws1 (x∗, z) =∞, lim

z→x∗B1ws


⇒ D1 = D2



Reconstruction methods

Reconstruction methods connected to the uniqueness proof ofKirsch, K.:

Singular source method of Potthast 2001Needle method of Ikehata 2000Use of ws(x , z)→∞, if x , z ∈ IR3 \ D → ∂D

z∗x∗

D

Sampling method of Colton, Kirsch 1996Use of ws(x , z)→∞ if x , z ∈ D → ∂D


Reconstruction methods

Reconstruction methods connected to the uniqueness proof ofKirsch, K.:

Singular source method of Potthast 2001Needle method of Ikehata 2000Use of ws(x , z)→∞, if x , z ∈ IR3 \ D → ∂D

z∗x∗

D

Sampling method of Colton, Kirsch 1996Use of ws(x , z)→∞ if x , z ∈ D → ∂D



��ν

XXXz -��

uius, u = ui + us

D

∆u + k2u = 0 in IR3 \ D

u = 0 on ∂D

∂us

∂r− ikus = o

(1r

), r = |x | → ∞

us(x) =eik |x |

|x |

{u∞

(x|x |

)+ O

(1|x |

)}, |x | → ∞

Given: Far field u∞ for one incident plane waveFind: Shape and location of scatterer DNonlinear and ill-posed


Example for nonlinearity

A priori information: D is ball of radius R centered at origin

Incident field: ui(x) =sin k |x ||x |

Scattered field: us(x) = −sin kReikR

eik |x |

|x |

Far-field pattern: u∞(x) = −sin kReikR

Nonlinear equation for the unknown radius R






eik |x |

|x |








eik |x |

|x |




Example for ill-posedness

Perturbed data: u∞(x) = −sin kReikR + εYn(x)

Total field:

u(x) =sin k(|x | − R)

eikR |x |+ ε k in+1 h(1)

n (k |x |)Yn

(x|x |

)

u(x) = ε k in+1 h(1)n (kR)Yn

(x|x |

), |x | = R

|u(x)| ≈ ε k(

2nekR

)n

Yn

(x|x |

), |x | = R

Small errors in data u∞ can cause large errors in solution,or solution may not exist anymore.




Total field:

u(x) =sin k(|x | − R)


n (k |x |)Yn

(x|x |

)


(x|x |

), |x | = R

|u(x)| ≈ ε k(

2nekR

)n

Yn

(x|x |

), |x | = R





Total field:

u(x) =sin k(|x | − R)


n (k |x |)Yn

(x|x |

)


(x|x |

), |x | = R

|u(x)| ≈ ε k(

2nekR

)n

Yn

(x|x |

), |x | = R





Total field:

u(x) =sin k(|x | − R)


n (k |x |)Yn

(x|x |

)


(x|x |

), |x | = R

|u(x)| ≈ ε k(

2nekR

)n

Yn

(x|x |

), |x | = R



Existence???

��ν

XXXz -��

uius,u∞, u = ui + us

D

Wrong question to ask: Would need to characterize far-fieldpatterns for which the corresponding total field vanishes on aclosed surface.

Main Task: Assuming correct data or perturbed correct data,design methods for a stable approximate solution


Iterative methods versus qualitative methods

Iterative methods: Reformulate inverse problem as nonlinearill-posed operator equation.Solve by iteration methods such as regularized Newtonmethods, Landweber iterations or conjugate gradient methods

Qualitative methods: Develop criterium in terms of behaviourof certain ill-posed linear integral equations that decide onwhether a point lies inside or outside the scatterer.Linear sampling, factorization, probe methods, etc


Iterative methods versus qualitative methods

Iterative methods: Reformulate inverse problem as nonlinearill-posed operator equation.Solve by iteration methods such as regularized Newtonmethods, Landweber iterations or conjugate gradient methods

Qualitative methods: Develop criterium in terms of behaviourof certain ill-posed linear integral equations that decide onwhether a point lies inside or outside the scatterer.Linear sampling, factorization, probe methods, etc


Iterative methods for boundary to far field map

XXXz -��

uiu∞

D

Interpret inverse problem as operator equation F (∂D) = u∞

For simplicity: ∂D = {p(x) : x ∈ S2}, p : S2 → IR3

Then F : C2(S2, IR3)→ L2(S2, IC), F : p → u∞

Inverse problem: Solve F (p) = u∞

Linearize: F (p + q) = F (p) + F ′(p; q) + o(q)

and, given an approximation p, solveF (p) + F ′(p; q) = u∞

for q to update p into p + q.

Regularization required


Iterative methods for boundary to far field map

XXXz -��

uiu∞

D

Interpret inverse problem as operator equation F (∂D) = u∞

For simplicity: ∂D = {p(x) : x ∈ S2}, p : S2 → IR3

Then F : C2(S2, IR3)→ L2(S2, IC), F : p → u∞

Inverse problem: Solve F (p) = u∞

Linearize: F (p + q) = F (p) + F ′(p; q) + o(q)

and, given an approximation p, solveF (p) + F ′(p; q) = u∞


Regularization requiredRainer Kress Inverse Obstacle Scattering

Fréchet derivative

TheoremFréchet derivative is given by

F ′(p; ·) : q → vq,∞

where vq,∞ is the far field ofradiating solution to

∆vq + k2vq = 0 in IR3 \ Dp

vq = −ν · q ∂u∂ν

on ∂Dp

��:��ν

qp

∂Dp

p + q

∂Dp+q

F (p + q)

= F (p) + F ′(p; q) + o(q)

Proof by hand waving

0 = u(∂Dp+q)|∂Dq+q ≈ u(∂Dp)|∂Dp︸︷︷︸= 0

+ [u′(∂D)q]|∂Dp︸︷︷︸= vq

+ grad u(∂Dp)|∂Dp︸︷︷︸= ∂ν u ν

·q


Fréchet derivative

TheoremFréchet derivative is given by

F ′(p; ·) : q → vq,∞

where vq,∞ is the far field ofradiating solution to

∆vq + k2vq = 0 in IR3 \ Dp

vq = −ν · q ∂u∂ν

on ∂Dp

��:��ν

qp

∂Dp

p + q

∂Dp+q

F (p + q)

= F (p) + F ′(p; q) + o(q)

Roger 1981, hand wavingKirsch, K. 1991, Hilbert space methods, domain derivativePotthast 1992, boundary integral equations


Newton iterations for boundary to far field map

Numerical examples:Hohage, Hettlich, Kirsch, K., Murch et al,Roger, Rundell, Tobocman, ... 1991-..., in 2DFarhat et al 2002, in 3DHarbrecht, Hohage 2005, in 3D

Pros:

Conceptually simpleVery good reconstructions

Contras:

Need efficient forward solver and good a priori informationConvergence not completely settled


Newton iterations for boundary to far field map

Numerical examples:Hohage, Hettlich, Kirsch, K., Murch et al,Roger, Rundell, Tobocman, ... 1991-..., in 2DFarhat et al 2002, in 3DHarbrecht, Hohage 2005, in 3D

Pros:

Conceptually simpleVery good reconstructions

Contras:

Need efficient forward solver and good a priori informationConvergence not completely settled


Huygens’ principle

us(x) = − 14π

∫∂D

eik |x−y |

|x − y |∂u∂ν

(y) ds(y), x ∈ IR3 \ D

Data equation

u∞(x) = − 14π

∫∂D

e−ik x ·y ∂u∂ν

(y) ds(y), x ∈ S2

Field equation

ui(x) =1

4π

∫∂D

eik |x−y |

|x − y |∂u∂ν

(y) ds(y), x ∈ ∂D

Two integral equations for two unknowns


Huygens’ principle

us(x) = − 14π

∫∂D

eik |x−y |

|x − y |∂u∂ν

(y) ds(y), x ∈ IR3 \ D

Data equation

u∞(x) = − 14π

∫∂D

e−ik x ·y ∂u∂ν

(y) ds(y), x ∈ S2

Field equation

ui(x) =1

4π

∫∂D

eik |x−y |

|x − y |∂u∂ν

(y) ds(y), x ∈ ∂D

Two integral equations for two unknowns


Parameterized equations

Recall ∂D = {p(x) : x ∈ S2}

Define A,A∞ : C2(S2, IR3)× L2(S2, IC)→ L2(S2, IC)by

A(p, ψ)(x) :=1

4π

∫S2

eik |p(x)−p(y)|

|p(x)− p(y)|ψ(y) ds(y), x ∈ S2

and

A∞(p, ψ)(x) :=1

4π

∫S2

e−ik x ·p(y)ψ(y) ds(y), x ∈ S2

Then setting ψ := J(p)∂u∂ν◦ p

Data equation A∞(p, ψ) = u∞Field equation A(p, ψ) = −ui ◦ p



Recall ∂D = {p(x) : x ∈ S2}Define A,A∞ : C2(S2, IR3)× L2(S2, IC)→ L2(S2, IC)by

A(p, ψ)(x) :=1

4π

∫S2

eik |p(x)−p(y)|

|p(x)− p(y)|ψ(y) ds(y), x ∈ S2

and

A∞(p, ψ)(x) :=1

4π

∫S2






Recall ∂D = {p(x) : x ∈ S2}Define A,A∞ : C2(S2, IR3)× L2(S2, IC)→ L2(S2, IC)by

A(p, ψ)(x) :=1

4π

∫S2

eik |p(x)−p(y)|

|p(x)− p(y)|ψ(y) ds(y), x ∈ S2

and

A∞(p, ψ)(x) :=1

4π

∫S2





Derivatives of operators

A(p, ψ)(x) :=1

4π

∫S2

eik |p(x)−p(y)|

|p(x)− p(y)|ψ(y) ds(y)

A′(p, ψ; q)(x) =1

4π

∫S2

gradeik |p(x)−p(y)|

|p(x)− p(y)|·[q(x)−q(y)]ψ(y) ds(y)

andA∞(p, ψ)(x) :=

14π

∫S2

e−ik x ·p(y)ψ(y) ds(y)

A′∞(p, ψ; q)(x) = − ik4π

∫S2

e−ik x ·p(y) x · q(y)ψ(y) ds(y)

Linearizations in the sense

‖A(p + q, ψ)− A(p, ψ)− A′(p, ψ; q)‖L2(S2) = o(‖q‖C2(S2))


Linearization of the data equation

If k2 is not a Dirichlet eigenvalue of the negative Laplacian in D,then A(p, ·) : H−1/2(S2)→ H1/2(S2) is a homeomorphism.

Given an approximation for p solve the field equation

A(p, ψ) = −ui ◦ p

for the density ψ, that is, ψ = −[A(p, ·)]−1(ui ◦ p)Keeping ψ fixed, linearize the data equation

A∞(p, ψ) = u∞

to obtainA′∞(p, ψ; q) = u∞ − A∞(p, ψ)


Johansson, Sleeman 2007





A(p, ψ) = −ui ◦ p


A∞(p, ψ) = u∞








A(p, ψ) = −ui ◦ p


A∞(p, ψ) = u∞






Recall boundary to far field operator F : p → u∞.

Can represent

F (p) = −A∞(p, [A(p, ·)]−1(ui ◦ p))

with derivative

F ′(p; q) = −A′∞(p, [A(p, ·)]−1(ui ◦ p); q)

+A∞(p, [A(p, ·)]−1A′(p, [A(p, ·)]−1(ui ◦ p); q))

−A∞(p, [A(p, ·)]−1((grad ui) ◦ p) · q)

Linearization of the data equation corresponds to Newtoniteration for

F (p) = u∞

with the derivate of F approximated through the first term



Recall boundary to far field operator F : p → u∞.Can represent

F (p) = −A∞(p, [A(p, ·)]−1(ui ◦ p))

with derivative

F ′(p; q) = −A′∞(p, [A(p, ·)]−1(ui ◦ p); q)

+A∞(p, [A(p, ·)]−1A′(p, [A(p, ·)]−1(ui ◦ p); q))

−A∞(p, [A(p, ·)]−1((grad ui) ◦ p) · q)


F (p) = u∞

with the derivate of F approximated through the first term



Recall boundary to far field operator F : p → u∞.Can represent

F (p) = −A∞(p, [A(p, ·)]−1(ui ◦ p))

with derivative

F ′(p; q) = −A′∞(p, [A(p, ·)]−1(ui ◦ p); q)

+A∞(p, [A(p, ·)]−1A′(p, [A(p, ·)]−1(ui ◦ p); q))

−A∞(p, [A(p, ·)]−1((grad ui) ◦ p) · q)


F (p) = u∞

with the derivate of F approximated through the first termRainer Kress Inverse Obstacle Scattering

Simultaneous linearization of both equations

Given approximations p and ψ linearize both equations toobtain

A′∞(p, ψ; q) + A∞(p, χ) = −A∞(p, ψ) + u∞

and

A′(p, ψ; q) + ((grad ui) ◦ p) · q + A(p, χ) = −A(p, ψ)− ui ◦ p

to be solved for q and χ to update p and ψ into p + q and ψ + χ

K., Rundell 2005 Laplace equationIvanyshyn, K. 2006,. . . Helmholtz equation



Given approximations p and ψ linearize both equations toobtain

A′∞(p, ψ; q) + A∞(p, χ) = −A∞(p, ψ) + u∞

and


to be solved for q and χ to update p and ψ into p + q and ψ + χ

K., Rundell 2005 Laplace equationIvanyshyn, K. 2006,. . . Helmholtz equation



Theorem (Ivanyshyn, Kress 2008)

Assume that k2 is not a Dirichlet eigenvalue of the negativeLaplacian in D and set ψ := −[A(p, ·)]−1(ui ◦ p).

Provided q satisfies linearized boundary to far field equation

F ′(p; q) = u∞ − F (p)

then q and χ := −[A(p, ·)]−1(A′(p, ψ; q) + ((grad ui) ◦ p) · q)satisfy linearized integral equations

A′∞(p, ψ; q) + A∞(p, χ) = −A∞(p, ψ) + u∞

and


and vice versa.



Theorem (Ivanyshyn, Kress 2008)

Assume that k2 is not a Dirichlet eigenvalue of the negativeLaplacian in D and set ψ := −[A(p, ·)]−1(ui ◦ p).Provided q satisfies linearized boundary to far field equation

F ′(p; q) = u∞ − F (p)

then q and χ := −[A(p, ·)]−1(A′(p, ψ; q) + ((grad ui) ◦ p) · q)satisfy linearized integral equations

A′∞(p, ψ; q) + A∞(p, χ) = −A∞(p, ψ) + u∞

and


and vice versa.


Linearization of the field equation

If k2 is not a Dirichlet eigenvalue of the negative Laplacian in D,then A∞(p, ·) : L2(S2)→ L2(S2) is injective and has denserange.

Given an approximation for p, find a regularized solution ψ ofthe data equation

A∞(p, ψ) = u∞.

Keeping ψ fixed, linearize the field equation

A(p, ψ) = −ui ◦ p

to obtain

A′(p, ψ; q) + ((grad ui) ◦ p) · q = −A(p, ψ)− ui ◦ p


Slight modification leads to the hybrid decomposition method ofK., Serranho 2003,. . .





A∞(p, ψ) = u∞.


A(p, ψ) = −ui ◦ p

to obtain








A∞(p, ψ) = u∞.


A(p, ψ) = −ui ◦ p

to obtain





Decomposition methods

Decompose the inverse problem into a linear ill-posed partand a nonlinear part.

Step 1.Reconstruct scattered field us from the given far field patternu∞.

Step 2.Find unknown boundary ∂D as location where the boundarycondition ui + us = 0 for the total field is satisfied.

For example in first step represent

us(x) =1

4π

∫Γ

eik |x−y |

|x − y |ϕ(y) ds(y)

for some surface Γ ⊂ D.Kirsch, K. 1986



Decompose the inverse problem into a linear ill-posed partand a nonlinear part.

Step 1.Reconstruct scattered field us from the given far field patternu∞.

Step 2.Find unknown boundary ∂D as location where the boundarycondition ui + us = 0 for the total field is satisfied.

For example in first step represent

us(x) =1

4π

∫Γ

eik |x−y |

|x − y |ϕ(y) ds(y)

for some surface Γ ⊂ D.Kirsch, K. 1986




A∞(p, ψ) = u∞and define the single-layer potential

u(x) = ui(x) +1

4π

∫S2

eik |x−p(y)|

|x − p(y)|ψ(y) ds(y).

Find an update p + q by linearizing the boundary condition

u ◦ (p + q) = 0,

that is, by solving the linear equation

u ◦ p + ((grad u) ◦ p) · q = 0

for q. Equivalent to linearizing the field equationA(p, ψ) = −ui ◦ p only with respect to the evaluation point.

K., Serranho 2003,. . .





u(x) = ui(x) +1

4π

∫S2

eik |x−p(y)|

|x − p(y)|ψ(y) ds(y).


u ◦ (p + q) = 0,


u ◦ p + ((grad u) ◦ p) · q = 0

for q.

Equivalent to linearizing the field equationA(p, ψ) = −ui ◦ p only with respect to the evaluation point.

K., Serranho 2003,. . .





u(x) = ui(x) +1

4π

∫S2

eik |x−p(y)|

|x − p(y)|ψ(y) ds(y).


u ◦ (p + q) = 0,


u ◦ p + ((grad u) ◦ p) · q = 0

for q. Equivalent to linearizing the field equationA(p, ψ) = −ui ◦ p only with respect to the evaluation point.

K., Serranho 2003,. . .Rainer Kress Inverse Obstacle Scattering

Implementation: Hybrid method

1. Given an approximation p, solve ill-posed integral equation

A∞(p, ψ) = u∞,

that is, ∫S2

e−ik x ·p(y)ψ(y) ds(y) = u∞(x), x ∈ S2

Smooth integrand, for example, Gauss-trapezoidal ruleSevere ill-posedness requires regularization,for example, via Tikhonov regularization



1. Given an approximation p, solve ill-posed integral equation

A∞(p, ψ) = u∞,

that is, ∫S2

e−ik x ·p(y)ψ(y) ds(y) = u∞(x), x ∈ S2

Smooth integrand, for example, Gauss-trapezoidal ruleSevere ill-posedness requires regularization,for example, via Tikhonov regularization



2. Approximate

u(x) = ui(x) +1

4π

∫S2

eik |x−p(y)|

|x − p(y)|ψ(y) ds(y)

and evaluate boundary values and normal derivatives on∂D = {p(x) : x ∈ S2} by jump relations.Use spectral quadrature rules ofWienert 1990, Ganesh, Graham, Sloan 2002–2004

Find update q by solving

u ◦ p + ((grad u) ◦ p) · q = 0

Insert q(x) = r(x)x with

r =M∑

m=0

m∑n=−m

amnYmn,

collocate at L points on ∂D and solve L× (M + 1)2 linearsystem for amn by penalized least squares.



2. Approximate

u(x) = ui(x) +1

4π

∫S2

eik |x−p(y)|

|x − p(y)|ψ(y) ds(y)

and evaluate boundary values and normal derivatives on∂D = {p(x) : x ∈ S2} by jump relations.Use spectral quadrature rules ofWienert 1990, Ganesh, Graham, Sloan 2002–2004Find update q by solving

u ◦ p + ((grad u) ◦ p) · q = 0


r =M∑

m=0

m∑n=−m

amnYmn,

collocate at L points on ∂D and solve L× (M + 1)2 linearsystem for amn by penalized least squares.


Implementation: Linearize data equation

1. Given an approximation p, solve well-posed integral equation

A(p, ψ) = u∞,

that is,

14π

∫S2

eik |p(x)−p(y)|

|p(x)− p(y)|ψ(y) ds(y) = −ui(p(x)), x ∈ S2

Use spectral quadrature rules ofWienert 1990, Ganesh, Graham, Sloan 2002–2004and collocation.



1. Given an approximation p, solve well-posed integral equation

A(p, ψ) = u∞,

that is,

14π

∫S2

eik |p(x)−p(y)|

|p(x)− p(y)|ψ(y) ds(y) = −ui(p(x)), x ∈ S2

Use spectral quadrature rules ofWienert 1990, Ganesh, Graham, Sloan 2002–2004and collocation.



2. Find update q by solving ill-posed linearized data equation,that is,

A′∞(p, ψ; q) = u∞ − A∞(p, ψ)

by Tikhonov regularization (smooth integrands again).


r =M∑

m=0

m∑n=−m

amnYmn,

collocate at L points on ∂D and solve L× (M + 1)2 linearsystem for amn by Tikhonov regularization.



2. Find update q by solving ill-posed linearized data equation,that is,

A′∞(p, ψ; q) = u∞ − A∞(p, ψ)

by Tikhonov regularization (smooth integrands again).


r =M∑

m=0

m∑n=−m

amnYmn,

collocate at L points on ∂D and solve L× (M + 1)2 linearsystem for amn by Tikhonov regularization.




Sampling and probe methods

Develop criterium in terms of behaviour of certain ill-posedlinear integral equations that decide on whether a point z liesinside or outside the scatterer D.

D

Evaluate the criterium numerically for a grid of points

Need full data, i.e, u∞(x ,d) for all x ,d ∈ S2



Develop criterium in terms of behaviour of certain ill-posedlinear integral equations that decide on whether a point z liesinside or outside the scatterer D.

D

Evaluate the criterium numerically for a grid of points

Need full data, i.e, u∞(x ,d) for all x ,d ∈ S2


Linear sampling method

Define far field operator F : L2(S2)→ L2(S2) by

Fg(x) :=

∫S2

u∞(x ,d)g(d) ds(d), x ∈ S2

Recall point source w i(x , z) =eik |x−z|

|x − z|= incident field

ws(x , z) = scattered field, w∞(x , z) = far fieldw i∞(x , z) = e−ik z·x = far field of incident field

Consider ill-posed linear integral equation

Fg(· , z) = w i∞(· , z)

for arbitrary source locations z




Fg(x) :=

∫S2

u∞(x ,d)g(d) ds(d), x ∈ S2





Fg(· , z) = w i∞(· , z)





Fg(x) :=

∫S2

u∞(x ,d)g(d) ds(d), x ∈ S2





Fg(· , z) = w i∞(· , z)




Let z ∈ D and g be a solution of∫S2

u∞(x ,d)g(d , z) ds(d) = e−ik z·x , x ∈ S2

⇒∫

S2us(x ,d)g(d , z) ds(d) = w i(x , z), x ∈ IR3 \ D

⇒∫

S2eik x ·dg(d , z) ds(d)︸︷︷︸

Hg(· , z)(x)

= −eik |x−z|

|x − z|, x ∈ ∂D

⇒ ‖g(· , z)‖L2(S2) →∞, ‖Hg(· , z)‖L2(∂D) →∞, z → ∂D

Bad news: Integral equation, in general, not solvable

Good news: Method works welland can be justified by approximation arguments





⇒∫


⇒∫


Hg(· , z)(x)

= −eik |x−z|

|x − z|, x ∈ ∂D

⇒ ‖g(· , z)‖L2(S2) →∞, ‖Hg(· , z)‖L2(∂D) →∞, z → ∂D







⇒∫


⇒∫


Hg(· , z)(x)

= −eik |x−z|

|x − z|, x ∈ ∂D

⇒ ‖g(· , z)‖L2(S2) →∞, ‖Hg(· , z)‖L2(∂D) →∞, z → ∂D







⇒∫


⇒∫


Hg(· , z)(x)

= −eik |x−z|

|x − z|, x ∈ ∂D

⇒ ‖g(· , z)‖L2(S2) →∞, ‖Hg(· , z)‖L2(∂D) →∞, z → ∂D







⇒∫


⇒∫


Hg(· , z)(x)

= −eik |x−z|

|x − z|, x ∈ ∂D

⇒ ‖g(· , z)‖L2(S2) →∞, ‖Hg(· , z)‖L2(∂D) →∞, z → ∂D







⇒∫


⇒∫


Hg(· , z)(x)

= −eik |x−z|

|x − z|, x ∈ ∂D

⇒ ‖g(· , z)‖L2(S2) →∞, ‖Hg(· , z)‖L2(∂D) →∞, z → ∂D





Theorem (Colton, Kirsch 1996)

For every ε > 0 and z ∈ D there exists g(· , z) ∈ L2(S2) suchthat

‖Fg(· , z)− w i∞(· , z)‖L2(S2) ≤ ε

and

‖g(· , z)‖L2(S2) →∞, ‖Hg(· , z)‖L2(∂D) →∞, z → ∂D

Arens 2003 Why linear sampling works?


Factorization method

Recall far field operator F : L2(S2)→ L2(S2) with

Fg(x) :=

∫S2

u∞(x ,d)g(d) ds(d), x ∈ S2

and consider ill-posed linear integral equation

(F ∗F )1/4g(· , z) = w i∞(· , z)

Theorem (Kirsch 1998)

The (F ∗F )1/4 equation is solvable if and only if z ∈ D.


Factorization method

Recall far field operator F : L2(S2)→ L2(S2) with

Fg(x) :=

∫S2

u∞(x ,d)g(d) ds(d), x ∈ S2

and consider ill-posed linear integral equation

(F ∗F )1/4g(· , z) = w i∞(· , z)

Theorem (Kirsch 1998)

The (F ∗F )1/4 equation is solvable if and only if z ∈ D.


Singular source method

Exploit the uniqueness proof and characterize the boundary ofthe scatterer D by the points z where

ws(z, z)

becomes large.Potthast 2001, pointwiseIkehata 2000, in energy norm

∂Dz∗

Λ

Approximate incident point source fieldw i(· , z) by linear combination of plane waves.



Exploit the uniqueness proof and characterize the boundary ofthe scatterer D by the points z where

ws(z, z)

becomes large.Potthast 2001, pointwiseIkehata 2000, in energy norm

∂Dz∗

Λ

Approximate incident point source fieldw i(· , z) by linear combination of plane waves.



Let z ∈ IR3 \ D and g be a solution of∫S2

eik x ·dg(d , z) ds(d) =eik |x−z|

|x − z|, x ∈ Λ

⇒∫

S2us(x ,d)g(d , z) ds(d) = ws(x , z), x ∈ IR3 \ D

⇒∫

S2u∞(x ,d)g(d , z) ds(d) = w∞(x , z)

= us(z,−x), x ∈ S2

⇒ ws(z, z) =

∫S2

u∞(−x ,d) g(x , z) g(d , z) ds(x) ds(d)







|x − z|, x ∈ Λ

⇒∫


⇒∫

S2u∞(x ,d)g(d , z) ds(d) = w∞(x , z)

= us(z,−x), x ∈ S2

⇒ ws(z, z) =

∫S2








|x − z|, x ∈ Λ

⇒∫


⇒∫

S2u∞(x ,d)g(d , z) ds(d) = w∞(x , z)

= us(z,−x), x ∈ S2

⇒ ws(z, z) =

∫S2








|x − z|, x ∈ Λ

⇒∫


⇒∫

S2u∞(x ,d)g(d , z) ds(d) = w∞(x , z)

= us(z,−x), x ∈ S2

⇒ ws(z, z) =

∫S2








|x − z|, x ∈ Λ

⇒∫


⇒∫

S2u∞(x ,d)g(d , z) ds(d) = w∞(x , z)

= us(z,−x), x ∈ S2

⇒ ws(z, z) =

∫S2








|x − z|, x ∈ Λ

⇒∫


⇒∫

S2u∞(x ,d)g(d , z) ds(d) = w∞(x , z)

= us(z,−x), x ∈ S2

⇒ ws(z, z) =

∫S2






Pros:Nice mathematicsSimple implementationNo a priori information needed

Contras:Need a lot of dataNo sharp boundaries (∞ = ?)Sensitive to noise


References

Serranho, P.A hybrid method for inverse scattering for sound-softobstacles in IR3.Inverse Problems and Imaging, 1, 691–712 (2007).

Ivanyshyn, O., Kress, R. and Serranho, P.Huygens’ principle and iterative methods in inverseobstacle scattering.Advances in Computational Mathematics 33, 413–429(2010).

Ivanyshyn, O. and Kress, R.Identification of sound-soft 3D obstacles fromphaseless data.Inverse Problems and Imaging 4, 111–130 (2010).

Colton, D. and Kress, R.Inverse scattering.In: Handbook of Mathematical Methods in Imaging(Scherzer, O., ed.)Springer-Verlag, pp. 551–598 (2011).


References






References






Documents

AIP Kress - Texas A&M University