Volume 2013, Issue 1 125139
Research Article
Open Access

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

Yuan Li

Yuan Li

College of Mathematics and Information Science, Wenzhou University, Wenzhou 325035, China wzu.edu.cn

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Rong An

Corresponding Author

Rong An

College of Mathematics and Information Science, Wenzhou University, Wenzhou 325035, China wzu.edu.cn

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First published: 16 July 2013
Citations: 4
Academic Editor: Stanislaw Migorski

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 size H 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.

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)
where Ω is a bounded domain in 2 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   div u = 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)
where , 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) = σσnn, 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 aL2(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 [24], 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 [1017]. 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 as the penalty finite element approximation solution to (u, p)∈(H3(Ω) 2, H2(Ω)). The error estimate derived in [19] is
(3)
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)
where k+ 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)
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 are of the following error estimates:
(6)

Throughout this paper, we will use c 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. Let ψ 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 aL2(S) 2 if and only if
(7)

In what follows, we employ the standard notation Hl(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)
The space V is equipped with the norm
(9)
It is well known that | | v | |V is equivalent to | | v | |1 due to Poincare′s inequality. Introduce two bilinear forms
(10)
and a trilinear form
(11)
It is easy to verify that this trilinear form satisfies the following important properties [12, 23]:
(12)
(13)
(14)
for all u, v, wV, and
(15)
for all uV,   vH2(Ω) 2, and wL2(Ω) 2, where N > 0 depends only on Ω.
Given fL2(Ω) 2 and gL2(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)
where j(η) = ∫Sg | η | ds for all ηL2(S) 2. Saito in [8] showed that there exists some positive β > 0 such that
(17)
then the variational inequality (16) is equivalent to the following: find uVσ such that for all vVσ
(18)
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)
then there exists a unique solution uVσ to the variational inequality problem (18) such that
(20)
where κ1 > 0 satisfies
(21)

3. Iteration Penalty Finite Element Approximation

Suppose that Ω is a convex and polygon domain. Let 𝒯h 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 K𝒯h, 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)
Denote V0h = V0 ∩ Wh. It is well known that V0h and Mh satisfy the Babuška-Brezzi condition [24, 25]:
(23)
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)
(25)
It follows from the trace inequality [26] that
(26)

Let ε > 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 such that for all (vh, qh)∈(Vh, Mh)

(27)

Step 2. For k = 1,2, …, find such that for all (vh, qh) ∈ (Vh, Mh)

(28)

First, we give the a priori estimate of the solution to the problem (28).

Theorem 2. Suppose that is the solution to the problem (28); then it satisfies

(29)

Proof. Setting vh = 0, in (27), using (12) and Young′s inequality, it yields that

(30)
Then we have
(31)
For k = 1,2, …, setting vh = 0, in (28), it yields that
(32)
Thus, we obtain
(33)

The next theorem gives the error estimate between the solutions (u, p) and to the problems (16) and (28), respectively. The proof can be found in [20].

Theorem 3. Let (u, p) ∈ H3(Ω) 2V × H2(Ω)∩M and be the solutions to the problems (16) and (28), respectively; then they satisfy

(34)

Next, we will show the error estimate 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.

Given zL2(Ω) 2, find (w, π)∈(V, M) such that for all (v, q)∈(V, M)
(35)
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)
Let Ih be the L2 orthogonal projections onto V0h and satisfy
(37)

Theorem 4. Let (u, p) ∈ H3(Ω) 2V × H2(Ω)∩M and be the solutions to the problems (16) and (28), respectively; then they satisfy

(38)

Proof. Setting and in the first equation of (35), we get

(39)
Taking v = u ± Ihw,  q = Qhπ in (16) and in (28), respectively, we obtain
(40)
Subtracting them, we get
(41)
Substituting the previous equation into (39), it yields that
(42)
Using (34), (36), and (37), J1 is estimated by
(43)
Similarly, using (25), (34), (36), and (37), J3 is estimated by
(44)
We rewrite J2 as
(45)
Then, from (13), (20), (29), (34), (36), and (37), it holds that
(46)
Finally, we estimate J4 by
(47)
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 𝒯H is made as in Section 3. And a fine mesh triangulation 𝒯h is generated by a mesh refinement process to 𝒯H. The conforming finite element space pairs (Vh, Mh) and (VH, MH)⊂(Vh, Mh) corresponding to the triangulations 𝒯h and 𝒯H, 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 Steps 1 and 2, we solve (27) and (28) on the coarse mesh, as in the follwing.

Step 1. Find such that for all (vH, qH)∈(VH, MH)

(48)

Step 2. For k = 1,2, …, find such that for all (vH, qH)∈(VH, MH)

(49)

In Step 3, 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)

As a direct consequence of Theorem 2, the solution to the problem (49) satisfies

(51)
(52)
Next, we estimate uεh. Taking vh = 0, qh = pεh in (50), it yields
(53)
That is,
(54)
Suppose that the initial data satisfies
(55)
then using (51)-(52), we can estimate uεh by
(56)
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 Theorems 3 and 4 that is of the following error estimates:
(57)
(58)
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) 2V × H2(Ω) ∩ M and (uεh, pεh)∈(Vh, Mh) be the solutions to the problems (16) and (50), respectively; then they satisfy

(59)

Proof. Define a generalized bilinear form on (Vh, Mh)×(Vh, Mh) by

(60)
Taking vh = Rhu,  qh = pεhQhp in (50), we have
(61)
Let v = uεh and v = 2uRhu in the first inequality of (16); then
(62)
Adding the above two inequalities gives
(63)
Substituting the above inequality into (61), it yields that
(64)
It follows from Hölder′s inequality and Young′s inequality that
(65)
where η > 0 is some small constant determined later. We rewrite I2 as
(66)
Then using (13), (15), and Young′s inequality, we can estimate I2 by
(67)
It is easy to show that
(68)
Finally, from triangle inequality, I5 is estimated by
(69)
Substituting (65)–(69) into (64), it yields that
(70)
where we use (24)–(26) and (57)-(58). Next, we estimate | | pεhQhp||. For all whV0h, let v = u ± wh in (16) and vh = uεh ± wh in (50), respectively. Then we get
(71)
Subtracting them and using (13), (15), we obtain
(72)
Therefore, it follows from (23) that | | Qhppεh|| can be estimated by
(73)
If we choose such that , then substituting (73) into (70), we show
(74)
From (73), again, we obtain
(75)
Thus, we complete the proof of (59).

4.2. Two-Level Oseen Iteration Penalty Method

In Steps 1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.

Step 1. Find by (48).

Step 2. For k = 1,2, …, find by (49).

In Step 3, 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)

From (12), it is easy to show that the solution (uεh, pεh) to the problem (76) satisfies
(77)
Suppose that the initial data satisfies
(78)
then using (52), we can estimate uεh by
(79)

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)

Proof. Proceeding as in the proof of (64), we can get

(81)
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)
Then substituting (65), (68), (69), and (82) into (81), it yields that
(83)
In (83), we use (24)–(26) and (57)-(58). Next, we estimate | | pεhQhp||. For all whV0h, proceeding as in the proof of (72), from (51) and (78), we can show
(84)
It follows from (23) and (84) that
(85)
If we choose such that , then substituting (85) into (83), we show
(86)
From (85), again, we obtain
(87)
Thus, we complete the proof of (80).

4.3. Two-Level Newton Iteration Penalty Method

In Steps 1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.

Step 1. Find by (48).

Step 2. For k = 1,2, …, find by (49).

In Step 3, 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)

In this section, we will suppose that the initial data satisfies
(89)
Then from (51), satisfies . Let vh = 0,  qh = pεh in (88). Using (12), we obtain
(90)
Since
(91)
then
(92)

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)

Proof. Proceeding as the in proof of (64), we can get

(94)

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)
From (13), (20), and Young′s inequality, I8 is estimated by
(96)
Using (13) and (24), we estimate I9 by
(97)
where C1 > 0 is independent of h, H, and ε. Similarly, it follows from (13), (24), and (57) that
(98)
where C2 > 0 is independent of h,   H, and ε. Finally, we can estimate I11 by
(99)
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)
For all whV0h, proceeding as in the proof of (72), we can show
(101)
Since
(102)
where C3 > 0 is independent of h,   H, and ε, then from (23) we have
(103)
Thus, for sufficiently small h,   H,   ε, and η such that
(104)
substituting (103) into (100), we obtain
(105)
From (103), again, we obtain
(106)
Thus, we complete the proof of (93).

Remark 9. In terms of Theorems 6, 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)

4.4. An Improved Scheme

In this section, we will propose a scheme to improve the error estimates derived in Theorems 68, which is described as follows.

In Steps 1 and 2, we solve (48) and (49) on the coarse mesh, as in the following.

Step 1. Find by (48).

Step 2. For k = 1,2, …, find by (49).

At Step 3, 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 Step 4, we solve a Newton correction of (uεh, pεh) on the fine mesh in terms of Newton iteration, as in the following.

Step 4. Find such that for all (vh, qh)∈(Vh, Mh)

(108)
First, we show the following theorem.

Theorem 10. Let and be the solutions of (28) and (108), respectively. Then there holds that

(109)
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 | |Vμ/2N. Taking in (28) and in (108) and adding them yield

(110)
Using (13), Hölder′s inequality, and Young′s inequality, we obtain
(111)
That is,
(112)
For all whV0h, taking in the first inequality of (28) and in the first inequality of (108), it yields that
(113)
Subtracting them and using (23), (112), we obtain
(114)

From (34) and Theorems 610, we get the following error estimates.

Theorem 11. Let (u, p) ∈ H3(Ω) 2V × H2(Ω)∩M and be the solutions to the problems (16) and (108), respectively. Then for the two-level Stokes or Oseen iteration penalty methods, they satisfy

(115)
And for the two-level Newton iteration penalty method, they satisfy
(116)

Remark 12. If we choose H = 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)

5. Numerical Results

In this section, we will give numerical results to confirm the error analysis obtained in Section 4. 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)
where λ ∈ Λ = {γL2(S) :    | γ(x)| ≤ 1  a.e. on  S}. In this case, we can solve the problem (16) by the following Uzawa iteration scheme:
(119)
then λn is known; we compute (un, pn) and λn+1 by
(120)
where
(121)
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)
Details are in the caption following the image
Domain Ω.
It is easy to verify that the exact solution u 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)
On the other hand, from the nonlinear slip boundary conditions (2), there holds that
(124)
then the function g can be chosen as g = −στ ≥ 0 on S1 and S2.
In all experiments, we choose μ = 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)
Then we choose ε = O(H),   H = O(h5/9),  k = 2 such that
(126)
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 between 1/H and 1/h.
1/H 4 6 8 10 12 14 16 18
1/h 12.125 25.157 42.224 63.095 87.604 115.619 147.033 181.756

Setting ε = ε0H, the comparison of relative error | | uuεh | |V/| | u | |V and | | ppεh||/| | 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 with H = 1/14 and h = 1/115.
ε0 0.01 0.001 0.0001 0.00001
Stokes 1.04374e − 03   1.84283e − 04   1.52977e − 04   1.52631e − 04 
Oseen  1.04345e − 03   1.78281e − 04   1.45638e − 04   1.45275e − 04 
Table 3. Numerical relative error for pressure with H = 1/14 and h = 1/115.
ε0 0.01 0.001 0.0001 0.00001
Stokes  1.85222e − 04   6.20827e − 05   5.96578e − 05   5.96412e − 05 
Oseen  1.85112e − 04   6.13519e − 05   5.88604e − 05   5.88398e − 05 

Thus, set ε = 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 ∥p − pεh∥/∥p Iteration CPU(s)
4 12  1.51517e − 02   5.41171e − 03  2 0.281
6 25  3.30825e − 03   1.25062e − 03  2 0.889
8 42  1.15862e − 03   4.44992e − 04  2 2.302
10 63  5.25908e − 04   1.99251e − 04  2 5.038
12 87  2.91391e − 04   1.05862e − 04  2 9.806
14 115  1.84283e − 04   6.20827e − 05  2 17.621
16 147 1.30602e − 04   3.96161e − 05  2 40.549
18 181 1.01944e − 04  2.77123e − 05 2 53.445
  
Order 1.727 1.913
Table 5. Numerical relative error for Oseen method.
1/H 1/h ppεh∥/∥p Iteration CPU(s)
4 12  1.49411e − 02   5.38449e − 03  2 0.316
6 25  3.23052e − 03   1.24245e − 03  2 1.006
8 42  1.12567e − 03   4.41332e − 04  2 2.613
10 63  5.08693e − 04   1.93710e − 04  2 5.728
12 87  2.81455e − 04  1.04715e − 04  2 11.213
14 115  1.78281e − 04   6.13519e − 05 2 20.045
16 147 1.26872e − 04   3.91207e − 05  2 42.242
18 181 9.95916e − 05  2.73676e − 05 2 59.391
  
Order 1.728 1.915
Next, we give the numerical results by using the two-level Newton iteration penalty method. In terms of Theorem 8, there holds that
(127)
Then we choose ε = 0.01H5/4 and 1/h = (1/H) 2 such that
(128)
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 ∥p − pεh∥/∥p Iteration CPU(s)
4 16  8.10332e − 03   3.02474e − 03  2 0.535
6 36  1.51165e − 03   5.98186e − 04  2 2.256
8 64  4.76536e − 04   1.89983e − 04  2 6.991
10 100 2.08297e − 04   8.02333e − 05  2 16.977
12 144 1.10901e − 04  3.89471e − 05  2 37.804
14 196  7.20875e − 05   2.21184e − 05 2 78.811
  
Order 1.811 1.947

Figures 2, 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.

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Streamline of flow and pressure contour for exact solution.
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Streamline of flow and pressure contour by Stokes method.
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Streamline of flow and pressure contour by Oseen method.
Details are in the caption following the image
Streamline of flow and pressure contour by Newton method.

Acknowledgments

This work is supported by the National Natural Science Foundation of China under Grant nos. 10901122, 11001205 and by Zhejiang Provincial Natural Science Foundation of China under Grant no. LY12A01015.

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