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HAL Id: hal-00939283

https://hal.archives-ouvertes.fr/hal-00939283

Submitted on 30 Jan 2014

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Riemann–Hilbert problem approach for two-dimensional flow inverse scattering

Alexey Agaltsov, Roman Novikov

To cite this version:

Alexey Agaltsov, Roman Novikov. Riemann–Hilbert problem approach for two-dimensional flow in-

verse scattering. Journal of Mathematical Physics, American Institute of Physics (AIP), 2014, 55 (10),

pp.103502. �hal-00939283�

(2)

two-dimensional ow inverse sattering

A.D.Agaltsov 1

,R.G.Novikov 2

Abstrat

We onsider inverse sattering for the time-harmoni wave equa-

tionwithrst-order perturbation intwodimensions. This problem

arisesinpartiularin theaoustitomographyofmovinguid. We

onsider linearized and nonlinearizedreonstrution algorithms for

this problem of inverse sattering. Our nonlinearized reonstru-

tionalgorithm isbasedon thenon-loal RiemannHilbert problem

approah. Comparisonswithpreeding resultsaregiven.

Keywords: aoustitomography,movinguid,waveequationwith

rst-order perturbation, inverse sattering, RiemannHilbert prob-

lem.

1 Introdution

Weonsider theequation

− ∆ψ − 2iA(x) ∇ ψ + V (x)ψ = Eψ, x = (x

1

, x

2

) ∈ R

2

, E > 0,

(1.1)

where

∆ = ∂

x21

+ ∂

x22,

∇ = (∂

x1

, ∂

x2

)

,

xk

= ∂/∂x

k,

k = 1

,

2

,and

A = (A

1

, A

2

)

and

V

are vetor and salar potentials on

R

2, respetively. In addition we assumethat

A

1,

A

2 and

V

are suiently regular funtions on

R

2

with suientdeayatinnity.

(1.2)

Equation (1.1) anbe onsidered asa model equation for the time-harmoni

exp( − iωt)

aoustipressure

ψ

inatwo-dimensionalmovinguid,seee.g.[RW℄, [RE℄.Inthissetting

E = ω

c

0

2

, A(x) = ω c

0

u(x), V (x) = 1 − n

2

(x) ω c

0

2

,

(1.3)

where

c

0 isareferenesoundspeed,

n(x)

isasalarindexofrefration,

u(x)

is

anormalizeduidveloityvetor.

ThemainpartoftheworkwasfullledduringthestageoftherstauthorintheCentre

deMathematiquesAppliqueesofEolePolytehniqueinOtoberDeember2013

1

119991,LomonosovMosowStateUniversity,FaultyofComputationalMathematisand

Cybernetis,Mosow,Russia;email: agaletsgmail.om

2

CNRS(UMR7641),CentredeMathematiques Appliquees,EolePolytehnique, 91128

Palaiseau, Frane; IEPTRAS, 117997, Mosow, Russia; Mosow Institute of Physis and

Tehnology,Dolgoprudny,Russia;email: novikovmap.polytehnique.fr

(3)

equationatxed energy

E

withmagnetipotential

A

andeletripotential

v

,

where

V (x) = A

2

(x) − i div A(x) + v(x)

= A

21

(x) + A

22

(x) − i

∂A

1

(x)

∂x

1

+ ∂A

2

(x)

∂x

2

+ v(x),

(1.4)

seee.g. [HN1℄,[ER2℄.

Forequation(1.1)weonsiderthelassialsatteringsolutions

ψ

+ontinu-

ousandboundedon

R

2withtheirrstderivativesandspeiedbythefollowing asymptotisas

| x | → ∞

:

ψ

+

(x, k) = e

ikx

+ C( | k | ) e

i|k||x|

| x |

1/2

f

k, | k | x

| x |

+ o 1

| x |

1/2

,

(1.5)

x ∈ R

2

, k ∈ R

2

, k

2

= E, C( | k | ) = − πi √

2πe

iπ/4

| k |

1/2

,

where apriori unknownfuntion

f = f (k, l)

,

k

,

l ∈ R

2,

k

2

= l

2

= E

, arising

in(1.5)isthelassialsatteringamplitudefor(1.1).

Given potentials

A

,

V

, to determine

ψ

+ and

f

one an use the following

LippmannShwingerintegralequation(1.6)andformula(1.8)(seee.g. [HN1℄):

ψ

+

(x, k) = e

ikx

+ Z

R2

G

+

(x − y, k) ×

× − 2iA(y) ∇

y

ψ

+

(y, k) + V (y)ψ

+

(y, k) dy,

(1.6)

G

+

(x, k) = − (2π)

2

Z

R2

e

iξx

ξ

2

− k

2

− i0 = − i

4 H

01

( | x || k | ),

(1.7)

where

x ∈ R

2,

k ∈ R

2,

k

2

= E

,

H

01 istheHankelfuntionoftherstkind;

f (k, l) = (2π)

2

Z

R2

e

ily

− 2iA(y) ∇

y

ψ

+

(y, k) + V (y)ψ

+

(y, k)

dy,

(1.8)

where

k ∈ R

2,

l ∈ R

2,

k

2

= l

2

= E

. Atually, we onsider (1.6) and its

dierentiatedversion,where

isappliedtobothsidedof(1.6),asasystemof

linearintegralequationsfor

ψ

+and

∇ ψ

+.

Oneanseethatthesatteringamplitude

f

forequation(1.1)at xed

E

is

denedon

M

E

=

k ∈ R

2

, l ∈ R

2

: k

2

= l

2

= E , E > 0.

(1.9)

Notethat

f

on

M

E isinvariantwithrespettothegaugetransformations

A → A + ∇ ϕ,

V → V − i∆ϕ + ( ∇ ϕ)

2

+ 2A ∇ ϕ,

(1.10)

where

ϕ

isasuientlyregularfuntion on

R

2withsuientdeayatinnity, seee.g. [HN2℄,[ER2℄. Inaddition,

ψ

+ istransformedas

ψ

+

→ e

ψ

+ (1.11)

(4)

Inthis work weonsider thefollowinginverse satteringproblem forequa-

tion(1.1)underassumptions(1.2):

Problem 1.1. Given satteringamplitude

f

on

M

E at xed

E > 0

, nd po-

tentials

A

and

V

on

R

2 (atleastapproximately).

Problem 1.1 for the ase when

A ≡ 0

wasstudied, in partiular, in [N1℄,

[GM℄,[N3℄,[GN2℄,[N4℄, [BBMRS℄,[BAR℄andin [N2℄,[Buk℄.

Problem1.1 forthegeneralasewasonsidered,inpartiular, onpage457

of[N3℄andin[X℄.

Problem1.1isalsorelatedwithseveralotherinverseproblemsfortheShrodinger

equationinmagnetield(andforthetime-harmoniwaveequationwithrst-

orderperturbation)in dimension

d ≥ 2

. Conerning these otherinverseprob-

lemssee[DKN℄,[Sh℄,[HN2℄,[Nor℄,[RW℄,[RE℄,[BBS℄,[ER1℄,[NSU℄,[A℄,[ER2℄,

[Ni℄,[WY℄,[GT℄,[IY℄andreferenestherein.

Notethatapproximatending

A

and

V

inProblem1.1means,inpartiular,

ndingmodulotransformations(1.10) . However,forreal-valued

A

and

V

there

is no gauge nonuniqueness (1.10) in Problem 1.1! In addition,

A

and

V

of

formulas(1.3)(ofmovinguidaoustis)arerealif

n

isreal.

Inthisworkwearemainlymotivatedbyappliationstotheaoustitomog-

raphyofmovinguiddisribedin [Nor℄,[RW℄,[RE℄, [BBS℄.Notethat intheir

reonstrution results works [Nor℄, [RW℄, [RE℄, [BBS℄ proeed from near-eld

sattering data (e.g. from some near eld information on

ψ

+) instead of the

satteringamplitude

f

. Butitisalsoknownthatnear-eldsatteringdataan

betransformedintovaluesof

f

,see e.g. [Ber℄,[BBS℄.

Resultsofthepresentworkanbedesribedasfollows:

InSetion 2wegiveformulasforsolvingProblem1.1 inthe Bornapproxi-

mation. Toourknowledgethese formulaswerenotyet givenexpliitely in the

literaturefortheasewhen

A 6≡ 0

. Theseformulasareprovedin Setion5.

In Setion 3wegive anonlinearizedreonstrution algorithm for Problem

1.1. For the ase when

A ≡ 0

this algorithm is redued to the algorithm of

[N4℄and wasimplementednumerially in [BAR℄. Forthegeneral asethis al-

gorithmanbealsoregardedassimpliationanddevelopmentofthealgorithm

mentioned on page 457 of [N3℄ and based on the RiemannHilbertManakov

problemapproahof[GN1℄,[N1℄. Aderivationofthereonstrutionalgorithm

ofSetion3isgivenin Setion6.

InSetion4weshowthatin theBornapproximationthealgorithm ofSe-

tion3isreduedtoformulasofSetion2. RelatedproofsaregiveninSetion7.

Inasimilarwaywithresultsof[NS℄thereonstrutionalgorithmofSetion3

anbegeneralizedto themulti-hannelase. Thisgeneralizationwill begiven

inasubsequentwork.

InthepresentworkwearefousedonapproximatereonstrutionsforProb-

lem 1.1, admitting stable numerial implementation. Issues related with the-

orems of uniqueness and examples of nonuniqueness for Problem 1.1 will be

onsideredinasubsequentwork.

(5)

If

A = (A

1

, A

2

)

and

V

satisfy(1.2)andare suientlysmall, thenproeeding

from(1.6),(1.8)wehavethefollowingBornapproximationformulasfordiret

sattering:

ψ

+

(x, k) ≈ e

ikx

, ∇ ψ

+

(x, k) ≈ e

ikx

ik, f (k, l) ≈ f

lin

(k, l),

(2.1)

f

lin

(k, l) == (2π)

def 2

Z

R2

e

i(kl)x

2kA(x) + V (x)

dx,

(2.2)

where

x

,

k

,

l ∈ R

2,

k

2

= l

2

= E

. Inpartiular, formulas(2.1)anbespeied

as(2.14).

Notethat

f

linon

M

E isinvariantwithrespettothegaugetransformations

A → A + ∇ ϕ, V → V − i∆ϕ,

(2.3)

where

ϕ

isasuientlyregularfuntion on

R

2withsuientdeayatinnity.

This invariane follows from (2.2), (2.3), integrating by parts and using that

k

2

− l

2

= 0

. Weonsider(2.3)asalinearizationof (1.10)forsmall

A

,

V

and

ϕ

.

Inthissetionweonsiderthefollowinglinearizedinversesatteringproblem

forequation(1.1)under assumptions(1.2) :

Problem2.1. Givenlinearizedsatteringamplitude

f

linon

M

Eatxed

E > 0

,

ndpotentials

A

and

V

on

R

2 (atleastapproximately).

Notethatapproximatending

A

and

V

inProblem2.1means,inpartiular,

ndingmodulotransformations(2.3). However,inasimilarwaywithProblem

1.1, thereis nogaugenonuniqueness (2.3)in Problem 2.1 forthease ofreal-

valued

A

and

V

.

Problem2.1isalinearizationofProblem1.1.

TostudyProblem 2.1 itis onvenientto introdue

ϕ

div,

A

div,0,

V

div,0 and

ϕ

±,

A

±,0,

V

±,0suhthat:

∆ϕ

div

(x) = − div A(x), ϕ

div

(x) → 0

as

| x | → ∞ , A

div,0

(x) = A(x) + ∇ ϕ

div

(x), V

div,0

(x) = V (x) − i∆ϕ

div

(x),

(2.4)

where

x ∈ R

2;

z

ϕ

(x) = − 1

2 (A

1

(x) − iA

2

(x)), ϕ

(x) → 0

as

| x | → ∞ , A

,0

(x) = A(x) + ∇ ϕ

(x), V

,0

(x) = V (x) − i∆ϕ

(x),

(2.5)

¯z

ϕ

+

(x) = − 1

2 (A

1

(x) + iA

2

(x)), ϕ

+

(x) → 0

as

| x | → ∞ , A

+,0

(x) = A(x) + ∇ ϕ

+

(x), V

+,0

(x) = V (x) − i∆ϕ

+

(x),

(2.6)

(6)

z

= 1

2 (∂

x1

− i∂

x2

), ∂

= 1

2 (∂

x1

+ i∂

x2

), x = (x

1

, x

2

) ∈ R

2

.

(2.7)

Oneanseethat

div A

div,0

(x) = 0, A

1,0

(x) − iA

2,0

(x) = 0, A

+,01

(x) + iA

+,02

(x) = 0,

where

x ∈ R

2,

A

±,0

= (A

±1,0

, A

±2,0

)

.

Itisalsoonvenienttotransformformula(2.2)to theform

f

lin

(k, l) − f

lin

( − l, − k) = 2(k + l) A(k b − l),

f

lin

(k, l) + f

lin

( − l, − k) = 2(k − l) A(k b − l) + 2 V b (k − l),

(2.8)

A(p) = (2π) b

2

Z

R2

e

ipx

A(x) dx, V b (p) = (2π)

2

Z

R2

e

ipx

V (x) dx,

(2.9)

where

(k, l) ∈ M

E,

p ∈ R

2. Notethat

(k, l) ∈ M

E

= ⇒ k − l ∈ B

2E

,

p ∈ B

2E

= ⇒ p = k − l

forsome

(k, l) ∈ M

E

,

(2.10)

where

B

r

= { p ∈ R

2

: | p | ≤ r } , r > 0.

(2.11)

Wedene

C

N,σ

( R

2

) =

u ∈ C

N

( R

2

) : k u k

N,σ

< + ∞ , k u k

N,σ

= max

|n|6N

sup

x∈R2

(1 + | x |

2

)

σ/2

| ∂

n

u(x) | , N ∈ N ∪ { 0 } , σ > 0,

(2.12)

where

C

N

( R

2

)

isthespaeof

N

-timesontinouslydierentiableomplex-valued funtionson

R

2,

n

= ∂

xn11

xn22

, n = (n

1

, n

2

) ∈ N ∪ { 0 }

2

, | n | = n

1

+ n

2

.

(2.13)

Notethatif

A

1,

A

2,

V ∈ C

0,σ

( R

2

)

forsome

σ > 2

and

k A

j

k

0,σ

≤ q

,

j = 1

,

2

,

k V k

0,σ

≤ q

, then

ψ

+

(x, k) = e

ikx

+ O(q), ∇ ψ

+

(x, k) = e

ikx

ik + O(q), f (k, l) = f

lin

(k, l) + O(q

2

)

as

q → +0,

(2.14)

uniformlywithrepsetto

x

,

k

,

l ∈ R

2,

k

2

= l

2

= E

,atxed

E > 0

.

Theorem 2.1. Suppose that

A

1,

A

2,

V

are real-valued and

A

1,

A

2,

V ∈

C

N,σ

( R

2

)

for some

N ≥ 3

,

σ > 2

. Then the following formulas for solving

(7)

A(k b − l) = f

lin

(k, l) − f

lin

(l, k) 2

k − l

| k − l |

2

+ f

lin

(k, l) − f

lin

( − l, − k) 2

k + l

| k + l |

2

, V b (k − l) = f

lin

(l, k) + f

lin

( − l, − k)

2 ,

(2.15)

where

A b

,

V b

are denedby (2.9)and

(k, l) ∈ M

E;

A(x) = A

appr

(x, E) + A

err

(x, E), x ∈ R

2

, E > 0, A

appr

(x, E) =

Z

|p|≤2√ E

e

ipx

A(p) b dp, A

err

(x, E) = Z

|p|≥2√ E

e

ipx

A(p) b dp,

(2.16)

V (x) = V

appr

(x, E) + V

err

(x, E), x ∈ R

2

, E > 0,

V

appr

(x, E) = Z

|p|≤2√ E

e

ipx

V b (p) dp, V

err

(x, E) = Z

|p|≥2√ E

e

ipx

V b (p) dp,

(2.17)

where

A(p) b

and

V b (p)

for

| p | ≤ 2 √

E

aregiven intermsof

f

lin on

M

E aording

to (2.10), (2.15)and

| A

err,j

(x, E) | ≤ c

1

(N, σ) k A

j

k

N,σ

E

N22

,

(2.18)

| V

err

(x, E) | ≤ c

1

(N, σ) k V k

N,σ

E

N22

,

(2.19)

where

x ∈ R

2,

j = 1

,

2

,

A

err

= (A

err,1

, A

err,2

)

,

E ≥

14 and

c

1

(N, σ) = 4

(N − 2)(σ − 2) .

(2.20)

Theorem2.2. Supposethat

A

1,

A

2,

V ∈ C

N,σ

( R

2

)

forsome

N ≥ 4

and

σ > 2

.

Let

A

div,0,

V

div,0 bedenedaordingto (2.4). Thenthe followingformulas for

solvingProblem 2.1arevalid:

b

A

div,0

(k − l) = f

lin

(k, l) − f

lin

( − l, − k) 2

k + l

| k + l |

2

, V b

div,0

(k − l) = f

lin

(k, l) + f

lin

( − l, − k)

2 ,

(2.21)

where

A b

div,0,

V b

div,0 aretheFouriertransformsof

A

div,0,

V

div,0 (see (2.9))and

(k, l) ∈ M

E;

A

div,0

(x) = A

div,0appr

(x, E) + A

div,0err

(x, E), x ∈ R

2

, E > 0,

(2.22)

A

div,0appr

(x, E) =

Z

|p|≤2√ E

e

ipx

A b

div,0

(p) dp, A

div,0err

(x, E) = Z

|p|≥2√ E

e

ipx

A b

div,0

(p) dp, V

div,0

(x) = V

apprdiv,0

(x, E) + V

errdiv,0

(x, E), x ∈ R

2

, E > 0,

(2.23)

V

apprdiv,0

(x, E) =

Z

|p|≤2√ E

e

ipx

V b

div,0

(p) dp, V

errdiv,0

(x, E) = Z

|p|≥2√ E

e

ipx

V b

div,0

(p) dp,

(8)

where

A b

div,0

(p)

and

V b

div,0

(p)

for

| p | ≤ 2 √

E

are given in termsof

f

lin on

M

E

aording to(2.10) , (2.21)and

| A

div,0err,j

(x, E) | ≤ (1 + √

2)c

1

(N, σ) k A k

N,σ

E

N22

,

(2.24)

| V

errdiv,0

(x, E) | ≤ c

1

(N, σ)

k V k

N,σ

E

N22

+ √

2 k A k

N,σ

E

N23

,

(2.25)

k A k

N,σ

= max k A

1

k

N,σ

, k A

2

k

N,σ

,

(2.26)

where

x ∈ R

2,

j = 1

,

2

,

E ≥

14,

A

div,0err

= (A

div,0err,1

, A

div,0err,2

)

and

c

1

(N, σ)

isdened

by (2.20). Furthermore, if

div A = 0

then

A

div,0

= A

,

V

div,0

= V

.

Theorem2.3. Supposethat

A

1,

A

2,

V ∈ C

N,σ

( R

2

)

forsome

N ≥ 4

and

σ > 2

.

Let

A

±,0,

V

±,0 bedenedaording to(2.5)(2.6). Then thefollowingformulas forsolving Problem 2.1arevalid:

A b

±1,0

(k − l) = 1 2

f (k, l) − f ( − l, − k)

k

1

+ l

1

± i(k

2

+ l

2

) , A b

±2,0

(k − l) = ± i A b

±1,0

(k − l), V b

±,0

(k − l) = (l

1

± il

2

)f (k, l) + (k

1

± ik

2

)f ( − l, − k)

k

1

+ l

1

± i(k

2

+ l

2

) ,

(2.27)

where

A b

±,0,

V b

±,0 are the Fourier transforms of

A

±,0,

V

±,0 (see (2.9)) and

(k, l) ∈ M

E;

A

±,0

(x) = A

±appr,0

(x, E) + A

±err,0

(x, E), x ∈ R

2

, E > 0,

(2.28)

A

±appr,0

(x, E) =

Z

|p|≤2√ E

e

ipx

A b

±,0

(p) dp, A

±err,0

(x, E) = Z

|p|≥2√ E

e

ipx

A b

±,0

(p) dp, V

±,0

(x) = V

appr±,0

(x, E) + V

err±,0

(x, E), x ∈ R

2

, E > 0,

(2.29)

V

appr±,0

(x, E) =

Z

|p|≤2√ E

e

ipx

V b

±,0

(p) dp, V

err±,0

(x, E) = Z

|p|≥2√ E

e

ipx

V b

±,0

(p) dp,

where

A b

±,0

(p)

and

V b

±,0

(p)

for

| p | ≤ 2 √

E

are given in terms of

f

lin on

M

E

aording to(2.10) , (2.27)and

| A

±err,j,0

(x, E) | ≤ (1 + √

2)c

1

(N, σ) k A k

N,σ

E

N22

,

(2.30)

| V

err±,0

| ≤ c

1

(N, σ)

k V k

N,σ

E

N22

+ √

2 k A k

N,σ

E

N23

,

(2.31)

where

x ∈ R

2,

j = 1

,

2

,

A

±err,0

= (A

±err,1,0

, A

±err,2,0

)

,

k A k

N,σ is dened by (2.23)

and

c

1

(N, σ)

isgiven by (2.20). Furthermore, if

A

1

± iA

2

= 0

then

A = A

±,0,

V = V

±,0.

Theorems2.12.3areprovedin Setion5.

(9)

3.1. Some notations. Tostudy Problem1.1 itis onvenienttointrodue

ϕ

div,

A

div,

V

divand

ϕ

±,

A

±,

V

±,where

ϕ

divand

ϕ

± aredenedaordingto(2.4)

(2.6)and

A

div

= A + ∇ ϕ

div

, V

div

= V − i∆ϕ

div

+ ( ∇ ϕ

div

)

2

+ 2A ∇ ϕ

div

, A

±

= A + ∇ ϕ

±

, V

±

= V − i∆ϕ

±

+ ( ∇ ϕ

±

)

2

+ 2A ∇ ϕ

±

.

(3.1)

Inthis setion wegiveanonlinearizedalgorithm for approximate nding

A

±,

V

± and

A

div,

V

div on

R

2 from

f

on

M

E. This algorithm takesinto aount

multiplesatteringeetsandanberegardedasanonlinearversionofformulas

for

A

±appr,0 ,

V

appr±,0

,

A

div,0appr ,

V

apprdiv,0

of (2.28),(2.29),(2.22),(2.23) .

Itisonvenienttousethefollowingnotations:

z = x

1

+ ix

2

, z ¯ = x

1

− ix

2

,

(3.2)

λ = E

1/2

(k

1

+ ik

2

), λ

= E

1/2

(l

1

+ il

2

),

(3.3)

where

x = (x

1

, x

2

) ∈ R

2,

k = (k

1

, k

2

) ∈ Σ

E,

l = (l

1

, l

2

) ∈ Σ

E,

Σ

E

=

m = (m

1

, m

2

) ∈ C

2

: m

21

+ m

22

= E , E > 0.

(3.4)

Inthese notations

k

1

= 1

2 E

1/2

(λ + λ

1

), k

2

= i

2 E

1/2

1

− λ),

(3.5)

l

1

= 1

2 E

1/2

+ λ

′−1

), l

2

= i

2 E

1/2

′−1

− λ

),

(3.6)

exp(ikx) = exp

i

2 E

1/2

(λ¯ z + λ

1

z)

,

(3.7)

where

λ

,

λ

∈ C \ { 0 }

,

z ∈ C

2,

k

,

l ∈ Σ

E.

Inaddition,usingformulas(1.9),(3.3),(3.4),(3.5),(3.6)oneansee that

Σ

E

∼ = C \ { 0 } , Σ

E

∩ R

2

= S

1

E

∼ = T, M

E

∼ = T × T,

(3.8)

where

S

1r

=

m ∈ R

2

: | m | = r , r > 0, T =

λ ∈ C : | λ | = 1 .

(3.9)

Inaddition,thefuntions

ψ

+,

f

of(1.5)(1.8)anbewritten as

ψ

+

= ψ

+

(z, λ, E), f = f (λ, λ

, E),

(3.10)

where

λ

,

λ

∈ T

,

z ∈ C

,

E > 0

.

(10)

nding

A

±,

V

± and

A

div,

V

div on

R

2 from

f

on

M

E hasthefollowingsheme

f −→ h

±

−→ µ

+

−→ µ

±

−→ A

±appr

, V

appr±

−→ A

divappr

, V

apprdiv

(3.11)

andonsistsofthefollowingsteps:

Step1. Find funtions

h

±

(λ, λ

, E)

,

λ

,

λ

∈ T

, from thefollowing linearintegral

equations:

h

±

(λ, λ

, E) − πi Z

T

h

±

(λ, λ

′′

, E)χ

± i λ

λ

′′

− λ

′′

λ

×

× f (λ

′′

, λ

, E) | dλ

′′

| = f (λ, λ

, E),

(3.12)

where

χ(s) =

1

for

s ≥ 0,

0

for

s < 0

. (3.13)

Step2. Solvethefollowinglinearintegralequationfor

µ

+

(z, λ, E)

,

z ∈ C

,

λ ∈ T

,

E > 0

:

µ

+

(z, λ, E) + Z

T

B(λ, λ

, z, E)µ

+

(z, λ

, E) | dλ

| = 1,

(3.14)

where

B(λ, λ

, z, E) = 1 2

Z

T

h

(ζ, λ

, z, E)χ

− i ζ

λ

− λ

ζ

dζ ζ − λ(1 − 0) −

− 1 2

Z

T

h

+

(ζ, λ

, z, E)χ

i ζ

λ

− λ

ζ

ζ − λ(1 + 0) ,

(3.15)

h

±

(λ, λ

, z, E) ==

def

h

±

(λ, λ

, E) ×

× exp

− i

√ E

2 (λ − λ

)¯ z + (λ

1

− λ

′−1

)z ,

(3.16)

and

λ

,

λ

∈ T

,

z ∈ C

,

E > 0

.

Step3. Denefuntions

µ

±

(z, λ, E)

,

z ∈ C

,

λ ∈ T

,

E > 0

,byformulas

µ

±

(z, λ, E) = µ

+

(z, λ, E) + πi

Z

T

h

±

(λ, λ

, z, E) ×

× χ

± i λ

λ

− λ

λ

µ

+

(z, λ

, E) | dλ

| ,

(3.17)

where funtions

h

±

(λ, λ

, z, E)

are given by (3.16) and

χ

is dened by

(3.13).

(11)

Step4. Funtions

A

±appr,j

(x, E)

,

V

appr±

(x, E)

,

x ∈ R

2,

j = 1

,

2

,

E > 0

, aredened

byformulas

A

appr,1

(x, E) = i

4 a

z

(z, E ), A

appr,2

(x, E) = 1

4 a

z

(z, E), a

z

(z, E) = 4∂

ln

Z

T

µ

+

(z, ζ, E) | dζ | ,

V

appr

(x, E) =

√ E π

Z

T

z

µ

(z, ζ, E) dζ,

(3.18)

and

A

+appr,1

(x, E) = i

4 a

+z¯

(z, E), A

+appr,2

(x, E) = − 1

4 a

+¯z

(z, E), a

+z¯

(z, E) = − 4∂

z

ln

Z

T

µ

+

(z, ζ, E) | dζ | ,

V

appr+

(z, E ) = 2i √ E∂

Z

T

µ

+

(z, ζ, E) dζ ζ

2

Z

T

µ

+

(z, ζ, E) dζ ζ

,

(3.19)

where

z

isgivenby(3.2).

Step5. Find

A

divappr,j

(x, E)

,

V

apprdiv

(x, E)

,

x ∈ R

2,

j = 1

,

2

,

E > 0

,from formulas

A

divappr,1

(x, E) = i

8 (a

z

(z, E ) + a

+¯z

(z, E)), A

divappr,2

(x, E) = 1

8 (a

z

(z, E ) − a

+¯z

(z, E)),

(3.20)

V

apprdiv

(x, E) = 1

2 V

appr

(x, E) + V

appr+

(z, E)

− 1

8 a

z

(z, E)a

+¯z

(z, E ),

where

z

isdenedby(3.2)andfuntions

a

z,

a

+,

V

appr±

aredenedin(3.18),

(3.19).

Aderivationofthisreonstrutionalgorithmisbasedonthemethod ofthe

RiemannHilbert problem and on the

∂ ¯

-method. This derivation is given in

Setion6.

For theasewhen

A ≡ 0

thisalgorithm isreduedtothealgorithmof[N4℄

forapproximatiending

V

on

R

2from

f

on

M

E. Thealgorithmof[N4℄onsists

ofthesameaforementionedsteps1,2,3andtheformula

V

appr

= V

appr ,where

V

appr

isdenedin (3.18). Thisalgorithmof[N4℄wasimplementednumerially

in[BAR℄.

Forthegeneralasethisalgorithmanbealsoregardedassimpliationand

development of the algorithm mentioned (in few lines) on page 457 of [N3℄.

Atually, in [N3℄ the part of the algorithm onsisting in nding

µ

± from

h

±

(12)

mentionedfortheasewhen

A

1

= A

1

, A

2

= A

2

, − 2i div A + V = V,

(3.21)

i.e. for theself-adjointase, whereasthis assumption is not neessaryfor the

algorithm.

3.3. Properties ofthe algorithm. Let

k u

1

k

L2(T)

= Z

T

| u

1

(λ) |

2

| dλ |

1/2

,

k u

2

k

L2(T2)

= Z

T2

| u

2

(λ, λ

) |

2

| dλ | | dλ

|

1/2

, T

2

= T × T,

(3.22)

where

u

1and

u

2aretestfuntions on

T

and

T

2,respetively.

Proposition 3.1. Let

E > 0

be xed. Supposethat

f ∈ L

2

(T

2

), k f k

L2(T2)

< 1

π ,

(3.23)

where

f = f (λ, λ

, E)

.Thenequation(3.12)isuniquelysolvablefor

h

±

∈ L

2

(T

2

)

and

k h

±

k

L2(T2)

< k f k

L2(T2)

1 − π k f k

L2(T2)

,

(3.24)

k B k

L2(T2)

< 2π k f k

L2(T2)

1 − π k f k

L2(T2)

,

(3.25)

where

B

isdenedby(3.15), (3.16) (atxed

z

,

E

). In addition, if

k f k

L2(T2)

< 1

3π ,

(3.26)

then

k B k

L2(T2)

< 1

, equation (3.14), at xed

z

,

E

, is uniquely solvable for

µ

+

∈ L

2

(T )

and

k µ

+

k

L2(T)

< (2π)

1/2

1 − k B k

L2(T2)

, k µ

+

− 1 k

L2(T)

< (2π)

1/2

k B k

L2(T2)

1 − k B k

L2(T2)

,

(3.27)

k µ

±

− 1 k

L2(T)

< 3π(2π)

1/2

k f k

L2(T2)

1 − 3π k f k

L2(T2)

,

(3.28)

where

µ

± are denedby (3.17). Inaddition, atleast, if

k f k

L2(T2)

< 1

6π ,

(3.29)

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