Abstract and Applied Analysis

Volume 2013 (2013), Article ID 125139, 17 pages

http://dx.doi.org/10.1155/2013/125139

Research Article

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

Yuan Li and Rong An

College of Mathematics and Information Science, Wenzhou University, Wenzhou 325035, China

Received 2 April 2013; Revised 12 June 2013; Accepted 19 June 2013

Academic Editor: Stanislaw Migorski

Copyright © 2013 Yuan Li and Rong An. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 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 . The error estimate obtained in this paper shows that if , , and 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:

where is a bounded domain in assumed to have a Lipschitz continuous boundary , represents the viscous coefficient, denotes the velocity vector, the pressure and the prescribed body force vector. The solenoidal condition 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:

where , and is a scalar function; and are the normal and tangential components of the velocity, where stands for the unit vector of the external normal to ; , independent of , is the tangential components of the stress vector which is defined by with . The set denotes a subdifferential of the function at , 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 , 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 as the penalty finite element approximation solution to . The error estimate derived in [19] is

where is the penalty parameter. However, the condition number of the numerical discretization for the penalty methods is , which will result in an ill-conditioned problem when mesh size . In order to avoid the choice of the small parameter , Dai et al. [20] have studied the iteration penalty finite element method and derive

where 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 are of the following error estimate:

Finally, we propose an improved correction iteration scheme for in terms of the Newton iteration method. We prove that the correction finite element solutions are of the following error estimates:

Throughout this paper, we will use to denote a positive constant whose value may change from place to place but that remains independent of , , and and that may depend on , and the norms of ,,,   and  .

2. Preliminary

First, we give the definition of the subdifferential property. Let be a given function possessing the properties of convexity and weak semicontinuity from below. We say that the set is a subdifferential of the function at if and only if

In what follows, we employ the standard notation (or ) and , , for the Sobolev spaces of all functions having square integrable derivatives up to order in and the standard Sobolev norm. When , we write (or ) and instead of (or ) and , respectively.

For the mathematical setting, we introduce the following spaces:

The space is equipped with the norm

It is well known that is equivalent to due to Poincare's inequality. Introduce two bilinear forms

and a trilinear form

It is easy to verify that this trilinear form satisfies the following important properties [12, 23]:

for all , and

for all , and , where depends only on .

Given and with on , under the above notation, the variational formulation of the problem (1)-(2) reads as follows: find such that for all

where for all . Saito in [8] showed that there exists some positive such that

then the variational inequality (16) is equivalent to the following: find such that for all

The existence and uniqueness theorem of the solution to the problem (18) has been shown in [19]. Here, we only recall it.

Theorem 1. If the following uniqueness condition holds

then there exists a unique solution to the variational inequality problem (18) such that

where satisfies

3. Iteration Penalty Finite Element Approximation

Suppose that is a convex and polygon domain. Let be a family of quasi-uniform triangular partition of . The corresponding ordered triangles are denoted by . Let , and . For every , let denote the space of the polynomials on of degree at most . For simplicity, we consider the conforming finite element spaces and defined by

Denote . It is well known that and satisfy the Babuška-Brezzi condition [24, 25]:

where is a constant independent of . Denote and as the orthogonal projections onto and , respectively, which satisfy

It follows from the trace inequality [26] that

Let 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

Step 2. For , find such that for all

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

Proof. Setting , in (27), using (12) and Young's inequality, it yields that

Then we have

For , setting , in (28), it yields that

Thus, we obtain

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

Theorem 3. Let and be the solutions to the problems (16) and (28), respectively; then they satisfy

Next, we will show the error estimate for the penalty finite element approximation (28). This error analysis is based on the regularity assumption that the following linearized problem (35) is regular.

Given , find such that for all

According to (12) and (20), it is easy to verify that there exists a unique solution to the problem (35). The assumption that (35) is regular means that also belongs to and the following inequality holds:

Let be the orthogonal projections onto and satisfy

Theorem 4. Let and be the solutions to the problems (16) and (28), respectively; then they satisfy

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

Taking in (16) and in (28), respectively, we obtain

Subtracting them, we get

Substituting the previous equation into (39), it yields that

Using (34), (36), and (37), is estimated by

Similarly, using (25), (34), (36), and (37), is estimated by

We rewrite as

Then, from (13), (20), (29), (34), (36), and (37), it holds that

Finally, we estimate by

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, and with are two real positive parameters. The coarse mesh triangulation is made as in Section 3. And a fine mesh triangulation is generated by a mesh refinement process to . The conforming finite element space pairs and corresponding to the triangulations and , 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

Step 2. For , find such that for all

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 such that for all

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

Next, we estimate . Taking in (50), it yields

That is,

Suppose that the initial data satisfies

then using (51)-(52), we can estimate by

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 to the problem (50). Moreover, satisfy (56).

It follows from Theorems 3 and 4 that is of the following error estimates:

Next, we begin to prove the following error estimate for the solution to the problem (50).

Theorem 6. Suppose that the uniqueness condition (55) holds. Let and be the solutions to the problems (16) and (50), respectively; then they satisfy

Proof. Define a generalized bilinear form on by

Taking in (50), we have

Let and in the first inequality of (16); then

Adding the above two inequalities gives

Substituting the above inequality into (61), it yields that

It follows from Hölder's inequality and Young's inequality that

where is some small constant determined later. We rewrite as

Then using (13), (15), and Young's inequality, we can estimate by

It is easy to show that

Finally, from triangle inequality, is estimated by

Substituting (65)–(69) into (64), it yields that

where we use (24)–(26) and (57)-(58). Next, we estimate . For all , let in (16) and in (50), respectively. Then we get

Subtracting them and using (13), (15), we obtain

Therefore, it follows from (23) that can be estimated by

If we choose such that , then substituting (73) into (70), we show

From (73), again, we obtain

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 , 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 such that for all

From (12), it is easy to show that the solution to the problem (76) satisfies

Suppose that the initial data satisfies

then using (52), we can estimate by

For two-level Oseen iteration penalty method, the solution is of the following error estimate.

Theorem 7. Suppose that the uniqueness condition (78) holds. Let and be the solutions to the problems (16) and (76), respectively; then they satisfy

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

In the above equation, , , , and have been estimated in the proof of Theorem 6. Here, we only estimate . Using (12), (13), (15), and Young's inequality, we have

Then substituting (65), (68), (69), and (82) into (81), it yields that

In (83), we use (24)–(26) and (57)-(58). Next, we estimate . For all , proceeding as in the proof of (72), from (51) and (78), we can show

It follows from (23) and (84) that

If we choose such that , then substituting (85) into (83), we show

From (85), again, we obtain

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 , 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 such that for all

In this section, we will suppose that the initial data satisfies

Then from (51), satisfies . Let in (88). Using (12), we obtain

Since

then

Theorem 8. Suppose that the uniqueness condition (89) holds. Let and be the solutions to the problems (16) and (88), respectively; then they satisfy

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

In the above equation, , , , and have been estimated in the proof of Theorem 6. Here, we only estimate . We rewrite as

From (13), (20), and Young's inequality, is estimated by

Using (13) and (24), we estimate by

where is independent of , and . Similarly, it follows from (13), (24), and (57) that

where is independent of , and . Finally, we can estimate by

For sufficiently small , and such that , substituting (65), (68), (69), and (95)–(99) into (94), it yields that

For all , proceeding as in the proof of (72), we can show

Since

where is independent of , and , then from (23) we have

Thus, for sufficiently small , and such that

substituting (103) into (100), we obtain

From (103), again, we obtain

Thus, we complete the proof of (93).

Remark 9. In terms of Theorems 6, 7, and 8, if we choose for the two-level Stokes or Oseen iteration penalty methods and for the two-level Newton iteration penalty method, then

4.4. An Improved Scheme

In this section, we will propose a scheme to improve the error estimates derived in Theorems 6–8, 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 , 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 by (50) or (76) or (88).

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

Step 4. Find such that for all

First, we show the following theorem.

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

where is the solution to the problem (50) or (76) or (88).

Proof. Under the uniqueness condition (89), the solution satisfies . Taking in (28) and in (108) and adding them yield

Using (13), Hölder's inequality, and Young's inequality, we obtain

That is,

For all , taking in the first inequality of (28) and in the first inequality of (108), it yields that

Subtracting them and using (23), (112), we obtain

From (34) and Theorems 6–10, we get the following error estimates.

Theorem 11. Let and be the solutions to the problems (16) and (108), respectively. Then for the two-level Stokes or Oseen iteration penalty methods, they satisfy

And for the two-level Newton iteration penalty method, they satisfy

Remark 12. If we choose , in (115) for two-level Stokes or Oseen iteration penalty methods and , in (116) for the two-level Newton iteration penalty method, then we obtain

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 such that the variational inequality problem (16) is equivalent to the following variational identity problem:

where . In this case, we can solve the problem (16) by the following Uzawa iteration scheme:

then is known; we compute and by

where

Consider the problems (1)-(2) in the fixed square domain (see Figure 1). Let . The external force is chosen such that the exact solution is

Figure 1: Domain .

It is easy to verify that the exact solution satisfies on , on and on . Moreover, the tangential vector on and are and . Thus, we have

On the other hand, from the nonlinear slip boundary conditions (2), there holds that

then the function can be chosen as on and .

In all experiments, we choose , iteration initial value , and . In terms of Theorems 6 and 7, for the two-level Stokes/Oseen penalty iteration methods, there holds that

Then we choose such that

We pick eight coarse mesh size values; that is, . In Table 1, the scaling between and is given.

Table 1: Comparison of the scaling between and .

4
6
8
10
12
14
16
18

12.125
25.157
42.224
63.095
87.604
115.619
147.033
181.756

Setting , the comparison of relative error and for different is shown in Tables 2 and 3 when we use the two-level Stokes/Oseen penalty iteration methods with and . We can see that, for our present testing case, it suffices to set if it is hoped to be as large as possible.

Table 2: Numerical relative error for velocity with and .

0.01
0.001
0.0001
0.00001

Stokes

Oseen

Table 3: Numerical relative error for pressure with and .

0.01
0.001
0.0001
0.00001

Stokes

Oseen

Thus, set and . Tables 4 and 5 display the relative errors of the velocity and the relative 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 for both velocity and pressure, in - and -norms, respectively, as shown in (126).

Table 4: Numerical relative error for Stokes method.

Iteration
CPU(s)

4
12

2
0.281

6
25

2
0.889

8
42

2
2.302

10
63

2
5.038

12
87

2
9.806

14
115

2
17.621

16
147

2
40.549

18
181

2
53.445

Order

1.727
1.913

Table 5: Numerical relative error for Oseen method.

Iteration
CPU(s)

4
12

2
0.316

6
25

2
1.006

8
42

2
2.613

10
63

2
5.728

12
87

2
11.213

14
115

2
20.045

16
147

2
42.242

18
181

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

Then we choose and such that

Because when and , 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, . Table 6 displays the relative errors of the velocity and the relative 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 for both velocity and pressure, in - and -norms, respectively, as shown in (128).

Table 6: Numerical relative error for Newton method.

Iteration
CPU(s)

4
16

2
0.535

6
36

2
2.256

8
64

2
6.991

10
100

2
16.977

12
144

2
37.804

14
196

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.

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.

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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