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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
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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
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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
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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
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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
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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
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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=
1k (u|k)(v|k) + u|kv|kdx=K
k=
1k f|kv|kdx,Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 5 / 31
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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=1k (u|k)(v|k) + u|kv|kdx=K
k=1k f|kv|kdx,Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 5 / 31
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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,}
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 7 / 31
O
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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
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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)
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C ti 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)
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C ti 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)
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C ti 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)
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Continuous problem
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Continuous problem.Robin interface conditions
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)
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Continuous problem
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Continuous problem.Robin interface conditions
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)
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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)
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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)
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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)
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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)
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Continuous problem
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Continuous problem.Analysis of the convergence
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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Continuous problem.Analysis of the convergence
Let us now introduce
En =Kk=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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Continuous problem.Analysis of the convergence
Let us now introduce
En =Kk=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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Continuous problem.Analysis of the convergence
Let us now introduce
En =Kk=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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Continuous problem.Analysis of the convergence
Let us now introduce
En =Kk=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=1
En 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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pAnalysis of the convergence
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=1
En 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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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 .
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Outline
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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
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The triangulation and the discrete local spaces
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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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The triangulation and the discrete local spaces
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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}.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 15 / 31
The triangulation and the discrete local spaces
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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}.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 15 / 31
The triangulation and the discrete local spaces
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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}.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 15 / 31
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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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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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, =.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 16 / 31
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, =.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 16 / 31
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, =.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 16 / 31
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, =.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 16 / 31
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)
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 17 / 31
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Outline
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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
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 18 / 31
The iterative scheme
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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)
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 19 / 31
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
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 20 / 31
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)
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 21 / 31
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)
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 22 / 31
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,)
.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 23 / 31
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.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 24 / 31
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)|).
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 25 / 31
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.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 26 / 31
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.
Y. Maday (Paris 6 & Brown) Symetric Mortar EFEF07 27 / 31
Numerical results
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