# Two-Level Iteration Penalty Methods for the Navier-Stokes Equations with Friction Boundary Conditions

Authors: Yuan Li; Rong An
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

This paper presents two-level iteration penalty finite element methods to approximate the solution of the Navier-Stokes equations with friction boundary conditions. The basic idea is to solve the Navier-Stokes type variational inequality problem on a coarse mesh with mesh sizeH in combining with solving a Stokes, Oseen, or linearized Navier-Stokes type variational inequality problem for Stokes, Oseen, or Newton iteration on a fine mesh with mesh size h. The error estimate obtained in this paper shows that if H, h, and ε can be chosen appropriately, then these two-level iteration penalty methods are of the same convergence orders as the usual one-level iteration penalty method.


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## Body

## 1. Introduction

In this paper, we consider a two-level iteration penalty method for the incompressible flows which are governed by the incompressible Navier-Stokes equations:(1)-μΔu+(u·∇)u-∇p=f,inΩ,divu=0,inΩ,
where Ω is a bounded domain in ℝ2 assumed to have a Lipschitz continuous boundary ∂Ω, μ>0 represents the viscous coefficient, u=(u1(x),u2(x)) denotes the velocity vector, p=p(x) the pressure and f=(f1(x),f2(x)) the prescribed body force vector. The solenoidal condition divu=0 means that the flows are incompressible.Instead of the classical whole homogeneous boundary conditions, here we consider the following slip boundary conditions with friction type:(2)u=0,onΓ,un=0,-στ(u)∈g∂|uτ|onS,
where Γ∩S=∅,Γ∪S¯=∂Ω, and g is a scalar function; un=u·n and uτ=u-unn are the normal and tangential components of the velocity, where n stands for the unit vector of the external normal to S; στ(u)=σ-σnn, independent of p, is the tangential components of the stress vector σ which is defined by σi=σi(u,p)=(μeij(u)-pδij)nj with eij(u)=(∂ui/∂xj)+(∂uj/∂xi),i,j=1,2. The set ∂ψ(a) denotes a subdifferential of the function ψ at a∈L2(S)2, whose definition will be given in the next section.This type of boundary condition is firstly introduced by Fujita [1] where some problems in hydrodynamics are studied. Some theoretical problems are also studied by many scholars, such as Fujita in [2–4], Y. Li and K. Li [5, 6], and Saito and Fujita [7, 8] and references cited in their work.The development of appropriate mixed finite element approximations is a key component in the search for efficient techniques for solving the problem (1) quickly and efficiently. Roughly speaking, there exist two main difficulties. One is the nonlinear term (u·∇)u, which can be processed by the linearization method such as the Newton iteration method, Stokes iteration method, Oseen iteration method [9], or the two-level methods [10–17]. The other is that the velocity and the pressure are coupled by the solenoidal condition. The popular technique to overcome the second difficulty is to relax the solenoidal condition in an appropriate method and to result in a pseudocompressible system, such as the penalty method and the artificial compressible method [18]. Recently, using the Taylor-Hood element, the authors [19] study the penalty finite element method for the problem (1)-(2). Denote (uεh,pεh) as the penalty finite element approximation solution to (u,p)∈(H3(Ω)2,H2(Ω)). The error estimate derived in [19] is
(3)∥u-uεh∥1+∥p-pεh∥≤c(ε+h5/4),
where ε>0 is the penalty parameter. However, the condition number of the numerical discretization for the penalty methods is O(ε-1h-2), which will result in an ill-conditioned problem when mesh size h→0. In order to avoid the choice of the small parameter ε, Dai et al. [20] have studied the iteration penalty finite element method and derive
(4)∥u-uεhk∥1+∥p-pεhk∥≤c(εk+1+h5/4),
where k∈ℕ+ is the iteration step number.In this paper, we combine the iteration penalty method with the two-level method to approximate the solution of the problem (1)-(2). The iterative penalty method was first introduced by Cheng and Shaikh [21] for the Stokes equations and further used to solve the pure Neumann problem [22]. This iteration penalty method can be considered as the time discretization of the artificial compressible method [23]. The two-level iteration penalty methods studied in this paper can be described as follows. The first step and the second step are required to solve a small Navier-Stokes equations on the coarse mesh in terms of the iteration penalty method [20, 21]. The third step is required to solve a large linearization problem on the fine mesh in terms of the Stokes iteration, Oseen iteration, or Newtonian iteration, respectively. We prove that these two-level iteration penalty finite element solutions (uεh,pεh) are of the following error estimate:
(5)∥u-uεh∥1+∥p-pεh∥≤{c(h5/4+H9/4+εH5/4+εk+1),(StokesandOseeniteration),c(h5/4+H5/2+εH5/4+εk+2),(Newtonianiteration)
Finally, we propose an improved correction iteration scheme for (uεh,pεh) in terms of the Newton iteration method. We prove that the correction finite element solutions (uεh⋆,pεh⋆) are of the following error estimates:
(6)∥u-uεh⋆∥1+∥p-pεh⋆∥≤{c(h5/4+H9/2+ε3/2H5/4+ε1/2H9/4+εk+1/2),(StokesandOseeniteration),c(h5/4+H5+ε3/2H5/4+ε1/2H5/2+εk+1/2),(Newtonianiteration).Throughout this paper, we will usec to denote a positive constant whose value may change from place to place but that remains independent of h, H, and ε and that may depend on μ, Ω and the norms of u,p,f,   and  g.


## 2. Preliminary

First, we give the definition of the subdifferential property. Letψ be a given function possessing the properties of convexity and weak semicontinuity from below. We say that the set ∂ψ(a) is a subdifferential of the function ψ at a∈L2(S)2 if and only if
(7)∂ψ(a)={b∈L2(S)2:ψ(h)-ψ(a)≥b·(h-a),∀h∈L2(S)2}.In what follows, we employ the standard notationHl(Ω) (or Hl(Ω)2) and ||·||l, l≥0, for the Sobolev spaces of all functions having square integrable derivatives up to order l in Ω and the standard Sobolev norm. When l=0, we write L2(Ω) (or L2(Ω)2) and ||·|| instead of H0(Ω) (or H0(Ω)2) and ||·||0, respectively.For the mathematical setting, we introduce the following spaces:(8)V={u∈H1(Ω)2,u|Γ=0,u·n|S=0},V0=H01(Ω)2,Vσ={u∈V,divu=0},M=L02(Ω)={q∈L2(Ω),∫Ω‍qdx=0}.The spaceV is equipped with the norm
(9)∥v∥V=(∫Ω‍|∇v|2dx)1/2.
It is well known that ||v||V is equivalent to ||v||1 due to Poincare's inequality. Introduce two bilinear forms
(10)a(u,v)=μ∫Ω‍∇u·∇vdx,∀u,v∈V,d(v,q)=∫Ω‍qdivvdx,∀v∈V,q∈M,
and a trilinear form
(11)b(u,v,w)=∫Ω‍(u·∇)v·wdx-12∫Ω‍divuv·wdx=12∫Ω‍(u·∇)v·wdx-12∫Ω‍(u·∇)w·vdx.
It is easy to verify that this trilinear form satisfies the following important properties [12, 23]:
(12)b(u,v,w)=-b(u,w,v),(13)b(u,v,w)≤N∥u∥V∥v∥V∥w∥V,(14)b(u,v,w)≤N2∥u∥1/2∥u∥V1/2×(∥v∥V∥w∥1/2∥w∥V1/2+∥w∥V∥v∥1/2∥v∥V1/2),
for all u,v,w∈V, and
(15)|b(u,v,w)|+|b(v,u,w)|+|b(w,u,v)|≤N∥u∥V∥v∥2∥w∥,
for all u∈V,v∈H2(Ω)2, and w∈L2(Ω)2, where N>0 depends only on Ω.Givenf∈L2(Ω)2 and g∈L2(S) with g>0 on S, under the above notation, the variational formulation of the problem (1)-(2) reads as follows: find (u,p)∈(V,M) such that for all (v,q)∈(V,M)(16)a(u,v-u)+b(u,u,v-u)+j(vτ)-j(uτ)-d(v-u,p)≥(f,v-u),d(u,q)=0,
where j(η)=∫S‍g|η|ds for all η∈L2(S)2. Saito in [8] showed that there exists some positive β>0 such that
(17)β∥q∥≤supv∈Vd(v,q)∥v∥V;
then the variational inequality (16) is equivalent to the following: find u∈Vσ such that for all v∈Vσ(18)a(u,v-u)+b(u,u,v-u)+j(vτ)-j(uτ)≥(f,v-u).
The existence and uniqueness theorem of the solution u to the problem (18) has been shown in [19]. Here, we only recall it.Theorem 1.
If the following uniqueness condition holds(19)4κ1N(∥f∥+∥g∥L2(S))μ2<1,
then there exists a unique solution u∈Vσ to the variational inequality problem (18) such that
(20)∥u∥V≤2κ1μ(∥f∥+∥g∥L2(S))<μ2N,
where κ1>0 satisfies
(21)|(f,v)-j(vτ)|≤κ1(∥f∥+∥g∥L2(S))∥v∥V,∀v∈Vσ.


## 3. Iteration Penalty Finite Element Approximation

Suppose thatΩ is a convex and polygon domain. Let 𝒯h be a family of quasi-uniform triangular partition of Ω. The corresponding ordered triangles are denoted by K1,K2,…,Kn. Let hi=diam(Ki),i=1,…,n, and h=max{h1,h2,…,hn}. For every K∈𝒯h, let Pr(K) denote the space of the polynomials on K of degree at most r. For simplicity, we consider the conforming finite element spaces Vh and Mh defined by
(22)Vh=V∩WhwithWh={vh∈C(Ω¯)2,vh|K∈[P2(K)]2,∀K∈𝒯h,{(Ω¯)2}},Mh={qh∈C(Ω¯),qh|K∈P1(K),∀K∈𝒯h,∫Ω‍qhdx=0}.
Denote V0h=V0∩Wh. It is well known that V0h and Mh satisfy the Babuška-Brezzi condition [24, 25]:
(23)κ∥qh∥≤supwh∈V0hd(wh,qh)∥wh∥V,
where κ>0 is a constant independent of h. Denote Rh and Qh as the L2 orthogonal projections onto Vh and Mh, respectively, which satisfy
(24)∥v-Rhv∥+h∥v-Rhv∥V≤chi∥v∥i,∀v∈H3(Ω)2∩V,i=1,2,3,(25)∥q-Qhq∥≤chj∥q∥j,∀q∈H2(Ω)∩M,j=1,2.
It follows from the trace inequality ||v||L2(S)≤c||v||1/2||v||V1/2 [26] that
(26)∥v-Rhv∥L2(S)≤c∥v-Rhv∥1/2∥v-Rhv∥V1/2≤chi-1/2∥v∥i,∀v∈H3(Ω)2∩V,i=1,2,3.Letε>0 be some small parameter. The one-level iteration penalty finite element method for the problem (16) has been studied in [20], which can be described as follows.Step 1.
Find(uεh0,pεh0)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(27)a(uεh0,vh-uεh0)+b(uεh0,uεh0,vh-uεh0)+j(vhτ)-j(uεhτ0)-d(vh-uεh0,pεh0)≥(f,vh-uεh0),d(uεh0,qh)+ε(pεh0,qh)=0.Step 2.
Fork=1,2,…, find (uεhk,pεhk)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(28)a(uεhk,vh-uεhk)+b(uεhk,uεhk,vh-uεhk)+j(vhτ)-j(uεhτk)-d(vh-uεhk,pεhk)≥(f,vh-uεhk),d(uεhk,qh)+ε(pεhk,qh)=ε(pεhk-1,qh).
First, we give the a priori estimate of the solution(uεhk,pεhk) to the problem (28).Theorem 2.
Suppose that(uεhk,pεhk)∈(Vh,Mh) is the solution to the problem (28); then it satisfies
(29)μ∥uεhk∥V2+ε∥pεhk∥2≤(2k+1)κ122μ(∥f∥+∥g∥L2(S))2.Proof.
Settingvh=0, qh=pεh0 in (27), using (12) and Young's inequality, it yields that
(30)μ∥uεh0∥V2+ε∥pεh0∥2≤(f,uεh0)-j(uεhτ0)≤κ1(∥f∥+∥g∥L2(S))∥uεh0∥V≤μ2∥uεh0∥V2+κ122μ(∥f∥+∥g∥L2(S))2.
Then we have
(31)μ∥uεh0∥V2+2ε∥pεh0∥2≤κ12μ(∥f∥+∥g∥L2(S))2.
For k=1,2,…, setting vh=0, qh=pεhk in (28), it yields that
(32)μ∥uεhk∥V2+ε∥pεhk∥2≤(f,uεhk)-j(uεhτk)+ε(pεhk-1,pεhk)≤μ2∥uεhk∥V2+κ122μ(∥f∥+∥g∥L2(S))2+ε2∥pεhk∥2+ε2∥pεhk-1∥2.
Thus, we obtain
(33)μ∥uεhk∥V2+ε∥pεhk∥2≤κ12μ(∥f∥+∥g∥L2(S))2+ε∥pεhk-1∥2≤⋯≤kκ12μ(∥f∥+∥g∥L2(S))2+ε∥pεh0∥2≤(2k+1)κ122μ(∥f∥+∥g∥L2(S))2.The next theorem gives the error estimate between the solutions(u,p) and (uεhk,pεhk) to the problems (16) and (28), respectively. The proof can be found in [20].Theorem 3.
Let(u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεhk,pεhk)∈(Vh,Mh) be the solutions to the problems (16) and (28), respectively; then they satisfy
(34)∥u-uεhk∥V+∥p-pεhk∥≤c(h5/4+εk+1).Next, we will show the error estimate||u-uεhk|| for the penalty finite element approximation (28). This L2 error analysis is based on the regularity assumption that the following linearized problem (35) is (H2(Ω)2,H1(Ω)) regular.Givenz∈L2(Ω)2, find (w,π)∈(V,M) such that for all (v,q)∈(V,M)(35)a(w,v)+b(uεhk,v,w)+b(v,u,w)-d(v,π)=(z,v),d(w,q)=0.
According to (12) and (20), it is easy to verify that there exists a unique solution (w,π) to the problem (35). The assumption that (35) is (H2(Ω)2,H1(Ω)) regular means that (w,π) also belongs to (H2(Ω)2,H1(Ω)) and the following inequality holds:
(36)∥w∥2+∥π∥1≤c∥z∥.
Let Ih be the L2 orthogonal projections onto V0h and satisfy
(37)∥w-Ihw∥V≤ch∥w∥2.Theorem 4.
Let(u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεhk,pεhk)∈(Vh,Mh) be the solutions to the problems (16) and (28), respectively; then they satisfy
(38)∥u-uεhk∥≤c(h9/4+εh5/4+εk+1).Proof.
Settingz=u-uεhk and v=u-uεhk in the first equation of (35), we get
(39)∥u-uεhk∥2=a(w,u-uεhk)+b(uεhk,u-uεhk,w)+b(u-uεhk,u,w)-d(u-uεhk,π).
Taking v=u±Ihw,q=Qhπ in (16) and vh=uεhk±Ihw,qh=Qhπ in (28), respectively, we obtain
(40)a(u,Ihw)+b(u,u,Ihw)-d(Ihw,p)=(f,Ihw),d(u,Qhπ)=0,a(uεhk,Ihw)+b(uεhk,uεhk,Ihw)-d(Ihw,pεhk)=(f,Ihw),d(uεhk,Qhπ)+ε(pεhk,Qhπ)=ε(pεhk-1,Qhπ).
Subtracting them, we get
(41)a(u-uεhk,Ihw)+b(u,u,Ihw)-b(uεhk,uεhk,Ihw)-d(Ihw,p-pεhk)=0,d(u-uεhk,Qhπ)+ε(pεhk-1,Qhπ)-ε(pεhk,Qhπ)=0.
Substituting the previous equation into (39), it yields that
(42)∥u-uεhk∥2=a(w-Ihw,u-uεhk)︸J1+b(uεhk,u-uεhk,w)︸J2+b(u-uεhk,u,w)+b(uεhk,uεhk,Ihw)-b(u,u,Ihw)︸J2+d(Ihw-w,p-pεhk)-d(u-uεhk,π-Qhπ)︸J3+ε(pεhk-1,Qhπ)-ε(pεhk,Qhπ)︸J4.
Using (34), (36), and (37), J1 is estimated by
(43)J1=a(w-Ihw,u-uεhk)≤μ∥u-uεhk∥V∥w-Ihw∥V≤ch(h5/4+εk+1)∥w∥2≤ch(h5/4+εk+1)∥u-uεhk∥.
Similarly, using (25), (34), (36), and (37), J3 is estimated by
(44)J3=d(Ihw-w,p-pεhk)-d(u-uεhk,π-Qhπ)≤∥Ihw-w∥V∥p-pεhk∥+∥u-uεhk∥V∥π-Qhπ∥≤ch(h5/4+εk+1)(∥w∥2+∥π∥1)≤ch(h5/4+εk+1)∥u-uεhk∥.
We rewrite J2 as
(45)J2=b(uεhk,u-uεhk,w)+b(u-uεhk,u,w)+b(uεhk,uεhk,Ihw)-b(u,u,Ihw)=b(uεhk,u-uεhk,w)+b(u-uεhk,u,w)+b(uεhk,uεhk-u,Ihw)-b(u-uεhk,u,Ihw)=b(uεhk,u-uεhk,w-Ihw)+b(u-uεhk,u,w-Ihw).
Then, from (13), (20), (29), (34), (36), and (37), it holds that
(46)J2≤N(∥uεhk∥V+∥u∥V)∥u-uεhk∥V∥w-Rhw∥V≤ch(h5/4+εk+1)∥w∥2≤ch(h5/4+εk+1)∥u-uεhk∥.
Finally, we estimate J4 by
(47)J4=ε(pεhk-1,Qhπ)-ε(pεhk,Qhπ)=ε(pεhk-1-p,Qhπ)+ε(p-pεhk,Qhπ)≤ε(∥pεhk-1-p∥+∥p-pεhk∥)∥Qhπ∥≤cε(h5/4+εk)∥π∥1≤cε(h5/4+εk)∥u-uεhk∥.
Combining these estimates with (42), we conclude that (38) holds.


## 4. Two-Level Iteration Penalty Methods

In this section, based on the iteration penalty method described in the previous section, the two-level iteration penalty finite element methods for (16) are proposed in terms of the Stokes iteration, Oseen iteration, or Newtonian iteration. From now on, H and h with h<H are two real positive parameters. The coarse mesh triangulation 𝒯H is made as in Section 3. And a fine mesh triangulation 𝒯h is generated by a mesh refinement process to 𝒯H. The conforming finite element space pairs (Vh,Mh) and (VH,MH)⊂(Vh,Mh) corresponding to the triangulations 𝒯h and 𝒯H, respectively, are constructed as in Section 3. With the preavious notations, we propose the following two-level iteration finite element methods.


### 4.1. Two-Level Stokes Iteration Penalty Method

In Steps1 and 2, we solve (27) and (28) on the coarse mesh, as in the follwing.Step 1.
Find(uεH0,pεH0)∈(VH,MH) such that for all (vH,qH)∈(VH,MH)(48)a(uεH0,vH-uεH0)+b(uεH0,uεH0,vH-uεH0)+j(vHτ)-j(uεHτ0)-d(vH-uεH0,pεH0)≥(f,vH-uεH0),d(uεH0,qH)+ε(pεH0,qH)=0.Step 2.
Fork=1,2,…, find (uεHk,pεHk)∈(VH,MH) such that for all (vH,qH)∈(VH,MH)(49)a(uεHk,vH-uεHk)+b(uεHk,uεHk,vH-uεHk)+j(vHτ)-j(uεHτk)-d(vH-uεHk,pεHk)≥(f,vH-uεHk),d(uεHk,qH)+ε(pεHk,qH)=ε(pεHk-1,qh).
In Step3, we solve a Stokes-type variational inequality problem on the fine mesh in terms of the Stokes iteration, as in the following.Step 3.
Find(uεh,pεh)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(50)a(uεh,vh-uεh)+b(uεHk,uεHk,vh-uεh)+j(vhτ)-j(uεhτ)-d(vh-uεh,pεh)≥(f,vh-uεh),d(uεh,qh)+ε(pεh,qh)=ε(pεHk,qh).
As a direct consequence of Theorem2, the solution (uεHk,pεHk) to the problem (49) satisfies
(51)∥uεHk∥V≤2k+12κ1μ(∥f∥+∥g∥L2(S)),(52)ε∥pεHk∥2≤(2k+1)κ122μ(∥f∥+∥g∥L2(S))2.
Next, we estimate uεh. Taking vh=0,qh=pεh in (50), it yields
(53)μ∥uεh∥V2+ε∥pεh∥2≤(f,uεh)-j(uεhτ)-b(uεHk,uεHk,uεh)+ε(pεHk,pεh)≤κ1(∥f∥+∥g∥L2(S))∥uεh∥V+N∥uεHk∥V2∥uεh∥V+ε∥pεHk∥∥pεh∥≤μ2∥uεh∥V2+ε2∥pεh∥2+ε2∥pεHk∥2+κ12μ(∥f∥+∥g∥L2(S))2+N2μ∥uεHk∥V4.
That is,
(54)∥uεh∥V≤2κ1μ(∥f∥+∥g∥L2(S))+2Nμ∥uεHk∥V2+εμ∥pεHk∥.
Suppose that the initial data satisfies
(55)7Nκ1μ22k+12(∥f∥+∥g∥L2(S))<1;
then using (51)-(52), we can estimate uεh by
(56)∥uεh∥V≤2κ1μ(∥f∥+∥g∥L2(S))+27∥uεHk∥V+μ7N≤(2+2k+17)κ1μ(∥f∥+∥g∥L2(S))+μ12N≤(2+2k+17)2μ72k+1N+μ12N<(27+249+17)μN=23μ49N<μ2N.
By the classical existence theorem for the variational inequality problem of the second kind in the finite dimension [27], we have the following.Theorem 5.
Under the uniqueness condition (55), there exists a unique solution (uεh,pεh) to the problem (50). Moreover, uεh satisfy (56).It follows from Theorems3 and 4 that (uεHk,pεHk) is of the following error estimates:
(57)∥u-uεHk∥V+∥p-pεHk∥≤c(H5/4+εk+1),(58)∥u-uεHk∥≤c(H9/4+εH5/4+εk+1).
Next, we begin to prove the following error estimate for the solution (uεh,pεh) to the problem (50).Theorem 6.
Suppose that the uniqueness condition (55) holds. Let (u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεh,pεh)∈(Vh,Mh) be the solutions to the problems (16) and (50), respectively; then they satisfy
(59)∥u-uεh∥V+∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).Proof.
Define a generalized bilinear form on(Vh,Mh)×(Vh,Mh) by
(60)ℬεh(uh,ph;vh,qh)=a(uh,vh)-d(vh,ph)+d(uh,qh)+ε(ph,qh).
Taking vh=Rhu,qh=pεh-Qhp in (50), we have
(61)μ∥uεh-Rhu∥V2+ε∥pεh-Qhp∥2=ℬεh(uεh-Rhu,pεh-Qhp;uεh-Rhu,pεh-Qhp)=a(uεh,uεh-Rhu)-d(uεh-Rhu,pεh)+d(uεh,pεh-Qhp)+ε(pεh,pεh-Qhp)-ℬεh(Rhu,Qhp;uεh-Rhu,pεh-Qhp)≤(f,uεh-Rhu)+b(uεHk,uεHk,Rhu-uεh)+ε(pεHk,pεh-Qhp)+j(Rhuτ)-j(uεhτ)-ℬεh(Rhu,Qhp;uεh-Rhu,pεh-Qhp).
Let v=uεh and v=2u-Rhu in the first inequality of (16); then
(62)a(u,uεh-u)+b(u,u,uεh-u)+j(uεhτ)-j(uτ)-d(uεh-u,p)≥(f,uεh-u),a(u,u-Rhu)+b(u,u,u-Rhu)+j(2uτ-Rhuτ)-j(uτ)-d(u-Rhu,p)≥(f,u-Rhu).
Adding the above two inequalities gives
(63)(f,uεh-Rhu)≤a(u,uεh-Rhu)+b(u,u,uεh-Rhu)-d(uεh-Rhu,p)+j(2uτ-Rhuτ)+j(uεhτ)-2j(uτ).
Substituting the above inequality into (61), it yields that
(64)μ∥uεh-Rhu∥V2+ε∥pεh-Qhp∥2=a(u-Rhu,uεh-Rhu)︸I1+b(uεHk,uεHk,Rhu-uεh)-b(u,u,Rhu-uεh)︸I2-d(uεh-Rhu,p-Qhp)+d(u-Rhu,pεh-Qhp)︸I3+ε(pεHk-Qhp,pεh-Qhp)︸I4+j(2uτ-Rhuτ)-2j(uτ)+j(Rhuτ)︸I5.
It follows from Hölder's inequality and Young's inequality that
(65)I1=a(u-Rhu,uεh-Rhu)≤μ∥u-Rhu∥V∥uεh-Rhu∥V≤μ8∥uεh-Rhu∥V2+2μ∥u-Rhu∥V2,I3=d(u-Rhu,pεh-Qhp)-d(uεh-Rhu,p-Qhp)≤∥u-Rhu∥V∥pεh-Qhp∥+∥uεh-Rhu∥V∥p-Qhp∥≤μ8∥uεh-Rhu∥V2+η2∥pεh-Qhp∥2+1μ∥p-Qhp∥2+14η2∥u-Rhu∥V2,
where η>0 is some small constant determined later. We rewrite I2 as
(66)I2=b(uεHk,uεHk,Rhu-uεh)-b(u,u,Rhu-uεh)=b(uεHk-u,u,Rhu-uεh)+b(u,uεHk-u,Rhu-uεh)+b(uεHk-u,uεHk-u,Rhu-uεh).
Then using (13), (15), and Young's inequality, we can estimate I2 by
(67)I2≤N∥u∥2∥Rhu-uεh∥V∥uεHk-u∥+N∥uεHk-u∥V2∥Rhu-uεh∥V≤μ8∥Rhu-uεh∥V2+c(∥uεHk-u∥2+∥uεHk-u∥V4).
It is easy to show that
(68)I4=ε(pεHk-Qhp,pεh-Qhp)≤ε∥pεHk-Qhp∥∥pεh-Qhp∥≤η2∥pεh-Qhp∥2+ε24η2∥pεHk-Qhp∥2≤η2∥pεh-Qhp∥2+ε22η2(∥p-pεHk∥2+∥p-Qhp∥2).
Finally, from triangle inequality, I5 is estimated by
(69)I5=j(2uτ-Rhuτ)-2j(uτ)+j(Rhuτ)≤2∫S‍g|uτ-Rhuτ|ds≤2∥g∥L2(S)∥uτ-Rhuτ∥L2(S).
Substituting (65)–(69) into (64), it yields that
(70)∥u-uεh∥V≤∥u-Rhu∥V+∥uεh-Rhu∥V≤2ημ∥pεh-Qhp∥+c(∥u-Rhu∥V+∥p-Qhp∥+∥uτ-Rhuτ∥L2(S)1/2+ε∥p-pεHk∥+∥u-uεHk∥+∥u-uεHk∥V2)≤2ημ∥pεh-Qhp∥+c(h5/4+εH5/4+H9/4+εk+1),
where we use (24)–(26) and (57)-(58). Next, we estimate ||pεh-Qhp||. For all wh∈V0h, let v=u±wh in (16) and vh=uεh±wh in (50), respectively. Then we get
(71)a(u,wh)+b(u,u,wh)-d(wh,p)=(f,wh),a(uεh,wh)+b(uεHk,uεHk,wh)-d(wh,pεh)=(f,wh).
Subtracting them and using (13), (15), we obtain
(72)d(wh,p-pεh)=a(u-uεh,wh)+b(u,u,wh)-b(uεHk,uεHk,wh)=a(u-uεh,wh)+b(u,u-uεHk,wh)+b(u-uεHk,u,wh)-b(u-uεHk,u-uεHk,wh)≤({N∥u-uεHk∥V2}μ∥u-uεh∥V+N∥u∥2∥u-uεHk∥+N∥u-uεHk∥V2)∥wh∥V.
Therefore, it follows from (23) that ||Qhp-pεh|| can be estimated by
(73)κ∥Qhp-pεh∥≤supwh∈V0hd(wh,Qhp-pεh)∥wh∥V=supwh∈V0hd(wh,Qhp-p)+d(wh,p-pεh)∥wh∥V≤∥p-Qhp∥+μ∥u-uεh∥V+N∥u∥2∥u-uεHk∥+N∥u-uεHk∥V2≤μ∥u-uεh∥V+c(h5/4+εH5/4+H9/4+εk+1).
If we choose η=κ/4μ such that (2η/μ)·(μ/κ)=1/2, then substituting (73) into (70), we show
(74)∥u-uεh∥V≤c(h5/4+εH5/4+H9/4+εk+1).
From (73), again, we obtain
(75)∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).
Thus, we complete the proof of (59).


### 4.2. Two-Level Oseen Iteration Penalty Method

In Steps1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.Step 1.
Find(uεH0,pεH0)∈(VH,MH) by (48).Step 2.
Fork=1,2,…, find (uεHk,pεHk)∈(VH,MH) by (49).In Step3, we solve an Oseen type variational inequality problem on the fine mesh in terms of the Oseen iteration, as in the following.Step 3.
Find(uεh,pεh)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(76)a(uεh,vh-uεh)+b(uεHk,uεh,vh-uεh)+j(vhτ)-j(uεhτ)-d(vh-uεh,pεh)≥(f,vh-uεh),d(uεh,qh)+ε(pεh,qh)=ε(pεHk,qh).From (12), it is easy to show that the solution (uεh,pεh) to the problem (76) satisfies
(77)μ∥uεh∥V2+ε∥pεh∥2≤(f,uεh)-j(uεhτ)+ε(pεHk,pεh)≤κ1(∥f∥+∥g∥L2(S))∥uεh∥V+ε∥pεHk∥∥pεh∥≤μ2∥uεh∥V2+ε2∥pεh∥2+ε2∥pεHk∥2+κ122μ(∥f∥+∥g∥L2(S))2.
Suppose that the initial data satisfies
(78)24k+2Nκ1μ2(∥f∥+∥g∥L2(S))<1;
then using (52), we can estimate uεh by
(79)∥uεh∥V≤εμ∥pεHk∥+κ1μ(∥f∥+∥g∥L2(S))≤(2k+12+1)κ1μ(∥f∥+∥g∥L2(S))≤4k+2κ1μ(∥f∥+∥g∥L2(S))≤μ2N.For two-level Oseen iteration penalty method, the solution(uεh,pεh) is of the following error estimate.Theorem 7.
Suppose that the uniqueness condition (78) holds. Let (u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεh,pεh)∈(Vh,Mh) be the solutions to the problems (16) and (76), respectively; then they satisfy
(80)∥u-uεh∥V+∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).Proof.
Proceeding as in the proof of (64), we can get
(81)μ∥uεh-Rhu∥V2+ε∥pεh-Qhp∥2=a(u-Rhu,uεh-Rhu)︸I1+b(uεHk,uεh,Rhu-uεh)-b(u,u,Rhu-uεh)︸I6-d(uεh-Rhu,p-Qhp)+d(u-Rhu,pεh-Qhp)︸I3+ε(pεHk-Qhp,pεh-Qhp)︸I4+j(2uτ-Rhuτ)-2j(uτ)+j(Rhuτ)︸I5.
In the above equation, I1, I3, I4, and I5 have been estimated in the proof of Theorem 6. Here, we only estimate I6. Using (12), (13), (15), and Young's inequality, we have
(82)I6=b(uεHk,uεh,Rhu-uεh)-b(u,u,Rhu-uεh)=b(uεHk,uεh-Rhu,Rhu-uεh)+b(uεHk,Rhu-u,Rhu-uεh)+b(uεHk-u,u,Rhu-uεh)=b(uεHk,Rhu-u,Rhu-uεh)+b(uεHk-u,u,Rhu-uεh)≤N∥uεHk∥V∥Rhu-u∥V∥Rhu-uεh∥V+N∥u∥2∥Rhu-uεh∥V∥uεHk-u∥≤μ8∥Rhu-uεh∥V2+c(∥uεHk-u∥2+∥Rhu-u∥V2).
Then substituting (65), (68), (69), and (82) into (81), it yields that
(83)∥u-uεh∥V≤2ημ∥pεh-Qhp∥+c(h5/4+εH5/4+H9/4+εk+1).
In (83), we use (24)–(26) and (57)-(58). Next, we estimate ||pεh-Qhp||. For all wh∈V0h, proceeding as in the proof of (72), from (51) and (78), we can show
(84)d(wh,p-pεh)=a(u-uεh,wh)+b(u,u,wh)-b(uεHk,uεh,wh)=a(u-uεh,wh)+b(uεHk,u-uεh,wh)+b(u-uεHk,u,wh)≤(μ∥u-uεh∥V+N∥u∥2∥u-uεHk∥+N∥uεHk∥V∥u-uεh∥V)∥wh∥V≤(5μ4∥u-uεh∥V+N∥u∥2∥u-uεHk∥)∥wh∥V.
It follows from (23) and (84) that
(85)κ∥Qhp-pεh∥≤supwh∈V0hd(wh,Qhp-pεh)∥wh∥V=supwh∈V0hd(wh,Qhp-p)+d(wh,p-pεh)∥wh∥V≤∥p-Qhp∥+5μ4∥u-uεh∥V+N∥u∥2∥u-uεHk∥≤5μ4∥u-uεh∥V+c(h5/4+εH5/4+H9/4+εk+1).
If we choose η=κ/5μ such that (2η/μ)·(5μ/4κ)=1/2, then substituting (85) into (83), we show
(86)∥u-uεh∥V≤c(h5/4+εH5/4+H9/4+εk+1).
From (85), again, we obtain
(87)∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).
Thus, we complete the proof of (80).


### 4.3. Two-Level Newton Iteration Penalty Method

In Steps1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.Step 1.
Find(uεH0,pεH0)∈(VH,MH) by (48).Step 2.
Fork=1,2,…, find (uεHk,pεHk)∈(VH,MH) by (49).
In Step3, we solve a linearized Navier-Stokes type variational inequality problem on the fine mesh in terms of the Newton iteration, as in the following.Step 3.
Find(uεh,pεh)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(88)a(uεh,vh-uεh)+b(uεHk,uεh,vh-uεh)+b(uεh,uεHk,vh-uεh)+j(vhτ)-j(uεhτ)-d(vh-uεh,pεh)≥(f,vh-uεh)+b(uεHk,uεHk,vh-uεh),d(uεh,qh)+ε(pεh,qh)=ε(pεHk,qh).In this section, we will suppose that the initial data satisfies(89)8Nκ1μ22k+12(∥f∥+∥g∥L2(S))<1.
Then from (51), uεHk satisfies ||uεHk||V≤μ/8N. Let vh=0,qh=pεh in (88). Using (12), we obtain
(90)μ∥uεh∥V2-b(uεHk,uεh,uεh)+ε∥pεh∥2≤(f,uεh)-j(uεhτ)-b(uεHk,uεHk,uεh)+ε(pεHk,pεh)≤κ1(∥f∥+∥g∥L2(S))∥uεh∥V+N∥uεHk∥V2∥uεh∥V+ε∥pεHk∥∥pεh∥≤μ2∥uεh∥V2+ε2∥pεh∥2+ε2∥pεHk∥2+κ12μ(∥f∥+∥g∥L2(S))2+N2μ∥uεHk∥V4.
Since
(91)μ∥uεh∥V2-b(uεHk,uεh,uεh)≥μ∥uεh∥V2-N∥uεHk∥V∥uεh∥V2≥7μ8∥uεh∥V2,
then
(92)∥uεh∥V≤2κ13μ(∥f∥+∥g∥L2(S))+2ε3μ∥pεHk∥+2N3μ∥uεHk∥V2≤24312k+1μN+143∥uεHk∥V+μ43N<(14+155+16)μN=574μ1320N<μ2N.Theorem 8.
Suppose that the uniqueness condition (89) holds. Let (u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεh,pεh)∈(Vh,Mh) be the solutions to the problems (16) and (88), respectively; then they satisfy
(93)∥u-uεh∥V+∥p-pεh∥≤c(h5/4+εH5/4+H5/2+εk+2).Proof.
Proceeding as the in proof of (64), we can get (94)μ∥uεh-Rhu∥V2+ε∥pεh-Qhp∥2=a(u-Rhu,uεh-Rhu)︸I1+-d(uεh-Rhu,p-Qhp)+d(u-Rhu,pεh-Qhp)︸I3+ε(pεHk-Qhp,pεh-Qhp)︸I4+j(2uτ-Rhuτ)-2j(uτ)+j(Rhuτ)︸I5+b(uεHk,uεh,Rhu-uεh)+b(uεh,uεHk,Rhu-uεh)-b(u,u,Rhu-uεh)-b(uεHk,uεHk,Rhu-uεh)︸I7.
In the above equation,I1, I3, I4, and I5 have been estimated in the proof of Theorem 6. Here, we only estimate I7. We rewrite I7 as
(95)I7=b(uεHk,uεh,Rhu-uεh)+b(uεh,uεHk,Rhu-uεh)-b(u,u,Rhu-uεh)-b(uεHk,uεHk,Rhu-uεh)=b(uεh-u,u,Rhu-uεh)+b(uεh,uεh-u,Rhu-uεh)-b(uεh-uεHk,uεh-uεHk,Rhu-uεh)=b(uεh-Rhu,u,Rhu-uεh)+b(Rhu-u,u,Rhu-uεh)︸I8+b(uεh,Rhu-u,Rhu-uεh)︸I9+b(Rhu-uεh,Rhu-uεHk,Rhu-uεh)︸I10+b(uεHk-Rhu,Rhu-uεHk,Rhu-uεh)︸I11.
From (13), (20), and Young's inequality, I8 is estimated by
(96)I8=b(uεh-Rhu,u,Rhu-uεh)+b(Rhu-u,u,Rhu-uεh)≤N∥u∥V∥Rhu-uεh∥V2+N∥u∥V∥Rhu-uεh∥V∥Rhu-u∥V≤μ2∥Rhu-uεh∥V2+μ2∥Rhu-uεh∥V∥Rhu-u∥V≤μ2∥Rhu-uεh∥V2+μ32∥Rhu-uεh∥V2+4μ∥Rhu-u∥V2.
Using (13) and (24), we estimate I9 by
(97)I9=b(uεh,Rhu-u,Rhu-uεh)=b(uεh-Rhu,Rhu-u,Rhu-uεh)+b(Rhu,Rhu-u,Rhu-uεh)≤N∥Rhu-u∥V∥Rhu-uεh∥V2+N∥Rhu∥V∥Rhu-u∥V∥Rhu-uεh∥V≤C1h2∥Rhu-uεh∥V2+μ32∥Rhu-uεh∥V2+c∥Rhu-u∥V2,
where C1>0 is independent of h,H, and ε. Similarly, it follows from (13), (24), and (57) that
(98)I10=b(Rhu-uεh,Rhu-uεHk,Rhu-uεh)≤N∥Rhu-uεHk∥V∥Rhu-uεh∥V2≤N(∥Rhu-u∥V+∥u-uεHk∥V)∥Rhu-uεh∥V2≤C2(h2+H5/4+εk+1)∥Rhu-uεh∥V2,
where C2>0 is independent of h,H, and ε. Finally, we can estimate I11 by
(99)I11=b(uεHk-Rhu,Rhu-uεHk,Rhu-uεh)≤N∥Rhu-uεHk∥V2∥Rhu-uεh∥V≤μ32∥Rhu-uεh∥V2+c(∥Rhu-u∥V4+∥u-uεHk∥V4).
For sufficiently small h,H, and ε such that C1h2+C2(h2+H5/4+εk+1)=1/32, substituting (65), (68), (69), and (95)–(99) into (94), it yields that
(100)∥u-uεh∥V≤4ημ∥pεh-Qhp∥+c(h5/4+εH5/4+H5/2+εk+2).
For all wh∈V0h, proceeding as in the proof of (72), we can show
(101)d(vh,p-pεh)=a(u-uεh,vh)+b(u,u,vh)-b(uεHk,uεh,vh)-b(uεh,uεHk,vh)+b(uεHk,uεHk,vh).
Since
(102)b(u,u,vh)-b(uεHk,uεh,vh)-b(uεh,uεHk,vh)+b(uεHk,uεHk,vh)=b(u-uεh,u,vh)+b(u,u-Rhu,vh)-b(u-uεh,u-Rhu,vh)+b(uεh-uεHk,Rhu-uεHk,vh)-b(uεHk,uεh-Rhu,vh)≤(N∥u∥V∥u-uεh∥V+N∥u∥V∥u-Rhu∥V+∥u-Rhu∥V∥u-uεh∥V)∥vh∥V+(∥u-uεh∥V+∥u-uεHk∥V)×(∥u-Rhu∥V+∥u-uεHk∥V)∥vh∥V+N∥uεHk∥V(∥u-uεh∥V+∥u-Rhu∥V)∥vh∥V≤C3(1+h2+H2+εk+1)∥u-uεh∥V∥vh∥V+c(∥u-Rhu∥V+∥u-Rhu∥V2+∥u-uεHk∥V2)∥vh∥V,
where C3>0 is independent of h,H, and ε, then from (23) we have
(103)κ∥Qhp-pεh∥≤C3(1+h2+H2+εk+1)∥u-uεh∥V+c({∥u-uεHk∥V2}∥u-Rhu∥V+∥p-Qhp∥+∥u-Rhu∥V2+∥u-uεHk∥V2).
Thus, for sufficiently small h,H,ε, and η such that
(104)C3(1+h2+H2+εk+1)κ4ημ<12,
substituting (103) into (100), we obtain
(105)∥u-uεh∥V≤c(h5/4+εH5/4+H5/2+εk+2).
From (103), again, we obtain
(106)∥p-pεh∥≤c(h5/4+εH5/4+H5/2+εk+2).
Thus, we complete the proof of (93).Remark 9.
In terms of Theorems6, 7, and 8, if we choose ε=O(H),H=O(h5/9) for the two-level Stokes or Oseen iteration penalty methods and ε=O(H5/4),H=O(h1/2) for the two-level Newton iteration penalty method, then
(107)∥u-uεh∥V+∥p-pεh∥≤ch5/4.


### 4.4. An Improved Scheme

In this section, we will propose a scheme to improve the error estimates derived in Theorems6–8, which is described as follows.In Steps1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.Step 1.
Find(uεH0,pεH0)∈(VH,MH) by (48).Step 2.
Fork=1,2,…, find (uεHk,pεHk)∈(VH,MH) by (49).
At Step3, we solve a linearized problem (50) or (76) or (88) on the fine mesh in terms of Stokes iteration or Oseen iteration or Newton iteration, as in the following.Step 3.
Find(uεh,pεh)∈(Vh,Mh) by (50) or (76) or (88).
At Step4, we solve a Newton correction of (uεh,pεh) on the fine mesh in terms of Newton iteration, as in the following.Step 4.
Find(uεh⋆,pεh⋆)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(108)a(uεh⋆,vh-uεh⋆)+b(uεh,uεh⋆,vh-uεh⋆)+b(uεh⋆,uεh,vh-uεh⋆)+j(vhτ)-j(uεhτ⋆)-d(vh-uεh⋆,pεh⋆)≥(f,vh-uεh⋆)+b(uεh,uεh,vh-uεh⋆),d(uεh⋆,qh)+ε(pεh⋆,qh)=ε(pεh,qh).
First, we show the following theorem.Theorem 10.
Let(uεhk,pεhk) and (uεh⋆,pεh⋆) be the solutions of (28) and (108), respectively. Then there holds that
(109)∥uεhk-uεh⋆∥V+∥pεhk-pεh⋆∥≤c(∥uεhk-uεh∥V2+ε1/2∥pεh-pεhk-1∥),
where (uεh,pεh) is the solution to the problem (50) or (76) or (88).Proof.
Under the uniqueness condition (89), the solution uεh satisfies ||uεh||V≤μ/2N. Taking vh=uεh⋆,qh=pεh⋆-pεhk in (28) and vh=uεhk,qh=pεhk-pεh⋆ in (108) and adding them yield
(110)μ∥uεh⋆-uεhk∥V2+ε∥pεh⋆-pεhk∥2≤ε(pεh-pεhk-1,pεh⋆-pεhk)+b(uεhk,uεhk,uεh⋆-uεhk)-b(uεh,uεh⋆,uεh⋆-uεhk)-b(uεh⋆,uεh,uεh⋆-uεhk)+b(uεh,uεh,uεh⋆-uεhk)=ε(pεh-pεhk-1,pεh⋆)+b(uεh-uεhk,uεh-uεhk,uεh⋆-uεhk)+b(uεhk-uεh⋆,uεh,uεh⋆-uεhk).
Using (13), Hölder's inequality, and Young's inequality, we obtain
(111)μ∥uεh⋆-uεhk∥V2+ε∥pεh⋆-pεhk∥2≤ε∥pεh-pεhk-1∥∥pεh⋆-pεhk∥+N∥uεh-uεhk∥V2×∥uεh⋆-uεhk∥V+N∥uεh∥V∥uεh⋆-uεhk∥V2≤ε2∥pεh⋆-pεhk∥2+ε2∥pεh-pεhk-1∥2+μ2∥uεh⋆-uεhk∥V2+μ4∥uεh⋆-uεhk∥V2+N2μ∥uεh-uεhk∥V4.
That is,
(112)∥uεh⋆-uεhk∥V≤2εμ∥pεh-pεhk-1∥+2Nμ∥uεh-uεhk∥V2.
For all wh∈V0h, taking vh=uεhk±wh in the first inequality of (28) and vh=uεh⋆±wh in the first inequality of (108), it yields that
(113)a(uεhk,wh)+b(uεhk,uεhk,wh)-d(wh,pεhk)=(f,wh),a(uεh⋆,wh)+b(uεh,uεh⋆,wh)+b(uεh⋆,uεh,wh)-d(wh,pεh⋆)=(f,wh)+b(uεh,uεh,wh).
Subtracting them and using (23), (112), we obtain
(114)κ∥pεh⋆-pεhk∥≤supwh∈V0hd(wh,pεh⋆-pεhk)∥wh∥V=supwh∈V0ha(uεh⋆-uεhk,wh)+b(uεh,uεh⋆,wh)∥wh∥V+supwh∈V0hb(uεh⋆,uεh,wh)-b(uεhk,uεhk,wh)-b(uεh,uεh,wh)∥wh∥V=supwh∈V0ha(uεh⋆-uεhk,wh)+b(uεh⋆-uεhk,uεh,wh)∥wh∥V+supwh∈V0hb(uεh,uεh⋆-uεhk,wh)-b(uεhk-uεh,uεhk-uεh,wh)∥wh∥V≤μ∥uεh⋆-uεhk∥V+2N∥uεh∥V∥uεh⋆-uεhk∥V+N∥uεhk-uεh∥V2≤2μ∥uεh⋆-uεhk∥V+N∥uεhk-uεh∥V2≤22με∥pεh-pεhk-1∥+5N∥uεhk-uεh∥V2.From (34) and Theorems 6–10, we get the following error estimates.Theorem 11.
Let(u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεh⋆,pεh⋆)∈(Vh,Mh) be the solutions to the problems (16) and (108), respectively. Then for the two-level Stokes or Oseen iteration penalty methods, they satisfy
(115)∥u-uεh⋆∥V+∥p-pεh⋆∥≤c(h5/4+H9/2+ε3/2H5/4+ε1/2H9/4+εk+1/2).
And for the two-level Newton iteration penalty method, they satisfy
(116)∥u-uεh⋆∥V+∥p-pεh⋆∥≤c(h5/4+H5+ε3/2H5/4+ε1/2H5/2+εk+1/2).Remark 12.
If we chooseH=O(h5/18), ε=O(h5/4) in (115) for two-level Stokes or Oseen iteration penalty methods and H=O(h1/4), ε=O(h5/4) in (116) for the two-level Newton iteration penalty method, then we obtain
(117)∥u-uεh⋆∥V+∥p-pεh⋆∥≤ch5/4.


## 4.1. Two-Level Stokes Iteration Penalty Method

In Steps1 and 2, we solve (27) and (28) on the coarse mesh, as in the follwing.Step 1.
Find(uεH0,pεH0)∈(VH,MH) such that for all (vH,qH)∈(VH,MH)(48)a(uεH0,vH-uεH0)+b(uεH0,uεH0,vH-uεH0)+j(vHτ)-j(uεHτ0)-d(vH-uεH0,pεH0)≥(f,vH-uεH0),d(uεH0,qH)+ε(pεH0,qH)=0.Step 2.
Fork=1,2,…, find (uεHk,pεHk)∈(VH,MH) such that for all (vH,qH)∈(VH,MH)(49)a(uεHk,vH-uεHk)+b(uεHk,uεHk,vH-uεHk)+j(vHτ)-j(uεHτk)-d(vH-uεHk,pεHk)≥(f,vH-uεHk),d(uεHk,qH)+ε(pεHk,qH)=ε(pεHk-1,qh).
In Step3, we solve a Stokes-type variational inequality problem on the fine mesh in terms of the Stokes iteration, as in the following.Step 3.
Find(uεh,pεh)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(50)a(uεh,vh-uεh)+b(uεHk,uεHk,vh-uεh)+j(vhτ)-j(uεhτ)-d(vh-uεh,pεh)≥(f,vh-uεh),d(uεh,qh)+ε(pεh,qh)=ε(pεHk,qh).
As a direct consequence of Theorem2, the solution (uεHk,pεHk) to the problem (49) satisfies
(51)∥uεHk∥V≤2k+12κ1μ(∥f∥+∥g∥L2(S)),(52)ε∥pεHk∥2≤(2k+1)κ122μ(∥f∥+∥g∥L2(S))2.
Next, we estimate uεh. Taking vh=0,qh=pεh in (50), it yields
(53)μ∥uεh∥V2+ε∥pεh∥2≤(f,uεh)-j(uεhτ)-b(uεHk,uεHk,uεh)+ε(pεHk,pεh)≤κ1(∥f∥+∥g∥L2(S))∥uεh∥V+N∥uεHk∥V2∥uεh∥V+ε∥pεHk∥∥pεh∥≤μ2∥uεh∥V2+ε2∥pεh∥2+ε2∥pεHk∥2+κ12μ(∥f∥+∥g∥L2(S))2+N2μ∥uεHk∥V4.
That is,
(54)∥uεh∥V≤2κ1μ(∥f∥+∥g∥L2(S))+2Nμ∥uεHk∥V2+εμ∥pεHk∥.
Suppose that the initial data satisfies
(55)7Nκ1μ22k+12(∥f∥+∥g∥L2(S))<1;
then using (51)-(52), we can estimate uεh by
(56)∥uεh∥V≤2κ1μ(∥f∥+∥g∥L2(S))+27∥uεHk∥V+μ7N≤(2+2k+17)κ1μ(∥f∥+∥g∥L2(S))+μ12N≤(2+2k+17)2μ72k+1N+μ12N<(27+249+17)μN=23μ49N<μ2N.
By the classical existence theorem for the variational inequality problem of the second kind in the finite dimension [27], we have the following.Theorem 5.
Under the uniqueness condition (55), there exists a unique solution (uεh,pεh) to the problem (50). Moreover, uεh satisfy (56).It follows from Theorems3 and 4 that (uεHk,pεHk) is of the following error estimates:
(57)∥u-uεHk∥V+∥p-pεHk∥≤c(H5/4+εk+1),(58)∥u-uεHk∥≤c(H9/4+εH5/4+εk+1).
Next, we begin to prove the following error estimate for the solution (uεh,pεh) to the problem (50).Theorem 6.
Suppose that the uniqueness condition (55) holds. Let (u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεh,pεh)∈(Vh,Mh) be the solutions to the problems (16) and (50), respectively; then they satisfy
(59)∥u-uεh∥V+∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).Proof.
Define a generalized bilinear form on(Vh,Mh)×(Vh,Mh) by
(60)ℬεh(uh,ph;vh,qh)=a(uh,vh)-d(vh,ph)+d(uh,qh)+ε(ph,qh).
Taking vh=Rhu,qh=pεh-Qhp in (50), we have
(61)μ∥uεh-Rhu∥V2+ε∥pεh-Qhp∥2=ℬεh(uεh-Rhu,pεh-Qhp;uεh-Rhu,pεh-Qhp)=a(uεh,uεh-Rhu)-d(uεh-Rhu,pεh)+d(uεh,pεh-Qhp)+ε(pεh,pεh-Qhp)-ℬεh(Rhu,Qhp;uεh-Rhu,pεh-Qhp)≤(f,uεh-Rhu)+b(uεHk,uεHk,Rhu-uεh)+ε(pεHk,pεh-Qhp)+j(Rhuτ)-j(uεhτ)-ℬεh(Rhu,Qhp;uεh-Rhu,pεh-Qhp).
Let v=uεh and v=2u-Rhu in the first inequality of (16); then
(62)a(u,uεh-u)+b(u,u,uεh-u)+j(uεhτ)-j(uτ)-d(uεh-u,p)≥(f,uεh-u),a(u,u-Rhu)+b(u,u,u-Rhu)+j(2uτ-Rhuτ)-j(uτ)-d(u-Rhu,p)≥(f,u-Rhu).
Adding the above two inequalities gives
(63)(f,uεh-Rhu)≤a(u,uεh-Rhu)+b(u,u,uεh-Rhu)-d(uεh-Rhu,p)+j(2uτ-Rhuτ)+j(uεhτ)-2j(uτ).
Substituting the above inequality into (61), it yields that
(64)μ∥uεh-Rhu∥V2+ε∥pεh-Qhp∥2=a(u-Rhu,uεh-Rhu)︸I1+b(uεHk,uεHk,Rhu-uεh)-b(u,u,Rhu-uεh)︸I2-d(uεh-Rhu,p-Qhp)+d(u-Rhu,pεh-Qhp)︸I3+ε(pεHk-Qhp,pεh-Qhp)︸I4+j(2uτ-Rhuτ)-2j(uτ)+j(Rhuτ)︸I5.
It follows from Hölder's inequality and Young's inequality that
(65)I1=a(u-Rhu,uεh-Rhu)≤μ∥u-Rhu∥V∥uεh-Rhu∥V≤μ8∥uεh-Rhu∥V2+2μ∥u-Rhu∥V2,I3=d(u-Rhu,pεh-Qhp)-d(uεh-Rhu,p-Qhp)≤∥u-Rhu∥V∥pεh-Qhp∥+∥uεh-Rhu∥V∥p-Qhp∥≤μ8∥uεh-Rhu∥V2+η2∥pεh-Qhp∥2+1μ∥p-Qhp∥2+14η2∥u-Rhu∥V2,
where η>0 is some small constant determined later. We rewrite I2 as
(66)I2=b(uεHk,uεHk,Rhu-uεh)-b(u,u,Rhu-uεh)=b(uεHk-u,u,Rhu-uεh)+b(u,uεHk-u,Rhu-uεh)+b(uεHk-u,uεHk-u,Rhu-uεh).
Then using (13), (15), and Young's inequality, we can estimate I2 by
(67)I2≤N∥u∥2∥Rhu-uεh∥V∥uεHk-u∥+N∥uεHk-u∥V2∥Rhu-uεh∥V≤μ8∥Rhu-uεh∥V2+c(∥uεHk-u∥2+∥uεHk-u∥V4).
It is easy to show that
(68)I4=ε(pεHk-Qhp,pεh-Qhp)≤ε∥pεHk-Qhp∥∥pεh-Qhp∥≤η2∥pεh-Qhp∥2+ε24η2∥pεHk-Qhp∥2≤η2∥pεh-Qhp∥2+ε22η2(∥p-pεHk∥2+∥p-Qhp∥2).
Finally, from triangle inequality, I5 is estimated by
(69)I5=j(2uτ-Rhuτ)-2j(uτ)+j(Rhuτ)≤2∫S‍g|uτ-Rhuτ|ds≤2∥g∥L2(S)∥uτ-Rhuτ∥L2(S).
Substituting (65)–(69) into (64), it yields that
(70)∥u-uεh∥V≤∥u-Rhu∥V+∥uεh-Rhu∥V≤2ημ∥pεh-Qhp∥+c(∥u-Rhu∥V+∥p-Qhp∥+∥uτ-Rhuτ∥L2(S)1/2+ε∥p-pεHk∥+∥u-uεHk∥+∥u-uεHk∥V2)≤2ημ∥pεh-Qhp∥+c(h5/4+εH5/4+H9/4+εk+1),
where we use (24)–(26) and (57)-(58). Next, we estimate ||pεh-Qhp||. For all wh∈V0h, let v=u±wh in (16) and vh=uεh±wh in (50), respectively. Then we get
(71)a(u,wh)+b(u,u,wh)-d(wh,p)=(f,wh),a(uεh,wh)+b(uεHk,uεHk,wh)-d(wh,pεh)=(f,wh).
Subtracting them and using (13), (15), we obtain
(72)d(wh,p-pεh)=a(u-uεh,wh)+b(u,u,wh)-b(uεHk,uεHk,wh)=a(u-uεh,wh)+b(u,u-uεHk,wh)+b(u-uεHk,u,wh)-b(u-uεHk,u-uεHk,wh)≤({N∥u-uεHk∥V2}μ∥u-uεh∥V+N∥u∥2∥u-uεHk∥+N∥u-uεHk∥V2)∥wh∥V.
Therefore, it follows from (23) that ||Qhp-pεh|| can be estimated by
(73)κ∥Qhp-pεh∥≤supwh∈V0hd(wh,Qhp-pεh)∥wh∥V=supwh∈V0hd(wh,Qhp-p)+d(wh,p-pεh)∥wh∥V≤∥p-Qhp∥+μ∥u-uεh∥V+N∥u∥2∥u-uεHk∥+N∥u-uεHk∥V2≤μ∥u-uεh∥V+c(h5/4+εH5/4+H9/4+εk+1).
If we choose η=κ/4μ such that (2η/μ)·(μ/κ)=1/2, then substituting (73) into (70), we show
(74)∥u-uεh∥V≤c(h5/4+εH5/4+H9/4+εk+1).
From (73), again, we obtain
(75)∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).
Thus, we complete the proof of (59).


## 4.2. Two-Level Oseen Iteration Penalty Method

In Steps1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.Step 1.
Find(uεH0,pεH0)∈(VH,MH) by (48).Step 2.
Fork=1,2,…, find (uεHk,pεHk)∈(VH,MH) by (49).In Step3, we solve an Oseen type variational inequality problem on the fine mesh in terms of the Oseen iteration, as in the following.Step 3.
Find(uεh,pεh)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(76)a(uεh,vh-uεh)+b(uεHk,uεh,vh-uεh)+j(vhτ)-j(uεhτ)-d(vh-uεh,pεh)≥(f,vh-uεh),d(uεh,qh)+ε(pεh,qh)=ε(pεHk,qh).From (12), it is easy to show that the solution (uεh,pεh) to the problem (76) satisfies
(77)μ∥uεh∥V2+ε∥pεh∥2≤(f,uεh)-j(uεhτ)+ε(pεHk,pεh)≤κ1(∥f∥+∥g∥L2(S))∥uεh∥V+ε∥pεHk∥∥pεh∥≤μ2∥uεh∥V2+ε2∥pεh∥2+ε2∥pεHk∥2+κ122μ(∥f∥+∥g∥L2(S))2.
Suppose that the initial data satisfies
(78)24k+2Nκ1μ2(∥f∥+∥g∥L2(S))<1;
then using (52), we can estimate uεh by
(79)∥uεh∥V≤εμ∥pεHk∥+κ1μ(∥f∥+∥g∥L2(S))≤(2k+12+1)κ1μ(∥f∥+∥g∥L2(S))≤4k+2κ1μ(∥f∥+∥g∥L2(S))≤μ2N.For two-level Oseen iteration penalty method, the solution(uεh,pεh) is of the following error estimate.Theorem 7.
Suppose that the uniqueness condition (78) holds. Let (u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεh,pεh)∈(Vh,Mh) be the solutions to the problems (16) and (76), respectively; then they satisfy
(80)∥u-uεh∥V+∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).Proof.
Proceeding as in the proof of (64), we can get
(81)μ∥uεh-Rhu∥V2+ε∥pεh-Qhp∥2=a(u-Rhu,uεh-Rhu)︸I1+b(uεHk,uεh,Rhu-uεh)-b(u,u,Rhu-uεh)︸I6-d(uεh-Rhu,p-Qhp)+d(u-Rhu,pεh-Qhp)︸I3+ε(pεHk-Qhp,pεh-Qhp)︸I4+j(2uτ-Rhuτ)-2j(uτ)+j(Rhuτ)︸I5.
In the above equation, I1, I3, I4, and I5 have been estimated in the proof of Theorem 6. Here, we only estimate I6. Using (12), (13), (15), and Young's inequality, we have
(82)I6=b(uεHk,uεh,Rhu-uεh)-b(u,u,Rhu-uεh)=b(uεHk,uεh-Rhu,Rhu-uεh)+b(uεHk,Rhu-u,Rhu-uεh)+b(uεHk-u,u,Rhu-uεh)=b(uεHk,Rhu-u,Rhu-uεh)+b(uεHk-u,u,Rhu-uεh)≤N∥uεHk∥V∥Rhu-u∥V∥Rhu-uεh∥V+N∥u∥2∥Rhu-uεh∥V∥uεHk-u∥≤μ8∥Rhu-uεh∥V2+c(∥uεHk-u∥2+∥Rhu-u∥V2).
Then substituting (65), (68), (69), and (82) into (81), it yields that
(83)∥u-uεh∥V≤2ημ∥pεh-Qhp∥+c(h5/4+εH5/4+H9/4+εk+1).
In (83), we use (24)–(26) and (57)-(58). Next, we estimate ||pεh-Qhp||. For all wh∈V0h, proceeding as in the proof of (72), from (51) and (78), we can show
(84)d(wh,p-pεh)=a(u-uεh,wh)+b(u,u,wh)-b(uεHk,uεh,wh)=a(u-uεh,wh)+b(uεHk,u-uεh,wh)+b(u-uεHk,u,wh)≤(μ∥u-uεh∥V+N∥u∥2∥u-uεHk∥+N∥uεHk∥V∥u-uεh∥V)∥wh∥V≤(5μ4∥u-uεh∥V+N∥u∥2∥u-uεHk∥)∥wh∥V.
It follows from (23) and (84) that
(85)κ∥Qhp-pεh∥≤supwh∈V0hd(wh,Qhp-pεh)∥wh∥V=supwh∈V0hd(wh,Qhp-p)+d(wh,p-pεh)∥wh∥V≤∥p-Qhp∥+5μ4∥u-uεh∥V+N∥u∥2∥u-uεHk∥≤5μ4∥u-uεh∥V+c(h5/4+εH5/4+H9/4+εk+1).
If we choose η=κ/5μ such that (2η/μ)·(5μ/4κ)=1/2, then substituting (85) into (83), we show
(86)∥u-uεh∥V≤c(h5/4+εH5/4+H9/4+εk+1).
From (85), again, we obtain
(87)∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).
Thus, we complete the proof of (80).


## 4.3. Two-Level Newton Iteration Penalty Method

In Steps1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.Step 1.
Find(uεH0,pεH0)∈(VH,MH) by (48).Step 2.
Fork=1,2,…, find (uεHk,pεHk)∈(VH,MH) by (49).
In Step3, we solve a linearized Navier-Stokes type variational inequality problem on the fine mesh in terms of the Newton iteration, as in the following.Step 3.
Find(uεh,pεh)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(88)a(uεh,vh-uεh)+b(uεHk,uεh,vh-uεh)+b(uεh,uεHk,vh-uεh)+j(vhτ)-j(uεhτ)-d(vh-uεh,pεh)≥(f,vh-uεh)+b(uεHk,uεHk,vh-uεh),d(uεh,qh)+ε(pεh,qh)=ε(pεHk,qh).In this section, we will suppose that the initial data satisfies(89)8Nκ1μ22k+12(∥f∥+∥g∥L2(S))<1.
Then from (51), uεHk satisfies ||uεHk||V≤μ/8N. Let vh=0,qh=pεh in (88). Using (12), we obtain
(90)μ∥uεh∥V2-b(uεHk,uεh,uεh)+ε∥pεh∥2≤(f,uεh)-j(uεhτ)-b(uεHk,uεHk,uεh)+ε(pεHk,pεh)≤κ1(∥f∥+∥g∥L2(S))∥uεh∥V+N∥uεHk∥V2∥uεh∥V+ε∥pεHk∥∥pεh∥≤μ2∥uεh∥V2+ε2∥pεh∥2+ε2∥pεHk∥2+κ12μ(∥f∥+∥g∥L2(S))2+N2μ∥uεHk∥V4.
Since
(91)μ∥uεh∥V2-b(uεHk,uεh,uεh)≥μ∥uεh∥V2-N∥uεHk∥V∥uεh∥V2≥7μ8∥uεh∥V2,
then
(92)∥uεh∥V≤2κ13μ(∥f∥+∥g∥L2(S))+2ε3μ∥pεHk∥+2N3μ∥uεHk∥V2≤24312k+1μN+143∥uεHk∥V+μ43N<(14+155+16)μN=574μ1320N<μ2N.Theorem 8.
Suppose that the uniqueness condition (89) holds. Let (u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεh,pεh)∈(Vh,Mh) be the solutions to the problems (16) and (88), respectively; then they satisfy
(93)∥u-uεh∥V+∥p-pεh∥≤c(h5/4+εH5/4+H5/2+εk+2).Proof.
Proceeding as the in proof of (64), we can get (94)μ∥uεh-Rhu∥V2+ε∥pεh-Qhp∥2=a(u-Rhu,uεh-Rhu)︸I1+-d(uεh-Rhu,p-Qhp)+d(u-Rhu,pεh-Qhp)︸I3+ε(pεHk-Qhp,pεh-Qhp)︸I4+j(2uτ-Rhuτ)-2j(uτ)+j(Rhuτ)︸I5+b(uεHk,uεh,Rhu-uεh)+b(uεh,uεHk,Rhu-uεh)-b(u,u,Rhu-uεh)-b(uεHk,uεHk,Rhu-uεh)︸I7.
In the above equation,I1, I3, I4, and I5 have been estimated in the proof of Theorem 6. Here, we only estimate I7. We rewrite I7 as
(95)I7=b(uεHk,uεh,Rhu-uεh)+b(uεh,uεHk,Rhu-uεh)-b(u,u,Rhu-uεh)-b(uεHk,uεHk,Rhu-uεh)=b(uεh-u,u,Rhu-uεh)+b(uεh,uεh-u,Rhu-uεh)-b(uεh-uεHk,uεh-uεHk,Rhu-uεh)=b(uεh-Rhu,u,Rhu-uεh)+b(Rhu-u,u,Rhu-uεh)︸I8+b(uεh,Rhu-u,Rhu-uεh)︸I9+b(Rhu-uεh,Rhu-uεHk,Rhu-uεh)︸I10+b(uεHk-Rhu,Rhu-uεHk,Rhu-uεh)︸I11.
From (13), (20), and Young's inequality, I8 is estimated by
(96)I8=b(uεh-Rhu,u,Rhu-uεh)+b(Rhu-u,u,Rhu-uεh)≤N∥u∥V∥Rhu-uεh∥V2+N∥u∥V∥Rhu-uεh∥V∥Rhu-u∥V≤μ2∥Rhu-uεh∥V2+μ2∥Rhu-uεh∥V∥Rhu-u∥V≤μ2∥Rhu-uεh∥V2+μ32∥Rhu-uεh∥V2+4μ∥Rhu-u∥V2.
Using (13) and (24), we estimate I9 by
(97)I9=b(uεh,Rhu-u,Rhu-uεh)=b(uεh-Rhu,Rhu-u,Rhu-uεh)+b(Rhu,Rhu-u,Rhu-uεh)≤N∥Rhu-u∥V∥Rhu-uεh∥V2+N∥Rhu∥V∥Rhu-u∥V∥Rhu-uεh∥V≤C1h2∥Rhu-uεh∥V2+μ32∥Rhu-uεh∥V2+c∥Rhu-u∥V2,
where C1>0 is independent of h,H, and ε. Similarly, it follows from (13), (24), and (57) that
(98)I10=b(Rhu-uεh,Rhu-uεHk,Rhu-uεh)≤N∥Rhu-uεHk∥V∥Rhu-uεh∥V2≤N(∥Rhu-u∥V+∥u-uεHk∥V)∥Rhu-uεh∥V2≤C2(h2+H5/4+εk+1)∥Rhu-uεh∥V2,
where C2>0 is independent of h,H, and ε. Finally, we can estimate I11 by
(99)I11=b(uεHk-Rhu,Rhu-uεHk,Rhu-uεh)≤N∥Rhu-uεHk∥V2∥Rhu-uεh∥V≤μ32∥Rhu-uεh∥V2+c(∥Rhu-u∥V4+∥u-uεHk∥V4).
For sufficiently small h,H, and ε such that C1h2+C2(h2+H5/4+εk+1)=1/32, substituting (65), (68), (69), and (95)–(99) into (94), it yields that
(100)∥u-uεh∥V≤4ημ∥pεh-Qhp∥+c(h5/4+εH5/4+H5/2+εk+2).
For all wh∈V0h, proceeding as in the proof of (72), we can show
(101)d(vh,p-pεh)=a(u-uεh,vh)+b(u,u,vh)-b(uεHk,uεh,vh)-b(uεh,uεHk,vh)+b(uεHk,uεHk,vh).
Since
(102)b(u,u,vh)-b(uεHk,uεh,vh)-b(uεh,uεHk,vh)+b(uεHk,uεHk,vh)=b(u-uεh,u,vh)+b(u,u-Rhu,vh)-b(u-uεh,u-Rhu,vh)+b(uεh-uεHk,Rhu-uεHk,vh)-b(uεHk,uεh-Rhu,vh)≤(N∥u∥V∥u-uεh∥V+N∥u∥V∥u-Rhu∥V+∥u-Rhu∥V∥u-uεh∥V)∥vh∥V+(∥u-uεh∥V+∥u-uεHk∥V)×(∥u-Rhu∥V+∥u-uεHk∥V)∥vh∥V+N∥uεHk∥V(∥u-uεh∥V+∥u-Rhu∥V)∥vh∥V≤C3(1+h2+H2+εk+1)∥u-uεh∥V∥vh∥V+c(∥u-Rhu∥V+∥u-Rhu∥V2+∥u-uεHk∥V2)∥vh∥V,
where C3>0 is independent of h,H, and ε, then from (23) we have
(103)κ∥Qhp-pεh∥≤C3(1+h2+H2+εk+1)∥u-uεh∥V+c({∥u-uεHk∥V2}∥u-Rhu∥V+∥p-Qhp∥+∥u-Rhu∥V2+∥u-uεHk∥V2).
Thus, for sufficiently small h,H,ε, and η such that
(104)C3(1+h2+H2+εk+1)κ4ημ<12,
substituting (103) into (100), we obtain
(105)∥u-uεh∥V≤c(h5/4+εH5/4+H5/2+εk+2).
From (103), again, we obtain
(106)∥p-pεh∥≤c(h5/4+εH5/4+H5/2+εk+2).
Thus, we complete the proof of (93).Remark 9.
In terms of Theorems6, 7, and 8, if we choose ε=O(H),H=O(h5/9) for the two-level Stokes or Oseen iteration penalty methods and ε=O(H5/4),H=O(h1/2) for the two-level Newton iteration penalty method, then
(107)∥u-uεh∥V+∥p-pεh∥≤ch5/4.


## 4.4. An Improved Scheme

In this section, we will propose a scheme to improve the error estimates derived in Theorems6–8, which is described as follows.In Steps1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.Step 1.
Find(uεH0,pεH0)∈(VH,MH) by (48).Step 2.
Fork=1,2,…, find (uεHk,pεHk)∈(VH,MH) by (49).
At Step3, we solve a linearized problem (50) or (76) or (88) on the fine mesh in terms of Stokes iteration or Oseen iteration or Newton iteration, as in the following.Step 3.
Find(uεh,pεh)∈(Vh,Mh) by (50) or (76) or (88).
At Step4, we solve a Newton correction of (uεh,pεh) on the fine mesh in terms of Newton iteration, as in the following.Step 4.
Find(uεh⋆,pεh⋆)∈(Vh,Mh) such that for all (vh,qh)∈(Vh,Mh)(108)a(uεh⋆,vh-uεh⋆)+b(uεh,uεh⋆,vh-uεh⋆)+b(uεh⋆,uεh,vh-uεh⋆)+j(vhτ)-j(uεhτ⋆)-d(vh-uεh⋆,pεh⋆)≥(f,vh-uεh⋆)+b(uεh,uεh,vh-uεh⋆),d(uεh⋆,qh)+ε(pεh⋆,qh)=ε(pεh,qh).
First, we show the following theorem.Theorem 10.
Let(uεhk,pεhk) and (uεh⋆,pεh⋆) be the solutions of (28) and (108), respectively. Then there holds that
(109)∥uεhk-uεh⋆∥V+∥pεhk-pεh⋆∥≤c(∥uεhk-uεh∥V2+ε1/2∥pεh-pεhk-1∥),
where (uεh,pεh) is the solution to the problem (50) or (76) or (88).Proof.
Under the uniqueness condition (89), the solution uεh satisfies ||uεh||V≤μ/2N. Taking vh=uεh⋆,qh=pεh⋆-pεhk in (28) and vh=uεhk,qh=pεhk-pεh⋆ in (108) and adding them yield
(110)μ∥uεh⋆-uεhk∥V2+ε∥pεh⋆-pεhk∥2≤ε(pεh-pεhk-1,pεh⋆-pεhk)+b(uεhk,uεhk,uεh⋆-uεhk)-b(uεh,uεh⋆,uεh⋆-uεhk)-b(uεh⋆,uεh,uεh⋆-uεhk)+b(uεh,uεh,uεh⋆-uεhk)=ε(pεh-pεhk-1,pεh⋆)+b(uεh-uεhk,uεh-uεhk,uεh⋆-uεhk)+b(uεhk-uεh⋆,uεh,uεh⋆-uεhk).
Using (13), Hölder's inequality, and Young's inequality, we obtain
(111)μ∥uεh⋆-uεhk∥V2+ε∥pεh⋆-pεhk∥2≤ε∥pεh-pεhk-1∥∥pεh⋆-pεhk∥+N∥uεh-uεhk∥V2×∥uεh⋆-uεhk∥V+N∥uεh∥V∥uεh⋆-uεhk∥V2≤ε2∥pεh⋆-pεhk∥2+ε2∥pεh-pεhk-1∥2+μ2∥uεh⋆-uεhk∥V2+μ4∥uεh⋆-uεhk∥V2+N2μ∥uεh-uεhk∥V4.
That is,
(112)∥uεh⋆-uεhk∥V≤2εμ∥pεh-pεhk-1∥+2Nμ∥uεh-uεhk∥V2.
For all wh∈V0h, taking vh=uεhk±wh in the first inequality of (28) and vh=uεh⋆±wh in the first inequality of (108), it yields that
(113)a(uεhk,wh)+b(uεhk,uεhk,wh)-d(wh,pεhk)=(f,wh),a(uεh⋆,wh)+b(uεh,uεh⋆,wh)+b(uεh⋆,uεh,wh)-d(wh,pεh⋆)=(f,wh)+b(uεh,uεh,wh).
Subtracting them and using (23), (112), we obtain
(114)κ∥pεh⋆-pεhk∥≤supwh∈V0hd(wh,pεh⋆-pεhk)∥wh∥V=supwh∈V0ha(uεh⋆-uεhk,wh)+b(uεh,uεh⋆,wh)∥wh∥V+supwh∈V0hb(uεh⋆,uεh,wh)-b(uεhk,uεhk,wh)-b(uεh,uεh,wh)∥wh∥V=supwh∈V0ha(uεh⋆-uεhk,wh)+b(uεh⋆-uεhk,uεh,wh)∥wh∥V+supwh∈V0hb(uεh,uεh⋆-uεhk,wh)-b(uεhk-uεh,uεhk-uεh,wh)∥wh∥V≤μ∥uεh⋆-uεhk∥V+2N∥uεh∥V∥uεh⋆-uεhk∥V+N∥uεhk-uεh∥V2≤2μ∥uεh⋆-uεhk∥V+N∥uεhk-uεh∥V2≤22με∥pεh-pεhk-1∥+5N∥uεhk-uεh∥V2.From (34) and Theorems 6–10, we get the following error estimates.Theorem 11.
Let(u,p)∈H3(Ω)2∩V×H2(Ω)∩M and (uεh⋆,pεh⋆)∈(Vh,Mh) be the solutions to the problems (16) and (108), respectively. Then for the two-level Stokes or Oseen iteration penalty methods, they satisfy
(115)∥u-uεh⋆∥V+∥p-pεh⋆∥≤c(h5/4+H9/2+ε3/2H5/4+ε1/2H9/4+εk+1/2).
And for the two-level Newton iteration penalty method, they satisfy
(116)∥u-uεh⋆∥V+∥p-pεh⋆∥≤c(h5/4+H5+ε3/2H5/4+ε1/2H5/2+εk+1/2).Remark 12.
If we chooseH=O(h5/18), ε=O(h5/4) in (115) for two-level Stokes or Oseen iteration penalty methods and H=O(h1/4), ε=O(h5/4) in (116) for the two-level Newton iteration penalty method, then we obtain
(117)∥u-uεh⋆∥V+∥p-pεh⋆∥≤ch5/4.


## 5. Numerical Results

In this section, we will give numerical results to confirm the error analysis obtained in Section4. Since these two-level Stokes/Oseen/Newton iteration penalty methods are given in the form of the variational inequality problems which are not directly solved, the appropriate iteration algorithm must be constructed. Here we use the Uzawa iteration algorithm introduced in [28].For simplicity, we only give the Uzawa iteration method for solving the variational inequality problem (16). Similar schemes can be used to solve the two-level Stokes/Oseen/Newton iteration penalty schemes in Section 4. First, there exists a multiplier λ∈Λ such that the variational inequality problem (16) is equivalent to the following variational identity problem:
(118)a(u,v)+b(u,u,v)-d(v,p)+∫S‍λgvτds=(f,v),∀v∈V,d(u,q)=0,∀q∈M,λuτ=|uτ|,a.e.onS,
where λ∈Λ={γ∈L2(S):|γ(x)|≤1a.e.  onS}. In this case, we can solve the problem (16) by the following Uzawa iteration scheme:
(119)λ0∈Λisgiven;
then λn is known; we compute (un,pn) and λn+1 by
(120)a(un,v)+b(un,un,v)-d(v,pn)=(f,v)-∫S‍λngvτds,∀v∈V,d(un,q)=0,∀q∈M,λn+1=PΛ(λn+ρguτn),ρ>0,
where
(121)PΛ(γ)=sup(-1,inf(1,γ)),∀γ∈L2(S).Consider the problems (1)-(2) in the fixed square domain (0,1)×(0,1) (see Figure 1). Let μ=0.1. The external force f is chosen such that the exact solution (u,p) is
(122)u(x,y)=(u1(x,y),u2(x,y)),p(x,y)=(2x-1)(2y-1),u1(x,y)=-x2y(x-1)(3y-2),u2(x,y)=xy2(y-1)(3x-2).Figure 1
DomainΩ.It is easy to verify that the exact solutionu satisfies u=0 on Γ, u·n=u1=0,u2≠0 on S1 and u1≠0,u·n=u2=0 on S2. Moreover, the tangential vector τ on S1 and S2 are (0,1) and (-1,0). Thus, we have
(123)στ=4μy2(y-1)onS1,στ=4μx2(x-1)onS2.
On the other hand, from the nonlinear slip boundary conditions (2), there holds that
(124)|στ|≤g;
then the function g can be chosen as g=-στ≥0 on S1 and S2.In all experiments, we chooseμ=0.1, iteration initial value λ0=1, and ρ=μ/2. In terms of Theorems 6 and 7, for the two-level Stokes/Oseen penalty iteration methods, there holds that
(125)∥u-uεh∥V+∥p-pεh∥≤c(h5/4+εH5/4+H9/4+εk+1).
Then we choose ε=O(H),H=O(h5/9),k=2 such that
(126)∥u-uεh∥V+∥p-pεh∥≤ch5/4.
We pick eight coarse mesh size values; that is, H=1/4,1/6,1/8,…,1/18. In Table 1, the scaling between 1/H and 1/h=(1/H)9/5 is given.Table 1
Comparison of the scaling between1/H and 1/h.
1
/
H
4
6
8
10
12
14
16
18
1
/
h
12.125
25.157
42.224
63.095
87.604
115.619
147.033
181.756Settingε=ε0H, the comparison of relative error ||u-uεh||V/||u||V and ||p-pεh||/||p|| for different ε0>0 is shown in Tables 2 and 3 when we use the two-level Stokes/Oseen penalty iteration methods with 1/H=14 and 1/h=115. We can see that, for our present testing case, it suffices to set ε=0.001H if it is hoped to be as large as possible.Table 2
Numerical relative error for velocity withH=1/14 and h=1/115.
ε
0
0.01
0.001
0.0001
0.00001
Stokes
1.04374
e
-
03
1.84283
e
-
04
1.52977
e
-
04
1.52631
e
-
04
Oseen
1.04345
e
-
03
1.78281
e
-
04
1.45638
e
-
04
1.45275
e
-
04Table 3
Numerical relative error for pressure withH=1/14 and h=1/115.
ε
0
0.01
0.001
0.0001
0.00001
Stokes
1.85222
e
-
04
6.20827
e
-
05
5.96578
e
-
05
5.96412
e
-
05
Oseen
1.85112
e
-
04
6.13519
e
-
05
5.88604
e
-
05
5.88398
e
-
05Thus, setε=0.001H and 1/h≈(1/H)9/5. Tables 4 and 5 display the relative H1 errors of the velocity and the relative L2 errors of the pressure and their average convergence orders and CPU time when we use the two-level Stokes iteration penalty method and two-level Oseen iteration penalty method, respectively. Based on Tables 4 and 5, the two-level Stokes/Oseen iteration penalty methods can reach the convergence orders of O(h5/4) for both velocity and pressure, in H1- and L2-norms, respectively, as shown in (126).Table 4
Numerical relative error for Stokes method.
1
/
H
1
/
h
∥
u
-
u
ε
h
∥
V
/
∥
u
∥
V
∥
p
-
p
ε
h
∥
/
∥
p
∥
Iteration
CPU(s)
4
12
1.51517
e
-
02
5.41171
e
-
03
2
0.281
6
25
3.30825
e
-
03
1.25062
e
-
03
2
0.889
8
42
1.15862
e
-
03
4.44992
e
-
04
2
2.302
10
63
5.25908
e
-
04
1.99251
e
-
04
2
5.038
12
87
2.91391
e
-
04
1.05862
e
-
04
2
9.806
14
115
1.84283
e
-
04
6.20827
e
-
05
2
17.621
16
147
1.30602
e
-
04
3.96161
e
-
05
2
40.549
18
181
1.01944
e
-
04
2.77123
e
-
05
2
53.445
Order
1.727
1.913Table 5
Numerical relative error for Oseen method.
1
/
H
1
/
h
∥
u
-
u
ε
h
∥
V
/
∥
u
∥
V
∥
p
-
p
ε
h
∥
/
∥
p
∥
Iteration
CPU(s)
4
12
1.49411
e
-
02
5.38449
e
-
03
2
0.316
6
25
3.23052
e
-
03
1.24245
e
-
03
2
1.006
8
42
1.12567
e
-
03
4.41332
e
-
04
2
2.613
10
63
5.08693
e
-
04
1.93710
e
-
04
2
5.728
12
87
2.81455
e
-
04
1.04715
e
-
04
2
11.213
14
115
1.78281
e
-
04
6.13519
e
-
05
2
20.045
16
147
1.26872
e
-
04
3.91207
e
-
05
2
42.242
18
181
9.95916
e
-
05
2.73676
e
-
05
2
59.391
Order
1.728
1.915Next, we give the numerical results by using the two-level Newton iteration penalty method. In terms of Theorem8, there holds that
(127)∥u-uεh∥V+∥p-pεh∥≤c(h5/4+εH5/4+H5/2+εk+2).
Then we choose ε=0.01H5/4 and 1/h=(1/H)2 such that
(128)∥u-uεh∥V+∥p-pεh∥≤ch5/4.
Because when H=1/16 and h=1/256, this method does not work and the computer displays “out of memory”. Thus, in this experiment, we pick six coarse mesh size values; that is, H=1/4,1/6,…,1/14. Table 6 displays the relative H1 errors of the velocity and the relative L2 errors of the pressure and their average convergence orders and CPU time when we use the two-level Newton iteration penalty method. Based on Tables 4 and 5, we can see that the two-level Newton iteration penalty method also reaches the convergence orders of O(h5/4) for both velocity and pressure, in H1- and L2-norms, respectively, as shown in (128).Table 6
Numerical relative error for Newton method.
1
/
H
1
/
h
∥
u
-
u
ε
h
∥
V
/
∥
u
∥
V
∥
p
-
p
ε
h
∥
/
∥
p
∥
Iteration
CPU(s)
4
16
8.10332
e
-
03
3.02474
e
-
03
2
0.535
6
36
1.51165
e
-
03
5.98186
e
-
04
2
2.256
8
64
4.76536
e
-
04
1.89983
e
-
04
2
6.991
10
100
2.08297
e
-
04
8.02333
e
-
05
2
16.977
12
144
1.10901
e
-
04
3.89471
e
-
05
2
37.804
14
196
7.20875
e
-
05
2.21184
e
-
05
2
78.811
Order
1.811
1.947Figures2, 3, 4, and 5 show the streamline of flow and the pressure contour of the numerical solution by the two-level Stokes/Oseen/Newton iteration penalty methods and the exact solution, respectively.Figure 2
Streamline of flow and pressure contour for exact solution.Figure 3
Streamline of flow and pressure contour by Stokes method.Figure 4
Streamline of flow and pressure contour by Oseen method.Figure 5
Streamline of flow and pressure contour by Newton method.



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