Maday-efef.pdf

8/13/2019 Maday-efef.pdf

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A Symmetric mortar element method:with its Robin iterative procedure

Y. Maday1

1Laboratoire Jacques-Louis LionsUniversit Pierre et Marie Curie, Paris, France

and Brown Univ.

European Finite Element Fair

Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 1 / 31
http://find/


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Collaboration with

Caroline Japhet

Frdric Nataf

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Outline

1 IntroductionHybrid formulation

2 Iterative method

Continuous problem.

3 The discrete case

The triangulation and the discrete local spaces

The iterative scheme

4 Numerical results

http://find/


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Outline


2 Iterative method

Continuous problem.

3 The discrete case



4 Numerical results

http://find/


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Hybrid formulation.Domain decomposition

Consider the problem

(Id )u = f in, u=0 on, (1)

written under the following variational formulation : find uH10 ()such that

(uv+ uv) dx=

fvdx, vH10 (). (2)

taking into account the domain decomposition

=K

k=1

k

. (3)the problem can be written finduH10 ()such thatvH

10 ().

K

k=1

k (u|k)(v|k) + u|kv|kdx=K

k=1

k f|kv|kdx,Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 5 / 31
http://goforward/http://find/http://goback/


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(Id )u = f in, u=0 on, (1)


(uv+ uv) dx=

fvdx, vH10 (). (2)


=K

k=1

k


10 ().

K

k=1


k=1

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(Id )u = f in, u=0 on, (1)


(uv+ uv) dx=

fvdx, vH10 (). (2)


=K

k=1

k


10 ().

K

k=1


k=1

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(Id )u = f in, u=0 on, (1)


(uv+ uv) dx=

fvdx, vH10 (). (2)


=K

k=1

k


10 ().

K

k=

1k (u|k)(v|k) + u|kv|kdx=K

k=

1k f|kv|kdx,Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 5 / 31
http://find/


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(Id )u = f in, u=0 on, (1)


(uv+ uv) dx=

fvdx, vH10 (). (2)


=K

k=1

k


10 ().

K

k=1k (u|k)(v|k) + u|kv|kdx=K

k=1k f|kv|kdx,Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 5 / 31
http://find/http://goback/


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Hybrid formulation.Localisation

The spaceH10 ()can be identified with

V ={v= (v1, ..., vK)K

k=1

H1 (k),

k, , k=, 1 k, K, vk =v overk }.

or again

V ={v= (v1, ..., vK)K

k=1H1 (

k),

p(pk)K

k=1

H1/2(k)withpk=p overk,,

K

k=1

H1/2

(

k

)

H1/2(

k

)

= 0}.



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Hybrid formulation.Localisation

The spaceH10 ()can be identified with

V ={v= (v1, ..., vK)K

k=1

H1 (k),

k, , k=, 1 k, K, vk =v overk }.

or again

V ={v= (v1, ..., vK)K

k=1H1 (

k),

p(pk)K

k=1

H1/2(k)withpk=p overk,,

K

k=1

H1/2(k) H1/2(k) = 0}.

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Hybrid formulation.Localisation with Lagrange multipliers

The problem (2) is equivalent to the following one : Find (u, p) V

such thatv

Kk=1 H1 (k),

Kk=1

k

(ukvk+ ukvk) dxK

k=1

H1/2(k) H1/2(k)

=K

k=1

k

fkvkdx.

With the constrained space defined as follows

V ={(v, q)

Kk=1

H1 (k)

K

k=1

H1/2(k)

,

vk =v andqk=q overk,}

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Hybrid formulation.Localisation with Lagrange multipliers

The problem (2) is equivalent to the following one : Find (u, p) V

such thatv

Kk=1 H1 (k),

Kk=1

k

(ukvk+ ukvk) dxK

k=1

H1/2(k) H1/2(k)

=K

k=1

k

fkvkdx.

With the constrained space defined as follows

V ={(v, q)

Kk=1

H1 (k)

K

k=1

H1/2(k)

,

vk =v andqk=q overk,}


O
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Outline


2 Iterative method

Continuous problem.

3 The discrete case



4 Numerical results


C i bl
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Continuous problem.Robin interface conditions

Instead of imposing the equalities ofuandpwe write

pk+ uk=p+ u overk,

whereis a given positive real number, and introduce the following

algorithm : let(un

k

, pn

k

) H1

(k) H1/2(k)be an approximation

of(u, p)ink at stepn, then,(un+1k , pn+1k )is the solution in

H1 (k) H1/2(k)of

k un+1k vk+ u

n+1k vkdxH1/2(k) H1/2(k)

=

k

fkvkdx, vkH1 (

k) (4)

k,=k, , vkH

1/200 (

k,) (5)


C ti bl
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pk+ uk=p+ u overk,


algorithm : let(un

k

, pn

k

) H1 (k) H1/2(k)be an approximation


H1 (k) H1/2(k)of

k un+1k vk+ u


=

k

fkvkdx, vkH1 (

k) (4)

k,=k, , vkH

1/200 (

k,) (5)


C ti bl
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pk+ uk=p+ u overk,


algorithm : let(un

k

, pn

k



H1 (k) H1/2(k)of

k un+1k vk+ u


=

k

fkvkdx, vkH1 (

k) (4)

k,=k, , vkH

1/200 (

k,) (5)


C ti bl
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pk+ uk=p+ u overk,


algorithm : let(un

k

, pn

k



H1 (k) H1/2(k)of

k un+1k vk+ u


=

k

fkvkdx, vkH1 (

k) (4)

k,=k, , vkH

1/200 (

k,) (5)


Continuous problem
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It is obvious to remark that this series of equations results in

uncoupled problems set on everyk whenever f L2()we have

un+1

k

+ un+1

k

=fk overk

un+1knk

+ un+1k =pn + u

n over

k,

pn+1k = un+1k

nk

overk (6)


Continuous problem
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It is obvious to remark that this series of equations results in

uncoupled problems set on everyk whenever f L2()we have

un+1

k

+ un+1

k

=fk overk

un+1knk

+ un+1k =pn + u

n over

k,

pn+1k = un+1k

nkoverk (6)


Continuous problem
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Continuous problem.Analysis of the convergence

As in [Lions] and [Despres] by using an energy estimate (done with

f =0) k

|un+1k |

2 + |un+1k |2

dx=

k

pn+1k un+1k ds

=

1

4k,

(p

n+1

k + u

n+1

k )

2

(p

n+1

k u

n+1

k )

2ds

By using the interface conditions (5) we obtain

k

|u

n+1

k |

2

+ |u

n+1

k |

2dx+

1

4

k,(p

n+1

k u

n+1

k )

2

ds

= 1

4

k,

(pn + un )

2ds (7)


Continuous problem
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f =0) k

|un+1k |

2 + |un+1k |2

dx=

k

pn+1k un+1k ds

=

1

4k,

(p

n+1

k + u

n+1

k )

2

(p

n+1

k u

n+1

k )

2ds


k

|u

n+1

k |

2

+ |u

n+1

k |

2dx+

1

4

k,(p

n+1

k u

n+1

k )

2

ds

= 1

4

k,

(pn + un )

2ds (7)


Continuous problem


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f =0) k

|un+1k |

2 + |un+1k |2

dx=

k

pn+1k un+1k ds

=

1

4k,

(p

n+1

k + u

n+1

k )

2

(p

n+1

k u

n+1

k )

2ds


k

|u

n+1

k |

2

+ |u

n+1

k |

2dx+

1

4

k,(p

n+1

k u

n+1

k )

2

ds

= 1

4

k,

(pn + un )

2ds (7)


Continuous problem
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f =0) k

|un+1k |

2 + |un+1k |2

dx=

k

pn+1k un+1k ds

=

1

4k,

(p

n+1

k +

u

n+1

k )

2

(p

n+1

k

u

n+1

k )

2ds


k

|u

n+1

k |

2

+ |u

n+1

k |

2dx+

1

4

k,(p

n+1

k u

n+1

k )

2

ds

= 1

4

k,

(pn + un )

2ds (7)


Continuous problem
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Let us now introduce

En =K

k=1

k

|unk|

2 + |unk|2

and Bn = 14

Kk=1

=k

k,

(pnk unk)

2ds.

By summing up the previous estimates overk=1, ..., K, we have

En+1

+Bn+1

Bn,

so that, by summing up these inequalities, now overn, we obtain :

n=1En B0.

We thus havelim

nEn =0

so that

limn

pnkH1/2(k)=0, fork=1, ..., K,

which proves the convergenceof the continuous algorithm.Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 12 / 31

Continuous problem
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En =Kk=1

k

|unk|

2 + |unk|2

and Bn = 14

Kk=1

=k

k,

(pnk unk)

2ds.


En+1 + Bn+1 Bn,


n=1En B0.

We thus havelim

nEn =0

so that

limn

pnkH1/2(k)=0, fork=1, ..., K,


Continuous problem
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En =Kk=1

k

|unk|

2 + |unk|2

and Bn = 14

Kk=1

=k

k,

(pnk unk)

2ds.


En+1 + Bn+1 Bn,


n=1En B0.

We thus havelim

nEn =0

so that

limn

pnkH1/2(k)=0, fork=1, ..., K,


Continuous problem.
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En =Kk=1

k

|unk|

2 + |unk|2

and Bn = 14

Kk=1

=k

k,

(pnk unk)

2ds.


En+1 + Bn+1 Bn,


n=1En B0.

We thus havelim

nEn =0

so that

limn

pnkH1/2(k)=0, fork=1, ..., K,


Continuous problem.
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En =Kk=1

k

|unk|

2 + |unk|2

and Bn = 14

Kk=1

=k

k,

(pnk unk)

2ds.


En+1 + Bn+1 Bn,


n=1

En B0.

We thus havelim

nEn =0

so that

limn

pnkH1/2(k)=0, fork=1, ..., K,


Continuous problem.
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pAnalysis of the convergence


En =K

k=1

k

|unk|

2 + |unk|2

and Bn = 14

Kk=1

=k

k,

(pnk unk)

2ds.


En+1 + Bn+1 Bn,


n=1

En B0.

We thus havelim

nEn =0

so that

limn

pnkH1/2(k)=0, fork=1, ..., K,


Continuous problem.
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pAnalysis of the convergence

We have just proven that

Theorem

Assume that f is in L2()and(p0k)1kK H1/2(k,). Then, thealgorithm (4)-(5) converges in the sense that

limn

unk ukH1(k)+ p

nk pkH1/2(k)

=0, for1 kK,

where ukis the restriction tok of the solution u to (1), and pk =

uk

n

koverk, 1 kK .


Outline
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2 Iterative method

Continuous problem.

3 The discrete case



4 Numerical results


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g p

Eachk is provided with its own meshTkh , 1 kK, such that

k =TTkh

T.

As is standard we introduce the discretization parameter

h= max1kK

hk, with hk= maxTTkh

hT.

We assume thatTkh is uniformly regular Then, we define over eachsubdomain two conforming spacesYkh andX

kh by :

Ykh = {vh,k C0(k), vh,k|T PM(T), T Tkh},

Xkh = {vh,kYkh, vh,k|k=0}.


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g p


k =TTkh

T.


h= max1kK


hT.


kh by :




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k =TTkh

T.


h= max1kK


hT.


kh by :




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k =TTkh

T.


h= max1kK


hT.


kh by :




The interface spaces
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Over eachk, we consider the space of traces of elements of Ykh . It is

denoted byY

k,

h .k, Ykh = Y

k,h

over eachsuch thatk, =. Mortar subspace

Wk,

h

Yk,

h

Example in 2D, over each (possibly curved) side k, of piecewisepolynomials of degree M.W

k,

h

composed of polynomials of degree M 1 over both[xk,

0

, xk,

1

]

and[xk,n1, xk,n ].

Wkhis the product space of theWk,h over eachsuch that

k, =.




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denoted byY

k,

h .k, Ykh = Y

k,h


Wk,

h

Yk,

h


k,

h


0

, xk,

1

]

and[xk,n1, xk,n ].


k, =.




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denoted byY

k,

h .k, Ykh = Y

k,h


Wk,

h

Yk,

h


k,

h


0

, xk,

1

]

and[xk,n1, xk,n ].


k, =.




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denoted byY

k,

h .k, Ykh = Y

k,h


Wk,

h

Yk,

h


k,

h


0

, xk,

1

]

and[xk,n1, xk,n ].


k, =.




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denoted byY

k,

h .k, Ykh = Y

k,h


Wk,

h

Yk,

h


k,

h


0

, xk,

1

]

and[xk,n1, xk,n ].


k, =.




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denoted byY

k,

h .k, Ykh = Y

k,h


Wk,

h

Yk,

h


k,

h


0

, xk,

1

]

and[xk,n1, xk,n ].


k, =.


The global spaces and the problem
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The global discrete constrained space is then

Vh={(uh, ph) K

k=1

Xkh

K

k=1

Wkh

,

k,

((ph,k+ uh,k) (ph,+ uh,))h,k,=0, h,k, Wk,h },

and the discrete problem is : Find (uh, ph) Vhsuch that

vh= (vh,1,...vh,K)K

k=1 Xkh,

K

k=1

kuh,kvh,k+ uh,kvh,kdx

K

k=1

k

ph,kvh,kds

=K

k=1

k

fkvh,kdx. (8)

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Outline


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1

IntroductionHybrid formulation

2 Iterative method

Continuous problem.

3 The discrete case



4 Numerical results


http://find/


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We also have the discrete algorithm : let (un

h,k, pn

h,k) Xk

h W

k

h then,(un+1h,k , p

n+1h,k )is the solution in X

kh

Wkh of

k un+1h,k vh,k+ u

n+1h,k vh,k

dx

kpn+1h,k vh,kds

=k

fkvh,kdx, vh,kXkh (9)

k,(pn+1h,k + u

n+1h,k )h,k,=

k,

(pnh,+ unh,)h,k,,

h,k, Wk,h . (10)


Convergence for the iterative scheme
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We can prove the convergence of the iterative scheme

Theorem

Let us assume thathc, for some constant c small enough. Then,

the discrete problem (8) has a unique solution(uh,

ph) Vh. Thealgorithm (9)-(10) is well posed and converges in the sense that

limn

u

nh,k uh,kH1(k)+ =k p

nh,k, ph,k,

H

12

(k,)

=0, for1 k


Convergence for the iterative scheme
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Based on a stability result for the Lagrange multipliers, proven in 2D by

Ben Belgacem, and valid also in 3D

Lemma

There exists a constant c such that, for any ph,k, inWk,h , there exists

an element wh,k, in Xkhthat vanishes overk \ k, and satisfies

k,ph,k,w

h,k, ph,k,2

H

12

(k,)

(11)

with a bounded norm

wh,k,H1(k)cph,k,H

12

(k,)

. (12)


Analysis of the discrete problem


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Lemma

There exists a constant >0independent of h such that

(uh, ph) Vh, vhK

k=1

Xkh,

a((uh, ph), vh)) (uh+ ph 12 ,)vh. (13)

Moreover, we have the continuity argument : there exists a constant

c>0such that

(uh, ph) Vh, vhK

k=1

Xkh,

a((uh, ph), vh)) c(uh+ ph 12)(vh). (14)


Convergence of the discrete scheme -1-
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Theorem

Assume that the solution u of (1)-(1) is in H2() H10 (), anduk =u|k H

2+m(k), with M 1 m0, and let pk,= unk

over

eachk,. Then, there exists a constant c independent of h andsuch

that

uh u+ ph p 12 ,c(h2+m + h1+m)

Kk=1

uH2+m(k)

+c( h

m

+ h1+m)

Kk=1

pk,H

12+m(k,)

.


Convergence of the discrete scheme -2-
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Theorem

Assume that the solution u of (1)-(1) is in H2() H10 (),

uk =u|k H2+m(k), and pk,=

unk

is in H32+m(k,)with

M 1 m0. Then there exists a constant c independent of h andsuch that

uh u+ ph p 12 ,c(h2+m + h1+m)

Kk=1

uH2+m(k)

+c(h1+m

+ h2+m)(log h)(m)

K

k=1

pk,

H

3

2 +m(k,)

with(m) =0if mM 2and(m) =1if m=M 1.


What aboutP1elements

AK H2(k ) d i t t i d d t f h th
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AssumeuK

k=1 H2 (

k)and is a constant independent ofhthen

uh ucK

i=1

uH2(k)

which is not optimal.

In order to improveshould depend onh, or assume more regularity

onu, or at least overpk,= unk :

If the solutionuK

k=1 H2 (

k)and = ch, then

uh uchK

i=1

uH2(k)

If the solutionuK

k=1 H2 (

k),pk,H32 (k,)andis a

constant independent ofhthen

uh u=O(h| log(h)|).


Numerical results
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0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x

y

Mesh

Figure:Lebesgue constant on the interval.


Choice of the Robin parameter


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The Robin parameter is either an arbitrary constant or takes thefollowing values :

min= [(2 + 1)((

hmin)2 + 1)]

14

mean= [(2 + 1)(( hmean

)2 + 1)] 14

max= [(2 + 1)((

hmax)2 + 1)]

14

wherehmin,hmeanandhmaxstands respectively for the smallestmeanest or highest step size on the interface.


Numerical results
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