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Mixed finite element methods
Lucia Gastaldi
DICATAM - University of Brescia http://dm.ing.unibs.it/gastaldi/
Abstract setting
Let V and Q be Hilbert spaces.
a : V × V → R and b : V × Q → R continuous bilinear forms with
a(u, v ) ≤ kakkukV kv kV ∀u, v ∈ V
b(v , q) ≤ kbkkv kV kqkQ ∀v ∈ V , ∀q ∈ Q
Saddle point problem
Given f ∈ V 0 and g ∈ Q 0 , find (u, p) ∈ V × Q such that
a(u, v ) + b(v , p) = hf , v i ∀v ∈ V
(SP)
b(u, q) = hg , qi
∀q ∈ Q
The problem in operator form
Continuous operators
A : V → V 0 defined by hAu, v i = a(u, v )
B : V → Q 0 and B t : Q → V 0 defined by
hBv , qiQ 0 ×Q = b(v , q) = hv , B t qiV ×V 0
Then problem (SP) can be written as
Au + B t p = f
Bu = g
in V 0
in Q 0
The algebraic case
A ∈ Rn×n , B ∈ Rm×n , f ∈ Rn , g ∈ Rm , x ∈ Rn , y ∈ Rm
Ax + B t y = f
Bx = g
Notation
KerB = {x ∈ Rn : Bx = 0}
ImB = {z ∈ Rm : z = Bx for some x ∈ Rn }
KerB t = {y ∈ Rm : B t y = 0}
ImB t = {w ∈ Rn : w = B t y for some y ∈ Rm }
Prop. 1 The restriction of B to (KerB)⊥ is a one to one mapping
from (KerB)⊥ to ImB.
Theorem 1 Let B ∈ Rm×n . Then
KerB t = (ImB)⊥
ImB = (KerB t )⊥
KerB = (ImB t )⊥
ImB t = (KerB)⊥
Theorem 1 Let B ∈ Rm×n . Then
KerB t = (ImB)⊥
ImB = (KerB t )⊥
KerB = (ImB t )⊥
ImB t = (KerB)⊥
Theorem 2 A ∈ Rn×n , B ∈ Rm×n . Let K = KerB.
The matrix
A Bt
M=
B 0
is nonsingular if and only if
I A : K → K is surjective (and injective)
I B : Rn → Rm is surjective.
The infinite dimensional case
Au + B t p = f
Bu = g
in V 0
in Q 0
Notation
K = KerB ={v ∈ V : Bv = 0}
={v ∈ V : b(v , q) = 0 ∀q ∈ Q}
={v ∈ V : hv , B t qi = 0 ∀q ∈ Q}
Definition
V Hilbert space, S subspace of V , polar set S o ⊆ V 0 such that
S o = {L ∈ V 0 : hL, v i = 0 ∀v ∈ S}.
We have:
I S o is closed;
I (S o )o = S if and only if S is closed;
I S o = {0} if and only if S is dense.
Remark
(ImB t )o = {v ∈ V : hv , B t qi = 0 ∀q ∈ Q} = KerB
Theorem 2 Let V and Q be Hilbert spaces, B : V → Q 0 a
continuous linear operator, then
(KerB)o = ImB t
(KerB t )o = ImB
Remark
(ImB t )o = {v ∈ V : hv , B t qi = 0 ∀q ∈ Q} = KerB
Theorem 2 Let V and Q be Hilbert spaces, B : V → Q 0 a
continuous linear operator, then
(KerB)o = ImB t
(KerB t )o = ImB
Theorem 3 Let V and Q be Hilbert spaces, B : V → Q 0 a
continuous linear operator. The following properties are equivalent:
a. ImB is closed in Q 0 ;
b. ImB = (KerB t )o ;
c. ∃LB : ImB → (KerB t )⊥ continuous linear operator and β > 0
such that BLB g = g for all g ∈ ImB and
βkLB g kV ≤ kg kQ 0 .
Lemma The following properties are equivalent.
a. There exists β > 0 such that
(IS)
sup
v ∈V
b(v , q)
≥ βkqk
kv kV
∀q ∈ Q;
b. B t is an isomorphism from Q to K o and
kB t qkV 0 ≥ βkqkQ
∀q ∈ Q;
c. B is an isomorphism from K ⊥ to Q 0 and
kBv kQ 0 ≥ βkv kV
∀v ∈ K ⊥ .
An equivalent formulation of Problem (SP)
Let K (g ) = {v ∈ V : Bv = g }.
Problem 1 Find u ∈ K (g ) such that
a(u, v ) = hf , v i
∀v ∈ K .
Proposition 2 If b satisfies the condition (IS) then Problem (SP)
and Problem 1 are equivalent, that is, there exists a unique solution
of (SP) if and only if there exists unique solution of Problem 1.
Lemma 2 If there exists α > 0 such that
a(u, v )
≥ αkukV
v ∈K kv kV
a(u, v )
sup
≥ αkv kV
u∈K kukV
sup
(ISa )
∀u ∈ K
∀v ∈ K
then for any f ∈ K 0 there exists w ∈ K such that
a(w , v ) = hf , v i
and
kw kV ≤
∀v ∈ K
1
kf kV 0 .
α
Theorem 4 Assume that a satisfies the inf-sup conditions (ISa ),
and that b satisfies (IS), then there exists a unique solution
(u, p) ∈ V × Q of Problem (SP) and
1
1
kak
0
kukV ≤ kf kV +
1+
kg kQ 0
α
β
α
1
kpkQ ≤
β
kak
kak
kak
0
1+
kf kV + 2 1 +
kg kQ 0 .
α
β
α
The symmetric case
Assume that a : V × V → R is symmetric.
a satisfies the ellipticity on the kernel condition if there exists
α > 0 such that
(EK)
a(v , v ) ≥ αkv k2V
∀v ∈ K .
If a satisfies the ellipticity on the kernel condition then it satisfies
also (ISa ).
Sufficient and necessary condition
Theorem 5 Let Λ : V × Q → V 0 × Q 0 given by
Λ(v , q) = (Av + B t q, Bv ).
The linear operator Λ is an isomorphism from V × Q → V 0 × Q 0 if
and only if
I πA is an isomorphism from K to K’;
I there exists β > 0 such that kB t qkV 0 ≥ βkqkQ for all q ∈ Q.
Here π represents the orthogonal projection from V 0 → K 0 .
Finite element discretization
Let Vh and Qh be finite dimensional subspaces of V and Q
respectively.
Discrete saddle point problem
Given f ∈ V 0 and g ∈ Q 0 , find (uh , ph ) ∈ Vh × Qh such that
a(uh , v ) + b(v , ph ) = hf , v i ∀v ∈ Vh
(SP)
b(uh , q) = hg , qi
∀q ∈ Qh
Notation
We introduce the discrete version Bh of the operator B.
Bh : Vh → Qh0
such that hBh v , qi = b(v , q)
∀v ∈ Vh , ∀q ∈ Qh .
Kh = KerBh = {v ∈ Vh : b(v , q) = 0 ∀q ∈ Qh }
Kh (g ) = {v ∈ Vh : Bh v = g } for g ∈ Qh0
Notice that Kh 6⊂ K
Main assumptions
Discrete inf-sup condition for a There exists α∗ > 0 such that
sup
(DISa )
v ∈Kh
sup
u∈Kh
a(u, v )
≥ α∗ kukV
kv kV
∀u ∈ Kh
a(u, v )
≥ α∗ kv kV
kukV
∀v ∈ Kh
Discrete inf-sup condition for b There exists β ∗ > 0 such that
(DISb )
sup
v ∈Vh
b(u, q)
≥ β ∗ kqkQ
kv kV
∀q ∈ Qh
Approximation theorem
Theorem 6 Assume that a and b satisfy the discrete inf-sup
conditions (DISa ) and (DISb ). Then there exists a constant C
depending on α∗ , β∗ , kak and kbk such that the following error
estiamte holds true
ku − uh kV + kp − ph kQ ≤ C inf ku − v kV + inf kp − qkQ .
v ∈Vh
q∈Qh
Approximation theorem
Theorem 6 Assume that a and b satisfy the discrete inf-sup
conditions (DISa ) and (DISb ). Then there exists a constant C
depending on α∗ , β∗ , kak and kbk such that the following error
estiamte holds true
ku − uh kV + kp − ph kQ ≤ C inf ku − v kV + inf kp − qkQ .
v ∈Vh
q∈Qh
Moreover, if α∗ and β ∗ do not depend on h also C is independent
of h.
Approximation theorem
Theorem 6 Assume that a and b satisfy the discrete inf-sup
conditions (DISa ) and (DISb ). Then there exists a constant C
depending on α∗ , β∗ , kak and kbk such that the following error
estiamte holds true
ku − uh kV + kp − ph kQ ≤ C inf ku − v kV + inf kp − qkQ .
v ∈Vh
q∈Qh
Moreover, if α∗ and β ∗ do not depend on h also C is independent
of h.
If Kh ⊆ K then
ku − uh kV ≤ C inf ku − v kV .
v ∈Vh
The symmetric case
Assume a : V × V symmetric and the following discrete ellipticity
on the kernel condition: there exists α∗ > 0 such that
(DEK)
a(v , v ) ≥ α∗ kv k2V
∀v ∈ Kh .
Then the discrete inf-sup conditions (DISa ) are satisfied.
Fortin operator
Theorem 7 Assume that the continuous inf-sup condition (IS)
holds true. The discrete inf-sup condition for b (DISb ) holds true
with β ∗ independent of h if and only if there exists an operator
Πh : V → Vh such that
b(v − Πh v , q) = 0 ∀q ∈ Qh , ∀v ∈ V
kΠh v kV ≤ C kv kV ∀v ∈ V
with C independent of h.