The Factorization Method for Inverse Scattering Problemsaipc.tamu.edu/speakers/kirsch.pdfD. Colton, R. Kress: Inverse Acoustic and Electromagnetic Scattering Theory. 2nd edition, Springer,

Karlsruhe Institute of Technology

KIT – University of the State of Baden-Wurttemberg and National Large-scale Research Center of the Helmholtz Association www.kit.edu

The Factorization Method forInverse Scattering ProblemsAndreas Kirsch AIP 2011, College Station

Department of Mathematics

Karlsruhe Institute of TechnologyOutline of the Talk

Introduction

The Direct Scattering Problem

The Inverse Scattering Problem

Reconstruction Techniques

Factorization of the Far Field Operator and Range Identity

Explicit Form of Characteristic Function

Connections to Linear Sampling Method and Time Reversal

Some Numerical Simulations

Final Remarks

The Factorization Method for Inverse Scattering Problems 2/30

Karlsruhe Institute of TechnologyIntroduction

uinc

us

D

General situation in scattering theory:

A given (“incident”) wave uinc

is disturbed by a medium Dand produces a scattered field us

Total field: u = uinc + us


Karlsruhe Institute of TechnologyIntroductionNecessary specification of:

model of wave propagation (scalar wave equation, Maxwell’sequations, Helmholtz equation, Navier’s equation, ...)incident wave uinc

form of scattering medium (impenetrable, penetrable, index ofrefraction)

In this talk fields uinc ,us,u : Rd → C are scalar functions (d = 2 ord = 3). Incident field uinc satisfies Helmholtz equation (reducedwave equation)

∆uinc(x) + k2uinc(x) = 0 in Rd

with wave number k > 0.Scattering medium is given by index of refraction n ∈ L∞(Rd) withn = 1 outside of bounded domain D ⊂ Rd . Total field u satisfies

∆u(x) + k2n(x) u(x) = 0 in Rd .


Karlsruhe Institute of TechnologyIntroduction

This model appears

as TM-mode for the time harmonic Maxwell system with d = 2and wave number k = ω

√ε0µ0 with frequency ω > 0 and index

of refraction n(x) = 1ε0

[ε(x) + i σ(x)/ω

]in acoustic wave progation with d = 2 or d = 3 and wavenumber k = ω/c0 where ω > 0 is frequency and c0 is speed ofsound in vacuum and index of refractionn(x) =

c20

c(x)2

[1 + i γ(x)/ω

]where γ is a damping coefficient.

This models penetrable media. Also possible: impenetrable media:

Helmholtz equation ∆u + k2u = 0 in Rd \ D and boundaryconditions as, e.g., u = 0 on ∂D or ∂u/∂ν = 0 on ∂D or∂u/∂ν + iλu = 0 on ∂D.

In addition, a radiation condition for the scattered field us = u − uinc

is necessary:


Karlsruhe Institute of TechnologyThe Direct Scattering ProblemGiven: wave number k > 0, incident field uinc , and refractive indexn ∈ L∞(Rd), determine u ∈ H1

loc(Rd) as a solution of

∆u(x) + k2n(x) u(x) = 0 in Rd ,

us = u − uinc satisfies Sommerfeld’s radiation condition (SRC):

∂us(x)

∂r− ik us(x) = O

(r−(d+1)/2) , r = |x | → ∞ ,

uniformly with respect to x = x/|x | ∈ Sd−1.Examples of incident waves:

(a) Plane wave of direction θ ∈ Sd−1: uinc(x) = eik θ·x , x ∈ Rd .

(b) Spherical wave with source point y ∈ Rd :

uinc(x) = Φ(x , y) :=

i4 H(1)

0 (k |x − y |) , d = 2,

exp(ik |x − y |)4π|x − y | , d = 3,

x 6= y .


Karlsruhe Institute of TechnologyThe Direct Scattering ProblemWriting equation for scattered wave yields

∆us + k2n us = −∆uinc − k2n uinc = −k2(n − 1)uinc in Rd ,

and us satisfies the SRC. This is special case of

∆v + k2n v = −f in Rd , and v satisfies SRC.

First uniqueness: Assume f = 0. Then:

o(1) =

∫|x |=R

∣∣∣∣∂v∂r− ikv

∣∣∣∣2 ds

=

∫|x |=R

∣∣∣∣∂v∂r

∣∣∣∣2 + k2|v |2ds − 2k Im∫

|x |=R

v∂v∂r

ds


Karlsruhe Institute of TechnologyThe Direct Scattering Problem

Green’s theorem yields∫|x |=R

v∂v∂r

ds =

∫∫|x |<R

[|∇v |2 + v ∆v

]dx =

∫∫|x |<R

[|∇v |2 − k2n|v |2

]dx

and thus

o(1) =

∫|x |=R

∣∣∣∣∂v∂r

∣∣∣∣2 + k2|v |2ds + 2k3∫∫D

(Im n)︸︷︷︸≥ 0

|v |2 dx

Therefore,∫|x |=R |v |

2ds −→ 0 as R →∞. Rellich’s Lemma yieldsv = 0 outside ball. Unique continuation yields v = 0 everywhere.


Karlsruhe Institute of TechnologyThe Direct Scattering ProblemExistence by integral equation approach:

Theorem: v ∈ H1loc(R

d) is solution of

∆v + k2n v = −f in Rd , and v satisfies SRC

if, and only if, v ∈ L2(D) solves Lippmann-Schwinger equation

v(x) =

∫∫D

[k2q(y) v(y) + f (y)

]Φ(x , y) dy , x ∈ D ,

where q = n − 1 is contrast. Asymptotic behavior of Φ(x , y) as|x | → ∞ yields:

v(x) = γdexp(ik |x |)|x |(d−1)/2

[v∞(x) + O(1/|x |)

], |x | → ∞ ,

uniformly with respect to x := x/|x | ∈ Sd−1 with far field pattern v∞.The Factorization Method for Inverse Scattering Problems 9/30

Karlsruhe Institute of TechnologyThe Inverse Scattering ProblemIf, in particular, us = us(x , θ) is scattered field corresponding toincident plane wave uinc(x) = exp(ik θ · x) then

us(x , θ) = γdexp(ik |x |)|x |(d−1)/2

[u∞(x , θ) + O(1/|x |)

], |x | → ∞ ,

with far field pattern u∞ = u∞(x , θ).Inverse scattering problem: u∞(x , θ) is known for all x , θ ∈ Sd−1,medium n or only its support D has to be determined.Example: Which domain D ⊂ R2 corresponds to the following farfields u∞(φ, θ), φ, θ ∈ [0,2π]?

Re u∞ Im u∞ Re u∞ Im u∞The Factorization Method for Inverse Scattering Problems 10/30

Karlsruhe Institute of TechnologyThe Inverse Scattering ProblemLeft example simple:Theorem of Karp: If u∞(x , θ

)= ψ(x · θ) for all x , θ ∈ Sd−1, then D

is a ball in Rd−1.This follows from uniqueness of inverse problem:Theorem (Nachman, Novikov, Ramm (all 1988, d = 3), Bukhgeim(2007, d = 2)) The far field patterns u∞(x , θ) determine nuniquely; that is, if nj ↔ u∞j (x , θ) for j = 1,2, then:

u∞1 (x , θ) = u∞2 (x , θ) for all x , θ ∈ Sd−1 =⇒ n1 = n2 .

Some literature on time harmonic (inverse) scattering theory:F. Cakoni, D. Colton: Qualitative Methods in Inverse ScatteringTheory. An Introduction. Springer, 2006.D. Colton, R. Kress: Inverse Acoustic and ElectromagneticScattering Theory. 2nd edition, Springer, 1998.A. Kirsch: Introduction to the Mathematical Theory of InverseProblems. Springer 1996, 2011.


Karlsruhe Institute of TechnologyReconstruction Techniques(A) Linearization, e.g. Born approximation: Let n = 1 + q

∆u + k2(1 + q)u = 0 , u = uinc + us , us SRC

∆u + k2u = −k2qu i.e. ∆us + k2us = −k2qu ≈ −k2quinc

usB(x , θ) = k2

∫∫D

q(y) uinc(y , θ) Φ(x , y) dy , x ∈ Rd ,

u∞B (x , θ) = k2∫∫

Dq(y) uinc(y , θ) e−ik x ·ydy

= k2∫∫

Dq(y) eik(θ−x)·ydy = k2 q

(k(x − θ)

).

x , θ ∈ Sd−1; that is, k(x − θ) ∈ {z ∈ Rd : |z| ≤ 2k}.The Factorization Method for Inverse Scattering Problems 12/30

Karlsruhe Institute of TechnologyReconstruction Techniques

(B) Iterative methods to determine contrast function n: Definemapping T : L∞(D)→ L2(Sd−1 × Sd−1), n 7→ u∞. Applyiterative method to solve T (n) = f for n where f = f (x , θ) is given(measured) far field pattern.

Possible methods: Newton-type methods, gradient-type methods,second order methods.

Derivative: T ′(n)h = v∞ where v is radiating solution of∆v + k2nv = −k2hu. Derivative T ′(n) is compact and one-to-one!

Advantages: Very general, accurate, incorporation of a prioriinformation possible.

Disadvantages: “Expensive”, only local convergence is expected, norigorous convergence result known.


Karlsruhe Institute of TechnologyThe Far Field Operator(C) Factorization Method determines only support of q = 1− n!Values of n ∈ L∞(Rd) do not have to be known in advance.

Define far field operator F : L2(Sd−1)→ L2(Sd−1) by

(Fg)(x) =

∫Sd−1

u∞(x , θ) g(θ) ds(θ) , x ∈ Sd−1 .

Properties of F :F is compact.If n is real-valued then F is normal; that is, F ∗F = F F ∗, andeven: S := I + ik

2π F is unitary (=scattering matrix).F is one-to-one if k2 is not an interior transmission eigenvalue;

that is, ∆u + k2n u = 0 , ∆w + k2w = 0 in D ,

u = w on ∂D , ∂u/∂ν = ∂w/∂ν on ∂D ,

implies u = w = 0 in D.The Factorization Method for Inverse Scattering Problems 14/30

Karlsruhe Institute of TechnologyThe Factorization MethodFirst: Born approximation with far field operator FB; that is,

(FBg)(x) =

∫Sd−1

u∞B (x , θ) g(θ) ds(θ)

= k2∫

Sd−1

∫∫D

q(y) g(θ) eik(θ−x)·ydy ds(θ)

= k2∫∫

Dq(y) e−ik x ·y

∫Sd−1

g(θ) eik θ·y ds(θ︸︷︷︸=: (Hg)(y)

) dy

= H∗ TB Hg with TBf = k2q f .

Therefore, FB = H∗ TB H .

This is factorization of FB. Note: FB is self-adjoint for real q while Fis only normal!


Karlsruhe Institute of TechnologyThe Factorization MethodNow general nonlinear case:

Recall: u∞(·, θ) ←→ uinc(·, θ)

superposition: Fg ←→∫

Sd−1uinc(·, θ) g(θ) ds(θ) =: vg

Theorem: F has factorization F = H∗T H with

(Hg)(x) =

∫Sd−1

g(θ) eik θ·x ds(θ) , x ∈ Rd ,

(H∗ϕ)(x) =

∫∫Dϕ(y) e−ik x ·y dy , x ∈ Sd−1 ,

and Tf = k2q (f + k2v) where v solves

∆v + k2(1 + q)v = −q f , v satisfies SRC.

Compare to FB = H∗TB H with TBf = k2q f .

This is factorization, what’s the method?The Factorization Method for Inverse Scattering Problems 16/30

Karlsruhe Institute of TechnologyThe Factorization MethodTheorem: Let Rd \ D be connected. For any z ∈ Rd defineφz ∈ L2(Sd−1) by

φz(x) = e−ik x ·z , x ∈ Sd−1 .

Then z ∈ D if, and only if, φz ∈ R(H∗).

Proof: (H∗ϕ)(x) =

∫∫Dϕ(y) e−ik x ·y dy ?

= e−ik x ·z , x ∈ Sd−1 .

This is equivalent to (because complement of D is connected)

(∗)∫∫

Dϕ(y) Φ(x , y) dy = Φ(x , z) , x /∈ (D ∪ {z}) .

z ∈ D: Choose Φ ∈ C∞(Rd) with Φ(x) = Φ(x , z) outside of D anddefine ϕ by

∫∫D ϕ(y) Φ(·, y) dy = Φ in Rd ; that is, −ϕ = ∆Φ + k2Φ.

z /∈ D: (∗) can not have a solution! (Left hand side bounded, righthand side unbounded for x → z.)


Karlsruhe Institute of TechnologyRange IdentityRecall: F = H∗ T H and z ∈ D ⇔ φz ∈ R(H∗) .

Goal: Express range R(H∗) by known operator F !General situation:

X X

Y Y-

-

?

6

H H∗

T

F

Theorem: If T : X → X is selfadjoint and coercive; that is,

〈ψ,Tϕ〉 = 〈Tψ,Tϕ〉 , 〈ϕ,Tϕ〉 ≥ c‖ϕ‖2 for all ψ,ϕ ∈ X ,

then R(H∗) = R(F 1/2) .


Karlsruhe Institute of TechnologyRange IdentityTheorem: Let F = H∗T H : Y → Y be one-to-one and such thatI + ir F is unitary for some r > 0. Furthermore, let T : X ∗ → X becomp. perturb. of s.a. and coercive operator and Im〈ϕ,Tϕ〉 6= 0 for

all ϕ ∈ closureR(H) with ϕ 6= 0. Then R(H∗) = R((F ∗F )1/4

).

Idea of proof: I + irF unitary implies F normal and thusFψj = λjψj . Then F = H∗T H = |F |1/2S|F |1/2 where

|F |1/2ψ =∑

j

√|λj | 〈ψ,ψj〉Y ψj ,

Sψ =∑

j

λj

|λj |〈ψ,ψj〉Y ψj .

|〈Sψ,ψ〉| =∣∣∣∣∑j

λj

|λj |∣∣〈ψ,ψj〉Y

∣∣2∣∣∣∣≥ c‖ψ‖2Y


Karlsruhe Institute of TechnologyApplication to Scattering ProblemLet k2 be no int. transm. eigenvalue, q real, q(x) ≥ q0 on D.Recall: F = H∗T H and F is one-to-one and I + ik

2πF is unitaryand T : L2(D)→ L2(D), f 7→ k2q(f + v) is compact perturbation ofcoercive operator and and Im〈ϕ,Tϕ〉 > 0 for all

ϕ ∈ closureR(H), ϕ 6= 0. Thus R(H∗) = R((F ∗F )1/4

).

Combination of previous theorems:Theorem: Let again φz(x) = exp(−ik x · z), x ∈ Sd−1.Under above assumptions:

z ∈ D ⇐⇒ φz ∈ R((F ∗F )1/4)

Let {λj : j ∈ N} ⊂ C be eigenvalues of (normal!) operator F withnormalized eigenfunctions ψj ∈ L2(Sd−1) for j ∈ N. Then:

z ∈ D ⇐⇒∑j∈N

|〈φz , ψj〉L2 |2

|λj |<∞ ⇐⇒

∑j∈N

|〈φz , ψj〉L2 |2

|λj |

−1

> 0 .


Karlsruhe Institute of TechnologyMedia with AbsorptionNow n = 1 + q ∈ L∞(Rd) complex valued, Im n ≥ 0. StillF = H∗TF but not normal anymore. Define

Re F =12(F + F ∗) = H∗(Re T ) H

Im F =12i

(F − F ∗) = H∗(Im T ) H

F# =def |Re F | + Im F = H∗T H

with coercive T .

Theorem: z ∈ D ⇐⇒ φz ∈ R(F 1/2# )

Let {λj : j ∈ N} ⊂ R be eigenvalues of (selfadjoint!) operator F# withnormalized eigenfunctions ψj ∈ L2(Sd−1) for j ∈ N. Then:

z ∈ D ⇐⇒∑j∈N

|〈φz , ψj〉L2 |2

λj<∞ ⇐⇒

∑j∈N

|〈φz , ψj〉L2 |2

λj

−1

> 0 .


Karlsruhe Institute of TechnologyConnection to Linear Sampling MethodRecall Factorization Method:

z ∈ D ⇐⇒ (F ∗F )1/4h = φz solvable in L2(Sd−1)

Linear Sampling Method:

Solve (approximately): Fg = φz in L2(Sd−1)

If Fg = φz solvable then (F ∗F )1/4S(F ∗F )1/4g = φz , thus z ∈ D.For example, apply Tikhonov regularization to Fg = φz :

(εI + F ∗F ) gz,ε = F ∗φz

for z ∈ Rd and ε > 0. Let vg be Herglotz function with kernel g.Then (Arens, Lechleiter 2004, 2007):

(a) x ∈ D: c ‖hz‖2L2(S2) ≤ limε→0

∣∣vgz,ε(z)∣∣ ≤ ‖hz‖2L2(S2)

(b) x /∈ D: limε→0|vgz,ε(z)| =∞


Karlsruhe Institute of TechnologyConnection to Time Reversal

Recall: Fg ←→ vg(x) =

∫Sd−1

g(θ) eikx ·θds(θ)

Define (Rg)(x) = g(−x). Then (Hazard, Ramdani 2004):

g F7→ Fg R7→ RFg F7→ FRFg R7→ RFR︸︷︷︸= F∗

Fg = F ∗Fg

Focusing: Choose (largest) eigenvalue λ of F and correspondingeigenfunction ψ and plot |vψ(z)|2 =

∣∣(ψ, φz)L2

∣∣2.

Note: vψ(z) =4πk2

λ

∫∫D

q(y) uψ(y) j0(k |z − y |) dy , z ∈ R3 .

Factorization Method: Plot w(z) =

∑j∈N

∣∣(ψj , φz)L2

∣∣2|λj |

−1


Karlsruhe Institute of TechnologyNumerical SimulationsRecall:

z ∈ D ⇐⇒∑j∈N

|〈φz , ψj〉L2 |2

|λj |< ∞

⇐⇒ w(z) =

∑j∈N

|〈φz , ψj〉L2 |2

|λj |

−1

> 0 .

Therefore, sign(w) is the characteristic function of D!

The following examples show plots of

wN(z) =

N∑j=1

|〈φz , ψj〉|2

|λj |

−1

, z ∈ R2 :

for N = 32 or N = 36, respectively.The Factorization Method for Inverse Scattering Problems 24/30

Karlsruhe Institute of TechnologyNumerical Simulations

Dirichlet boundary conditions:



Dirichlet boundary conditions:



Scattering by an open arc:



Real data:


Karlsruhe Institute of TechnologyNumerical Simulations3D-Example (joint work with A, Kleefeld): Scattering underconductive transmission conitions

∆u + k2u = 0 in R3 \ ∂D ,

u+ = u− ,∂u+

∂ν− ∂u−

∂ν= λ u on ∂D .


Karlsruhe Institute of TechnologyFinal Remarks

Rigorous justification of Factorization Method for:

Scattering by anisotropic media: ∇ · (A∇u) + k2u = 0

Scattering of electromagnetic and elastic waves

Scattering by arcs, periodic structures, in wave guides

Nonlinear Helmholtz equation (Lechleiter, Minisymposium M04)

Tomography (Bruhl, Hanke, Hyvonen, Lechleiter, von Harrach,Griesmaier)

References:

A. Kirsch, N. Grinberg: The Factorization Method for InverseProblems. Oxford University Press, 2008.

A. Kirsch: Introduction to the Mathematical Theory of InverseProblems. Springer, 2nd edition, 2011.

Thank you for your attention!


Documents

The Factorization Method for Inverse Scattering Problemsaipc.tamu.edu/speakers/kirsch.pdfD. Colton, R. Kress: Inverse Acoustic and Electromagnetic Scattering Theory. 2nd edition, Springer,