Domain decomposition methods in a geometrical multi-scale domain using finite volume schemes.

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Domain decomposition methods in a geometrical

multi-scale domain using finite volume schemes.

Marie-Claude Viallon

To cite this version:

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multi-s ale domain using 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. The heat equation is solved by using a nite volume dis retization in a domain

that onsists of a two-dimensional (2D) entral node and several one-dimensional (1D)

out-going bran hes. Several interfa e onne tion options to mat h the submodels with or without

ontinuity are looked at. Forea h of them, monolithi s hemes are dened, and existen e and

uniqueness of solutions is proved. New s hemes are dedu ed from them, whi h are obtained

through domain de omposition methods in the form of interfa e systems, with one or two

interfa e unknowns. A omparativesystemati study is arried out froman algebrai and

nu-meri alpointof viewtolearnmore about a ura yof theses hemes a ording tothe interfa e

onditions : Diri hlet, Neumann or Robin, onsidering espe ially the impa t of the Robin

pa-rameter. It is shown that some of the properties of lassi al iterative substru turing methods

remain validin the multi-s ale ase.

Résumé. On appro he la solution de l'équation de la haleur ave une dis rétisation de type

volumes nis dans un domaine onstitué d'un noeud bi-dimensionnel (2D) d'où partent

plu-sieurs bran hes uni-dimensionnelles (1D). Plusieurs types de ra ordements, ontinus ou non,

sontenvisagés auxinterfa esentre lessous-modèles1Det2D.Pour ha un,dess hémas

mono-lithiquessont onstruitsetl'existen eetl'uni itéd'unesolutionest démontrée.Onendéduitde

nouveauxs hémasobtenuspardé ompositiondedomainesouslaformedesystèmesd'interfa e,

ave une oudeux in onnues par interfa e.Uneétude omparativesystématiqueest menéed'un

point de vue algébrique et numérique pour onnaître la pré ision de es s hémas en fon tion

des onditions d'interfa e : Diri hlet, Neumannou Robin, en regardant notamment l'inuen e

du paramètre de Robin. On montre que ertaines propriétés des méthodes itératives lassiques

sont en ore vraies dans le ontexte multi-é helle.

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

Mots lés : nite volume s heme, paraboli problem, multi-s ale domain, domain

de omposi-tion, stability and onvergen e of numeri almethods, Robin interfa e ondition.

1 Introdu tion

The aim of the present paper is to use a domain de omposition approa h to dene,

ana-lyze and ompare, new nite volume s hemes to solve a model problem set in a geometri al

multi-s aledomain.This kindofdomainsderive forinstan e fromanite rodstru ture, whi h

is a onne ted union of ylinders (re tangles in the two-dimensional (2D) ase). Arterial trees

in the ardiovas ular system (see [15℄), systems of pipes in industrial installations, or anal

systems, are lassi al examples of rod stru tures. Partial dierential equation solving in su h

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multi-s aledomainwheretheoriginalspa es aleoftheproblemiskeptintheregionsofinterest.

The main appli ation area is omputational hemodynami s. Dimensionally-heterogeneous

modelling in that ontext an be found for instan e in [2,3, 4, 5, 6, 7,13, 14, 25,30℄. In [19℄,

the authorsdealwithdimensionally-heterogeneoushydrauli networks.Coupling of1Dand 2D

CFD odes is dis ussed in [17℄ (Euler system). This work has been extended re ently in [9℄to

the oupling of a density-based 3D Euler ode to a 1D version of the ode (for instan e, an

appli ation is the simulation of diesel inje tors). In [2℄, the appli ations are the heat transfer

problem with purediusion phenomena and the linear elasti ity problem.

The modelproblem,in this paper, is the heat equation in a 1D-2D domain.Three options

are onsidered for the transmission onditions : the solution and the mean value of the ux

(normalderivative)are ontinuous, the uxandthe meanvalueof thesolutionare ontinuous,

and the mean value of the solution or of the ux is ontinuous supplemented by a Robin

ondition. A nitevolume s hemeof hybrid dimensionfor solving the modelproblemwith the

ontinuity of the solution onthe interfa es wasstudied in[24℄. The problemwas fully oupled

and treated as a monolithi entity, it was solved by using a dire t method, an error estimate

wasobtained. Sofar,however, fewtheoreti al resultsfordimensionaly-heterogeneous problems

have been produ ed. For instan e, in [2℄, the authorswork in a generaltheoreti al framework

and prove existen e and uniqueness of the solution for both the ontinuous problem and its

nite elementapproximationbywritinganaugmented variationalformulationwhere Lagrange

multipliers derive from the oupling onditions. On the ontrary, in this paper, we deal

ex- lusively with nite volume approximations. Hybrid dimension monolithi s hemes are built,

followingthe samestru tureas[24℄,toapproximatethesolutionofthemodelproblemwiththe

two other kinds of transmission onditions. Existen e and uniqueness of the solution of these

s hemes isproved. Thenadomainde ompositionmethodisused (asin[2,3, 13℄for instan e).

Thispartitionedapproa h,inwhi hthesubmodelsaresolvedseparately,wasnot overedin[24℄.

We use the iterativestrong ouplingmethodbe ausethis isa de omposition strategy that

workswith subdomainswhi hdonot overlap.The methodologyisdeveloped in[2,3,4,18,19,

32℄. This te hnique an be understood as a domain de omposition approa h where the

parti-tioning takes pla e at the interfa es of models of dierent geometri dimensions, the interfa e

system is solved by using a Krylov method. The pro edure involves two times more interfa e

unknowns than lassi al methods but it is not a drawba k here sin e there is a smallnumber

of interfa e unknowns. In this paper, a systemati review is arried out by hoosing Diri hlet,

Neumann or Robin interfa e onditions in pairs, for the three kinds of interfa e onditions

mat hing the submodels. Ea h time, it is shown how to write restri tion operators, with or

without the mean value of the solution or its normal derivative, for s hemes to be onsistent

with transmission onditions. All s hemes are analyzed from an algebrai point of view. We

ompare the numeri al results with the ones obtained by solving the monolithi s hemes with

a dire t method.In parti ular, the way the Robin parameter inuen es the a ura yis looked

at. We showthat approximationsare not equallya urate for all interfa e onditions.

Classi aldomainde ompositionmethodsaregenerallyamenabletopro edures foran

inter-fa e problem, for instan e the S hur omplement system or the ux equation, by eliminating

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geome-transmission onditions. Another pro edure refered to as "the Robin equation" is dened in

the paper, whi h an be onsidered asa dual method inthe same way as the ux equation.

Then,we look atsome iterative substru turing methods. Many lassi al domain

de ompo-sitionmethodsarepre onditioned Ri hardsonmethodsfortheS hur omplementsystemorfor

the ux equation (see [29℄ or [10℄ for instan e). We explain the extent towhi h this property

remainstruefortheDiri hlet-NeumannmethodandfortheNeumann-Diri hletmethod

(adap-tation to the geometri al multi-s ale ase). The relationship with the Gauss-Seidel algorithm

is highlighted, as wesee inthe lassi al ase for instan e in [28℄.

Thisworkisorganizedasfollows.Se tion2presentsthemodelproblemanditsvariants.The

hybrid monolithi s heme developed in [24℄ is re alled inSe tion 3 and two other versions are

dened that allowdis ontinuity ofthe solution a rossthe interfa es. Existen e anduniqueness

ofthe solutionof ea hs heme isproved. InSe tion4,the iterativestrong ouplingpro edureis

applied,some lassi aldomainde omposition strategies are adapted and ompared,the Robin

equation is introdu ed. We give numeri al experiments in Se tion 5 illustrating the dieren e

between the three monolithi s hemes and givingthe error between the approximations

obtai-ned by ea h monolithi s heme and its asso iated interfa e systems deriving from the strong

ouplingmethod,testing many dierent interfa e onditions.

2 The model problems

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

The geometri al multi-s ale 1D-2D domain on whi h our model problems are set derives

from a nite rod stru ture, whi h onsists of one node and

p

bran hes. This onstru tion is

done in [23℄ and is reminded below.

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 [23℄). 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

by removing the orners. We assumethat

ω

ε

0 \ ∪

p

j=1

B

j

ε

is not too large. Let

Ω

ε

=

∪

p

j=1

B

j

ε

∪ ω

0 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

(5)

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

.

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 the initialdomain

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 problems

The rst boundary value problem in the domain

D

ε

, whi h is onsidered in this paper, is

the following :











∂v

j

∂t

(x

e

j

_{, t)}

₋

∂

2 _v

j

∂x

e

j

2 (x

e

j

_{, t) = f}

j

(x

e

j

, t), x

e

j

∈ (δ, l

j

),

j = 1, ..., p, t

_{∈ (0, T ) (a)}

v

j

(x

e

j

, 0) = 0, x

e

j

∈ (δ, l

j

), j = 1, ..., p

v

j

(l

j

, t) = 0, t

∈ (0, T )

∂u

∂t

(x, y, t)

− △u(x, y, t) = f(x, y, t), (x, y) ∈ Ω(0), t ∈ (0, T ) (b)

u(x, y, 0) = 0, (x, y)

_{∈ Ω(0)}

∂u

∂n

(x, y, t) = 0, (x, y)

∈ ∂Ω(0)\(∪

p

j=1

γ

j

), t

∈ (0, T )

u(x, y, t) = 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 =

∂v

j

∂x

e

j

(δ, t), j = 1, ..., p, t

∈ (0, T ) (c)

(1)

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, 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 ontinuoustillthe ordertwo.Thisallowsustoobtain

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More pre isely, we dene aglobal solution

u

d

of (1)by letting

u

d

(x, y, t) =

u(x, y, t)

if

(x, y)

∈ Ω(0), t ∈ (0, T )

v

j

(x

e

j

, t)

if

(x, y)

∈ B

ε

j

, x

e

j

∈ (δ, l

j

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

∈ (0, T )

(2) Thesolution

u

d

isdenedin

Ω

ε

but

u

d

_{(x, y, t)}

doesnotdependon

y

e

j

when

(x, y)

∈ B

ε

j

\B

′ε

j

.

Deningthe solutionon

Ω

ε

allowsustousestandard normsina2Ddomain,and morebroadly

to easily ompare

u

d

with

u

ε

(see below).

Remark 1 Problem (1) was introdu ed in [24℄ in the framework of the method of asymptoti

partial domainde omposition (MAPDD) (see [22℄). The followingresult was proved in [24℄.

For any

J > 0

, there is

M

, independent of

ε

, su h that if

δ = Mε

|

ln

ε

|,

then

ku

ε

− u

d

k

H

1 _(Ω

ε

×(0,T ))

= O(ε

J

_),

where

u

ε

isthe unique solution of the following paraboli linear model equation











∂u

ε

∂t

− △u

ε

= f,

in

Ω

ε

, t

∈ (0, T )

u

ε

= 0,

on

β

ε

j

, j = 1, ..., p, t

∈ (0, T )

∂u

ε

∂n

= 0,

on

∂Ω

ε

\(∪

p

j=1

β

j

ε

), t

∈ (0, T )

u

ε

(x, y, 0) = 0, (x, y)

∈ Ω

ε

, t

∈ (0, T )

(3)

where

f

is a smooth fun tion dened in

Ω

ε

× [0, T ]

su h that

f (x, y, t) = f

j

(x

e

j

_{, t)}

, if

(x, y)

∈

B

ε

j

, j = 1, ..., p

,

t

∈ (0, T )

; where

f

j

are independent of

ε C

J+5

₋

smooth fun tions,

J

≥ 5

, and

they vanish in some neighborhood of the node

O

and verti es

O

j

, j = 1, ..., p

. We assume that

f

j

(., t) = 0

for

t

≤ t

0 , t

0 > 0

.

Of ourse, in reality, the problem whi h arises in the rst instan e is (3). Then, some

assumptions are madethat givethe problem(1). Forinstan e, the 1D simpliedmodel an be

viewed as aredu tion of the omplete 2D modelby means of kinemati alrestri tions (see [3℄)

introdu ed on apart of the bran hes. Then,boundary onditions based uponmean quantities

are pres ribed on the oupling interfa es. Therefore, it makes sense to require the ontinuity

of the solution and its normal derivative inan average sense (asin [18℄) a ross

γ

j

, j = 1, ..., p

,

that is tosay torequire that











1 θ

j

ε

Z

γ

j

u(., t)dγ = v

j

(δ, t), j = 1, ..., p, t

∈ (0, T )

1 θ

j

ε

Z

γ

j

∂u

∂n

(., t)dγ =

∂v

j

∂x

e

j

(δ, t), j = 1, ..., p, t

∈ (0, T )

(4) instead of







u(x, y, t) = 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 =

∂v

j

∂x

e

j

(δ, t), j = 1, ..., p, t

∈ (0, T )

(5)

However, the problem (1) with (4) instead of (5) is not well posed : some supplementary

onditions are needed. Itis learthat (1)wasobtained by supplementing (4)by the ontinuity

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alternative,whi h is obtained when the problem(1) is hanged su h that (5)is repla ed by











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 )

(6)

that is tosay (4)is supplemented by the ontinuity of the ux onthe interfa es.

Otheroptions an be onsidered.The problem (1)"rsa",toindi atea Robin sele ted

alter-native, isobtained when the problem(1)is hanged su h that (5) isrepla ed by











1 θ

j

ε

Z

γ

j

u(., t)dγ = v

j

(δ, t), j = 1, ..., p, t

∈ (0, T )

∂u

∂n

(x, y, t) + 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 )

(7) or











1 θ

j

ε

Z

γ

j

∂u

∂n

(., t)dγ =

∂v

j

∂x

e

j

(δ, t), j = 1, ..., p, t

∈ (0, T )

∂u

∂n

(x, y, t) + 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 )

(8)

thatistosay(4)issupplementedbyRobin onditionsattheinterfa es.Here,

q

j

, j = 1, ..., p,

are

any non-negative values. The onditions (7) and (8)are equivalent if

q

j

6= 0

for all

j = 1, ..., p

.

If

q

j

= 0

for all

j = 1, ..., p

,then the problem(1)"rsa"with (7)is equivalent to(1)"dsa", while

(1)"rsa" with (8)isnot wellposed assoonasonly one value of

q

j

isnil.

Itisworth notingthatthe solutionofthe problem(1)is ontinuous whereasthe solutionof

the problems (1)"dsa" or (1)"rsa" may be dis ontinuous. Furthermore, all the versions whi h

may be imagined by using average values in the Robin ondition lead to problems whi h are

notwelldenedortoproblemswhi harealreadydenedabove.Tomentionbutafewexamples:











1 θ

j

ε

Z

γ

j

∂u

∂n

(., t)dγ =

∂v

j

∂x

e

j

(δ, t), j = 1, ..., p, t

∈ (0, T )

1 θ

j

ε

Z

γ

j

∂u

∂n

(., t)dγ + q

j

1 θ

j

ε

Z

γ

j

u(., t)dγ =

∂v

j

∂x

e

j

(δ, t) + q

j

v

j

(δ, t), (x, y)

∈ γ

j

,

j = 1, ..., p, t

_{∈ (0, T )}

(9)

is not well dened.

(8)

Remark 2 An error estimate proved in [24℄ is reminded. Let

ω

ε

0,

poly be a polygonal domain in ludedin

ω

ε

0

, and

B

ε,

poly

=

∪

p

j=1

B

j

ε

∪ ω

0,

ε

poly

. It isstatedthat, if

u

ε

isthe solutionof theheat

equation (3) and

u

d,T ,k

is the approximate nite volume solution of the problem (1),if

h

is the

size of the mesh and

k

the time step, then the following estimate holds under the assumptions

of Remark 1 :

ku

ε

− u

d,T ,k

k

L

2 _(B

ε,

poly

×(0,T ))

= O

h

√

ε

+ k

+ O(ε

J

)

Thetheoreti alstudyoftheproblems(1)"dsa"and(1)"rsa"isnot onsideredinthispaper.Here

the fo us is on only one aspe t : domain de omposition and a ura y with whi h the solutions

of the monolithi s hemes an be approximated.

3 Monolithi numeri al s hemes

Were allherethemonolithi numeri als hemewhi his onstru tedin[24℄toapproa hthe

solution of (1). Other monolithi s hemes are dened to solve (1)"dsa" and (1)"rsa". Impli it

methodsare used whether itis for (1), (1)"dsa"or (1)"rsa".

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

.

Thenan admissiblemesh over

Ω(0)

, denoted by

T

, is onstru ted. Weassume inthe

follo-wingthat

Ω(0)

ispolygonal.Were all(seethe denitionin[12℄)that su hamesh onsistsina

family of open polygonal onvex subsets

K

of

Ω(0)

alled ontrolvolumes, afamily of edges

σ

of the ontrol volumes denoted by

E

, and afamily of points

x

K

hosen in ea h ontrol volume

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.

(9)

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

u(x

K

, t

n

), K

∈ T , n ∈ {0, ..., N

k

+ 1

}

.











h

e

j

i

v

n+1

j,i

− 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)

(13)

with the initial ondition

u

0 K

= 0,

∀K ∈ T

v

0

(10)

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.

Indeed (13 ) an be rewrittenin the followingway :

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

(14)

Forthe sake of simpli ity,in(13) and (14), the summationis donefor

σ

⊂ γ

j

, and forea h

of them,

K

is the ontrolvolumesu h that

σ

∈ E

K

.

Letus deneon

Ω

ε

the fun tion

u

n

d,T

for

n

∈ {0, ..., N

k

+ 1

}

by

u

n

_d,T

(x, y) =







u

n

K

, (x, y)

∈ K, K ∈ T

v

n

j,i

, (x, y)

∈ B

j

ε

, x

e

j

∈ (x

e

j

i−1/2

, x

e

j

i+1/2

), i = 1, ..., N

j

,

j = 1, ..., p.

(15)

The approximate solutionof (1)is dened on

Ω

ε

× (0, (N

k

+ 1)k)

by

u

d,T ,k

(x, y, t) = u

n+1

d,T

(x, y), (x, y)

∈ Ω

ε

, t

∈ (nk, (n + 1)k), n ∈ {0, ..., N

k

}

The s heme (13) leads to a linear system in the form of

AU

n+1

_{= BU}

n

_{+ F}

n

in whi 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 is proved in[24℄ that there isa unique solution

(U

n+1

₎

to equations(13),

(U

n

₎

being given.

To solve (1), we assume that

ε

and

δ

are xed given parameters. Letus remark that a

nu-meri al approximation of the solution of (1) is also a numeri alapproximation of the solution

of (3) under some onditions about

ε

and

δ

. Indeed, in view of Remark 1, (1) is a reasonable

approximationfor (3)if

ε

issmalland

δ

of order

ε

ln

ε

.

Let us dene now the nite volume s hemes orresponding with the alternatives (1)"dsa"

and (1)"rsa". In both ases, the value of the solution is not assumed to be onstant on the

interfa es, andthe approximate values ofthe solutionontheedge

σ

ontheinterfa e

γ

j

attime

t

n

is alled

u

n

σ

, σ

⊂ γ

j

, j = 1, ..., p

.For both s hemes, the equation (13d) isrepla ed by

F

_K,σ

n

=

₋

m(σ)

d

σ

(u

n

_σ

_{− u}

n

_K

),

_{∀σ ⊂ γ}

j

, σ

∈ E

K

, j = 1, ..., p

(16)

Then, toobtain the s heme for (1)"dsa", the equation(13 ) is repla ed by

(11)

Itispointed outthat

v

n

j,0

remainsa onvex ombinationofthe approximatedvaluesof the

solu-tiononea hsideofthe interfa e(thisisalsothe asefor(1)"rsa"below).Thes heme resulting

fromtheequations(13) with(16)and (17)inpla eof(13 )and(13d)isthe monolithi s heme

that is used to solve (1)"dsa".

Lemma 3 There is a unique solution

(U

n+1

₎

to the linear system that orresponds to the

mo-nolithi s heme to solve (1)"dsa",

(U

n

₎

being given.

Proof. For a given

n

∈ {0, ..., N

k

}

, set

f

e

j

,n

i

= 0, v

j,i

n

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

j

and

f

_K

n

= 0, u

n

_K

= 0, K

∈ T

,then the system be omesin the formof

AU

n+1

_{= 0}

.Let usprove that

U

n+1

_{= 0}

.

Letusmultiply(13a)by

v

n+1

j,i

and sum over

i

, then multiplyby

θ

j

ε

and sum over

j

. In the

same way, letusmultiply(13b) by

u

n+1

K

andsum over K.Reordering the summationsover the

set of edges in

Ω(0)

, onsidering separately the interior edges and the edges on

γ

j

, j = 1, ..., p

,

we get

p

X

j=1

θ

j

ε

N

j

X

i=1

h

e

j

i

k

(v

n+1

j,i

)

2 +

p

X

j=1

θ

j

ε





N

j

X

i=1

F

_i+1/2

j,n+1

v

n+1

j,i

−

N

j

−1

X

i=0

F

_i+1/2

j,n+1

v

n+1

j,i+1





+

X

K∈T

m(K)

k

(u

n+1

K

)

2 +

X

σ∈E

int

,σ=σ

K|L

F

_K,σ

n+1

(u

n+1

_K

− u

n+1

L

)

+

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

F

_K,σ

n+1

u

n+1

_K

= 0

Repla ingthe numeri aluxes by their values results in

p

X

j=1

θ

j

ε

N

j

X

i=1

h

e

j

i

k

(v

n+1

j,i

)

2 +

X

K∈T

m(K)

k

(u

n+1

K

)

2 +

p

X

j=1

θ

j

ε





N

j

X

i=1

(v

_j,i+1

n+1

− v

n+1

j,i

)

2 h

e

j

i+1/2

+

v

n+1

j,1

− v

j,0

n+1

h

e

j

1/2

v

_j,1

n+1





+

X

σ∈E

int

,σ=σ

K|L

m(σ)

d

σ

(u

n+1

_K

_{− u}

n+1

_L

)

2 ₋

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1

_σ

_{− u}

n+1

_K

)u

n+1

_K

= 0

(19)

It follows from(17) that

(12)

p

X

j=1

θ

j

ε

N

j

X

i=1

h

e

j

i

k

(v

n+1

j,i

)

2 +

X

K∈T

m(K)

k

(u

n+1

K

)

2 +

p

X

j=1

θ

j

ε

N

j

X

i=0

(v

_j,i+1

n+1

_{− v}

n+1

_j,i

)

2 h

e

j

i+1/2

+

X

σ∈E

int

,σ=σ

K|L

m(σ)

d

σ

(u

n+1

_K

_{− u}

n+1

_L

)

2 +

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1

_σ

− u

n+1

K

)

2 = 0

(20)

From (20), it easily follows that all the omponents of

U

n+1

are nil. Thereby

A

is a regular

matrix.

For

n

∈ {0, N

k

+ 1

}

, let us onsider the fun tion

u

n

d,T

dened as the restri tion of

u

n

d,T

(see

(15)) to

D

ε

. It isan interesting point that (20) an be written as

1 k

ku

n+1

d,T

k

2 2,T

+

| u

n+1

d,T

|

2 1,T

= 0

.

See belowthe denitions of

k.k

2,T

and

| . |

1,T

forthe problem(1)"dsa". The sameresult o urs

for the problem (1).

Andnally,toobtainthe s heme for(1)"rsa"with the onditions (7), the equation(13 )is

repla ed by











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

+ q

j

u

n

σ

=

v

n

j,1

− v

j,0

n

h

e

j

1/2

+ q

j

v

j,0

n

,

∀σ ⊂ γ

j

, σ

∈ E

K

, j = 1, ..., p,

(r)

(21) that gives

v

_j,0

n

=



(

X

σ⊂γ

j

m(σ)d

σ

1 + q

j

d

σ

)v

_j,1

n

+ h

e

j

1/2

X

σ⊂γ

j

m(σ)

1 + q

j

d

σ

u

n

_K







(

X

σ⊂γ

j

m(σ)d

σ

1 + q

j

d

σ

)(1

_{− q}

j

h

e

_1/2

j

) + θ

j

εh

e

_1/2

j





−1

(22)

If

q

j

= 0

for all

j = 1, ..., p

, (21) is brought ba k to (17) and (22) to (18). This was exa tly

what was expe ted, sin e the problem(1)"rsa" with (7)is equivalent to (1)"dsa"inthis ase.

The problem (1)"rsa" an be written with the onditions (7) or (8), but ea h ondition

yields a spe i s heme. The equation (21r) des ribes both. The equation (21d) orresponds

only to(7). A s heme whi h dis retizes (8)is writtenby keeping (21r) supplemented by

1 θ

j

ε

X

σ⊂γ

j

,σ∈E

K

m(σ)

u

n

σ

− u

n

K

d

σ

=

v

n

j,1

− v

n

j,0

h

e

j

1/2

, j = 1, ..., p

(23)

and this time

(13)

It has been said that (1)"rsa" with (8) is not well posed if there is

j

∈ {1, ..., p}

su h that

q

j

= 0

. That isfound in (24) whose denominator an els inthis ase.

The s heme resulting from the equations(13) with (16) and (21r), supplemented by (21d)

to des ribe (7) or (23) to des ribe (8), in pla e of (13 ) and (13d), is the monolithi s heme

that isused tosolved (1)"rsa".If

q

j

6= 0

forall

j = 1, ..., p

, then(21) is equivalentto(21r) and

(23).

Lemma 4 There is a unique solution

(U

n+1

₎

to the linear system that orresponds to the

mo-nolithi s heme to solve (1)"rsa",

(U

n

₎

being given.

Proof. Let usassume that

q

j

6= 0

for all

j = 1, ..., p

to use inter hangeably (21d) and (23)

(when it omes to dealingwith

j

∈ {1, ..., p}

su h that

q

j

= 0,

we pro eed as for the problem

(1)"dsa" toaddress the asso iatedterm). The proof is almost the same asabove. Considering

again (19), it follows from (23) that

−

p

X

j=1

θ

j

ε

v

_j,1

n+1

_{− v}

_j,0

n+1

h

e

j

1/2

v

_j,0

n+1

+

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1

_σ

_{− u}

n+1

_K

)v

_j,0

n+1

= 0.

Writing that

v

n+1

j,0

= u

n+1

σ

+ (v

n+1

j,0

− u

σ

n+1

)

and adding tothe aboveequality leads to

p

X

j=1

θ

j

ε

N

j

X

i=1

h

e

j

i

k

(v

n+1

j,i

)

2 +

X

K∈T

m(K)

k

(u

n+1

K

)

2 +

p

X

j=1

θ

j

ε

N

j

X

i=0

(v

_j,i+1

n+1

_{− v}

n+1

_j,i

)

2 h

e

j

i+1/2

+

X

σ∈E

int

,σ=σ

K|L

m(σ)

d

σ

(u

n+1

_K

− u

n+1

L

)

2 +

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1

_σ

− u

n+1

K

)

2 +

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1

_σ

_{− u}

n+1

_K

)(v

_j,0

n+1

_{− u}

n+1

_σ

) = 0

(25)

To derive from (25) that all the omponents of

U

n+1

are nil, letus prove by using (21r) that

the last term isnon-negative :

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

(u

n+1

σ

− u

n+1

K

)(v

n+1

j,0

− u

n+1

σ

)

=

p

X

j=1

X

σ⊂γ

j

,σ∈E

K

m(σ)

v

n+1

j,1

− v

n+1

j,0

h

e

j

1/2

+ q

j

(v

j,0

n+1

− u

n+1

σ

)

!

(v

_j,0

n+1

_{− u}

n+1

_σ

)

=

p

X

j=1

q

j

X

σ⊂γ

j

,σ∈E

K

m(σ)(v

_j,0

n+1

− u

n+1

σ

)

2 ≥ 0

by meansof (21d). Thereby

A

is a regularmatrix.

In ontrast, in the present "rsa" ase, (25) an be written as

1 k

ku

n+1

d,T

k

2 2,T

+

| u

n+1

d,T

|

2 1,T

+

"interfa e term"

= 0

. See below the denition of

| . |

1,T

for the problem (1)"rsa". This

"interfa e term"measures the gapbetween the limitvaluesof thesolutionontheleftand right

(14)

Three norms are dened in the three spa es of pie ewise onstant fun tions on

D

ε

satis-fying thesame dis reteboundary andinterfa e onditions asthe solutionsof(1), (1)"dsa"and

(1)"rsa"respe tively.The dis rete

L

2

and

H

1

norms forsolutions of(1)are previously dened

in [31℄.

Denition 5 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

K

, (x, y)

∈ K, K ∈ T

W

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 and we denote the fun tions

W

7→ 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

and

W

_{7→| W |}

1,T

=





X

σ∈E

int

,σ⊂∪

p

j=1

γ

j

m(σ)d

σ

D

σ

W

d

σ

2 +

p

X

j=1

θ

j

ε

N

j

X

i=0

(W

j,i+1

− W

j,i

)

2 h

e

j

i+1/2





1/2

where, for the problem (1),

D

σ

W =

| W

K

− W

L

|, σ ∈ E

int

, σ = σ

K|L

| W

K

− W

j,0

|, σ ⊂ γ

j

, σ

∈ E

K

, j = 1, ..., p

and

W

j,0

=





W

j,1

h

e

j

1/2

+

1 θ

j

ε

X

σ⊂γ

j

,σ∈E

K

m(σ)

d

σ

W

K









1 h

e

j

1/2

+

1 θ

j

ε

X

σ⊂γ

j

m(σ)

d

σ





−1

and

W

j,N

j

+1

= 0, j

∈ {1, ..., p}

.

For the problem (1)"dsa",

D

σ

W

d

σ

=

| W

j,1

− W

j,0

|

h

e

j

1/2

, σ

⊂ γ

j

, σ

∈ E

K

, j = 1, ..., p

and

W

j,0

is given by (18).

For the problem (1)"rsa",

D

σ

W =

| W

σ

− W

K

|, σ ⊂ γ

j

, σ

∈ E

K

, j = 1, ..., p

where

W

σ

is

dedu ed from (21r) and

W

j,0

is given by (22) or (24).

The fun tion

| . |

1,T

is a norm in the dis rete spa e related to the solutions of (1). It is a

semi-norm in the dis retespa es related tothe solutions of (1)"dsa"and (1)"rsa".

4 Strong oupling

Tosolvetheheatequation,a lassi alapproa hisused:youdis retizestime(impli itEuler)

toobtainasequen e ofsteady problemstowhi hthedomainde ompositionstrategyisapplied

(see[8℄,[21℄).Indeed,awaveformrelaxationalgorithmexists(see[16℄),wherethede omposition

methodisdire tlyformulatedfortheoriginalproblemwithouttimedis retization,butthiskind

of algorithm is useful when the physi al problems that are solved in the subdomains are not

(15)

sition strategy espe iallytailored tosolve dimensionallyheterogeneous problems. The method

is explained in 4.1. From 4.2, s hemes dedi ated to the three kind of transmission onditions

are dened. All s hemes are analyzed from an algebrai point of view. The interfa e systems

obtainedby thestrong ouplingmethodhave

2p

unknowns.Throughsubstitution, lassi al

me-thods with

p

unknowns are found, su h asthe S hur omplement system orthe uxequation,

adjustedtothespe i ontext,pre onditionedornot.Onthesamelevel,theRobinequationis

introdu ed. The onne tionis nallymade with lassi al domainde omposition methodssu h

as Diri hlet-Neumann orNeumann-Diri hlet method in4.5.

4.1 The method

The methodology proposed in [3℄,[4℄,[18℄,[19℄ and [32℄, is applied now to solve (1). Let us

begin by writing (1) as a set of segregated s hemes and by performing an Euler's impli ite

time-dis retization(see the beginningof Se tion 3.2).











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

j

0 (x

e

j

) = 0, x

e

j

∈ (δ, l

j

), j = 1, ..., p

v

_j

n+1

(l

j

) = 0

v

n+1

j

(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)

(26)











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)

(27)

The basi idea from the previous referen es is to onsider the numeri al resolution of the 2D

problem(27)ononehand,andofthe1Dproblems(26)ontheotherhand,asbla k-boxeswhi 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

onverselyifthe problemiswelldened).Forinstan e, ifwe hoose that(26)re eivestheinput

(β

_j

n+1

, j = 1, ..., p)

and (27) re eives the input (

α

n+1

j

, j = 1, ..., p

), this an beexpressed by

α

n+1

j

=

F

1D,j

n+1

(β

j

n+1

), j = 1, ..., p

β

_j

n+1

=

F

n+1

2D,j

(α

1 n+1

, ..., α

n+1

p

), j = 1, ..., p

(28) where

F

n+1

1D,j

givesthe value of the solutionof (26-option (b)) at the point

δ

on the axis

j

, and

F

2D,j

n+1

givesthemeanvalueontheinterfa e

γ

j

ofthesolutionof(27-option(a))atthetime

t

n+1

,

where (26-option (b)) is the system (26) with its penultimate equation (a) removed, and

(27-option (a)) is the system (27) with its lastequation (b) removed. The options(a) and (b) are

(16)

(

α

n+1

j

, β

j

n+1

, j = 1, ..., p

)isobtained,whi hisfullydis retizedandsolvedbyaniterativemethod.

On the ontrary, in the previous paper [24℄, a ount is taken of the equations (26-option

(a)) and (27-option (a)) in additionto the other equations (26b) and (27b). They are related

by (1- ) whi h iswritten below atthe time

t

n+1

:

∂v

n+1

_j

∂x

e

j

(δ) =

1 θ

j

ε

Z

γ

j

∂u

n+1

∂n

dγ, j = 1, ..., p

(29)

Therefore the 2D and 1D problems are not understood as separated bla k-boxes. The values

β

j

n+1

, j = 1, ..., p

, are no more unknowns and the only interfa e unknowns that are kept are

α

_j

n+1

, j = 1, ..., p

. The fully dis retization leads to the monolithi s heme (13) whi h is solved

by adire t method (fullmatrix method). Letus remark that

α

n+1

j

= v

n+1

j,0

, j = 1, ..., p

, in(13).

4.2 The numeri al s heme

4.2.1 Dis retization

The fully dis retes heme orresponding with (28) tosolve(1) isthe following :

(17)











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)

(31)

As above,let us denote(30-option (b)) (resp. (31-option (a))) the system (30) (resp. (31))

with its last equation (30a) (resp. (31b)) removed. The fully-dis retized version of (28) is the

system onstituted by (30a) and (31b) (ex ept that (30a) should be re-written

α

n+1

j

= ...

). Of

ourse,

v

n+1

j,1

in (30a) isan impli itefun tionof

β

n+1

j

sin e

(v

n+1

j,i

)

i

is the solutionof (30-option

(b)). The same thing about the value of

(u

n+1

K

)

K

on

γ

j

in (31b) is true sin e

(u

n+1

K

)

K

is the

solution of (31-option (a)),so a fun tionof

(α

n+1

j

)

j=1,...,p

.

The system onstituted by (30a) and (31b) will be named the interfa e system. Interior

unknowns are eliminatedby the means of (30-option (b)) and (31-option (a)).

4.2.2 Formulation for the interfa e system with

2p

unknowns

The 1D dis rete operator in (30a) onsists to solve the 1D Neumann problem (30-option

(b)) andtogiveanapproximate valueof the solutionatthe point

δ

ontheaxis

j

.Itis denoted

by

S

N D,n+1

1D,j

below,

j = 1, ..., p

.

The 2D dis rete operator in (31b) onsists to solve the 2D Diri hlet problem (31-option

(a)) and to give a dis rete mean value of the normal derivative on

γ

j

, j = 1, ..., p

. It an be

des ribed by introdu ing general 2D problems, at the ontinuous level, with a Diri hlet or

Neumann boundary value whi h is not ompulsory onstant. Below we give the semi-dis rete

version of the general Diri hlet problem (option (a)) and of the general Neumann problem