Schwarz method in a geometrical multi-scale domain using finite volume schemes.

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Schwarz method in a geometrical multi-scale domain

using finite volume schemes.

Marie-Claude Viallon

To cite this version:

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nite volume s hemes.

Marie-Claude Viallon

Univ Lyon, UJM-Saint-Étienne, CNRS UMR 5208, Institut CamilleJordan, 10rue Trélerie,

CS 82301, F-42023Saint-Etienne Cedex 2,Fran e

Abstra t. Domainde omposition methodswith anitevolumedis retizationtosolvea

para-boli linear partial dierential equation in a geometri al multi-s ale domain are studied. The

model domain onsists of a two-dimensional (2D) entral node, and several one-dimensional

(1D) outgoing bran hes. The interfa e onditions between the node and the bran hes are

ei-ther ontinuity of the solution orthe ux. The optimized S hwarz methodis adjusted and the

onvergen e isstatedinea h ase.The methodisimplementedinthe formofinterfa e system.

The results are ompared with those froma monolithi s heme, a ura y is studiedas a

fun -tion of the Robin parameter. Several variants are developed, anew e ientpre onditioning of

the interfa e system isdened.

Résumé. Nous étudions des méthodes de dé omposition de domaine ave une dis rétisation

de type volumes nis pour résoudre une équation aux dérivées partielles parabolique linéaire

dans une stru ture géométriquement multi-é helle. La stru ture est onstituée d'un noeud

bi-dimensionnel (2D) d'où partent plusieurs bran hes uni-dimensionnelles (1D). Les onditions

d'interfa e entre le noeud etles bran hes sont la ontinuité ou bien de la solution,ou bien du

ux. La méthode de S hwarz optimisée est adaptée et la onvergen e est établie dans haque

as. La méthode est programmée sous laformed'un système d'interfa e.Les résultatsobtenus

sont omparésave euxdonnésparuns hémamonolithique:uneétudedelapré isionde

l'ap-proximationen fon tiondu paramètre de Robinest faite.Plusieursvariantes sontdéveloppées.

On dénitnotammentun nouveau pré onditionnementtrès e a e du système d'interfa e.

Mathemati s Subje t Classi ation: 35K05, 74S10,65F10, 65N08, 65N12, 65N55.

Mots lés:nitevolumes heme,paraboli problem,multi-s aledomain,domainde omposition,

stability and onvergen e of numeri al methods, optimized S hwarz methods, Robin interfa e

ondition.

1 Introdu tion

The model problem that is onsidered here is set in a geometri al multi-s ale domain

de-riving from a nite rod stru ture, whi h is a onne ted nite union of re tangles in our ase.

Arterial treesinthe ardiovas ularsystem(see [?℄),systems ofpipesinindustrialinstallations,

or anal systems, are lassi alexamples of rodstru tures. When the knowledge of the solution

is needed in the whole domain, the rods are generally onsidered as one-dimensional domains

at some distan e from the jun tions to redu e the omputational osts. This therefore leads

to solve the problem in a geometri al multi-s ale domain (a single numeri al model with

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Themodelproblem,whi his onsideredhere,istheheatequationsetina1D-2Dmulti-s ale

domain. Two options are provided for the 1D-2D oupling interfa e onditions : the solution

andthemeanvalueoftheuxare ontinuous,ortheuxandthemeanvalueofthesolutionare

ontinuous. A monolithi nite volume numeri al s heme of hybrid dimension for solving the

modelproblemwas dened in[?℄ and[?℄ ineither ase. Aniterativeapproa hby solving

sepa-rately thesubproblemswasadopted in[?℄,following[?℄ or[?℄,interfa e systemswere obtained,

whi hweresolvedby usingaKrylovmethod.Comparisonofthe resultswiththeones obtained

bysolvingthemonolithi s hemeswithadire tmethodweredone.However,in[?℄,the oupling

interfa e onditionsthat are hosenforthe separated2D problemwere justpointwiseDiri hlet

onditions tosolvethe modelproblemwhen the solutionis ontinuous, orpointwiseNeumann

onditions to solve the model problem when the ux is ontinuous. In the present paper, we

lookatnon-standard2Dproblems by using forinstan e boundary onditions inrelationtothe

mean value of the ux. This 2D problem is not well dened as su h, but assuming the

solu-tion is onstant along the interfa e makes it well-posed, it is used to solve the model problem

whose solution is ontinuous. Using boundary ondition in relation to the mean value of the

solution and assumingthe ux is onstant along the interfa e gives the possibilityto solve the

modelproblemwhose uxis ontinuous.In the same vein, spe i Robin boundary onditions

alongtheinterfa esareproposedforthe2DproblemandleadtotheoptimizedS hwarzmethod.

There are dierent versions of the S hwarz method (see for instan e [?℄ and the referen es

ontained therein). We look at optimized versions of the method be ause they are onvergent

without overlap, and this is the only kind of domain de omposition methods we are

inter-essed in. The original optimized S hwarz method, whi h was rst introdu ed in [?℄ at the

ontinuouslevel,repla esthe lassi alDiri hletinterfa e onditionswithRobininterfa e

ondi-tions.OtheroptimizedS hwarz methodsuse even moregeneralinterfa e onditionsthanRobin

ones. It was rst suggested in [?℄ to use nonlo al transmission onditions to lead to

opti-mal results in terms of iteration ounts in domain de omposition. Numerous approximations

of these onditions were developed for dierent problems by many authors (see for instan e

[?℄,[? ℄,[?℄,[?℄,[?℄,[?℄,[?℄,[? ℄,[?℄,[?℄,[?℄,[?℄). Nevertheless, the problem that is solved in this paper

has a very fewnumber of unknowns sin e itis set in ageometri al multi-s aledomain.

Conse-quently, the number of iterationsneeded to a hieve onvergen e by using the Krylov method

(GMRES)isverylow.Thatiswhythereisnoneedtooptimizethetransmission onditions

(es-pe ially sin ethe optimizationis performedfor the onvergen e fa tor of the iterativemethod

andnotforaparti ularKrylovmethodappliedtotheinterfa esystem).Consequently,onlythe

version proposed in [?℄ with Robin interfa e onditions will be used in this paper. Regarding

the dis reteversion ofthe method,several dis retizationswere studiedin[?℄,[?℄,[? ℄,[?℄.In[?℄,a

nite volume s heme for onve tion diusion equation onnon mat hing grids is proposed and

studied. Inthepresentpaper,themethodisextended togeometri allymulti-s aledomainsand

anitevolumes hemeisproposedtoapproa hthemodelproblemforea htransmission

ondi-tion. The onvergen e of the numeri al solution towards the one given by the orresponding

monolithi s hemeisproved ineither ase.Itiswellknown thattheoptimizedS hwarz method

leadstoaJa obimethodtosolveaninterfa e problemandthe pro edureisusuallya elerated

by using a Krylov method instead (see [?℄), it is shown here that this property is still true.

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blems. The iterativestrong ouplingmethodis onsideredbe ause itis espe iallydedi atedto

geometri almulti-s alemodels.Themethodologyisproposedin[?℄anddeveloped in[?℄,[?℄,[?℄.

The pro edureinvolvestwotimesmoreinterfa eunknowns than lassi almethodsbut itisnot

adrawba k sin ethereisasmallnumberofinterfa eunknowns inthis kindof problems.Along

the same lines, inorder tohave anoverall vision, Diri hlet or Neumann, ouplingwith Robin,

boundary onditions are alsoused to solve the separated problems, and this leads to study a

new family of s hemes.

Domain de ompositionmethodsare generallyamenable to pro edures for aninterfa e

pro-blem,whi h an orrespondtotheS hur omplementsystemortheuxequation(see[?℄,[?℄,[?℄).

Several denitionsoftheS hur omplementsystem,theuxequation,ortheRobinequationin

the geometri almulti-s ale ase are proposed in[?℄, whi hare obtainedfromdierentinterfa e

onditions. These lastpro edures giveinterfa e systems with as many unknowns as interfa es.

In the present paper, su h pro edures deriving from the new s hemes are also onsidered. All

s hemes are ompared and analyzedfrom analgebrai and numeri alpoint of view.

This work is organized asfollows. Se tion 2 presents the model problemwith two possible

transmission onditions. The hybrid monolithi s heme developed previously in either ase is

re alled inSe tion 3.In Se tion 4,the S hwarz algorithmis dened and studied: onvergen e

is proved, the iterative strong oupling pro edure is applied, some lassi al domain

de ompo-sition strategies givinginterfa e systems with as many unknowns as interfa es are onsidered.

Other s hemes involving not only Robin interfa e onditions, but also Diri hlet or Neumann

values are onsideredinSe tion5.Wegivenumeri alexperimentsinSe tion6,givingtheerror

between the approximations obtained by ea h monolithi s heme and the asso iated interfa e

systems deriving from the domainde omposition methods.

2 The model problem

2.1 Des ription of the geometri al multi-s ale 1D-2D domain

The geometri almulti-s ale1D-2D domainon whi h our modelproblem isset was dened

in [?℄.It onsists of one node and

p

bran hes. The onstru tionis remindedbelowfor the sake of ompleteness.

Let

e

j

= [O, O

j

], j = 1, ..., p,

be

p

losed segments in

IR

2

, having a ommon end point

denoted by

O

, with length

l

j

= OO

j

, j = 1, ..., p

.

Let

(x, y)

denotethe oordinates inthe anoni albasis of

IR

2

,and

(x

e

j

_{, y}

e

j

₎

denotethelo al

oordinates asso iated with the segment

e

j

, j = 1, ..., p

. This lo al system is orthonormal and su h that

x

e

j

is the oordinatein the dire tion

e

j

.

Let

ε > 0.

Let

θ

1 , ..., θ

n

be positive numbers independent of

ε

.

Let

B

ε

j

= {(x, y) | x

e

j

∈ (0, l

j

), y

e

j

∈ (−

εθ

₂

j

,

εθ

₂

j

)},

and

β

ε

j

= {(x, y) | x

e

j

= l

j

, y

e

j

∈

(−

εθ

j

2 ,

εθ

j

2 )}.

Let

ω

0

be a bounded domain in

IR

2

with smooth boundary ontaining

O

(see [?℄). Let

ω

ε

0 = {(x, y) |

(x,y)−O

ε

∈ ω

0 }

. Weassume that

B

ε

j

\ ω

ε

0 ∩ B

i

ε

\ ω

0 ε

= ∅, i 6= j

.The domain

ω

ε

0

(see the dottedlineinFigure1(a))is addedinordertosmooth theboundaryof thenal stru ture

ω

ε

(5)

Let

Ω

ε

= ∪

p

j=1

B

j

ε

∪ ω

0 ε

.

x

y

A

x

y

l

O

Ω

ε

ω

0 δ

δ

5 O

l

₃

l

₂

4 l

β

1 ε

θ

ε

1

1 e

e

x

e

1

2

3

4

5 e1

1 ε

x

y

A

x

y

δ

4 l

e

x

e

2

3

4

5 e1

1 γ

γ

1

2 γ

₃

4

5 δ

δ

D

_ε

1 l

2

3 l

5 l

S

1

2

3

4

5 S

O

Ω(0)

Figure 1 (a)The rod stru ture

Ω

ε

and (b) The geometri almulti-s aledomain

D

ε

.

Now, let us des ribe the 1D-2D domain under onsideration. Let

δ > 0

, su h that

δ <

min

{l

j

, j = 1, ..., p}

and su h that

ω

ε

0

is inthe ballof enter

O

and radius

δ

. Denote

B

′ε

j

= B

j

ε

∩{(x, y) | x

e

j

∈ (0, δ)}, j = 1, ..., p

.Denote

Ω(0) = ∪

p

j=1

B

′ε

j

∪ω

0 ε

.

Therefore

Ω(0)

is atrun ated part of

Ω

ε

. Let

S

j

= {(x, y) | y

e

j

_{= 0, x}

e

j

_{∈ (δ, l}

j

)}, j = 1, ..., p,

besegments su h that

S

j

⊂ e

j

.

We denoteby

γ

j

= {(x, y) | x

e

j

_{= δ, y}

e

j

_{∈ (−}

εθ

j

2 ,

εθ

j

2 )}, j = 1, ..., p,

the 1D-2D interfa es. Let us dene

D

ε

= Ω(0) ∪ ∪

p

j=1

S

j

. The set

D

ε

is the so- alled geometri al multi-s ale domain.

2.2 The model problem

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









1 θ

j

ε

Z

γ

j

u(., t)dγ = v

j

(δ, t), j = 1, ..., p, t ∈ (0, T )

(d)

∂u

∂n

(x, y, t) =

∂v

j

∂x

e

j

(δ, t), (x, y) ∈ γ

j

, j = 1, ..., p, t ∈ (0, T )

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that is tosay ontinuity of the uxon the interfa es or ontinuity of the fun tion itself.

Weassumethatthefun tions

f

j

(respe tively

f

)aresmoothandindependentof

ε

,are onstant with respe t to

(x, y)

in some neighborhoodof

O

j

, j = 1, ..., p

(respe tively

O

),and vanish for

t ≤ t

0 , t

0 > 0

. In addition, for (??)-(??), we assume they are su h that the partial derivatives

of

u

in

Ω(0)

and

v

j

in

[δ, l

j

], j = 1, ..., p,

existand are ontinuous tillthe ordertwo.This allows

us toobtain the error estimatein [?℄.

Inthefollowing,theproblemsabovearereferredtoas(??)" sa"and(??)"dsa"respe tively,

to indi ate a ontinuous or dis ontinuous sele ted alternative, as the solution of the problem

(??)-(??)is ontinuous whereas the solutionof the problem(??)-(??) may be dis ontinuous.

3 Monolithi numeri al s hemes

We re all here the monolithi numeri al s hemes whi h are onstru ted in [?℄ and [?℄ to

approa h the solutionof (??)" sa" and (??)"dsa".

3.1 The meshes

First, a mesh of

S

j

on the axis

Ox

e

j

_{, j = 1, ..., p}

, is dened. For ea h value of

j

, we hoose

N

j

∈ IN

∗

,

and

N

j

+ 1

distin t and in reasing values

x

e

j

i+1/2

, i = 0, ..., N

j

, su h that

x

e

j

1/2

= δ, x

e

j

N

j

+1/2

= l

j

. Denote

I

e

j

i

= (x

e

j

i−1/2

, x

e

j

i+1/2

)

,and

h

e

j

i

= x

e

j

i+1/2

− x

e

j

i−1/2

, i = 1, . . . , N

j

. Set

h

e

j

₌

max

{h

e

j

i

, i = 1, ..., N

j

}

the size of the mesh of the interval

(δ, l

j

)

.

Then we hoose

N

j

points

x

e

j

i

, i = 1, ..., N

j

,

su h that

x

e

j

i

∈ I

e

j

i

. Set

x

e

j

0 = δ, x

e

j

N

j

+1

= l

j

, and

h

e

j

i+1/2

= x

e

j

i+1

− x

e

j

i

, i = 0, ..., N

j

.

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K

denoted by

P

. The mesh

T

satises the following properties 1)The losure of the union of allthe ontrolvolumes is

Ω(0).

2)

For any

K ∈ T ,

there isa subset

E

K

of

E

su h that

∂K =

[

σ∈E

K

σ,

and

[

K∈T

E

K

= E.

3)

For any

(K, L) ∈ T

2 _{, K 6= L,}

one of three followingassertions holds:

either

K ∩ L = ∅,

or

K ∩ L

is a ommon vertex of Kand L,

or

K ∩ L = σ, σ

being a ommon edge of Kand L denoted by

σ

K/L

.

4)The family

P = (x

K

)

K∈T

is su h that for any

K ∈ T , x

K

∈ K.

For any

(K, L) ∈ T

2 _{, K 6= L,}

it is assumed that

x

K

6= x

L

and thatthe straight linegoing through

x

K

and

x

L

is orthogonal to

σ

K/L

.

5)

For any

σ ∈ E,

if

σ ⊂ ∂Ω(0), σ ∈ E

K

and

x

K

∈ σ,

/

the orthogonal

proje tion of

x

K

onthe straightline ontainingthe edge

σ,

belongs to

σ.

(4)

Let

E

int

= {σ ∈ E, σ 6⊂ ∂Ω(0)}

.

For any

(K, L) ∈ T

2 _{, K 6= L}

, if

σ = σ

K/L

, let

d

σ

be the distan e between

x

K

and

x

L

. For any

K ∈ T

,if

σ ∈ E

K

and if

σ ⊂ ∂Ω(0),

let

d

σ

bethe distan e between

x

K

and

σ

.

Weassume that for any

σ ∈ E, d

σ

6= 0.

For any

K ∈ T ,

let

m(K)

be the area of

K

. For any

σ ∈ E,

let

m(σ)

be the length of

σ

. Let

h

0

be the size ofthe mesh

T

,

h

0 =

max

{

diam

(K), K ∈ T }

,wherediam isthe abbreviation for

diameter.

Thenumberofedgesof

γ

j

is alled

n

σ,j

, j = 1, ..., p

,andthe numberofedgesof

∪

p

j=1

γ

j

is alled

n

σ

=

P

p

_j=1

n

σ,j

.

We denote by

T S

the global 1D-2D mesh of

D

ε

. Let

h

be the size of the 1D-2D mesh of

D

ε

:

h =

max

{h

0 , h

e

j

_{, j = 1, ..., p}}

.

3.2 The monolithi s hemes

A onstant time step

k ∈ (0, T )

is merely hosen for the time dis retization.Let

N

k

∈ IN

∗

su h that

N

k

=

max

{n ∈ IN, nk < T }

. Let

t

n

= nk

, for

n ∈ {0, ..., N

k

+ 1}

. The value

v

n

j,i

is an approximation of

v

j

(x

e

j

i

, t

n

), i = 0, ..., N

j

+ 1

. The value

u

n

K

is an approximation of

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









h

e

j

i

v

_j,i

n+1

− v

n

j,i

k

+ F

j,n+1

i+1/2

− F

j,n+1

i−1/2

= h

e

j

i

f

e

j

,n

i

, i = 1, ..., N

j

, j = 1, ..., p,

n ∈ {0, ..., N

k

} (a)

F

_i+1/2

j,n

= −

v

n

j,i+1

− v

n

j,i

h

e

j

i+1/2

, i = 0, . . . , N

j

, v

j,N

n

j

+1

= 0, j = 1, ..., p,

n ∈ {1, ..., N

k

+ 1}

f

e

j

,n

i

=

1 h

e

j

i

Z

x

ej

_i+1/2

x

ej

_i−1/2

f

j

(x

1 , t

n+1

)dx

1 , i = 1, . . . , N

j

, j = 1, ..., p,

n ∈ {0, ..., N

k

}

m(K)

u

n+1

K

− u

n

K

k

+

X

σ∈E

K

F

_K,σ

n+1

= m(K)f

_K

n

, ∀K ∈ T , n ∈ {0, ..., N

k

} (b)

F

n

K,σ

= −

m(σ)

d

σ

(u

n

L

− u

n

K

), ∀σ ∈ E

int

,

if

σ = σ

K/L

, n ∈ {1, ..., N

k

+ 1}

F

n

K,σ

=







−

m(σ)

d

σ

(v

_j,0

n

− u

n

_K

) , ∀σ ⊂ γ

j

, σ ∈ E

K

, j = 1, ..., p

(d)

0 , ∀σ ⊂ ∂Ω(0)\(∪

p

_j=1

γ

j

)

f

n

K

=

1 m(K)

Z

K

f (x, t

n+1

)dx, ∀K ∈ T , n ∈ {0, ..., N

k

}

v

n

j,1

− v

n

j,0

h

e

j

1/2

=

1 θ

j

ε

X

σ⊂γ

j

σ∈E

K

m(σ)

d

σ

(v

_j,0

n

− u

n

_K

), j = 1, ..., p,

n ∈ {1, ..., N

k

+ 1} (c)

(5)

with the initial ondition

u

0 K

= 0, ∀K ∈ T

v

0 j,i

= 0, i = 1, ..., N

j

, j = 1, ..., p

Let us re all that

v

n

j,0

is a onvex ombination of the approximated values of the solution

on ea h side of

γ

j

, j = 1, ..., p

:

v

_j,0

n

=





v

n

j,1

h

e

j

1/2

+

1 θ

j

ε

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

u

n

_K









1 h

e

j

1/2

+

1 θ

j

ε

X

σ⊂γ

j

m(σ)

d

σ





−

1

(6)

Forthe sakeofsimpli ity,in(?? )and(??),thesummationisdonefor

σ ⊂ γ

j

,andforea h of them,

K

is the ontrolvolumesu h that

σ ∈ E

K

.

We remind now the nite volume s hemes orresponding with the alternative (??)"dsa".

The value ofthe solutionisnot assumed tobe onstantonthe interfa es,and theapproximate

value ofthesolutionontheedge

σ

ontheinterfa e

γ

j

attime

t

n

is alled

u

n

σ

, σ ⊂ γ

j

, j = 1, ..., p

.

The equation (??d) is repla ed by

F

_K,σ

n

= −

m(σ)

d

σ

(9)











1 θ

j

ε

X

σ⊂γ

j

σ∈E

K

m(σ)u

n

σ

= v

j,0

n

, j = 1, ..., p,

(d)

u

n

σ

− u

n

K

d

σ

=

v

n

j,1

− v

j,0

n

h

e

j

1/2

, ∀σ ⊂ γ

j

, σ ∈ E

K

, j = 1, ..., p

(8) that gives

v

_j,0

n

=



(

X

σ⊂γ

j

m(σ)d

σ

)v

n

j,1

+ h

e

j

1/2

X

σ⊂γ

j

,σ∈E

K

m(σ)u

n

_K









X

σ⊂γ

j

m(σ)d

σ

+ θ

j

εh

e

_1/2

j





−

₁

(9)

The s heme resulting from the equations (??) with (??)and (??) in pla e of (?? ) and (??d)

is the monolithi s heme that is used tosolve(??)"dsa".

Themonolithi s hemeslead toalinearsysteminthe formof

AU

n+1

_{= BU}

n

_{+ F}

n

inwhi h

U

n

is known and

U

n+1

is unknown,

n ∈ {0, ..., N

k

}

, the initialvalue

U

0

being given;

(U

n

₎

T

_{= {{v}

n

j,i

, i = 1, ..., N

j

}, j = 1, ..., p}, {u

n

K

, K ∈ T }

.

It isproved in[?℄ and[?℄ thatthereisaunique solution

(U

n+1

₎

toea hofthese linearsystems,

(U

n

₎

being given. The orresponding pie ewise onstant fun tions on

D

ε

are referred to as monolithi solutionsof (??)" sa" and (??)"dsa".

3.3 The spe i norm

A dis rete

L

2

norm is previously dened in [?℄ and [? ℄ in the spa e of pie ewise onstant

fun tions on

D

ε

.

Denition 1 We dene

X(T S)

theset of thefun tions from

D

ε

to

IR

whi hare onstantover ea h ontrol volume of

T S

. Let

W ∈ X(T S)

,

W (x, y) =

W

_W

K

, (x, y) ∈ K, K ∈ T

j,i

, (x, y) ∈ S

j

, x

e

j

∈ (x

_i−1/2

e

j

, x

e

_i+1/2

j

), i = 1, ..., N

j

, j = 1, ..., p.

We dene

kW k

2,T

=





X

K∈T

m(K)W

_K

2 +

p

X

j=1

θ

j

ε

N

j

X

i=1

h

e

j

i

W

j,i

2 



1

2

.

4 Domain de omposition strategy

Impli it Euler method is used for the time dis retization, then the domain de omposition

(10)

is illustrated by anexample.For instan e, tosolve (??)" sa", wewrite the segregated s hemes











v

n+1

_j

(x

e

j

_{) − v}

n

j

(x

e

j

)

k

−

∂

2 _v

n+1

j

∂x

e

j

2 (x

e

j

_{) = f}

j

(x

e

j

, t

n+1

), x

e

j

∈ (δ, l

j

),

j = 1, ..., p

v

0 j

(x

e

j

) = 0, x

e

j

∈ (δ, l

j

), j = 1, ..., p

v

_j

n+1

(l

j

) = 0

v

_j

n+1

(δ) = α

n+1

_j

, j = 1, ..., p

(a)

∂v

n+1

_j

∂x

e

j

(δ) = β

n+1

j

, j = 1, ..., p

(b)

(10)











u

n+1

_{(x, y) − u}

n

_{(x, y)}

k

− △u

n+1

_{(x, y) = f (x, y, t}

n+1

), (x, y) ∈ Ω(0)

u

0 _{(x, y) = 0, (x, y) ∈ Ω(0)}

∂u

n+1

∂n

(x, y) = 0, (x, y) ∈ ∂Ω(0)\(∪

p

j=1

γ

j

)

u

n+1

_{(x, y) = α}

n+1

j

, (x, y) ∈ γ

j

, j = 1, ..., p

(a)

1 θ

j

ε

Z

γ

j

∂u

n+1

∂n

dγ = β

n+1

j

, j = 1, ..., p

(b)

(11)

then the 1D problems (??) and the 2D problem (??) are onsidered as bla k-boxes whi h

re eive the input data (

α

n+1

j

, j = 1, ..., p

) and give ba k

(β

n+1

j

, j = 1, ..., p)

as output data (or

onversely if the problemis well dened, for a total of four possibilities).The options (a) and

(b) areex lusive.Forinstan e, letus hoose that(??)re eivestheinput

(β

n+1

j

, j = 1, ..., p)

and

(??) re eivesthe input(

α

n+1

j

, j = 1, ..., p

).

The fully dis retes heme orresponding with (??) and (??)is the following:

(11)











m(K)

u

n+1

K

− u

n

K

k

+

X

σ∈E

K

F

_K,σ

n+1

= m(K)f

_K

n

, ∀K ∈ T , n ∈ {0, ..., N

k

}

F

_K,σ

n+1

= −

m(σ)

d

σ

(u

n+1

_L

− u

n+1

K

), ∀σ ∈ E

int

,

if

σ = σ

K/L

, n ∈ {0, ..., N

k

}

F

_K,σ

n+1

=







−

m(σ)

d

σ

(α

n+1

_j

− u

n+1

_K

) , ∀σ ⊂ γ

j

, σ ∈ E

K

, j = 1, ..., p

(a)

0 , ∀σ ⊂ ∂Ω(0)\(∪

p

_j=1

γ

j

)

u

0 K

= 0, ∀K ∈ T ,

f

n

K

=

1 m(K)

Z

K

f (x, t

n+1

)dx, ∀K ∈ T , n ∈ {0, ..., N

k

}

1 θ

j

ε

X

σ⊂γ

j

σ∈E

K

m(σ)

d

σ

(α

n+1

_j

− u

n+1

K

) = β

j

n+1

, j = 1, ..., p, n ∈ {0, ..., N

k

}

(b)

(13)

Letus denote (??-option (b)) (resp. (??-option (a))) the system (??) (resp. (??)) with its

lastequation (??a)(resp. (??b)) removed.

It is lear that

v

n+1

j,i

, i = 1, ..., N

j

, u

n+1

K

, K ∈ T

, are solution of the monolithi s heme i

v

n+1

_j,i

, i = 1, ..., N

j

,aresolutionof(??-option(b)),

u

n+1

K

, K ∈ T ,

aresolutionof(??-option(a))),

and

α

1 , ..., α

p

, β

1 , ..., β

p

,

aresolutionoftheinterfa esystem onstitutedby(??a)and(??b)that

an be ondensed into

α

n+1

j

− S

N D,n+1

1D,j

(β

n+1

j

) = 0, j = 1, ..., p

β

_j

n+1

− R

j

S

_2D

DN,n+1

E(α

n+1

1 , ..., α

n+1

p

) = 0, j = 1, ..., p

(14)

The strong oupling method solves the interfa e system. The 1D dis rete operator

S

N D,n+1

1D,j

onsists tosolve the 1D Neumann problem(??-option (b)) and to give an approximate value

of the solution at the point

δ

on the axis

j

. The 2D dis rete operator

S

DN,n+1

2D

onsists to

solve a general2D Diri hlet problemlike (??-option (a)) but with a Diri hlet boundary value

whi h is not ompulsory onstant, and to give the value on

γ

j

of a dis rete normal derivative of the solution. The operator

R

j

is the mean value operator and

E

is the trivial extension operator. More information an be found in [?℄. The interfa e system (??) has

2p

unknowns

α

1 , ..., α

p

, β

1 , ..., β

p

,thatistosaytwi easmanyunknownsasinterfa es. Eventually,itissolved

by using the matrix-free GMRES method.It is alsothe GMRES methodthat is used to solve

any interfa e system throughout this paper.

Regarding the s heme naming, we use apital letters (D and N for instan e) to dene the

interfa e onditions, writing rstly the input/output values of the 2D problem, and se ondly

those of the 1D problems. For instan e, the s heme orresponding to (??) is referred to as

DmNND.Toprovidemeaningtothemulti-dimensionalinterfa e onditions,wewrite "mN"to

indi ate that the mean values of the normal derivatives of the solution on the interfa es are

used. IfRobin onditions are used, we write "mR". If this time, tosolve (??)"dsa", the mean

value ofthe solution isneeded, wewrite the Diri hlet ondition "mD", orthe Robin ondition

(12)

onsidered. Forinstan e, by using dire t elimination,(??) yields

S

_1D,j

N D,n+1

R

j

S

2D

DN,n+1

E(α

n+1

1 , ..., α

n+1

p

) − α

n+1

j

= 0, j = 1, ..., p

(15)

whi h is amulti-s aleS hur omplementsystem, with only

p

unknowns

α

1 , ..., α

p

.Indeed, the equation (??) is an adaptation of the Steklov-Poin aré equation by using mean values (with

pre onditioning). The ondition number of the interfa e system of the

p

unknowns s hemes is very small. Indeed, it is well known in the lassi al domain de omposition literature, when

there are two subdomains

A

and

B

of the same geometri dimension, that

S

N D,n+1

A

is a good pre onditioner for

S

DN,n+1

B

.The numeri alexperiments done both in the present paperand in [?℄,andalsowithRobin onditions, learlydemonstratethat the

p

unknowns s hemesaremore stable than the

2p

unknowns s hemes.

Choosing Robin onditions, insteadofDiri hletorNeumann onditions, asinputvaluesfor

the 2Dproblemwasnot overed in[?℄. Inthispaper, newRobinoptionsare oeredtodes ribe

the ontinuous problem (justtime-dis retized), that isto say (??) and (??).They are :

∂v

_j

n+1

∂x

e

j

(δ) + q

j

v

n+1

j

(δ) = ω

j

n+1

, j = 1, ..., p

(16) and

1 θ

j

ε

Z

γ

j

∂u

n+1

∂n

dγ + q

j

u

n+1

_{(x, y) = ω}

n+1

j

, (x, y) ∈ γ

j

, j = 1, ..., p.

(17) or

∂u

n+1

∂n

(x, y) + q

j

1 θ

j

ε

Z

γ

j

u

n+1

dγ = ω

_j

n+1

, (x, y) ∈ γ

j

, j = 1, ..., p.

(18)

where

q

j

is the Robin parameter, and

ω

n+1

j

is onstant,

j = 1, ..., p

. They are used to dene the S hwarz methodbelow.

5 The S hwarz algorithm

Herewe onsidertheLionsadaptationin[?℄oftheS hwarzalgorithm,su hthattheDiri hlet

interfa e onditionsare repla edbyRobininterfa e onditions.Thealgorithmhas alreadybeen

onsidered inthe ase of non mat hing nitevolume meshesin [?℄.Here weadapt the method

to the multi-s ale ase, prove the onvergen e of the algorithm and ompare with the strong

(13)

Letus begin with anEuler's impli itetime-dis retization











v

_j

n+1(l+1)

(x

e

j

_{) − v}

n

j

(x

e

j

)

k

−

∂

2 _v

n+1(l+1)

j

∂x

e

j

2 (x

e

j

_{) = f}

j

(x

e

j

, t

n+1

), x

e

j

∈ (δ, l

j

),

j = 1, ..., p

(a)

v

j

0 (x

e

j

) = 0, x

e

j

∈ (δ, l

j

), j = 1, ..., p

v

_j

n+1(l+1)

(l

j

) = 0

u

n+1(l+1)

_{(x, y) = u}

n+1(l+1)

γ,j

, (x, y) ∈ γ

j

, j = 1, ..., p,

(u)

−

∂v

n+1(l+1)

j

∂x

e

j

(δ) + q

j

v

n+1(l+1)

j

(δ) = −

1 θ

j

ε

Z

γ

j

∂u

n+1(l)

∂n

dγ + q

j

u

n+1(l)

γ,j

, j = 1, ..., p

(c)

u

n+1(l+1)

_{(x, y) − u}

n

_{(x, y)}

k

− △u

n+1(l+1)

_{(x, y) = f (x, y, t}

n+1

), (x, y) ∈ Ω(0)

(b)

u

0 _{(x, y) = 0, (x, y) ∈ Ω(0)}

∂u

n+1(l+1)

∂n

(x, y) = 0, (x, y) ∈ ∂Ω(0)\(∪

p

j=1

γ

j

)

1 θ

j

ε

Z

γ

j

∂u

n+1(l+1)

∂n

dγ + q

j

u

n+1(l+1)

γ,j

=

∂v

n+1(l)

_j

∂x

e

j

(δ) + q

j

v

n+1(l)

j

(δ), j = 1, ..., p

(d)

(19) with

n ∈ {0, ..., N

k

}

.

TheS hwarzmethodrequirestwokindsofunknownsontheinterfa ebetweentwosubdomains:

the values on the interfa e of the solution of the problem set in the left subdomain and the

values on the interfa e of the solution of the problem set in the right subdomain. Here, we

onsider that the leftsubdomain is

Ω(0)

and the right subdomain is

S

j

, j = 1, ..., p

. To ensure thattheS hwarz algorithmallowsustosolve(??)" sa",we hoosetowritetheRobin ondition

on the leftside with the mean value of the normal derivativeas below

1 θ

j

ε

Z

γ

j

∂u

n+1(l+1)

∂n

dγ + q

j

u

n+1(l+1)

_{(x, y) =}

∂v

n+1(l)

j

∂x

e

j

(δ) + q

j

v

n+1(l)

j

(δ), ∀(x, y) ∈ γ

j

, j = 1, ..., p

(20)

This relation involves that the value of the solution on the interfa e

γ

j

on the

Ω(0)

side is onstant, this value is named

u

n+1

γ,j

. Of ourse the interfa e value on the other side of

γ

j

is

(14)

(?? ) and (??d). It is assumed that

q

j

> 0, j = 1, ..., p

.The fully dis retized s heme is











h

e

j

i

v

_j,i

n+1(l+1)

− v

n

j,i

k

+ F

j,n+1(l+1)

i+1/2

− F

j,n+1(l+1)

i−1/2

= h

e

j

i

f

e

j

,n

i

, i = 1, ..., N

j

,

j = 1, ..., p, n ∈ {0, ..., N

k

}

(a)

F

_i+1/2

j,n(l+1)

= −

v

n(l+1)

j,i+1

− v

n(l+1)

j,i

h

e

j

i+1/2

, i = 0, . . . , N

j

, v

_j,N

n(l+1)

_j

₊₁

= 0, j = 1, ..., p,

n ∈ {1, ..., N

k

+ 1}

m(K)

u

n+1(l+1)

K

− u

n

K

k

+

X

σ∈E

K

F

_K,σ

n+1(l+1)

= m(K)f

_K

n

, ∀K ∈ T , n ∈ {0, ..., N

k

}

(b)

F

_K,σ

n(l+1)

= −

m(σ)

d

σ

(u

n(l+1)

_L

− u

n(l+1)

_K

), ∀σ ∈ E

int

,

if

σ = σ

K/L

, n ∈ {1, ..., N

k

+ 1}

F

_K,σ

n(l+1)

=







−

m(σ)

d

σ

(u

n(l+1)

_γ,j

− u

n(l+1)

_K

) , ∀σ ⊂ γ

j

, σ ∈ E

K

, j = 1, ..., p

0 , ∀σ ⊂ ∂Ω(0)\(∪

p

_j=1

γ

j

)

−

v

n+1(l+1)

j,1

− v

n+1(l+1)

j,0

h

e

j

1/2

+ q

j

v

j,0

n+1(l+1)

= −

1 θ

j

ε

X

σ⊂γ

j

σ∈E

K

m(σ)

d

σ

(u

n+1(l)

_γ,j

− u

n+1(l)

_K

) + q

j

u

n+1(l)

γ,j

,

j = 1, ..., p, n ∈ {0, ..., N

k

}

(c)

1 θ

j

ε

X

σ⊂γ

j

σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_γ,j

− u

n+1(l+1)

_K

) + q

j

u

n+1(l+1)

γ,j

=

v

_j,1

n+1(l)

− v

n+1(l)

_j,0

h

e

j

1/2

+ q

j

v

j,0

n+1(l)

,

j = 1, ..., p, n ∈ {0, ..., N

k

}

(d)

(21)

At iteration

(l)

, we assume that

v

n+1(l)

j,i

, j = 1, ..., p, i = 1, ..., N

j

, and

u

n+1(l)

K

, K ∈ T

, as wellas

v

n+1(l)

_j,0

(whi hrepresents

v

j

(δ)

),and

u

n+1(l)

γ,j

(whi hrepresentsthe value of

u

on

γ

j

),

j = 1, ..., p

, areknown,andwededu ethe orrespondingvaluesat

(l + 1)

.Weprovebelowthatthe problem has aunique solutionat

(l + 1)

, and thatthe algorithm onverges tothe solutionofthe hybrid s heme (??).

5.2 Convergen e of the algorithm

Let us re all that the s heme (??) leads to a linear system whose

U

n+1

is unknown,

n ∈ {0, ..., N

k

}

,

(U

n

₎

T

_{= {{v}

n

j,i

, i = 1, ..., N

j

}, j = 1, ..., p}, {u

n

K

, K ∈ T }

.

It is supplemented with the purpose of dening

(W

n

)

T

= {{v

n

j,i

, i = 1, ..., N

j

}, j = 1, ..., p}, {u

n

K

, K ∈ T }, {v

j,0

n

, j = 1, ..., p}, {v

j,0

n

, j = 1, ..., p}

where

v

n

j,0

is dened by (??). In the same way,let usdenote

(U

n(l)

)

T

=

{{v

n(l)

_j,i

, i = 1, ..., N

j

}, j = 1, ..., p}, {u

n(l)

K

, K ∈ T }, {v

n(l)

j,0

, j = 1, ..., p}

,

{u

n(l)

_γ,j

, j = 1, ..., p}

then

U

n+1(l+1)

isthe unknown ve torof alinearsysteminthe formof

AU

n+1(l+1)

_{= BU}

n+1(l)

₊

(15)

Lemma 2 1) The system (??) has a unique solution

U

n+1(l+1)

for a given value of

U

n+1(l)

.

2) The ve tor

U

n+1(l)

dened by the S hwarz algorithm onverges to the solution of the hybrid

s heme (??), that is the monolithi s heme to solve (??)" sa".

Proof. 1) To prove existen e and uniqueness, it issu ient to prove that

A

is regular.To dothis, fora given

n ∈ {0, ..., N

k

}

, letusprove that

U

n+1(l+1)

_{= 0}

if

f

e

j

,n

i

= 0, i = 1, ..., N

j

, j =

1, ..., p

,

f

n

K

= 0, K ∈ T , U

n

= 0, U

n+1(l)

= 0

.

Under this assumption, we multiply (??a) by

v

n+1(l+1)

j,i

and sum over

i

, then multiply by

θ

j

ε

and sum over

j

. Then we multiply (??b) by

u

n+1(l+1)

K

and sum over

K ∈ T

. Reordering

the summationsover theset of edgesin

Ω(0)

, onsideringseparatelythe interioredges and the edges on

γ

j

, j = 1, ..., p

, repla ing the numeri al uxes by their values result in

p

X

j=1

θ

j

ε

N

j

X

i=1

h

e

j

i

k

(v

n+1(l+1)

j,i

)

2 +

X

K∈T

m(K)

k

(u

n+1(l+1)

K

)

2 +

p

X

j=1

θ

j

ε





N

j

X

i=1

(v

n+1(l+1)

_j,i+1

− v

_j,i

n+1(l+1)

)

2 h

e

j

i+1/2

+

v

n+1(l+1)

j,1

− v

n+1(l+1)

j,0

h

e

j

1/2

v

_j,1

n+1(l+1)





+

X

σ∈E

int

,σ=σ

K|L

m(σ)

d

σ

(u

n+1(l+1)

_K

− u

n+1(l+1)

_L

)

2 −

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_γ,j

− u

n+1(l+1)

_K

)u

n+1(l+1)

_K

= 0

To get a positive term on the left-hand side of the equality, a same appropriate quantity is

added toboth sides. We obtain

p

X

j=1

θ

j

ε

N

j

X

i=1

h

e

j

i

k

(v

n+1(l+1)

j,i

)

2 +

X

K∈T

m(K)

k

(u

n+1(l+1)

K

)

2 +

p

X

j=1

θ

j

ε

N

j

X

i=0

(v

_j,i+1

n+1(l+1)

− v

_j,i

n+1(l+1)

)

2 h

e

j

i+1/2

+

X

σ∈E

int

,σ=σ

K|L

m(σ)

d

σ

(u

n+1(l+1)

_K

− u

n+1(l+1)

_L

)

2 +

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_γ,j

− u

n+1(l+1)

_K

)

2 = −

p

X

j=1

θ

j

ε

v

_j,1

n+1(l+1)

− v

_j,0

n+1(l+1)

h

e

j

1/2

v

n+1(l+1)

_j,0

+

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_γ,j

− u

n+1(l+1)

_K

)u

n+1(l+1)

_γ,j

(22)

By the means of (?? )and (??d) with aright-handside equalto zero,wehave

−

p

X

j=1

θ

j

ε

v

_j,1

n+1(l+1)

− v

_j,0

n+1(l+1)

h

e

j

1/2

v

n+1(l+1)

_j,0

+

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_γ,j

− u

n+1(l+1)

_K

)u

n+1(l+1)

_γ,j

= −

p

X

j=1

θ

j

εq

j

(v

_j,0

n+1(l+1)

)

2 + (u

n+1(l+1)

_γ,j

)

2

It follows that(??)may bewrittenasa sumof non-negativeterms thatisequaltozero, whi h

means that

U

n+1(l+1)

_{= 0}

(16)

to see that

U

n+1(l+1)

_{− W}

n+1

is solution of (??) with

f

e

j

,n

i

= 0, j = 1, ..., p, i = 1, ..., N

j

,

f

n

K

= 0, K ∈ T

, and

U

n

_{= 0}

. So it is su ient to prove that

U

n+1(l+1)

onverges towards

zero, when

f = 0

and

U

n

_{= 0}

, whi h is assumed below, to prove that the S hwarz algorithm

onverges to the solution of the hybrid s heme (??), and onsequently gives an approximate

solution of (??).Indeed, (?? )and (??d) lead to(?? ).

Onthe right-handside of (??),both terms are theprodu t of thequantities that appear in

(?? ) and (??d). Ex ept that these quantities are added in (??) whereas they are multiplied

in (??).We omplete the proofby following[?℄ inwhi hthe lassi al identity

4ab = (a + b)

2 ₋

(−a + b)

2

is used. Letus denote

L

l+1

the left-hand side of (??).We have

L

l+1

=

−

p

X

j=1

θ

j

ε

4q

j





v

_j,1

n+1(l+1)

− v

_j,0

n+1(l+1)

h

e

j

1/2

+ q

j

v

n+1(l+1)

j,0

!

2 −

−

v

n+1(l+1)

j,1

− v

n+1(l+1)

j,0

h

e

j

1/2

+ q

j

v

j,0

n+1(l+1)

!

2 



+

p

X

j=1

θ

j

ε

4q

j









1 θ

j

ε

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_γ,j

− u

n+1(l+1)

_K

) + q

j

u

n+1(l+1)

γ,j





2 −



−

1 θ

j

ε

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_γ,j

− u

n+1(l+1)

_K

) + q

j

u

n+1(l+1)

γ,j





2 



(23)

Thanks to (?? )and (??d), this an be written

L

l+1

= E

l

− E

l+1

with

E

l

=

p

X

j=1

θ

j

ε

4q

j

v

_j,1

n+1(l)

− v

_j,0

n+1(l)

h

e

j

1/2

+ q

j

v

n+1(l)

j,0

!

2 +

p

X

j=1

θ

j

ε

4q

j



−

1 θ

j

ε

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1(l)

_γ,j

− u

n+1(l)

_K

) + q

j

u

n+1(l)

γ,j





2

(24)

Weremark that the series

P

∞

l=1

L

l

is onvergentbe ause

L

l

≥ 0

and

P

N

l=1

L

l

= E

0 − E

N

≤ E

0

for all

N ≥ 1

, be ause

E

l

≥ 0

for all

l

. We dedu e that

L

l

tends ne essarily to

0

as

l

tends to

innity, and so that ea h omponent of

U

n+1(l+1)

tends to

0

as

l

tendsto innity.

5.3 Interfa e system

To begin, let us onsider the semi-dis rete version of the separated problems with Robin

(17)











u

n+1(l+1)

_{(x, y) − u}

n

_{(x, y)}

k

− △u

n+1(l+1)

_{(x, y) = f (x, y, t}

n+1

), (x, y) ∈ Ω(0)

u

0 _{(x, y) = 0, (x, y) ∈ Ω(0)}

∂u

n+1(l+1)

∂n

(x, y) = 0, (x, y) ∈ ∂Ω(0)\(∪

p

j=1

γ

j

)

u

n+1(l+1)

_{(x, y) = u}

n+1(l+1)

γ,j

, (x, y) ∈ γ

j

, j = 1, ..., p

1 θ

j

ε

Z

γ

j

∂u

n+1(l+1)

∂n

dγ + q

j

u

n+1(l+1)

γ,j

= ω

n+1(l)

j

, j = 1, ..., p,

(r)

(26)

What standsoutrstisthatthe option(r)in(??)isthe samerelationas(??).The rststepis

tointrodu e operatorswhi hgiveasimpledes riptionof theproblem.Theoperators

I

n+1

1D,j

and

J

_2D,j

n+1

aredenedby

I

n+1

1D,j

(ζ

n+1(l)

j

) =

∂v

n+1(l+1)

_j

∂x

e

j

(δ)+q

j

v

n+1(l+1)

j

(δ)

where

v

n+1(l+1)

j

isthe solution of (??); and

J

n+1

2D,j

(ω

n+1(l)

1 , ..., ω

n+1(l)

p

) = −

1 θ

j

ε

Z

γ

j

∂u

n+1(l+1)

∂n

dγ + q

j

u

n+1(l+1)

γ,j

where

u

n+1(l+1)

is the solution of (??). Wedene also











ω

n+1(l)

_j

=

∂v

n+1(l)

j

∂x

e

j

(δ) + q

j

v

n+1(l)

j

(δ), j = 1, ..., p

ζ

_j

n+1(l)

= −

1 θ

j

ε

Z

γ

j

∂u

n+1(l)

∂n

dγ + q

j

u

n+1(l)

γ,j

, j = 1, ..., p

(27)

The interfa esystem an then be expressed by

(

ω

_j

n+1(l+1)

= I

_1D,j

n+1

(ζ

_j

n+1(l)

), j = 1, ..., p

ζ

_j

n+1(l+1)

= J

_2D,j

n+1

(ω

n+1(l)

₁

, ..., ω

p

n+1(l)

), j = 1, ..., p

(28)

The fully-dis retized versions of (??)and (??) are

(18)

The 1Ddis reteoperatorthatdes ribes

I

n+1

1D,j

isnamed

S

R

−

R,n+1

1D,j

.It onsiststosolvethe Robin problem(??)andtogiveaRobinvalue(takingthistimestheoppositeofthenormalderivative)

of the solution on

γ

j

:

ω

n+1(l+1)

_j

=

v

n+1(l+1)

j,1

− v

n+1(l+1)

j,0

h

e

j

1/2

+ q

j

v

j,0

n+1(l+1)

, j = 1, ..., p

(31)

The notation for the Robin operator is

R

−

when the normal derivative is dire ted outwards fromthe1Ddomain,and

R

ifitisturnedinward.Itisexa tlythe oppositeforthe 2Dproblem. You an sum it up by saying that

R

is used when the normal derivative is dire ted from the 2D domain tothe 1D domain, and

R

−

inthe opposite dire tion.

For onsistent presentation and denition of

J

n+1

2D,j

, we introdu enow the general optionin the 2D problem(??)that is

1 θ

j

ε

Z

γ

j

∂u

n+1(l+1)

∂n

dγ + q

j

u

n+1(l+1)

_{(x, y) = ˜}

_ω

n+1(l)

j

(x, y), ∀(x, y) ∈ γ

j

, j = 1, ..., p,

(32)

whi h, unlike (??r), allows us to onsider that the value on the right-hand side, and thus the

value of

u

n+1(l+1)

on

γ

j

, is not ompulsory onstant. The fully-dis retized version of (??) with the option(??)is amodiedversionof (??)obtained by ex hanging

u

n+1(l+1)

γ,j

for

u

n+1(l+1)

σ

and

ω

_j

n+1(l)

for

ω

˜

n+1(l)

j,σ

, whi h approa hes

ω

˜

n+1(l)

j

on the edge

σ

,with the option

1 θ

j

ε

X

σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_σ

− u

n+1(l+1)

_K

) + q

j

u

n+1(l+1)

σ

= ˜

ω

n+1(l)

j,σ

, ∀σ ⊂ γ

j

, j = 1, ..., p

(33)

as Robininput values insteadof (??r).The dis reteoperator

S

mdRN,n+1

2D

omputes thedis rete

normal derivativesof the solutionon

γ

j

, whi h is

u

n+1(l+1)

σ

− u

n+1(l+1)

_K

d

σ

, σ ⊂ γ

j

, j = 1, ..., p

. The operator

S

mdRD,n+1

2D

gives

(u

n+1(l+1)

σ

)

σ⊂γ

j

,j=1,...,p

, thevaluesof the solutionon

γ

j

, j = 1, ..., p

.All that remains for usto do, nally,is to dene

S

mdRR

−

,n+1

2D

= −S

mdRN,n+1

2D

+ QS

mdRD,n+1

2D

(34)

where the linear operator

Q

isa blo k diagonalmatrix, ea h blo k equalto

q

j

I, j = 1, ..., p

.

We are now able togivethe fully-dis retizedversion of (??)

(

ω

_j

n+1(l+1)

= S

R

−

R,n+1

1D,j

(ζ

n+1(l)

j

), j = 1, ..., p

ζ

_j

n+1(l+1)

= R

j

S

2D

mdRR

−

,n+1

E(ω

n+1(l)

1 , ..., ω

n+1(l)

p

), j = 1, ..., p

(35)

that means with the average value forthe 2D problem

ζ

_j

n+1(l+1)

= −

1 θ

j

ε

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1(l+1)

_γ,j

− u

n+1(l+1)

_K

) + q

j

u

n+1(l+1)

γ,j

, j = 1, ..., p

(36)

The interfa esystem an be statedmore expli itly

(19)

with

(v

n+1(l+1)

j,i

)

i

solutionof (??),and

(u

n+1(l+1)

K

)

K

solution of (??),and where

sτ

j

=

1 θ

j

ε

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

, j = 1, ..., p

(38)

It isuseful topoint out that

R

j

S

2D

mdRR

−

,n+1

E(ω

n+1(l)

1 , ..., ω

p

n+1(l)

) = −R

j

S

2D

mdRN,n+1

E(ω

n+1(l)

1 , ..., ω

p

n+1(l)

) + q

j

u

n+1(l)

γ,j

(39)

sin e the value of the solution is onstant on the interfa es. Then, as "mean value" operator,

R

j

in(??) is a tually onlyappliedto the Neumann operator

S

mdRN,n+1

2D

. We observe that (??) isthe Ja obi method tosolve

(

ω

_j

n+1

= S

R

−

R,n+1

1D,j

(ζ

j

n+1

), j = 1, ..., p

ζ

_j

n+1

= R

j

S

2D

mdRR

−

,n+1

E(ω

n+1

1 , ..., ω

p

n+1

), j = 1, ..., p

(40)

As aboveit ispossible tosolve (??)by using the GMRES methodtoa elerate the resolution.

The system (??) is the fully-dis retized interfa e system. The s heme will be referred to as

mdRmdR

−

R

−

R.

5.4 Comparison with the strong oupling framework

Hereweprovethattheinterfa esystem(??)derivingfromS hwarz method anbeobtained

by using the iterative strong oupling method. The interfa e system (??) solves the problem

(??)" sa" by the means of Robin onditions su h as (??s) and (??r). Our related obje tive is

to fallba k on the same interfa e system by using the same tools.

First,we noti e that the onditions











1 θ

j

ε

Z

γ

j

∂u

∂n

(., t)dγ + q

j

u(x, y, t) =

∂v

j

∂x

e

j

(δ, t) + q

j

v

j

(δ, t), (x, y) ∈ γ

j

, j = 1, ..., p, t ∈ (0, T )

−

1 θ

j

ε

Z

γ

j

∂u

∂n

(., t)dγ + q

j

u(x, y, t) = −

∂v

j

∂x

e

j

(δ, t) + q

j

v

j

(δ, t), (x, y) ∈ γ

j

, j = 1, ..., p, t ∈ (0, T )

(41)

are equivalent to (??). Then, the problem (??)" sa" is rewritten with (??) instead of (??),

evenif bothareequivalent.Andnally,the separatedsemi-dis retized1Dand2Dproblems are

onsidered, whi hare just our originalproblems (??) and (??)with other options.