Constitutive law gap functionals to solve Cauchy problem for a linear elliptic PDE: a review

(1)

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problem for a linear elliptic PDE: a review

Thouraya Baranger, Stéphane Andrieux

To cite this version:

Thouraya Baranger, Stéphane Andrieux. Constitutive law gap functionals to solve Cauchy problem

for a linear elliptic PDE: a review. 2010. �hal-00489572v2�

(2)

problem for a linear ellipti PDE: a review

Thouraya N. Baranger

1

and Stéphane Andrieux

2

1

Universit deLyon,CNRS, UniversitLyon1,LaMCoSUMR5259,INSA-Lyon, F-69621,Villeurbanne,Fran e.

2

LaMSID,UMRCNRS-EDF2832,Clamart, Fran e.

Abstra t. Thispaperdes ribesageneralmethodbasedonminimizing onstitutive law gap fun tional in order to solve the Cau hy problem for a linear ellipti PDE. This fun tional measures the gap betweenthe solutionsof two well-posed problems. Ea h of these problems has one of the Cau hy data as known boundary ondition: Diri hlet or Neumann, and on the boundary where the data is la king, unknown Robinboundary onditions

η + ατ

and

η + βτ

areimposed,respe tively. Thedata

η

and

τ

haveto beidentied and(

α, β

)arepositivess alarsparameters ontrollingthe fun tional behavior. This approa h generalizes that presented in Andrieux et al [2℄ anden ompassesvariousmethodsproposedin theliterature. A ordingtothevalues of

α

and

β

when they tend toward

0

or

∞

, there are two groupsof methods: the rstgroupin ludesthose whi h depends ononlyoneunknowndata(

η

,

τ

or

η + ατ

). These ond groupin ludes thosewhi hdependontwounknown data

η

and

τ

. Then, theequivalen ebetweenEuler-Lagrange onditionsforthe onstitutivelawfun tionals and interfa ial operators usually used in the Domain De omposition eld is shown. UsingtheHadamardexampleweanalyseanalyti allythebehaviorof theseoperators asfun tionsoftheparameters(

α, β

). Then,thederivativesofthefun tionalaregiven usingadjointeldswhi hareparametrizedbythesameparameters. Finally,numeri al examplesaregiventoillustratethebehaviorofthesemethods,whi harenotfun tion oftheparameters(

α, β

)but alsooftheregularityoftheCau hydataandtheoverall geometryofthedomain.

(3)

1. Introdu tion

Consider a solid body

Ω

, given a ux

Φ

and the orresponding temperature

T

on

Γ

m

,

one wants to re over the orresponding ux and temperatureon the remaining part of

the boundary

Γ

u

, where

Γ

m

and

Γ

u

onstitute a partition of the whole boundary

∂Ω

.

The problemis therefore set as follows:

Find

(ϕ, T )

on

Γ

u

su h that there exists a eld

u

satisfying:











∇.k(x)∇u = 0

in

Ω

k(x)∇u.n = Φ

on

Γ

m

u = T

on

Γ

m

(1)

where the ondu tivity eld

k(x)

is real positive analyti in

L

∞

_(Ω)

. This problem is

known sin e Hadamard [16℄ to be ill-posed in the sense that the dependen e of

u

and

onsequently of

(ϕ, T )

on the data

(Φ, T )

isknown tobe not ontinuous.

We propose, in this paper, to identify the la king data

(ϕ, T )

by minimizing a

onstitutive lawgap fun tion whi h generalizes the one introdu edin [2℄. Then, Robin

(or Fourier)boundary onditions are dened on the

Γ

u

part of the boundary. The aim

istostudy ifbetternumeri albehavior anbeobservedwithspe ialvaluesoftheRobin

parameters. Werestri tourselveshereto ellipti operators althoughasimilarapproa h

an be applied to paraboli of hyperboli ones [1℄. Other ellipti operators des ribing

various physi alphenomena has been addressed in[5, 6,7, 13,14, 17℄.

Thispaperisorganizedasfollows: Inse tion2,afteraba kgroundontheliterature

dealing with Cau hy problem, two mixed well-posed problems are dened by splitting

the overspe ied data on

Γ

m

and Robin boundary onditions are introdu ed on the

boundary

Γ

u

. The latter are parametrized by two positive real onstants

α

and

β

.

Then, the boundary ondition identi ationproblem is dened asan optimization one

with onstraints, where the obje tive fun tional is a onstitutive law gap fun tion.

This fun tion quanties the energy gap between two elds solution of the well-posed

problems dened above, whi h onstitute the onstraints of the optimization problem.

Hen e, parti ular ases are outlined when

α

and

β

tend toward limit values

0

and

∞

. In se tion 3,we present anequivalent formulationbased on domainde omposition

strategy. Then, weshowfor allthe asesoutlinedinse tion 3,that the Euler-Lagrange

onditions for the onstitutive lawfun tionals and interfa ial operators are equivalent.

Hadamard example is presented in se tion 4, in order to illustrate the behavior of the

operators introdu ed above as fun tions of the Robin parameters

α

and

β

. In se tion

5,the evaluation ofthe derivatives ofthe onstitutive lawgap is given by usingadjoint

methods. Numeri al examples are presented in se tion 6, to illustrate the behavior

of the methods when the geometry or/and the boundary data on

Γ

m

and

Γ

u

present

(4)

2. Boundary onditions identi ation

ConsidertheaboveCau hy problem(1). Assumingthatthe data

(Φ, T )

are ompatible,

whi h means that this pair is indeed the tra e and normaltra e of a unique harmoni

fun tion

u

,extending the data means nding

(ϕ, T )

su h as:











∇.k(x)∇u = 0

in

Ω,

u = T, k(x)∇u.n = Φ

on

Γ

m

,

u = T , k(x)∇u.n = ϕ

on

Γ

u

(2)

The question now is how to re onstru t numeri ally the pair

(ϕ, T )

. In pra ti al

problems,dataisnotexpe tedtobe ompatible,sin edataerrors ano urfromerrors

inmeasurements. Theill-posednessinHadamard'ssenseshowsup-dramati ally-when

one tries toapproximate a given data

(Φ, T )

: it is possible to approa hit as losely as desired on

Γ

m

by tra es of a single harmoni fun tion, the "surprise" being a he ti

behavior of this fun tion onthe remainingpart of the boundary. This behavior an be

understoodbythe fa tthat the ompatibledata are dense inthe spa e of in ompatible

ones, whi h makes hopeless the natural idea of least square tting of the in ompatible

data by the ompatible ones. Regularizationpro edures are therefore required totreat

the data ompletion problem [22, 12, 11℄. There are several approa hes to regularize

su hill-posedness. Some of themtransform the ill-posedproblemintoawell-posed one

by adding a penalty term or by mollifying the data in order to avoid data os illations.

Tikhonov like methods use the penalty approa h. Another lass of rough but usually

e ientregularizingte hniquestrytosolvetheill-posedproblemiterativelyand hoosea

suitablestopping riteria,forinstan e L- urvebased riteria. Intheapproa hproposed

here, the introdu tion of two distin t elds, ea h of them meeting only one of the

over-spe ied data, turns out to avoid the need of a regularization pro edure for the

resolution of the data ompletionproblem,when the noise rate remainsreasonable, see

[5, 6, 7℄. Using separately the two boundary onditions on

Γ

m

has also been used in

the algorithmproposed by Kozlov et al [18℄ and analysed by Baumeister et al [8℄in a

general framework, where again no regularization pro edure is ast into the resolution

method.

We will restri t ourselves, throughout the paper for the setting to the ase where

the boundary

∂Ω

onsists of two losed manifoldsof lass

C

2

su h that

∂Ω = Γ

m

S

Γ

u

.

The following results remain true for less smooth boundaries and when

Γ

m

,

Γ

u

have

onta t points. However, for sake of simpli ity, we have hosen the above framework.

As already mentioned, the pairs of ompatible data are dense in the set of all possible

data pairs. Forthis known result werefertoFursikov [15℄and toapre eding paper[3℄,

where the mentionedproofs are adapted toour settings.

Lemma 2.1. (i) For a xed

T

in

H

1/2

_(Γ

m

)

, the set of data

Φ

for whi h there exists

a fun tion

u

in

H

1 _(Ω)

, satisfying the Cau hy problem (1) is everywhere dense in

H

−1/2

_(Γ

m

)

.

(5)

(ii) For a xed

Φ

in

H

−1/2

_(Γ

m

)

, the set of data

T

for whi h there exists a fun tion

u

in

H

1 _(Ω)

, satisfying the Cau hyproblem (1) is everywhere dense in

H

1/2

_(Γ

m

)

.

Observethat, whenthe omplete data are availableon

Γ

, wehaveanoverspe ied

boundary value problem











∇.k(x)∇u = 0

in

Ω,

u = T, k(x)∇u.n = Φ

on

Γ

m

,

u = T , k(x)∇u.n = ϕ

on

Γ

u

(3)

Theapproa hfollowedheregeneralizestheonegivenin[3℄. Itfollowstwosteps: onsider

foragivenpair

(η, τ ) ∈ H

−

1

2 (Γ

u

)×H

1

2 (Γ

u

)

thefollowingtwofamiliesofmixedwellposed problems











∇.k(x)∇u

1 = 0

in

Ω

u

1 = T

on

Γ

m

k(x)∇u

1 .n + αu

1 = η + ατ

on

Γ

u

(4)











∇.k(x)∇u

2 = 0

in

Ω

k(x)∇u

2 .n = Φ

on

Γ

m

k(x)∇u

2 .n + βu

2 = η + βτ

on

Γ

u

(5)

We denote by

α

and

β

two non-negative real oe ients. This ondition ensures that

problems (3) and (4) are well-posed. Using a

H

1

semi-norm the onstitutive law gap

fun tionalisdenedasameasureofthepseudo-energygapbetweenthetwoaboveelds

u

1

and

u

2

as follows:

E

αβ

(η, τ ) =

1

2 Z

Ω

k(x)(∇u

1 − ∇u

2 ).(∇u

1 − ∇u

2 )

(6)

This fun tional is positive and quadrati . Indeed,

u

1

and

u

2

are obviously equal

when the pair

(η, τ )

on the boundary

Γ

u

meets the a tual ompatible data pair

(ϕ, T ) ∈ H

−1/2

_{× H}

1/2

on the boundary

Γ

u

, then:

E

α,β

(η, τ ) = 0 = min

η,τ

E

αβ

(η, τ )

Thanks to the uniqueness of the Cau hy problem solution we an state that the data

ompletion problem an bea hieved through the minimization one:

(ϕ, T ) = arg min

η,τ

E

αβ

(η, τ )

(7)

UsingGreen theorem this fun tional an beexpressed asa boundary ontrol:

E

αβ

(η, τ ) =

1

2 Z

Γ

m

(k(x)∇u

1 .n − Φ)(T − u

2 )

(8)

+

1

2 Z

Γ

u

k(x)(∇u

1 − ∇u

2 ).n (u

1 − u

2 )

Notethattheboundary onditionsdened on

Γ

u

inthe problems(4)and(5)degenerate

(6)

we andenetwoapproa hesdieringbythenumberofunknowneldon

Γ

u

. Intherst

approa h there is onlyone unknown boundary eld, and, happens when

α = β

. Three

dierent single eld methods an be outlined: in the rst one, when

α = β = 0

, the

unknownboundarydataistheNeumannboundary ondition

η

,inthese ondapproa h,

when

α = β = ∞

, the unknown data is the Diri hlet boundary ondition

τ

and in the last one, when

0 < α = β < ∞

, the unknown data is the Robin boundary ondition

υ = η + ατ

. The se ond approa h is based on two unknown elds

(η, τ )

and in ludes the other ases where

α 6= β

.

Table1. Boundary onditionson

Γ

u

asfun tion oftheparameters

α

and

β

α = 0

α = ∞

0 < α < ∞

β = 0

∇u

1 .n = η

∇u

2 .n = η

u

1 = τ

∇u

2 .n = η

∇u

1 .n + αu

1 = η + ατ

∇u

2 .n = η

β = ∞

∇u

1 .n = η

u

2 = τ

u

1 = τ

u

2 = τ

∇u

1 .n + αu

1 = η + ατ

u

2 = τ

0 < β < ∞

∇u

1 .n = η

∇u

2 .n + βu

2 = η + βτ

u

1 = τ

∇u

2 .n + βu

2 = η + βτ

∇u

1 .n + αu

1 = η + ατ

∇u

2 .n + βu

2 = η + βτ

if

α = β

then:

∇u

1 .n + αu

1 = υ

∇u

2 .n + αu

2 = υ

2.1. A review

This general setting leads to an interesting interpretation for dierent values of

α

and

β

. In [3℄, we deal with the ase where

α = 0

and

β = +∞

. This two elds (i.e.

(η, τ ) ∈ H

−1/2

_(Γ

u

) × H

1/2

(Γ

u

)

)approa hhas been numeri allyexploredand turnedout

to be e ient and robust. Other authors, [4, 9,10℄ have explored mathemati ally and

numeri ally the ases with one single eld when

α = β = +∞

and its dual form when

α = β = 0

. One may,however, wantsto know whi h approa h ismore e ient. Noti e that when

α = β = +∞

, the rst optimality ondition of our optimization pro ess

lead tothe variationalformofthe well-known Steklov-Poin ar methodborrowed tothe

Domain De omposition eld, see [19, 21℄. This fa t has been already pointed out in

[3, 9℄. Moreover, when

α = β = 0

we nd the so- alled dual Steklov-Poin ar operator

method. An other alternative form with single eld formulation is the Alternating

Dire tion Iterative method, whi h onsists of solving two minimizationproblem where

ea hproblem depends ononlyone eld, su h that:

η

k

= arg min

η

E

αβ

(η, τ

k−1

))

(9) and

τ

k

= arg min

τ

E

αβ

(η

k

, τ ))

(10)

(7)

As it will be shown later, this last formulation turns out to be the KMF's (Kozlov,

Maz'ya and Fomin) pro ess des ribed in [3℄. Let us outline that, from the ontinuous

point of view, all these methods are equivalent. In fa t, they are all based on the

introdu tionoftwoelds

u

1

and

u

2

dealingseparatelywith the over-spe ieddata, and

the sear h for missing data by equalizingthe two elds. At the dis retelevel,all these

problems areof ourseill-posedbut someofthemare expe tedtobebetter onditioned

than others.

2.2. Two elds approa hes

Twoelds approa h an besetup if

0 ≤ α 6= β ≤ ∞

the onstitutivelawgap fun tional

an be expressed asfollows.

E

αβ

(η, τ ) =

1

2 Z

Γ

m

(k(x)∇u

1 .n − Φ)(T − u

2 )

(11)

+

1

2 Z

Γ

u

(α(τ − u

1 ) − β(τ − u

2 )) (u

1 − u

2 )

with,

u

1

and

u

2

arethe solutionof(4)and(5). Letusobservethat, for ompatibledata,

the followinglemma is straightforward.

Lemma 2.2. (Chara terization of the

u

1

and

u

2

elds at the minimum)

If

(Φ, T )

is a ompatible pair, there exists a pair

(ϕ, T )

solution of the Cau hy

problem su hthat:

(ϕ, T ) = arg min

η,τ

E

αβ

(η, τ ) and E

αβ

(ϕ, T ) = 0

Hen e,whenthefun tionalrea hesitsminimum,theelds

u

1

and

u

2

verify:

∇u

1 = ∇u

2

in

Ω

, whi h is equivalent to:

(

u

1 = u

2 + K

on

Γ

u

k(x)∇u

1 .n = k(x)∇u

2 .n on Γ

u

(12)

Where

K

is a real onstant. From (12) and the boundary onditionsdened in (4) and

(5),we an dedu e :

u = u

1 = τ +

β

α − β

K

(13)

∇u.n = η + α(τ − u) = η +

αβ

α − β

K

(14)

The general approa hbuilt with the energy and the Robin 's boundary onditions

an lead to many two elds methods by setting extreme values for

α

and

β

. These

methods are outlinedhereafter:

(i) if

α = 0

and

β = ∞

, the boundary onditions dened on

Γ

u

be omes Neumann

(8)

twowell-posedmixedproblems. Then,wedenotethefun tional

E

0∞

by

E

N D

whi h redu es to:

E

N D

(η, τ ) =

Z

Γ

u

(η − k(x)∇u

2 .n)(u

1 − τ ) +

Z

Γ

m

(k(x)∇u

1 .n − Φ)(T − u

2 )

(15)

When

E

N D

rea hes itsminimumwe have

u = u

1 = τ − K

and

∇u.n = η

. This ase

has been widely studiedin [2℄.

(ii) if

α = 0

and

β

is niteand non zero,the boundary ondition on

Γ

u

in(4)be omes

a Neumannone. Then we denotethe fun tional

E

0β

by

E

N β

whi hredu es to:

E

N β

(η, τ ) =

Z

Γ

u

β(u

2 − τ )(u

1 − u

2 ) +

Z

Γ

m

(k(x)∇u

1 .n − Φ)(T − u

2 )

(16)

When the fun tional

E

N β

rea hes its minimum we have

u = u

1 = τ − K

and

∇u.n = η

.

(iii) if

α = ∞

and

β

isniteandnonzero,the boundary onditionon

Γ

u

in(4)be omes

a Diri hlet one. Then we denotethe fun tional

E

∞β

by

E

Dβ

whi h redu es to:

E

Dβ

(η, τ ) =

Z

Γ

u

1 β

(k(x)∇u

2 .n−η)k(x)∇(u

1 −u

2 ).n+

Z

Γ

m

(k(x)∇u

1 .n−Φ)(T −u

2 )

(17)

When the fun tional

E

Dβ

rea hes itsminimum wehave

u = u

1 = τ

and

∇u.n = η

.

Similar results an be a hieved by swit hing

α

and

β

in the above ases. The only

dieren eliesintheadditional onditionsrequiredon

u

2

andthe ompatibility ondition

on the ux in the ase where (4) is a Neumann problem. In on lusion, to avoid

supplementary onstraints onthe elds

u

2

, the ases where

β = 0

should be avoided.

2.3. Single eldapproa hes:

α = β

The single eld approa h an be set up when

α = β

. Three dierent ases an be

distinguished:

(i) The Neumann approa h is obtained for

α = β = 0

. The unknown boundary data

isthe Neumannboundary ondition

η

. Wedenotethe fun tional

E

00

by

E

N

whi h

an beexpressed as boundary ontrolon

Γ

m

:

E

N

(η) =

Z

Γ

m

(k(x)∇u

1 .n − Φ)(T − u

2 )

(18)

(ii) The Diri hlet approa h is obtained for

α = β = ∞

. The unknown boundary data

istheDiri hletboundary ondition

τ

. Wedenotethefun tional

E

∞∞

by

E

D

whi h

an beexpressed as boundary ontrolon

Γ

m

:

E

D

(η) =

Z

Γ

m

(k(x)∇u

1 .n − Φ)(T − u

2 )

(19)

(iii) The Robin approa h is obtained for

0 < α = β < ∞

. The auxiliary unknown

boundary data is the Robin boundary ondition

υ = η + ατ

. We denote the

fun tional

E

αα

by

E

α

whi h an be expressed asfollows:

E

α

(υ) =

Z

Γ

m

(k(x)∇u

1 .n − Φ)(T − u

2 ) −

Z

Γ

u

k(x)

α

(∇u

1 .n − ∇u

2 .n)

2

(20)

(9)

or equivalently:

E

α

(υ) =

Z

Γ

m

(k(x)∇u

1 .n − Φ)(T − u

2 ) −

Z

Γ

u

α(u

1 − u

2 )

2

(21)

Here too, we observe that, for ompatible data, the following lemma is

straightforward.

Lemma 2.3. (Chara terization of

u

1

and

u

2

elds at the minimum)

If

(Φ, T )

is a ompatible pair, there exists a pair

(ϕ, T )

solution of the Cau hy

problem su hthat:

ϕ = arg min

η

E

N

(η) and E

N

(ϕ) = 0

T = arg min

τ

E

D

(τ ) and E

D

(T ) = 0

(ϕ + αT ) = arg min

υ

E

α

(υ) and E

α

(ϕ + αT ) = 0

Hen e, when these fun tionals rea h their minimum, the elds

u

1

and

u

2

verify:

∇u

1 = ∇u

2

in

Ω

, whi h is equivalent to:

(

u

1 = u

2 + K

on

Γ

u

k(x)∇u

1 .n = k(x)∇u

2 .n on Γ

u

(22)

Where

K

isa onstant. This onstantisundeterminedfor therst ase

(i)

, be ausethe

se ond problem (5) is Neumannproblem. For the se ond and third ases

K = 0

.

To highlight the properties of the dierentparametrization of the onstitutivelaw

fun tionals, we will now derive the rst optimalityor Euler-Lagrange onditions as an

interfa ial equation onthe boundary

Γ

u

where the data is unknown.

3. Euler-Lagrange onditions and interfa ial operators.

In the literature there are methods to equalize the two elds

u

1

and

u

2

on

Γ

u

using

onstraints onditions known as interfa e onditions. These methods are issued from

the domain de ompositioneld, see Quarteroni et al [21℄. However, it shouldbenoted

that,here there isonlyone domainandaboundarywhileinthe domainde omposition

methods, the interfa esare lo ated between subdomains. Nevertheless, inour problem,

Γ

u

plays the role of the interfa e.

We will prove hereafter that these pseudo interfa e onditions are equivalent to

the rst optimality ondition (12) of the minimization problem (6). First dierent

interfa e operators are dened on

Γ

u

,then for ea h ase previously emphasized aproof

isdeveloped. Tobeginwith, we onsider the followingmixed boundaryvalue problems:











∇.k(x)∇w

o

1 = 0

in

Ω

w

o

1 = 0

on

Γ

m

k(x)∇w

o

1 .n + αw

o

1 = υ

on

Γ

u

and











∇.k(x)∇w

∗

1 = 0

in

Ω

w

∗

1 = T

on

Γ

m

k(x)∇w

∗

1 .n + αw

∗

1 = 0

on

Γ

u

(23)

(10)









∇.k(x)∇w

o

2 = 0

in

Ω

k(x)∇w

o

2 .n = 0

on

Γ

m

k(x)∇w

o

2 .n + βw

2 o

= υ

on

Γ

u

and









∇.k(x)∇w

∗

2 = 0

in

Ω

k(x)∇w

∗

2 .n = Φ

on

Γ

m

k(x)∇w

∗

2 .n + βw

2 ∗

= 0

on

Γ

u

(24)

Given

υ

, for ea h

i = 1, 2, w

o

i

will be denoted by

H

i

(υ)

. We denoted also

R

1 (T )

and

R

2 (Φ)

insteadof

w

∗

1

and

w

∗

2

. Thanks to the linearityof the problems we an state

that:

u

1 = H

1 (υ) + R

1 (T ) and u

2 = H

2 (υ) + R

2 (Φ)

Now,forea h

i = 1, 2

wedenetheRobin toNeumannoperators

S

i

andthe Robin toDiri hlet operators

S

e

i

as follows:

S

i

: H

−1/2

(Γ

u

) −→ H

−1/2

(Γ

u

)

υ

7−→ k(x)∇(H

i

(υ)).n

(25)

e

S

i

: H

−1/2

(Γ

u

) −→ H

1/2

(Γ

u

)

υ

7−→ H

i

(υ)

(26)

The operators (

S

1 , e

S

1

) and (

S

2 , e

S

2

) depend onthe parameters

α

and

β

respe tively. In

the following these operators will be applied to

υ

elds expressed as

η + ατ

or

η + βτ

todeal with the problems (4) and (5)respe tively, with

(η, τ ) ∈ H(Γ

u

)

1/2

_{× H(Γ}

u

)

−1/2

.

Remark now, that for

α = 0

and

β = 0

, the Robin boundary onditions on

Γ

u

be ome

Neumann ones. Then, the operators

S

1

and

S

2

are the identity operator

I

and the

operators

S

e

1

and

S

e

2

arethe well-knownPoin ar-Steklovoperators. However, for

α = ∞

and

β = ∞

theRobinboundary onditionson

Γ

u

be omesDiri hletones. Theoperators

S

1

and

S

2

are redened as follows and are the lassi alSteklov-Poin ar operators:

S

1 : H

1/2

(Γ

u

) −→ H

−1/2

(Γ

u

)

τ

7−→ k(x)∇(H

1 (τ )).n

(27)

S

2 : H

1/2

(Γ

u

) −→ H

−1/2

(Γ

u

)

τ

7−→ k(x)∇(H

2 (τ )).n

(28)

while the operators

S

e

1

and

S

e

2

are the identity operator

I

.

Lemma 3.1. If

0 < α 6= β < ∞

, the interfa e onditionson

Γ

u

are:

u

1 = u

2 + K

(29)

k(x)∇u

1 .n = k(x)∇u

2 .n

(30)

Using the operators dened above, the interfa e onditions an be expressed as follows:

"

e

S

2 − e

S

1 β e

S

2 − α e

S

1 S

1 − S

2 αS

1 − βS

2 #

|

{z

}

S

αβ

(

η

τ

)

=

(

X

1 X

2 )

(31) where

X

1 = R

1 (T ) − R

2 (Φ) − K

(32)

X

2 = − k(x)∇R

1 (T ).n + k(x)∇R

2 (Φ).n

(33)

(11)

The interfa e onditions statedin the aboveLemmabe omefor the dierent ases

outlinedin the table 1:

(i) if

α = 0

and

β = ∞

then

u = τ − K

,

∇u.n = η

and:

"

− e

S

1 I

I

−S

2 #

|

{z

}

S

N D

(

η

τ

)

=

(

X

1 X

2 )

(ii) if

α = β = 0

then

u = u

1 = u

2 + K

,

∇u.n = η

and:

S

N

η = ( e

S

2 − e

S

1 )η = X

1

(iii) if

α = β = ∞

then

K = 0

,

u = u

1 = u

2 = τ

,

∇u.n = ∇u

1 .n

and:

S

D

τ = (S

1 − S

2 )τ = X

2

(iv) if

0 < α = β < ∞

then

υ = η + ατ

,

K = 0

,

u = u

1 = u

2 = τ

,

∇u.n = η

and:

S

α

υ =

e

S

2 − e

S

1 υ = X

1

(v) if

α = 0

and

0 < β < ∞

then

k(x)∇u.n = η

,

u = u

1 = τ − K

and:

"

e

S

2 − e

S

1 β e

S

2 I − S

2 −βS

2 #

|

{z

}

S

N β

(

η

τ

)

=

(

X

1 X

2 )

(vi) if

α = ∞

and

0 < β < ∞

then

K = 0

,

k(x)∇u.n = η

,

u = u

1 = u

2 = τ

and:

"

e

S

2 − I

β e

S

2 S

1 − S

2 −βS

2 #

|

{z

}

S

Dβ

(

η

τ

)

=

(

X

1 X

2 )

Remark. Noti e thatthe operator

S

αβ

is symetri .

3.1. Case

0 < α 6= β < +∞

In this se tion we show that the optimality onditions stated in the Lemma 2.2 are

equivalent to the interfa e onditions stated in the Lemma 3.1. We onsider the

fun tional dened by (6) and the elds

H

1

and

H

2

whi h are the solutions of the two

rst well posed problems dened by (23) and (24). These elds depend linearly on

η

and

τ

.

Lemma 3.2. The rst optimality ondition of the fun tional

E

αβ

reads:

e

S

2 − e

S

1 η +

β e

S

2 − α e

S

1 τ = X

1

(34)

(S

1 − S

2 ) η + (αS

1 − βS

2 ) τ

= X

2

(35)

(12)

Proof. Let us re all that

u

1

and

u

2

depend linearly on the variable

η

and

τ

, and

υ = η + ατ

for

H

1

and

υ = η + βτ

for

H

2

,then:

∂E

αβ

(η, τ )

∂η

.δη =

Z

Ω

k(x) (∇u

1 (η, τ ) − ∇u

2 (η, τ )) (∇H

1 (δη) − ∇H

2 (δη))

∂E

αβ

(η, τ )

∂τ

.δτ =

Z

Ω

k(x) (∇u

1 (η, τ ) − ∇u

2 (η, τ )) (α∇H

1 (δτ ) − β∇H

2 (δτ ))

Whi h an be writtenas follows:

∂E

αβ

(η, τ )

∂η

.δη =

Z

∂Ω

k(x) (∇u

1 (η, τ ) − ∇u

2 (η, τ )) .nH

1 (δη) − ∇H

2 (δη).n(u

1 (η, τ ) − u

2 (η, τ ))

∂E

αβ

(η, τ )

∂τ

.δτ =

Z

∂Ω

αk(x) (∇u

1 (η, τ ) − ∇u

2 (η, τ )) .nH

1 (δτ ) − β∇H

2 (δτ ).n(u

1 (η, τ ) − u

2 (η, τ ))

using the properties of the elds

H

1

and

H

2

we obtain:

∂E

αβ

(η, τ )

∂η

.δη =

Z

Γ

u

k(x)∇ (u

1 (η, τ ) − u

2 (η, τ )) .nH

1 (δη) − ∇H

2 (δη).n(u

1 (η, τ ) − u

2 (η, τ ))

∂E

αβ

(η, τ )

∂τ

.δτ =

Z

Γ

u

αk(x)∇ (u

1 (η, τ ) − u

2 (η, τ )) .nH

1 (δτ ) − β∇H

2 (δτ ).n(u

1 (η, τ ) − u

2 (η, τ ))

The stationarity ondition leads to:











∀ δη ∈ H

−1/2

R

Γ

u

k(x) (∇u

1 (η, τ ) − ∇u

2 (η, τ )) .nH

1 (δη) = 0

R

Γ

u

(u

1 (η, τ ) − u

2 (η, τ ))k(x)∇H

2 (δη).n = 0

∀ δτ ∈ H

1/2

R

Γ

u

αk(x) (∇u

1 (η, τ ) − ∇u

2 (η, τ )) .nH

1 (δτ ) = 0

R

Γ

u

β(u

1 (η, τ ) − u

2 (η, τ ))k(x)∇H

2 (δτ ).n = 0

(36)

Here we introdu ethe following Lemma:

Lemma 3.3. Consider two fun tions:

(ψ, ξ) ∈ H

1/2

_(Γ

u

) × H

−1/2

(Γ

u

)

with

R

Γ

u

ξ = 0

,

there exists a pair

(̟

ψ

, ̟

ξ

) ∈ H

−1/2

_(Γ

u

) × H

−1/2

(Γ

u

)

.











∇.k(x)∇w

1 = 0

in

Ω

w

1 = 0

on

Γ

m

k(x)∇w

1 .n + αw

1 = ̟

ψ

on

Γ

u

and











∇.k(x)∇w

2 = 0

in

Ω

k(x)∇w

2 .n = 0

on

Γ

m

k(x)∇w

2 .n + βw

2 = ̟

ξ

on

Γ

u

(37) With

w

1 = ψ

and

k(x)∇w

2 .n = ξ

.

The proof of this Lemma is detailed in the annexe. Using the above Lemma, we

denote by

ψ

and

ξ

respe tively

H

1 (δη)

and

∇H

2 (δτ ).n

, then we an state that:

Z

Γ

u

k(x) (∇u

1 (η, τ ) − ∇u

2 (η, τ )) .nψ = 0 ∀ ψ ∈ H

1/2

(Γ

u

)

(38)

Z

Γ

u

(u

1 (η, τ ) − u

2 (η, τ ))ξ

= 0 ∀ ξ ∈ H

−1/2

(Γ

u

)

(39) This leads to:

(13)

Then, using the operatorsdened above, weobtain:

( e

S

2 − e

S

1 )η + (β e

S

2 − α e

S

1 )τ = R

1 (T ) − R

2 (Φ) − K

and

(S

1 − S

2 )η + (αS

1 − βS

2 )τ = −k(x)∇R

1 (T ).n + k(x)∇R

2 (Φ).n

The rstoptimality onditions areexa tlythe interfa e onditionstatedinLemma

3.1.

"

e

S

2 − e

S

1 β e

S

2 − α e

S

1 S

1 − S

2 αS

1 − βS

2 # (

η

τ

)

=

(

X

1 X

2 )

(40) 3.2. Case:

α = 0

and

β = +∞

In this ase,

η

and

τ

denote the unknown values of

k(x)∇u.n

and the tra e of

u

on

Γ

u

.

The mixed boundary value problems dened by (23) and (24) be ome:











∇.k(x)∇w

o

1 = 0

in

Ω

w

o

1 = 0

on

Γ

m

k(x)∇w

o

1 .n = η

on

Γ

u

and











∇.k(x)∇w

∗

1 = 0

in

Ω

w

∗

1 = T

on

Γ

m

k(x)∇w

∗

1 .n = 0

on

Γ

u

(41)











∇.k(x)∇w

o

2 = 0

in

Ω

k(x)∇w

o

2 .n = 0

on

Γ

m

w

o

2 = τ

on

Γ

u

and











∇.k(x)∇w

∗

2 = 0

in

Ω

k(x)∇w

∗

2 .n = Φ

on

Γ

m

w

∗

2 = 0

on

Γ

u

(42)

The Cau hy problem (1) is solved, if and only if

u

1 = u

2 + K

and

∇u

1 (η).n =

∇u

2 (τ ).n

on

Γ

u

whi h are the interfa e onditions dened in (12).

E

N D

reads:

"

− e

S

1 I

I

−S

2 # (

η

τ

)

=

(

X

1 X

2 )

Proof. The prooffollows the same steps asthat of Lemma3.5and 3.6. In this ase the

fun tion depends onthe pair (

η

,

τ

):

E

N D

(η, τ ) =

1

2 Z

Ω

k(x) (∇u

1 (η) − ∇u

2 (τ )) (∇u

1 (η) − ∇u

2 (τ ))

Here, two optimality onditions have tobesatised:

∂E

N D

(η, τ )

∂η

.δη =

Z

Ω

k(x) (∇u

1 (η) − ∇u

2 (τ )) ∇H

1 (δη)

=

Z

Γ

u

(14)

∂E

N D

(η, τ )

∂η

.δη = 0 ∀δη =⇒

_Γ

_u

k(x) (∇u

1 (η) − ∇u

2 (τ )) .n = 0

=⇒ ∇u

1 (η).n = ∇u

2 (τ ).n

=⇒ Iη − S

2 τ = −k(x)∇R

1 (T ).n + k(x)∇R

2 (Φ).n

∂E

N D

(η, τ )

∂τ

.δτ = −

Z

Ω

k(x) (∇u

1 (η) − ∇u

2 (τ )) ∇H

2 (δτ )

= −

Z

Γ

u

(u

1 (η) − u

2 (τ )) .k(x)∇H

2 (δτ ).n because ∇H

2 (δτ ).n = 0 on Γ

m

∂E

N D

(η, τ )

∂τ

.δτ = 0 ∀δτ =⇒ u

1 (η) − u

2 (τ ) = K because

Z

Γ

u

k(x)∇H

2 (δτ ).n = 0

=⇒ ( e

S

1 η − Iτ ) = −R

1 (T ) + R

2 (Φ) + K

Therefore the optimality ondition isequivalentto the interfa ial equation dened

in the ase 1of the Lemma3.1:

"

− e

S

1 I

I

−S

2 # (

η

τ

)

=

(

X

1 X

2 )

3.3. The ase

α = β = +∞

This ase orrespond to the so- alled Cau hy-Poin ar-Steklov method, see Quarteroni

et al [21℄ and Ben Belga em et al [9℄. The unknown boundary ondition on

Γ

u

is the

Diri hlet one denoted by

τ

. The mixed boundary value problems dened by (23) and

(24) be ome:











∇.k(x)∇w

o

1 = 0

in

Ω

w

o

1 = 0

on

Γ

m

w

o

1 = τ

on

Γ

u

and











∇.k(x)∇w

∗

1 = 0

in

Ω,

w

∗

1 = T

on

Γ

m

,

w

∗

1 = 0

on

Γ

u

(43)











∇.k(x)∇w

o

2 = 0

in

Ω

k(x)∇w

o

2 .n = 0

on

Γ

m

w

o

2 = τ

on

Γ

u

and











∇.k(x)∇w

∗

2 = 0

in

Ω,

k(x)∇w

∗

2 .n = Φ

on

Γ

m

,

w

∗

2 = 0

on

Γ

u

(44)

Then

H

i

for

i = 1, 2

are fun tion of

τ

only. The Cau hy problem (1) is solved, if and only if

∇u

1 .n = ∇u

2 .n

on

Γ

u

whi h is the se ond optimality ondition dened in 2.2, the rst one issatised by the denition of

τ

.

E

D

reads:

(15)

Proof. Letus re all that

E

D

depends onlyon the variable

τ

:

E

D

(τ ) =

1

2 Z

Ω

k(x) (∇u

1 (τ ) − ∇u

2 (τ )) (∇u

1 (τ ) − ∇u

2 (τ ))

∂E

D

(τ )

∂τ

.δτ

=

Z

Ω

(∇u

1 (τ ) − ∇u

2 (τ )) (∇H

1 (δτ ) − ∇H

2 (δτ ))

(45)

=

Z

Ω

(∇u

1 (τ ) − ∇u

2 (τ )) .∇H

1 (δτ ) −

Z

Ω

(∇u

1 (λ) − ∇u

2 (λ)) .∇H

2 (δτ )

=

Z

∂Ω

(∇u

1 (τ ) − ∇u

2 (τ )) .n.H

1 (δτ ) −

Z

∂Ω

(u

1 (τ ) − u

2 (τ )) .∇H

2 (δτ ).n

one has:

Z

Γ

u

(u

1 (τ ) − u

2 (τ )) .∇H

2 (δτ ).n = 0 because u

1 − u

2 = 0

Z

Γ

m

(u

1 (τ ) − u

2 (τ )) .∇H

2 (δτ ).n = 0 because ∇H

2 (δτ ).n = 0

Z

Γ

m

(∇u

1 (τ ) − ∇u

2 (τ )) .n.H

1 (δτ ) = 0 because H

1 (δτ ) = 0

Then:

∂E

D

(τ )

∂τ

.δτ

=

Z

Γ

u

(∇u

1 (τ ) − ∇u

2 (τ )) .n.H

1 (δτ )

∂E

D

(τ )

∂τ

.δτ = 0 ∀δτ =⇒ (∇u

1 (τ ) − ∇u

2 (τ )) .n = 0

=⇒ ∇u

1 .n = ∇u

2 .n

=⇒ (S

1 − S

2 )τ = X

2 = −k(x)∇R

1 (T ).n + k(x)∇R

2 (Φ).n

(S

1 − S

2 )τ = X

2 .

3.4. The ase

α = β = 0

This ase orresponds tothe so- alledNeumann toNeumann Steklov-Poin ar method.

The unknown boundary ondition on

Γ

u

isthe Neumannone denoted by

η

. The mixed

boundary value problems dened by (23) and (24) be ome:











∇.k(x)∇w

o

1 = 0

in

Ω

w

o

1 = 0

on

Γ

m

k(x)∇w

o

1 .n = ¯

η

on

Γ

u

and











∇.k(x)∇w

∗

1 = 0

in

Ω

w

∗

1 = T

on

Γ

m

k(x)∇w

∗

1 .n = −

|Γ

1 u

|

R

Γ

m

Φ

on

Γ

u

(46)

(16)









∇.k(x)∇w

o

2 = 0

in

Ω

k(x)∇w

o

2 .n = 0

on

Γ

m

k(x)∇w

o

2 .n = ¯

η

on

Γ

u

and









∇.k(x)∇w

∗

2 = 0

in

Ω,

k(x)∇w

∗

2 .n = Φ

on

Γ

m

k(x)∇w

∗

2 .n = −

|Γ

1 u

|

R

Γ

m

Φ

on

Γ

u

(47) with

¯

η = η −

1 | Γ

u

|

Z

Γ

u

η

Here the elds

H

i

for

i = 1, 2

are fun tion of

η

¯

only. The following supplementary ondition is ne essary in this ase:

Z

Γ

m

Φ +

Z

Γ

u

η = 0

u

1 = u

2 + K

on

Γ

u

whi h is the rst

optimality ondition dened in 2.2, the se ond one is satisedby the denition of

η

.

E

N

reads:

Sη = ( e

S

2 − e

S

1 )η = X

1 = R

1 (T ) − R

2 (Φ) − K

Proof. In this ase the fun tiondepends only onthe variable

η

E

N

(η) =

1

2 Z

Ω

k(x) (∇u

1 (η) − ∇u

2 (η)) (∇u

1 (η) − ∇u

2 (η))

The optimality ondition is then:

∂E

N

(η)

∂η

.δη =

Z

Ω

k(x) (∇u

1 (η) − ∇u

2 (η)) (∇H

1 (δη) − ∇H

2 (δη))

(48)

=

Z

Ω

k(x) (∇u

1 (η) − ∇u

2 (η)) .∇H

1 (δη) −

Z

Ω

k(x) (∇u

1 (η) − ∇u

2 (η)) .∇H

2 (δη)

=

Z

∂Ω

k(x) (∇u

1 (η) − ∇u

2 (η)) .n.H

1 (δη) −

Z

∂Ω

(u

1 (η) − u

2 (η)) .k(x)∇H

2 (δη).n

Sin e

H

1 (δη) = 0

on

Γ

m

,

∇u

1 (η).n = ∇u

2 (η).n

on

Γ

u

and

k(x)∇H

2 (δη).n = 0 on Γ

m

:

∂E

N

(η)

∂η

.δη

=

−

Z

Γ

u

(u

1 (η) − u

2 (η)) k(x)∇H

2 (δη).n = −

Z

Γ

u

(u

1 (η) − u

2 (η)) .δ ¯

η

∂E

N

(η)

∂η

.δη = 0 ∀ δ ¯

η =⇒

Z

Γ

u

(u

1 (η) − u

2 (η)) .δ ¯

η = 0

(49)

=⇒ u

1 (η) − u

2 (η) = K because

Z

Γ

u

δ ¯

η = 0

=⇒ ( e

S

2 − e

S

1 )η = X

1 = R

1 (T ) − R

2 (Φ) − K

(50)

( e

S

2 − e

S

1 )η = X

1

(17)

3.5. The ase:

0 < α = β < ∞

This ase orrespond to the Diri hlet to Robin operator used in the domain

de omposition eld. Let

υ

be the unknown data on

Γ

u

.

The mixed boundary value

problems dened by (23) and (24) be ome:











∇.k(x)∇w

o

1 = 0

in

Ω

w

o

1 = 0

on

Γ

m

k(x)∇w

o

1 .n + αw

o

1 = υ

on

Γ

u

and











∇.k(x)∇w

∗

1 = 0

in

Ω

w

∗

1 = T

on

Γ

m

k(x)∇w

∗

1 .n + αw

∗

1 = 0

on

Γ

u











∇.k(x)∇w

o

2 = 0

in

Ω

k(x)∇w

o

2 .n = 0

on

Γ

m

k(x)∇w

o

2 .n + αw

o

2 = υ

on

Γ

u

and











∇.k(x)∇w

∗

2 = 0

in

Ω

k(x)∇w

∗

2 .n = Φ

on

Γ

m

k(x)∇w

∗

2 .n + αw

∗

2 = 0

on

Γ

u

1 = u

2 + K

on

Γ

u

whi h is the

rst optimality ondition dened in2.2, the se ond one is satisedby the denition of

υ

. Here

K = 0

.

E

α

reads::

e

S

2 − e

S

1 υ = X

1

Proof. The proof follows the same steps as that of above Lemma. In this ase the

fun tion depends on

υ = η + ατ

:

E

α

(υ) =

1

2 Z

Ω

k(x) (∇u

1 (υ) − ∇u

2 (υ)) (∇u

1 (υ) − ∇u

2 (υ))

∂E

α

(υ)

∂υ

.δυ

=

Z

Ω

k(x) (∇u

1 (υ) − ∇u

2 (υ)) (∇H

1 (δυ) − ∇H

2 (δυ))

=

Z

∂Ω

k(x) (∇u

1 (υ) − ∇u

2 (υ)) .n.H

1 (δυ) − (u

1 (υ) − u

2 (υ))k(x)∇H

2 (δυ).n

Sin e

H

1 (δυ)

and

∇H

2 (δυ).n = 0

on

Γ

m

then:

Z

Γ

m

k(x) (∇u

1 (υ) − ∇u

2 (υ)) .n.H

1 (δυ) − (u

1 (υ) − u

2 (υ))∇H

2 (δυ).n = 0

Then:

∂E

α

(υ)

∂υ

.δυ =

Z

Γ

u

k(x) (∇u

1 (υ) − ∇u

2 (υ)) .n.H

1 (δυ) − (u

1 (υ) − u

2 (υ))k(x)∇H

2 (δυ).n

Usingthe Robin boundary ondition:

k(x) (∇u

1 (υ) − ∇u

2 (υ)) .n = −α(u

1 (υ) − u

2 (υ))

the above optimality ondition be omes:

∂E

α

(υ)

∂υ

.δυ = −

Z

Γ

u

(18)

Using the Lemma 3.3 for

α = β

, we denote by

ψ

and

ξ

respe tively

H

1 (δυ)

and

∇H

2 (δυ).n

,then we an state that:

∂E

α

(υ)

∂υ

.δυ = 0 ∀ δυ =⇒ u

1 (υ) − u

2 (υ) = 0 on Γ

u

∀ (ψ, ξ) ∈ H

−1/2

_(Γ

u

) × H

1/2

(Γ

u

)

=⇒

S

e

2 − e

S

1 υ = R

1 (T ) − R

2 (Φ)

e

S

2 − e

S

1 υ = X

1

Remark that the ase

α = β = 0

is found by setting in the above optimality ondition

α = 0

.

3.6. Alternating Dire tion Iterative method

We onsider the Alternating Dire tion Iterative (ADI) method, whi h generates two

sequen es of tra es

υ

k

1

and

υ

k

2

build with the tra es of

u

k

1|Γ

u

and

u

k

2|Γ

u

respe tively.

Consider an initialguess

υ

0

2

; then, for

k ≥ 0

we look for

u

k+1

1

and then

u

k+1

2

su h that:











∇.k(x)∇u

k+1

₁

= 0

in

Ω

u

k+1

₁

= T

on

Γ

m

k(x)∇u

k+1

₁

.n + αu

k+1

₁

= υ

₁

k+1

on

Γ

u

(51) with

υ

k+1

1 = k(x)∇u

k

2 .n + αu

k

2

.











∇.k(x)∇u

k+1

₂

= 0

in

Ω

k(x)∇u

k+1

₂

.n = Φ

on

Γ

m

k(x)∇u

k+1

₂

.n + βu

k+1

₂

= υ

₂

k+1

on

Γ

u

(52) with

υ

k+1

2 = k(x)∇u

k+1

1 .n + βu

k+1

1

.

Using the operators

S

1

,

S

e

1

,

S

2

and

S

e

2

dened above, it is easy to show that it is

a xed-point iteration:

υ

k+1

2 = (S

1 + β e

S

1 )υ

k+1

1

, then using the expression of

υ

k+1

1

we obtain

υ

k+1

2 = (S

1 + β e

S

1 )(S

2 + α e

S

2 )υ

2 k

. Then:

υ

₂

k+1

= b

S

αβ

υ

2 k

, k ≥ 0

where the xed pointmap is given asfollows:

b

S

αβ

: H

1/2

(Γ

u

) −→ H

1/2

(Γ

u

)

υ

k

2 7−→

S

b

αβ

υ

k

2 = υ

2 k+1

with:

S

b

αβ

= (S

1 + β e

S

1 )(S

2 + α e

S

2 )

. The KMFiterativemethodoutlinedabovehappens when

α = 0

and

β = ∞

and then

S

b

N D

= e

S

1 S

2

. We an then on lude that the KMF

method an be interpreted as xed point resolution of aninterfa e problem

υ = b

S

αβ

υ

.

In this se tion, we showed that all the above methods are equivalent, from the

ontinuous point of view. However, we expe t that their numeri al behavior will be

(19)

4. Hadamard example

As pointed out previously, at the dis rete level the interfa ial operators outlined in

se tion 3 are expe ted to be dierently onditioned. The aimof this se tion is to give

ananalyti al taste towhat goes onfor the ondition numbers.

Let us onsider an annular domain with an outer

r

m

= 1

and aninner

r

u

≡ r < 1

radii,

k(x) = 1

and the polar oordinates system. The overspe ied data are available

on the external boundary

Γ

m

, whereas the la king data are on the inner boundary

Γ

u

.

The analyti al solution of the problems (4) and (5) take the general form of separate

variables fun tions:

H

1 (υ) =

∞

X

n=1

(r

n

− r

−n

)g

1 (nθ)

and

H

2 (υ) =

∞

X

n=1

(r

n

+ r

−n

)g

2 (nθ)

respe tively, whi h willbeusedto al ulatethe eigenvalues. Theeigenvalues of

S

n

1

,

S

n

2

,

e

S

n

1

and

S

e

n

2

are then given by the followingsequen e:

λ

n

1 =

n(r

2n

_{+ 1)}

n(r

2n

_{+ 1) − αr(r}

2n

_{− 1)}

(53)

λ

n

₂

=

n(r

2n

_{− 1)}

n(r

2n

_{− 1) − βr(r}

2n

_{+ 1)}

(54)

e

λ

n

₁

=

−r(r

2n

_{− 1)}

n(r

2n

_{+ 1) − αr(r}

2n

_{− 1)}

(55)

e

λ

n

₂

=

−r(r

2n

_{+ 1)}

n(r

2n

_{− 1) − βr(r}

2n

_{+ 1)}

(56)

orresponding to the eigenve tors

g

n

1 = cos(nθ)

and

g

n

2 = sin(nθ)

, respe tively. The

interfa ial operator

S

n

αβ

dened in the Lemma 3.1 an be expressed as follows for the

n

-rank:

S

_αβ

n

=

"

e

λ

n

2 − e

λ

n

1 βe

λ

n

2 − αe

λ

n

1 λ

n

1 − λ

n

2 αλ

n

1 − βλ

n

2 #

(57)

For the spe ial values of parameters

α

and

β

the above operator be omes:

S

_{N D}

n

=

"

_r

n

(

r

2n

₋₁

r

2 n

₊₁

)

1

1 n

_r

(

r

_r

2n

−1

₊₁

)

#

(58)

S

N β

n

=







r

n

4nr

2n

_+βr

4n+1

_−βr

(−nr

4n

_+βr

4n+1

_+2βr

2n+1

_+n+βr)

β

(

r

2n+1

_+r

₎

−r

2n

_n+n+βr

2n+1

_+βr

β

(

r

2n+1

_+r

₎

−r

2n

_n+n+βr

2n+1

_+βr

βn

(

r

2n

₋₁

₎

−r

2n

_n+n+βr

2n+1

_+βr







(59)

S

_Dβ

n

=







r

2n+1

_+r

−r

2n

_n+n+βr

2n+1

_+βr

n

(

r

2n

₋₁

₎

−r

2n

_n+n+βr

2 n+1

_+βr

n

(

r

2n

₋₁

₎

−r

2n

_n+n+βr

2n+1

_+βr

n

_r

(

−r

4n

_{n+4 βr}

2n+1

_+n

₎

(r

4n

_{n−2 r}

2n

_n−r

4n+1

_β+n+βr)







(60)

S

_D

n

= −

n

r

4r

2n

r

4n

_{− 1}

(61)

(20)

S

_N

n

= −

r

n

4r

2n

r

4n

_{− 1}

(62)

S

_α

n

= 4

r

2n+1

_n

−n

2 _r

4n

_{+ 2r}

4n+1

_{αn + n}

2 _{+ 2 nαr − α}

2 _r

4 n+2

_{+ α}

2 _r

2

(63)

Consider the one eld operators:

S

n

N

,

S

n

D

and

S

n

α

, their asymptoti development

when

n −→ ∞

shows that

S

n

N

≈ −

4 n

r

2n+1

,

S

n

D

≈ −4nr

2n−1

and

S

n

α

≈

n

4 r

2n+1

. Then,

one an dedu es that

S

n

D

and

S

n

α

, whi h have the same behavior, de rease faster than

S

n

D

. The operators whi h depend on two unknown elds, tend toward the following

expressions when

n −→ ∞

:

S

n

N D

≈

"

−

r

n

1

1 −

n

_r

#

; S

n

Dβ

≈

"

r

n

−1

n

_r

#

; S

n

N β

≈

"

r

2 n

2 _n

r

n

−1

#

(64)

From these expressions one an dedu es that the rst eigenvalue of ea h operator

vanishes qui kly. The se ond eigenvalues of

S

n

N D

and

S

n

Dβ

tend toward

n

r

and

−n

r

,

respe tively. However, the se ond eigenvalues of

S

n

N β

tend toward

−β

. the operators

S

n

N D

and

S

n

Dβ

have the same behavior when

n −→ ∞

. Then the operator

S

n

N β

has

the best behavior. The same results are obtained for the Hadamard example on the

square. The gures 1 and 2 show the evolution of the ondition number of the single

eld operatorsand the two eldsoperators, respe tively. They show the samebehavior

for high frequen ies for all these methods. These analysis give anidea on the behavior

ofea hoperator,butit isnotsu ient tode idewhi hmethodisbetter inanabsolute

way. In fa t, others parameters su h geometri and Cau hy data singularities, or the