Weak Imposition Of Boundary Conditions For
The Navier-Stokes Equations ∗
Atife C¸ a˘glar †
Ph.D. Student
Department of Mathematics
University of Pittsburgh
Pittsburgh, PA 15260, USA
November 23, 2002
Abstract
We prove the convergence of a finite element method for the Navier-
Stokes equations in which the no-slip condition, u·τ i = 0 on Γ for i = 1, 2
is imposed by a penalty method and the no-penetration condition, u·n = 0
on Γ, is imposed by Lagrange multipliers. This approach has been studied
for the Stokes problem in [2]. In most flows the Reynolds number is not ∗This work partially supported by NSF grants DMS9972622, INT9814115, and
INT9805563. †[email protected], http://www.math.pitt.edu/∼atife
1 negligable so the u · ∇u inertial effects are important. Thus the extension
beyond the Stokes problem to the Navier-Stokes equations is critical. We
show existence and uniqueness of the approximate solution and optimal
order of convergence can be achieved if the computational mesh follows
the real boundary. Our results for the (nonlinear) Navier-Stokes equations
improve known results for this approach for the Stokes problem.
Keywords. Navier-Stokes equations, Lagrange multiplier, slip with friction.
AMS Subject classifications. 65N30.
1 Introduction
The problem of predicting the equilibrium flow of a viscous incompressible fluid is one of solving approximately the stationary, incompressible Navier-Stokes equations:
−2Re−1∇ · (D(u)) + u · ∇u + ∇p = f in Ω,
u = 0 on Γ = ∂Ω, (1.1)
∇ · u = 0 in Ω, in a bounded polyhedral domain Ω ⊂ Rd d=2,3, where D is the deformation tensor given by
1 ∂u ∂u D (u) = i + j for 1 ≤ i, j ≤ d. i j 2 ∂x ∂x j i
The boundary Γ is assumed to be the union of k flat parts Γj :
k Γ = ∪j=1Γj .
2 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 3
The boundary condition u = 0 is decomposed of two separate conditions:
“no − penetration00 : u · n = 0 on Γ,
00 “no − slip : u · τ i = 0 on Γ,
where n be the unit normal vector on Γ and τ i, i = 1, . . . , d − 1, a system of orthonormal tangential vectors.
There are physical and computational reasons why these two boundary con- ditions should sometimes be separated and imposed using different techniques see, e.g. the discussions in [2]. For example, the Lagrange multiplier imple- mentation of the no-penetration condition allows slight, but locally balancing in and out flow capturing some aspects of surface roughness. For the no-slip condition, the penalty formulation (as we shall see) is equivalent to the Navier slip law. Thus, it is natural to use it for problems in which large tangential stress might occur. The numerical analysis of techniques for imposing essential boundary conditions weakly was begun in [3], see also [4], [5, 6], [7], [2] and the references therein. In many flows inertial effects are important and thus the extension beyond the Stokes problem, of [7,10] to the Navier-Stokes equations is important. The purpose of this report is to begin this extension. Our anal- ysis both extends the work of [2] and [7] from the (linear) Stokes problem to nonlinear Navier-Stokes problem and improves the basic results of [2] even in the linear case. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 4
2 The Continuous Problem
For our mathematical formulation we introduce the following spaces:
1 d 1 d X = [H (Ω)] , X0 = [H0 (Ω)] ,
2 V = {v ∈ X | h∇ · v, qiΩ = 0, ∀ q ∈ L0(Ω)},
2 V0 = v ∈ X0 | h∇ · v, qiΩ = 0, ∀ q ∈ L0(Ω) ,
2 2 Y = L0(Ω) = q ∈ L (Ω) | hq, 1iΩ = 0 , k −1/2 Z = H (Γj ), j=1 Y where k is the number of edges in 2d or faces in 3d of the boundary Γ. h., .iΩ is the usual L2 inner product, Hk(Ω) the usual W k,2(Ω) Sobolev space with
−k k norm k.kk,Ω, and the space H (Ω) is the dual of H0 (Ω), the space of functions in Hk(Ω) that vanish on Γ. The spaces H k−1/2(Γ) consist of the traces of all functions in Hk(Ω). Analogously, we denote by H −(k−1/2)(Γ) the dual space of
k−1/2 H (Γ) with h., .iΓ being the duality pairing.
k 1/2 A norm k.kΓ of a function ϕ ∈ j=1 H (Γj ) is defined by
Q 1/2 k 2 kϕkΓ = kϕk 1/2,Γj j=1 X ∗ 1/2 with the dual norm k.kΓ =: k.kZ. In addition, we set k.kΓj = k.kH (Γj ) and
∗ −1/2 k.kΓj = k.kH (Γj ).
The most common formulation of the Navier-Stokes equations in (X0, Y ) is WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 5
given by (see e.g. [8], [9]) : Find (u, p) ∈ (X0, Y ) such that:
a0(u, v) + a1(u; u, v) + a2(v, p) = hf, viΩ, ∀v ∈ X0,
a2(u, q) = 0, ∀q ∈ Y, where
−1 a0(u, v) = 2Re D(u) : D(v) dx, ZΩ
a1(u; v, w) = (u · ∇v) · w dx, ZΩ 1 b(u; v, w) = [a (u; v, w) − a (u; w, v)], 2 1 1
a2(u, p) = − p(∇ · u) dx. ZΩ
Lemma 2.1 For u ∈ V0 and v, w ∈ X0, b(u; v, w) = hu · ∇v, wiΩ. Since a1(u; v, w) = −a1(u; w, v) then b(u; v, v) = 0.
Proof It follows from integral by parts.
If we impose the boundary conditions weakly we must seek a formulation with velocities in X rather than X0. Using Green’s formula, the definition of defor- mation and stress tensor we arrive at the following weak formulation of (1.1) in
(X, Y ):
Find (u, p) ∈ (X, Y ) such that:
k
a0(u, v) + b(u; u, v) + a2(v, p) − hnj · =(u, p), viΓj = hf, viΩ, ∀v ∈ X, j=1 X
a2(u, q) = 0, ∀q ∈ Y. (2.1)
Here, the tensor =(., .) is defined by
1 = (u, p) = −pδ + 2Re−1D (u) + u u , for 1 ≤ i, k ≤ d, ik ik ik 2 i k WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 6 and for all v ∈ X, q ∈ Y.
Remark: =(., .) is a modification of the stress tensor where the pressure p
1 is replaced by the Bernoulli pressure pδik + 2 uiuk
Split the fourth term of the left hand side of equation (2.1) to its normal and tangential parts to obtain Lagrange multipliers:
ρ|Γj := −nj · =(u, p) · nj ,
τ (j) λ1|Γj := −nj · =(u, p) · 1 ,
τ (j) λ2|Γj := −nj · =(u, p) · 2 ,
Thus, ρ, λ1, λ2 are the individual components of the normal stress on Γj and
1 ≤ j ≤ k, where k is the number of boundary pieces of Γ. Then we get
1 1 2Re−1 D(u) : D(v) dx + (u · ∇u)v dx − (u · ∇v)u dx 2 2 ZΩ ZΩ ZΩ k k τ (j) − p(∇ · v) dx + ρ (v · nj ) ds + λ1 (v · 1 ) ds Ω j=1 Γj j=1 Γj Z X Z X Z k τ (j) + λ2 (v · 2 ) ds = f · v dx, (2.2) j=1 Γj Ω X Z Z − q(∇ · u) dx = 0, ZΩ Define c(.; ., ., ., .) on X × Y × Z × Z × Z as:
k 2 k τ (j) c(u; p, ρ, λ1, λ2) := hp, ∇ · uiΩ − hρ, u · nj iΓj − hλi, u · i iΓj . j=1 i=1 j=1 X X X The corresponding weak formulation of (2.1) is :
a(u; u, v) − c(v; p, ρ, λ1, λ2) = hf, viΩ, (2.3)
c(u; q, σ, χ1, χ2) = 0, WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 7
for all v ∈ X, q ∈ Y, σ, χ1, χ2 ∈ Z, where a(u; u, v) = a0(u; v) + b(u; u, v).
Define the space K by
K = {u ∈ X | c(u; p, ρ, λ1, λ2) = 0 ∀ p ∈ Y, ∀ ρ, λ1, λ2 ∈ Z} .
Lemma 2.2 The multi-linear form c(·; ·, ·, ·, ·) is continuous on X × Y × Z × Z.
Proof Define k|.k| as:
2 2 τ 2 τ 2 1/2 k|uk| := k∇uk + ku · nj kΓ + ku · 1kΓ + ku · 2kΓ which is equal to k.k1 on X, and apply Cauchy-Schwartz inequality to c(. : ., ., ., .) we obtain:
c(u; p, ρ, λ1, λ2)
k 2 k τ j ≤ kpkk∇ · uk + kρkZku · nj kΓj + kλikZ ku · 1kΓj j=1 i=1 j=1 X X X 1/2 1/2 k k ≤ kpkk∇uk + kρk2 ku · n k2 Z j Γj j=1 j=1 X X 1/2 1/2 2 k k + kλ k2 ku · τ j k2 i Z i Γj i=1 j=1 j=1 X X X
2 = kpkk∇uk + kρkZ ku · nj kΓ + kλikZ ku · τ ikΓ i=1 X
= h(kpk, kρkZ, kλ1kZ , kλ2kZ ), (k∇uk, ku · nj kΓ, ku · τ 1kΓ, ku · τ 2kΓ)i
2 1/2 2 1/2 2 2 2 2 2 τ 2 ≤ kpk + kρkZ + kλikZ k∇uk + ku · nj kΓ + ku · ikΓ " i=1 # " i=1 # X X 2 1/2 2 2 2 = kpk + kρkZ + kλikZ k|uk|. " i=1 # X WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 8
Lemma 2.3 The space K is a closed subspace of the Hilbert space X.
Proof This follows provided c(.; ., ., .) is continuous on X ×Y ×Z ×Z ×Z which was proven in Lemma 2.2
Lemma 2.4 For u ∈ K, Korn’s inequality
k∇uk ≤ CK (Ω)kD(u)k and the Poincar´e inequality
kuk ≤ CP (Ω)k∇uk hold.
Proof If u ∈ K then c(v, q, ρ, λ1, λ2) = 0 for all q ∈ Y, ρ, λ1, λ2 ∈ Z. If we pick
2 k τ ρ = 0, q = 0, then u satisfies i=1 j=1hλi, u · iiΓj = 0 for λ1, λ1 ∈ Z. This
1/2P P means that u · τ i = 0 in H (Γj ) for all j = 1, ..., k, and thus u · τ j = 0 a.e on
Γj . Similarly, u · n = 0 a.e on each Γj . These imply for all 1 ≤ j ≤ k so that u = 0 a.e on Γj with meas(Γj ) is strictly positive. By Korn’s inequalities see
[10], [11], [9] or [12] there exist CK > 0 such that
k∇uk ≤ CK (Ω)(γ(u) + kD(u)k) for all u ∈ X. In particular, for all u ∈ K, where γ(u) is a seminorm on L2(Ω) which is a norm on the constants. Since measure of Γj are strictly positive,
define γ(u) = kukΓj then, we get γ(u) = 0 if u is constant on Ω. The result thus follows. By the same argument, for the same γ(u) we get:
kuk ≤ CP (Ω)k∇(u)k WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 9 for all u ∈ K.
Corollary 2.1 The bilinear form a0(·, ·) is coercive on K:
2 αkuk1 ≤ a0(u, u)
−1 −2 −2 −2 with α = Re min{CK (Ω), CK (Ω)CP (Ω)}.
Proof This follows from the Korn and Poincar´e inequalities on K.
Also note that the problem associated with problem (2.3) has the following form in K. Find u ∈ K such that:
a(u; u, v) = hf, viΩ, ∀ v ∈ K. (2.4)
Remark: The problem (2.3) is different than the standard weak formulation of the Navier Stokes equations. Thus, we need to guarantee that the weak formulation is correct.
Lemma 2.5 (Existence) There exists a solution to the formulation (2.4) in
K.
Proof By the abstract theory developed in [8], the formulation (2.4) has a solution in K provided:
1) The form a(u; u, v) is coercive in K but since b(u; u, u) = 0 it is sufficient to show that a0(u, v) is coercive. This was done in the Corollary 1.1.
2) The space K is separable and u → a(u; u, v) is weakly continuous.The space K is separable since it is a closed subset of the Hilbert space X. Next, we prove that u → a(u; u, v) is weakly continuous in K. Let u be a function in K WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 10
and let um be a sequence in K so that um → u weakly as m → ∞. It is known by Rellich’s theorem that H 1(Ω) can be embedded compactly in L2(Ω). Then
2 d um converges strongly to u ∈ L (Ω) as m → ∞. Let v be in a dense subset L of K, L = {w ∈ D(Ω) ⊂ V : c(v, q, ρ, λ1, λ2) = 0, for all λ1, λ2, ρ ∈ Z}.
1 1 b(u ; u , v) = hu · ∇u , vi − hu · ∇v, u i . m m 2 m m Ω 2 m m Ω
Since um ∈ K so is in V . Thus, we have
hum · ∇um, viΩ = −hum · ∇v, umiΩ.
Hence we can actually write b(.; ., .) as:
d b(um; um, v) = −hum · ∇v, umiΩ = − umiumj vi,j dx. i,j=1 Ω X Z ∞ vi,j ∈ L (Ω) since v is infinitely many times differentiable. This implies that
1 lim umiumj = uiuj ∈ L (Ω). m→∞
Thus, d lim b(um; um, v) = − uiuj vi,j dx m→∞ i,j=1 Ω X Z = −b(u; v, u).
Since u ∈ K ⊂ V ,
−b(u; v, u) = hu; ∇u, viΩ.
2 The form a0(., .) is continuous and um → u in L (Ω) implies that
lim a0(um, v) = a0(u, v) m→∞
So we get
lim a(um; um, v) = a(u; u, v) m→∞ WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 11
for all v ∈ L. By density of L in K and continuity of the forms a0(., .) and b(.; ., .) the result follows.
In order to prove uniqueness we follow the approach of [8]. We define the
finite constant as b(u; v, w) N := sup . u,v,w∈K |u|1|v|1|w|1
Lemma 2.6 (Uniqueness) Suppose that the form a0(., .) is coercive and the form b(.; ., .) is locally Lipschitz continuous in K. Then, under the condition
1 hf, vi f ∗ f ∗ Ω 2 N| | < 1 where | | := sup α v∈K |v|1 problem (2.4) has a unique solution u ∈ K.
Proof Suppose there are two solutions u1 and u2 to the problem (2.4) then
a0(u1, v) + b(u1; u1, v) = hf, viΩ, ∀ v ∈ K and
a0(u2, v) + b(u2; u2, v) = hf, viΩ, ∀ v ∈ K.
Adding, subtracting the term b(u1; u2, v) and setting v = u1 − u2 we obtain:
a0(u1 −u2, u1 −u2)+b(u1; u1 −u2, u1 −u2)+b(u1 −u2; u2, u1 −u2) = 0. (2.7)
The second term in (2.7) vanishes by the skew-symmetric property of the form b(.; ., .). Thus
a0(u1 − u2, u1 − u2) ≤ |b(u1 − u2; u2, u1 − u2)|. (2.8)
Coercivity of the form a0(., .) and the definition of N imply that
2 2 α|u1 − u2|1 ≤ N|u2|1|u1 − u2|1. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 12
Since the form is coercive and continuous in K by Lax-Milgram lemma we have
1 |u | ≤ |f|∗. 2 1 α
Substituting (2.8) in (2.7) we obtain:
N α|u − u |2 ≤ |f|∗|u − u |2. 1 2 1 α 1 2 1
This implies that: N 2 (α − |f|∗)|u − u |2 ≤ 0. α 1 2 1
N ∗ Thus, u1 = u2 if 1 − α2 |f| > 0. We have shown that the problem (2.4) has a unique solution in K.
Theorem 2.1 Assume that the multilinear form c(.; ., ., .) satisfies the inf-sup condition. There is a constant β > 0 such that:
c(v; q, σ, χ1, χ2) inf sup 1/2 ≥ β. (2.9) χ1,χ2,σ∈Z v∈X |||v||| (kpk2 + kχ k2 + kχ k2 + kσk2 ) p∈Y 0 1 Z 2 Z Z
Then, for each solution u of problem (2.4), there exist a unique p ∈ Y, ρ, λ1, λ2 ∈
Z such that (u, p, ρ, λ1, λ2) is a solution of problem (2.3).
Proof It follows from [13]. For error analysis we need the following lemma.
Lemma 2.7 There is a constant β00 > 0 such that:
k (j) (j) hχ1, v · τ iΓ + hχ2, v · τ iΓ + hσ, v · nj iΓ j=1 1 j 2 j j 00 inf sup 1/2 ≥ β . σ,χ1,χ2∈Z v∈V 2 2 2 P kvk1 (kχ1kZ + kχ2kZ + kσkZ )
Proof See [2]. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 13
3 Penalty-Lagrange Multiplier Method
It has been observed by [14] that slip with friction boundary conditions match the experimental behavoir of real fluids at higher Reynolds number better than
τ (j) no-slip condition. Consequently, we propose to impose u· i |Γj = 0 for i = 1, 2 using a penalty technique that is equivalent to such a condition. We choose small, positive penalty parameters 1, 2 whose selection will be guided by the following analysis. We consider the following problem:
Find u ∈ V and ρ ∈ Z such that:
k −1 τ (j) a0(u, v) + b(u; u, v) + 1 h u, v · 1 iΓj j=1 X k k −1 τ (j) +2 hu, v · 2 iΓj + hρ, v · nj iΓj = hf, vi, ∀ v ∈ V, (3.1) j=1 j=1 X X k
hσ, u · nj iΓj = 0, ∀σ ∈ Z. j=1 X Suppose that instead of considering penalty methods to weaken the condition
τ (j) u · i |Γj = 0, we consider slip with non-linear resistance to slip, i.e.:
−1 −2Re ∇ · D(u) + u · ∇u + ∇p = f in Ω, (3.2)
∇ · u = 0 in Ω,
k u · nj ds = 0 on Γ = ∂Ω, j=1 Γj X Z 1 u · τ (j) + n · (D(u ) − |u |u ) · τ (j) = 0 on Γ , i = 1, 2, 1 ≤ j ≤ k. i i j 2 i j
Lemma 3.1 The variational formulation of equation (3.2) is the same as equa- tion (3.1). WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 14
Proof We multiply equation in (3.2) by v ∈ X, ζ ∈ Y, ρ ∈ Z, respectively, and integrate by parts yield:
−1 2Re hD(u) : D(v)iΩ + 1/2hu · ∇u, viΩ − 1/2hu · ∇v, uiΩ+
2 k −1 τ (j) τ (j) +hD(u) · nj , viΓj − hp, ∇ · viΩ + i h u · i , v · i iΓj + i=1 j=1 X X k
+ hρ, v · nj iΓj = hf, viΩ, ∀v ∈ X. j=1 X k
hσ, u · nj iΓj = 0, ∀σ ∈ Z. j=1 X hζ, ∇ · uiΩ = 0, ∀ζ ∈ Y.
The lemma implies that the penalty method captures the idea of slip with non- linear resistance to slip.To simplify the notation we define:
−1 τ (j) λ1 |Γj : = 1 u · 1 ,
−1 τ (j) λ2 |Γj : = 2 u · 2 ;
where λ1, λ2 ∈ Z. Then equation (3.1) takes the following form:
k τ (j) a0(u, v) + b(u; u, v)+ hλ1, v · 1 iΓj + j=1 X k k τ (j) + hλ2, v · 2 iΓj + hρ, v · nj iΓj = hf, vi, ∀ v ∈ V, j=1 j=1 X X k
hσ, u · nj iΓj = 0, ∀ σ ∈ Z, (3.3) j=1 X k k τ (j) hu · 1 , χ1iΓj = 1 hλ1, χ1iΓj , ∀ χ1 ∈ Z, j=1 j=1 X X k k τ (j) hu · 2 , χ2iΓj = 2 hλ2, χ2iΓj , ∀ χ2 ∈ Z. j=1 j=1 X X WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 15
Proposition 3.1 Let (u, λ1, λ2, ρ) be the solution of the Navier-Stokes equa-
tions (2.1) and let (u, λ1, λ2, ρ) be the solution to (3.2) Then:
2 2 2 2 1/2 e(λ1, λ2, ρ) + ku − uk1,Ω ≤ C[1kλ1kΓ + 2kλ2kΓ] , (3.4) where
2 2 2 1/2 e(λ1, λ2, ρ) := [kλ1 − λ1kZ + kλ2 − λ2kZ + kρ − ρkZ ] and C depends on Re, α, β00, and |f|∗.
Proof Subtracting equation (3.3) from equation (2.3) yields:
a0(u − u, v) + b(u − u; u, v) + b(u; u − u, v) + 2 k k τ (j) hλi − λi , v · i iΓj + hρ − ρ, v · nj iΓj = 0, ∀ v ∈ V, (3.5) i=1 j=1 j=1 X X X k
hσ, (u − u) · nj iΓj = 0, ∀ σ ∈ Z, j=1 X k k τ (j) h(u − u) · i , χiiΓj = −i hλi , χiiΓj , ∀ χi ∈ Z, j=1 j=1 X X
for i = 1, 2. From Lemma 2.7 given λ1, λ2, ρ ∈ Z we have:
00 β e(λ1, λ2, ρ) a (u − u , v) + b(u − u ; u , v) + b(u; u − u , v) ≤ sup 0 0=6 v∈V kvk1 2Re−1|u − u | |v| + M|u − u | |u | |v| + M|u − u | |v| |u| ≤ 1 1 1 1 1 1 1 1 kvk1
−1 −1 ∗ ≤ |u − u|1(2Re + 2Mα |f| )
−1 ≤ 2(Re + α)ku − uk1 (3.6) WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 16 where |b(u; v, w)| M := sup . 0=6 u,v,w∈V |u|1|v|1|w|1
In equation (3.5) we take v = u − u. Then,
a0(u − u, u − u) + b(u − u; u, u − u)
k k − i hλi , λi − λi iΓj + hρ − ρ, (u − u) · nj iΓj = 0. (3.7) j=1 j=1 X X Thus,
2 k a0(u − u, u − u) = b(u − u; u − u, u) + i hλi , λi − λi iΓj . i=1 j=1 X X 2 k Add and subtract the terms i=1 i j=1hλi, λi − λi iΓj , and dropping the pos-
2 k P P itive term i=1 i j=1hλi − λi , λi − λi iΓj we get:
P P 2 k a0(u − u, u − u) ≤ b(u − u; u − u, u) + i hλi, λi − λi iΓj . i=1 j=1 X X Applying Cauchy-Schwarz inequality and definition of M we get:
2 k a0(u − u, u − u) ≤ b(u − u; u, u − u) + i hλi, λi − λi iΓj i=1 j=1 X X −1 ∗ 2 ≤ Mα |f| |u − u|1 + ikλi − λi kZ kλikΓ.
By the arithmetic geometric mean inequality for n = 2, we have
a0(u − u, u − u) 2 1/2 2 1/2 −1 ∗ 2 2 2 2 ≤ Mα |f| |u − u|1 + kλi − λi kΓ i kλikΓ . i=1 ! i=1 ! X X
Using the coercivity of a0(., .) and inequality (3.6) we get
1/2 2(Re−1 + α) 2 ku − u k ≤ 2kλ k2 . 1 β00α(1 − Mα−2|f|∗) i i Γ i=1 ! X WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 17
Thus
2(Re−1 + α) e (λ , λ , ρ) + ku − u k ≤ ku − u k + ku − u k . 1 2 1 β00 1 1
Combining the latter inequalities we obtain:
2 1/2 2 2 e(λ1, λ2, ρ) + ku − uk1 ≤ C i kλikΓ . (3.8) i=1 ! X
4 Finite Element Spaces
The polyhedral domain Ω is subdivided into d-simplices with sides of length less than h with T h being the family of partitions. We will assume that T h satisfies the usual regularity assumptions, see e.g. [15] that:
1. Each vertex of Ω is a vertex of a T ∈ T h,
2. Each T ∈ T h has at least one vertex in the interior of Ω,
3. Any two d-simplices T, T 0 ∈ T h may meet in a vertex, a whole edge, or a
whole face,
The constants c0, c1 denote different constants which are independent of h.
h 1. Each T ∈ T contains a ball with radius c0h and is contained in a ball
with radius c1h.
h h Denote by Oj the partition of Γj which is induced by T .
Let Xh ⊂ X, Y h ⊂ Y, Zh ⊂ Z. Also let
h h h h X0 = u ∈ X | u = 0 on Γ . WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 18
The spaces Xh and Y h are assumed to satisfy the following properties:
I. There is a constant β˜ > 0 independent of h for which:
phdivuh dx Ω ˜ inf sup h h ≥ β, (4.1) 0=6 ph∈Y h uh h kp k0,Ωku k1,Ω 0=6 ∈X0 R
II.
h 1 inf kp − p k0,Ω ≤ chkpk1,Ω, ∀ p ∈ H (Ω), ph∈Y h
III. There exists a continuous linear operator Πh : H1(Ω)d → Xh for which:
h 1 d h Π (H0 (Ω) ) ⊂ X0 ,
h t−s t ku − Π uks,Ω ≤ ch kukt,Ω, ∀ u ∈ H (Ω) with s = 0, 1 and t = 1, 2,
h 1/2 ku − Π uk0,Γ ≤ ch kuk1,Ω,
k 1/2 where k.k0,Γ = ( j=1 k.k0,Γj ) . Assumption I balances the influence of the P constraint divu = 0 and also implies that the space:
h h h h h h h V0 = v ∈ X0 | hq , ∇ · v i = 0, ∀ q ∈ Y , h h h h h h h h V = v ∈ X | hq , ∇ · v i = 0, ∀ q ∈ Y ⊃ V0 , is also not empty. As usual V h is not a subset of V and in particular, the functions of V h are not divergence free. Hence we introduce anti-symmetric form b(.; ., ., .), h h h h b(u ; v , w ) N = sup h h h (4.2) 0=6 uh,vh,wh∈Xh |u |1|v |1|w |1 and hf, vhi f ∗ Ω k kh = sup h . vh∈V h |v |1 WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 19
With the above notation, the discrete analogue of problem (2.3) is:
h h h h h h h h Find u ∈ X , p ∈ Y , ρ , λ1 , λ2 ∈ Z such that:
h h h h h h h h h a(u ; u , v ) − c(v ; p , ρ , λ1 , λ2 ) = hf, v iΩ (4.3)
h h h h h c(u ; q , σ , χ1 , χ2 ) = 0
h h h h h h h h for all v ∈ X , q ∈ Y , σ , χ1 , χ2 ∈ Z .
By assumption I this is equivalent to the following problem in V h:
h h h h h h h h Find u ∈ X , p ∈ Y , ρ , λ1 , λ2 ∈ Z such that:
h h h h a(u , u , v ) = hf, v iΩ (4.4)
By the abstract theory developed in [8] the discrete problem will have unique solution provided:
−2 h ∗ α N |f|h < 1 (4.5) and, there is a constant β > 0 , independent of h such that
b c(vh; ph, ρh, λh, λh) inf sup 1 2 ≥ β. (4.6) h h h h 1/2 ρ ,λ1 ,λ2 ∈Z 0=6 vh∈Xh h h 2 h 2 h 2 h 2 ph∈Y h kv k1 kλ1 kZ + kλ2 kZ + kρ kZ + kp k0,Ω h i b Now we need to find right spaces X h, Y h, Zh so that they hold (4.6). An example of spaces Xh, Y h, Zh satisfying (4.6) and the classical inf-sup condition are given in [7].
Lemma 4.1 There exists a constant β > 0 independent of h, such that:
k h h τ (j) h h τ (j) h h j=1 hλ1 , v · 1 iΓj + hλ2 , v · 2 iΓj + hρ , v · nj iΓj inf sup ≥ β. h h 1/2 ρ ∈Z 0=6 vh∈Xh P kvhk kλhk2 + kλhk2 + kρhk2 h h h 1 1 Z 2 Z Z λ1 ,λ2 ∈Z (4.7) WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 20
Lemma 4.2 There is a constant β > 0 , independent of h such that:
bc(vh; ph, ρh, λh, λh) inf sup 1 2 ≥ β. (4.8) h h h h 1/2 ρ ,λ1 ,λ2 ∈Z 0=6 vh∈Xh h h 2 h 2 h 2 h 2 ph∈Y h kv k1 kλ1 kZ + kλ2 kZ + kρ kZ + kp k0,Ω h i b Proof The last two lemmas are proven in [2].
5 The Discrete Penalty-Lagrange Multiplier Method
We can now write the discrete analogue of the equation (3.1) as:
h h Find u ∈ X satisfying:
2 k h h h h h −1 h τ (j) h τ (j) a0(u , v ) + b(u ; u , v ) + i hu · i , v · i iΓj + i=1 j=1 X X k h h h h h + hρ , v · nj iΓj − hp , ∇ · v iΩ = hf, v i, (5.1) j=1 X h h hq , ∇ · u iΩ = 0,
k h h hσ , u · nj iΓj = 0, j=1 X h h h h h h h h for all v ∈ X , q ∈ Y , σ ∈ Z . This is equivalent to finding u ∈ V such that:
2 k h h h h h −1 h τ (j) h τ (j) a0(u , v ) + b(u ; u , v ) + i hu · i , v · i iΓj + (5.2) i=1 j=1 X X k h h h + hρ , v · nj iΓj = hf, v i, j=1 X k h h hσ , u · nj iΓj = 0, j=1 X for all vh ∈ V h, σh ∈ Zh. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 21
,h h Since λi for i = 1, 2 are not necessarily in Z equation (5.2) is not equiva- lent to (5.3). In fact
,h ˜h h τ (j) λi ∈ Zi := X · i |Γj , i = 1, 2, 1 ≤ j ≤ k.
2 k h h h h h ,h h τ (j) a0(u , v ) + b(u ; u , v ) + hλi , v · i iΓj + i=1 j=1 X X k h h h h h + hρ , v · nj iΓj = hf, v i, ∀ v ∈ V , j=1 X k h h h h hσ , u · nj iΓj = 0, ∀ σ ∈ Z , (5.3) j=1 X k k h τ (j) h ,h h h h hu · i , χi iΓj = i hλi , χi iΓj , ∀ χi ∈ Z , j=1 j=1 X X
for i = 1, 2. The latter problem will be equivalent if we make the following assumption: There exists β¨ > 0
k hλ,h, vh · τ (j)i + hλ,h, vh · τ (j)i + hρh, vh · n i j=1 1 1 Γj 2 2 Γj j Γj ¨ inf sup 1/2 ≥ β. h∈ ˜h 0=6 vh∈Xh ,h ,h λi Zi P vh 2 2 h 2 ρh∈Zh k k1 kλ1 kZ + kλ2 kZ + kρ kZ h i (5.4)
h We shall now study the error u − u in 1 − norm where (u, λ1, λ2, p, ρ) is the
h ,h ,h h h solution of (2.3) and (u , λ1 , λ2 , p , ρ ) is the solution of the finite element problem (5.3). We estimate the error in two cases. In the first case the computa- tional boundary Γh does not follow the fluid boundary Γ. For example, suppose the domain Ω has smooth boundary and we use quadratic elements for our ap- proximation. Then by approximating the fluid velocity u by penalized velocity
h u and u by discrete penalized velocity u . We will conclude that the error in WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 22
1 − norm has order h convergence. This requires that the penalty parameter should be scaled by h. In the second case, Γh follows Γ exactly. Our domain is polygonal (in 2d) with the choice of quadratic elements we approximate u by
h h h 2 discrete velocity u and u by u . This leads order h convergence in the error
1 − norm. This implies scaling by h2. Thus, it has improved the basic results of [2] even in the linear case.
h CASE I: In this case, the approximation u satisfies (5.3). We need similar equations for the continous u in the velocity space. For the reason we multiply by v ∈ V to obtain (3.1), instead we multiply by vh ∈ V h. It gives:
h h h h a0(u, v ) + b(u − u; u, v ) + hp − q , ∇ · v iΩ+
2 k k −1 τ (j) h τ (j) h h + i hu · i , v · i iΓj + hρ, v · nj iΓj = hf, v i, (5.5) i=1 j=1 j=1 X X X k h hσ , u · nj iΓj = 0, j=1 X for all vh ∈ V h and σh ∈ Zh, since Zh ⊂ Z. Subtracting (5.2) from (5.5) yields:
h h h h h h h h h a0(u − u , v ) + b(u ; u − u , v ) + b(u − u ; u, v ) − hp − q , ∇ · v iΩ+
2 k k −1 h τ (j) h τ (j) h h i h(u − u ) · i , v · i iΓj + hρ − ρ , v · nj iΓj = 0, (5.6) i=1 j=1 j=1 X X X for all vh ∈ V h. We write:
h h h h u − u = (u − v ) − (u − v ), (5.7)
h h with v being the best approximation of u in V . Let
h η := u − v ,
h h h φ := u − v . WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 23
Then equation (5.6) becomes:
h h h h h h h h h a0(η − φ , v ) + b(η − φ ; u, v ) + b(u ; η − φ , v ) − hp − q , ∇ · v iΩ
2 k k −1 h τ (j) h τ h h + i h(η − φ ) · i , v · iiΓj + hρ − ρ , v · nj iΓj = 0, (5.8) i=1 j=1 j=1 X X X for all vh ∈ V h. Equation (5.8) can be rewritten as:
2 k h h h h h h h −1 h τ h τ (j) a0(φ , v ) + b(φ ; u, v ) + b(u ; φ , v ) + i hφ · i, v · i iΓj i=1 j=1 X X h h h h h h = a0(η, v ) + b(η; u, v ) + b(u ; η, v ) − hp − q , ∇ · v iΩ (5.9)
k 2 k h h −1 τ (j) h τ (j) + hρ − ρ , v · nj iΓj + i hη · i , v · i iΓj j=1 i=1 j=1 X X X Set vh = φh. Then:
2 h 2 −1 h 2 h u h −1 h τ 2Re kD(φ )k0 + b(φ ; , φ ) + i φ · i 0,Γ = i=1 X
h h h h h h = a0(η, φ ) + b(u ; η, φ ) + b(η; u, φ ) − hp − q , ∇ · φ iΩ
2 k k −1 τ (j) h τ (j) h h + i hη · i , φ · i iΓj + hρ − ρ , φ · nj iΓj . (5.10) i=1 j=1 j=1 X X X h h h h But for each j, φ · nj ⊥Z since Z ⊂ Z. Thus we can replace ρ in (5.10) with any test function, say σh ∈ Zh. Then we have
2 −1 h 2 −1 h τ 2 2Re kD(φ )k + i kφ · ik0,Γ i=1 X 2 −1 ≤ 2Re−1|η| |φh| + N h|u h| |φh| |η| + N h|u ||φh| |η| + i kη · τ k2 1 1 1 1 1 1 1 2 i 0,Γ i=1 X 2 −1 + i kφh · τ k2 + kρ − σhk |φh| + kp − qhk|φh| (5.11) 2 i 0,Γ Z 1 1 i=1 X WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 24
2 −1 Re−1kD(φh)k2 + i kφh · τ k2 2 i 0,Γ i=1 X |φh|2 ≤ Re−1|η|2 + N h|φh| |η| (|u h| + |u| ) + kp − qhk2 + 1 1 1 1 1 1 4 2 −1 |φh|2 + i kη · τ k2 + kρ − σhk2 + 1 . (5.12) 2 i 0,Γ Z 4 i=1 X By Korn’s inequality we have:
2 −1 C kφhk2 ≤ Re−1kD(φh)k2 + i kφh · τ k2 1 1 2 i 0,Γ i=1 X h 2 −1 2 h ∗ −1 2 |φ |1 1 h 2 ≤ Re |η|1 + (N |f| α + |u|1)(|η|1 + ) + kp − q k 4 C1 2 −1 h 2 i τ 2 1 h 2 |φ |1 + kη · ik0,Γ + kρ − σ kZ + C1 . (5.13) 2 C1 2 i=1 X where C1 depends on Re, α, and
Since N|f|∗α−2 < 1, we can choose h sufficiently small so that:
h ∗ −1 N |f| hα < α.
Using this we obtain:
h 2 h 2 −1 2 |φ |1 1 h 2 C1kφ k1 ≤ Re |η|1 + (α + |u|1) + kp − q k 4 C1 2 −1 h 2 i τ 2 1 h 2 |φ |1 + kη · ik0,Γ + kρ − σ k0 + C1 . (5.14) 2 C1 2 i=1 X Thus we get:
2 −1 h 2 2 1 h 2 h 2 i τ 2 C2kφ k1 ≤ C3|η|1 + (kp − q k + kρ − σ k ) + kη · ik0,Γ (5.15) C1 2 i=1 X where C2 and C3 are depend on Re, , α and u but not on h. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 25
Adding and subtracting η in every norm of the left hand side of equation
(5.15) yields:
2 h 2 2 −1 τ 2 h 2 h 2 kη − φ k1 ≤ C4|η|1 + C5 i kη · ik0,Γ + C6(kρ − σ kZ + kp − q k ). i=1 X Applying the triangle inequality and taking infima we have:
h 2 ku − u k1 ≤
2 h 2 −1 h τ 2 ≤ C4 inf |u − v |1 + C5 i inf k(u − v ) · ik0,Γ+ 0=6 vh∈V h 0=6 vh∈V h i=1 X h 2 h 2 + C6( inf kρ − σ kZ + inf kp − q k ), 0=6 σh ∈Zh 0=6 qh∈Y h
where, for 1, 2 small. Now we have to derive a bound for the error in the
h h Lagrange multiplier. From Lemma 4.1 we take λ1 = λ2 = 0 :
k h h j=1hρ , v · nj iΓj inf sup h h = 0=6 ρh∈Zh h h v 0=6 v ∈X P k k1 kρ kZ k h h τ (j) h h τ (j) h h j=1hλ1 , v · 1 iΓj + hλ2 , v · 2 iΓj + hρ , v · nj iΓj inf sup 1/2 ≥ β. ρh∈Zh 0=6 vh∈Xh P kvhk kλhk2 + kλhk2 + kρhk2 h h h 1 1 Z 2 Z Z λ1 ,λ2 ∈Z (5.16)
Equation (5.16) implies that:
k h h vh n h h j=1hρ − σ , · j iΓj β ρ − σ Z ≤ sup h . (5.17) h h v v ∈X P k k1
WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 26
From (5.6) we have:
k h h h h h h h hρ − σ , v · nj iΓj = −a0(u − u , v ) − b(u − u ; u, v )− j=1 X h h h h h − b(u ; u − u , v ) + hp − q , ∇ · v iΩ+
2 k k −1 h τ (j) h τ (j) h h + i h(u − u ) · i , v · i iΓj + hρ − σ , v · nj iΓj i=1 j=1 j=1 X X X −1 u uh vh vh u uh h vh ≤ 2Re − 1 1 + 2α 1 − 1 + p − q 0 1 + 2 −1 h τ h h h + i k(u − u ) · ik0,Γ|v |1 + kρ − σ kZ |v |1 i=1 X 2 h h −1 h τ h h ≤ (C|u − u |1 + kp − q k + i k(u − u ) · ik0,Γ + kρ − σ kZ )|v |1 i=1 X (5.18) where C := 2(α + Re−1).
vh Thus dividing (5.18) by 1 , combining it with (5.17), applying the triangle inequality and taking infima yields:
h C h 1 h ρ − ρ ≤ u − u + inf p − q + Z β 1 β qh∈Y h 2 −1 i h τ 1 + β h + (u − u ) · i + inf ρ − σ . 0,Γ σh∈Zh Z i=1 β β X
(5.19)
Squaring both sides of (5.19) and bounding the mixed terms yields:
2 5C 2 5 2 h 2 ∗ u uh h kρ − ρ kZ ≤ 2 − 1 + 2 inf p − q + β β qh∈Y h
2 −2 5i h τ 2 10 h 2 + 2 (u − u ) · i + 2 inf kρ − σ kZ . 0,Γ σh∈Zh i=1 β β X
(5.20) WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 27
Combining (5.20) with (??) yields the following theorem for the total error.
Theorem 5.1 Assume that the discrete spaces satisfy condition (4.7). Then the error in the discrete solution of the penalty-Lagrange multiplier method is as follows:
2 h 2 h 2 −1 h τ 2 ku − u k1 ≤ C1 inf u − v + C2i inf u − v · i 0=6 vh∈V h 1 0=6 vh∈V h 0,Γ i=1 X
h 2 h 2 + C3 inf kρ − σ kZ + inf p − q . (5.21) 0=6 σh∈Zh 0=6 qh∈Y h 0
Incorporating (3.6) into (5.21) yields the following corollary for the total error for the penalty-Lagrange multiplier method:
Corollary 5.1 Assume that the discrete spaces satisfy condition (4.7). Then the error in the discrete solution of the penalty-Lagrange multiplier method is as follows:
2 h 2 h 2 C2 h τ 2 u − u ≤ C1 inf u − v + inf (u − v ) · i 1 vh∈V h 1,Ω vh∈V h 0,Γ i=1 i X
h 2 h 2 + C3 inf kρ − σ kZ + inf p − q σh∈Zh qh∈Y h 0 2 C + C |u − u |2 + 2 k(u − u ) · τ k2 + (5.22) 1 1,Ω i 0,Γ i=1 i X 2 2 2 2 2 C3kρ − ρkZ + kp − pk0 + C4 i kλikΓ . i=1 X Proof Squaring both sides of (3.4) yields the following:
2 2 2 2 2 2 2 ku − uk1,Ω + kλi − λi kZ + kρ − ρkZ ≤ C1 i kλikΓ . (5.23) i=1 i=1 X X
In (5.21) we replace the k.k0,Γ norm on the left hand side by k.kZ , add
(−u + u), (−p + p), (−ρ + ρ) in the appropriate terms of the right hand side and add it to (5.22) to get the stated result. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 28
Remark: The error bound that includes the difference in the value of the stress vector between the true and the discrete penalty-Lagrange multiplier so- lutions is as follows:
2 2 u uh ,h 2 h 2 − 1 + kλi − λi kZ + kρ − ρ kZ ≤ i=1 X 2 1 h 2 C2 h 2 ≤ C1 inf u − v + inf (u − v ) · τ i + min vh∈V h 1 vh∈V h 0,Γ i i ( i=1 i ) X
h 2 h 2 C3 min i + C0 inf kρ − σ kZ + inf p − q + i σh∈Zh qh∈Y h 0 2 2 2 C1 |u − u| + C3 min i + C0 kρ − ρkZ + kp − pk + 1 i 0 2 2 C 2 k(u − u ) · τ k2 + C 2 kλ k2 . i 0,Γ 4 i i Γ i=1 i i=1 X X
To obtain a bound on kp − pk0 we use (1.12) with χi = λi − λi , σ = ρ − ρ:
2 1/2 2 2 2 β kρ − ρkZ + kp − pk0 + kλi − λi kZ i=1 ! X 2 k (j) k hp − p, ∇ · vi − hλi − λ , v · τ iΓ − hρ − ρ, v · nj iΓ ≤ sup Ω i=1 j=1 i i j j=1 j v v ∈X P P k| |k P a0(u − u, v) + b(u − u; u, v) ≤ sup ≤ C |u − u|1 , v∈X k|v|k where C depends on Re and α. Thus:
C kp − p k2 ≤ |u − u |2 . (5.24) 0 β2 1,Ω
τ 2 τ (j) To get a bound on k(u − u) · ik0,Γ we set χi = (u − u) · i in (2.8):
τ k(u − u) · ik0,Γ ≤ i kλi k0,Γ . (5.25) WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 29
Add and subtract λi in the norm in the right hand side of (5.25) to yield:
τ k(u − u) · ik0,Γ ≤ i kλik0,Γ + kλi − λi k0,Γ ≤ i kλikΓ + kλi − λi k0,Γ , or 2 2 C 2 k(u − u ) · τ k2 ≤ C kλ k2 + kλ − λk2 . (5.26) i 0,Γ 2 i Γ i i 0,Γ i=1 i i=1 X X 2 Now we must bound kλi − λi k0,Γ. Take v = u − u in (3.7):
2 k a0(u − u, u − u) − b(u − u; u, u − u) − i hλi − λi , λi iΓj = 0. i=1 j=1 X X 2 k Add i=1 i j=1hλi − λi , λiiΓj to both sides:
2 P P 2 2 i kλi − λi k0,Γ ≤ i kλikΓ kλi − λi kZ +a0(u−u, u−u)+b(u−u; u, u−u). i=1 i=1 X X Definition of N and the Cauchy-Schwarz inequality:
2 2 kλ − λk2 ≤ i kλ k2 + kλ − λk2 . i i i 0,Γ 2 i 1/2,Γ i i Z i=1 i=1 X X Thus overall:
2 2 C 2 k(u − u ) · τ k2 ≤ C kλ k2 + kλ − λk2 . (5.27) i 0,Γ 2 i i Γ i i Z i=1 i i=1 X X Under the approximation assumption:
1/2 h h 2 h 2 inf u − v + inf p − q + inf kρ − ρ kZ ≤ vh∈V h 1 qh∈Y h 0 ρh∈Zh
k ≤ Ch max kfk−1 , kuk2 , (5.28) where k is the degree of the polynomial space we are using. If we let = 1 = 2 then corollary 5.1 indicates that proper choice of in (5.22) is = hk/2. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 30
Theorem 5.2 Assume that the discrete spaces satisfy condition (4.7) and let
= hk/2. Then the error in the solution of the penalty-Lagrange multiplier method and the discrete solution of the penalty-Lagrange multiplier method is as follows:
h ku − u k1 ≤ C.
h h h h h CASE II: Let (u, λ1, λ2, ρ, p) be the solution of (2.3) and (u , λ1 , λ2 , ρ , p ) be the finite element approximation of (2.3), that is (4.3) then substracting (4.3) from (2.3) we get:
h h h h h h h h h a0(u − u , v ) + b(u; u − u , v ) + b(u − u ; u , v ) − hp − p , ∇ · v iΩ
2 k k h τ (j) h τ (j) h h + h(u − u ) · i , v · i iΓj + hρ − ρ , v · nj iΓj = 0, (5.29) i=1 j=1 j=1 X X X for all vh ∈ V h. We write:
u − uh = (u − vh) − (uh − vh), with vh being the best approximation of u in V h. Let
η := u − vh,
φh := uh − vh.
Since Γh = Γ error terms which come from approximating boundary vanish, then equation (5.29) becomes:
h h h h h h h h h a0(η − φ , v ) + b(u; η − φ , v ) + b(η − φ ; u , v ) − hp − p , ∇ · v iΩ = 0,
(5.31) for all vh ∈ V h. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 31
Equation (5.31) can be rewritten as:
h h h h h h h h a0(φ , v ) + b(u; φ , v ) + b(φ ; u , v ) = a0(η, v )
h h h h h + b(u; η, v ) + b(η; u , v ) + hp − p , ∇ · v iΩ (5.32)
Setting vh = φh in (5.32), applying Cauchy-Schwartz inequality and definition of N h. We obtain:
−1 h 2 −1 h h ∗ −1 h 2Re kD(φ )k ≤ 2Re kD(η)kkD(φ )k + 2N |f|hα |φ |1|η|1+
h ∗ −1 h 2 h h N |f|hα |φ |1 + kp − p k|φ |1 (5.33)
By Korn’s inequality we have:
2Re−1kD(φh)k ≤ 2Re−1kD(η)k + 2αkD(η)k + αkD(φh)k + kp − phk
Thus we get:
h h C1kD(φ )k ≤ C2kD(η)k + kp − p k. (5.34)
∗ where C1 and C2 are depend on Re, α,and |f| . Adding and substracting η from the left hand side of (5.34) we get:
h h h kD(u − u )k ≤ (1 + C1/C2)kD(u − v )k + 1/C1kp − p k.
Taking infima we get:
h h h kD(u − u )k ≤ (1 + C1/C2) inf kD(u − v )k + 1/C1 inf kp − p k. 0=6 vh∈V h 0=6 ph∈Y h
(5.35)
h k Under the approximation assumption (5.30) we get ku − u k1 ∼ h . In order to
h h h bound ku − u k1 besides (5.35) we also need a bound on ku − u k1. For the WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 32
h h h reason we substract equation (5.3) from (4.3) then we set v = u − u and we obtain:
2 k h h h h h h h h h h ,h a0(u − u , u − u ) + b(u − u ; u , u − u ) + i hλi , λi − λi iΓj i=1 j=1 X X k h h h h + hρ − ρ , (u − u ) · nj iΓj = 0. (5.36) j=1 X From Lemma 3.1 we have:
−1 h ,h ,h h 2(Re + α) h h e (λ1 , λ2 , ρ ) ≤ u − u (5.37) β 1
where b
1/2 h ,h ,h h h ,h 2 h ,h 2 h h 2 e (λ1 , λ2 , ρ ) := kλ1 − λ1 kZ + kλ2 − λ2 kZ + kρ − ρ kZ . h i Since Γh = Γ adding and substracting
2 k k ,h h h i hλi, λi − λi iΓj , and hρ, (u − u ) · nj iΓj i=1 j=1 j=1 X X X in (5.36) gives:
h h h h h h h h h a0(u − u , u − u ) + b(u − u ; u , u − u ) (5.38)
2 k k ,h h h h + i hλi, λi − λi iΓj + hρ − ρ , (u − u ) · nj iΓj = 0. i=1 j=1 j=1 X X X By the same argument as in the proof of Proposition 2.1 we obtain:
h h 2 2 2 2 1/2 ku − u k1 ≤ C1 1kλ1kΓ + 2kλ2kΓ , (5.39) ∗ where C1 depends on Re, α, β, and |f| . Then combining (5.37) and (5.39).
We get: b
h h ku − u k1 ∼ WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 33
Thus from (5.35) and (5.38)
h k ku − u k1 ∼ h + . (5.40)
where C depends on C1 and kλikΓ for i = 1, 2. Thus (5.40) indicates that the proper choice of in (5.40) is = hk.
Theorem 5.3 Assume that the discrete spaces satisfy condition (4.7) and let
= hk. Then the error in the solution of the penalty-Lagrange multiplier method and the discrete solution of the penalty-Lagrange multiplier method is as follows:
h ku − u k1 ≤ C.
Conclusion. Weak imposition of essential boundary conditions by penalty-
Lagrange multipliers method lessens the ill-effects of domains with non-smooth boundaries. In this paper we have shown that the optimal order of convergence can be achieved if the computational boundary follows the real flow bound- ary exactly. We did not include numerical results because of the difficulty in implementation of the Lagrange multipliers method. Traditionally, Lagrange multipliers method is used as a model problem. However, we have done similar analysis by using the penalty- penalty method and verified our results numeri- cally in [20].
Acknowledgments. I would like to thank Prof. W. J. Layton and Dr.
T. Liakos for their valuable contributions to this paper. WEAK IMPOSITION OF B.C. IN NAVIER-STOKES 34
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