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Page 1: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

Two Iterative Methods

for Solving the Stokes Problem

Yuri Iliash Tuomo Rossi Jari Toivanen

Page 2: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

Iterative Methods

for Solving the Stokes Problem

Yuri Iliash

Tuomo Rossi

Jari Toivanen

Institute of Numerical Mathematics, Academy of Sciences of Russia,

Leninskij Prospekt 32-a, Moscow Russia.

Email [email protected].

University of Jyvaskyla, Laboratory of Scientic Computing,

P.O. Box 35, FIN-40351, Jyvaskyla Finland

Email [email protected]. (T. Rossi), [email protected]. (J.Toivanen)

Abstract

Two approaches to the numerical solution of the Stokes problem are consid-

ered. One of them uses the preconditioned Lanczos method of minimized itera-

tions which is based upon the principles of the multigrid domain decomposition

method (MGDD). The other approach is the classical Uzawa's algorithm, where

the Laplace equations are solved by using the ctitious domain method. Results

of computational experiments are given.

Key words: saddle point problem, Stokes problem, multigrid domain decomposi-

tion method, Uzawa's algorithm, ctitious domain method.

2

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Contents

1 Introduction 4

2 Dierential and variational formulation of the Stokes problem 5

3 Grids and approximation 5

3.1 FD approximation on staggered rectangular grids : : : : : : : : : : : : : : : 5

3.2 FE approximation on triangular grids : : : : : : : : : : : : : : : : : : : : : : 7

4 CG method via Uzawa's algorithm 9

4.1 Uzawa's algorithm : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9

4.2 Preconditioner for the Laplace operator based on the ctitious domain method 11

4.2.1 Fast direct Poisson solver in rectangle : : : : : : : : : : : : : : : : : : 11

4.2.2 Partial solution : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 13

4.2.3 The ctitious domain method for the Dirichlet boundary value problem 13

5 The Lanczos method of minimized iterations with a block-diagonal pre-

conditioner 17

5.1 Block-diagonal preconditioner : : : : : : : : : : : : : : : : : : : : : : : : : : 18

5.2 Multigrid domain decomposition method : : : : : : : : : : : : : : : : : : : : 21

5.3 The Lanczos method of minimized iterations : : : : : : : : : : : : : : : : : : 28

6 Numerical experiments 28

7 Acknowledgements 32

3

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1 Introduction

In the present paper we shall consider two iterative methods for solving the Stokes system

of partial dierential equations with Dirichlet boundary conditions for the velocity. Finite

dierence or nite element approximation of this system of equations leads the so-called

algebraic saddle point problem:

G

"

w

p

#

"

A B

B

T

0

# "

w

p

#

=

"

f

g

#

: (1:1)

Here A is a symmetric and positive denite 2N

v

2N

v

matrix, and G is an indenite singular

(2N

v

+N

p

) (2N

v

+N

p

) matrix.

There are some dierent approaches to the construction of iterative methods for solving

problem (1.1).

One of the approaches is to directly apply classical multigrid methods to the problem with

the matrix G [17, 33]. Bramble and Pasciak [9] suggested to reformulate the problem (1.1)

to obtain a problem with symmetric and positive denite matrix and apply the conjugate

gradient method to it. In [4] it is suggested to solve problem (1.1) using the following iterative

process

"

w

i+1

p

i+1

#

=

"

w

i

p

i

#

+

"

^

A B

B

T

^

C

#

1

"

f

g

#

"

A B

B

T

0

# "

w

i

p

i

#!

with

^

C = B

T

^

A

1

B

^

S, where operators

^

A and

^

S are in some sense close to the operators A

and B

T

^

A

1

B, respectively.

Two approaches to problem (1.1) are investigated in the present paper. One of them

is based on the generalized Lanczos method of minimized iterations with a block-diagonal

preconditioner of the form

~

G =

"

~

A 0

0

~

S

#

; (1:2)

where the matrix

~

A is spectrally equivalent to the matrix A, and the matrix

~

S to matrix

S B

T

A

1

B. The matrix

~

A is constructed by using a certain modication of the multigrid

domain decomposition (MGDD) method described in [22, 23, 24]. The use of the precondi-

tioner of the form (1.2) is theoretically justied in Section 5.1, the modication of the MGDD

method is described in Section 5.2, and the generalized Lanczos method of minimized itera-

tions is described in Section 5.3.

The other approach uses the Uzawa's algorithm (see, e.g., [10, 15, 16, 31]). We reformulate

the problem (1.1) to

(B

T

A

1

B)p = B

T

A

1

f g; w = A

1

f A

1

Bp; p ? kerB; (1:3)

which is solved with conjugate gradient method. In this approach we solve the systems with

the matrix A by using the ctitious domain method [1, 2, 13, 30, 32]. The asymptotic arith-

metical complexity estimates for this approach are worse than those for the block diagonal

preconditioner, but the numerical experimentations show that the Uzawa's algorithm works

faster in problems of practical size.

The results of computational experiments are given in Section 6. The computations were

made for the problem of \inverse stair".

4

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2 Dierential and variational formulation of the Stokes problem

Let be a simply-connected domain in R

2

with a piecewise-smooth boundary @. We

consider the Stokes system of equations in

(

k w +rp = 0

div w = 0

(2:1)

where w = [u; v] is the velocity of viscous incompressible uid, u is its x-component, v is its

y-component, p is the pressure and k is the kinematic viscosity of the uid. Without loss of

generality, hereafter we assume that k = 1.

On the boundary @ we pose the nonhomogeneous Dirichlet conditions for the velocity:

wj

@

=

0

: (2:2)

The function

0

should be chosen such that incoming and outgoing ows are equal, i. e.:

I

@

n

0

= 0: (2:3)

In this problem formulation no conditions are stated for the pressure p.

Along with the dierential formulation of the Stokes problem we consider the variational

(weak) formulation: nd ( w; p) 2

W

1

2

P :

D( w; z) (p; r z) = 0; 8z [z

1

; z

2

] 2

W

1

2

()

W

1

2

();

(r w; q) = 0; 8q 2 P; (2.4)

wj

@

=

0

:

Here (; ) denotes the inner product in L

2

(),

D(; ) is bilinear form dened by

D( w; z)

Z

ru rz

1

d +

Z

rv rz

2

d;

and

W

1

2

=W

1

2

()W

1

2

(); P =

n

q : q 2 L

2

(); (q; 1)

L

2

()

= 0

o

.

It is known [26] that if

0

2 W

1=2

2

(@) W

1=2

2

(@) and condition (2.3) holds then the

solution to problem (2.4) exists and is unique.

3 Grids and approximation

3.1 FD approximation on staggered rectangular grids

We approximate problem (2.1), (2.2) using nite dierences on uniform staggered grids.

These grids are constructed as follows [14]. In the domain we dene a basic square grid

with a step size h. Further, using this grid, we construct three other rectangular grids. Nodes

of the rst grid,

u

, are located at the midpoints of vertical (i. e., parallel to the y-axis) edges

of the basic grid (they are shown by in Fig. 3.1); the second grid,

v

, has its nodes at the

midpoints of horizontal (i. e., parallel to the x-axis) edges of the basic grid (they are shown

by in Fig. 3.1); nodes of the third grid,

p

, are at the centers of the cells of the basic grid

(they are shown by in Fig. 3.1).

5

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v

j;k1=2

v

j+1;k1=2

v

j+1;k+1=2

v

j;k+1=2

p

jk

p

j+1;k

u

j+3=2;k

u

j+1=2;k

u

j1=2;k

p

j+1;k1

p

j;k1

u

j+1=2;k1

Fig. 3.1. Staggered grids.

The rst equation of system (2.1) is discretized on the grid

u

. In order to approximate

the Laplace operator we use the standard ve-point scheme with

"

@

2

u

@x

2

#

j+1=2;k

u

j+3=2;k

+ 2u

j+1=2;k

u

j1=2;k

h

2

;

"

@

2

u

@y

2

#

j+1=2;k

u

j+1=2;k1

+ 2u

j+1=2;k

u

j+1=2;k+1

h

2

:

The pressure derivative is approximated using central dierences:

"

@p

@x

#

j+1=2;k

p

j+1;k

p

j;k

h

:

A similar discretization of the second equation of (2.1) uses the grid

v

. The discretization

of the equation for velocity divergence is performed using the grid

p

:

[div u]

j;k

u

j+1=2;k

u

j1=2;k

h

+

v

j;k+1=2

v

j;k1=2

h

:

For the approximation of the rst and the second equations at the nodes adjacent to the

boundary we employ the so-called re ection trick. It is based on constructing auxiliary

nodes outside the domain . For example, in order to obtain the discretization of the rst

equation at the node (3=2; 1) we introduce an auxiliary node (3=2; 0) which is symmetric

to the node (3=2; 1) regarding the boundary (see Fig. 3.2). Then we assume that u

3=2;1=2

=

1

2

u

3=2;0

+ u

3=2;1

, where the value of u

3=2;1=2

is taken from the boundary conditions. Thus,

we have u

3=2;0

= 2u

3=2;1=2

u

3=2;1

, and we may use the ve-point scheme to approximate

the Laplace operator at the node (3=2; 1). The second equation is approximated near the

boundary in a similar way.

6

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v

1;3=2

v

0;3=2

u

3=2;1

u

3=2;0

u

3=2;1=2

Fig. 3.2. Approximation at the nodes adjacent to the boundary.

As a result, we obtain a system of linear equations (1.1). Note that the matrix G in (1.1)

is singular: the dimension of kerG is equal to one, and kerG consists of vectors

"

0

c

#

, where

c = const

2

6

6

4

1

.

.

.

1

3

7

7

5

is m-dimensional vector.

The following important result can be proved (see, e. g., [21] ).

Lemma 3.1 The approximation of problem (2.1), (2.2) on the uniform staggered grids sat-

ises the Babuska-Brezzi-Ladyzhenskaya condition:

sup

w2R

2N

v

nf0g

(p; B

T

w)

kwk

A

c

0

kpk; 8p 2 R

N

p

x; p ? kerB;

here c

0

is a positive constant independent of h.

Using this lemma we can obtain the following result (see, e. g., [20]).

Theorem 3.1 For problem (1.1) arising from the approximation of the Stokes problem on

uniform staggered grids the matrix S B

T

A

1

B = S

T

is positive denite on the subspace

(kerB)

?

and

c

1

I S I; (3:1)

where a positive constant c

1

is independent of h; I is the identity matrix.

3.2 FE approximation on triangular grids

The nite element approximation is based on triangulated rectangular mesh. We wish to

use fast direct methods coupled with the ctitious domain approach, which requires that the

preconditioner is separable. This can be established by using so called P

1

isoP

2

=P

1

elements

[8]. (See Fig. 3.3.)

7

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@

@

@

@

@

@

@

@

@

@

@

@

@

@

@

@

@

@

@

@

x h x

x

h h

x

velocity and pressure

h

velocity

Fig. 3.3. P

1

isoP

2

=P

1

element.

The approximation of both of the velocity components and pressure is piece-wise linear.

We dene the nite element spaces

W

h

and P

h

by

W

h

= f w = [u; v] 2

W

1

2

W

1

2

j u and v are linear on each velocity triangleg: (3:2)

and

P

h

= fq 2 P j q is linear on each pressure triangleg: (3:3)

As a basis of

W

h

we use the basis f w

i

h

g

2N

v

i=1

dened by

w

i

h

2

W

h

; w

i

h

=

(

[

i

h

; 0] ; 1 i N

v

;

h

0;

iN

v

h

i

; N

v

+ 1 i 2N

v

:

(3:4)

Here

i

h

2

h

f 2

W

1

2

j is linear on each velocity triangleg; (3:5)

and

i

h

(x

j

) =

ij

; for each

i

h

; i = 1; :::; N

v

; and for each vertex x(j); j = 1; :::; N

v

of the

velocity mesh. For the pressure space P

h

we dene the Courant basis fq

i

h

g

N

p

i=1

by setting

q

i

h

2 P

h

; q

i

h

(y

j

) =

ij

; (3:6)

for each vertex y

j

; j = 1; :::; N

p

of the pressure triangulation and for each basis function q

i

h

;

i = 1; :::; N

p

: A sample triangulation for the \inverse stair" domain is illustrated in the gure

3.4.

It should be noted that the mesh must be generated in such a way that all the pressure

nodes of one element do not lie in the boundary of the domain. If this is taken into account

then it can be shown, that the Babuska-Brezzi-Ladyzhenskaya condition holds true for this

kind of discretization [8]. As a result we get the following linear system

"

A B

B

T

0

# "

w

p

#

=

"

f

g

#

; (3:7)

8

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Velocity triangle

Pressure triangle

Fig. 3.4. A sample triangulation of the inverse stair

domain with P

1

isoP

2

=P

1

elements

where w is the velocity components in the vertices of the velocity triangulation and p is the

pressure in the vertices of the pressure triangulation. The matrices A and B have the block

representation

A =

"

^

A 0

0

^

A

#

2 R

2N

v

2N

v

;

^

A 2 R

N

v

N

v

; B =

"

B

1

B

2

#

2 R

2N

v

N

p

; B

l

2 R

N

v

N

p

; l = 1; 2:

Analogously the vector block f has the form f = [f

t

1

f

t

2

]

t

: The block

^

A is computed by

^

A(i; j) =

Z

r

i

h

r

j

h

dx; i; j = 1; :::; N

v

: (3:8)

The matrix block B

l

is computed by

B

l

(i; j) =

Z

@

i

h

@x

l

q

j

h

dx; i = 1; :::; N

v

; j = 1; :::; N

p

; l = 1; 2: (3:9)

The force vector block f is computed by

f

l

(i) =

Z

r~u

l

r

i

h

; i = 1; :::; N

v

; l = 1; 2: (3:10)

The remaining vector block g is computed by

g(j) =

Z

r ~u

q

j

h

; j = 1; :::; N

p

: (3:11)

Here ~u

l

denotes the l:th component function of the nite element extension ~u

of

0

to the

domain : This extension ~u

is chosen in such a way that it equals to zero in the interior

nodes.

4 CG method via Uzawa's algorithm

4.1 Uzawa's algorithm

In this chapter we will construct so called Uzawa's algorithm for solving the linear system

(3.7).

9

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Lemma 4.1 The linear system (3.7) is equivalent to

(B

T

A

1

B)p = B

T

A

1

f g; w = A

1

f A

1

Bp; p ? kerB: (4:1)

Furthermore, Condj

kerB

?B

T

A

1

B c; where c is a positive constant independent of h:

Proof: Let us rst eliminate the velocity block w from the rst equation. This can be done

because the matrix block A is invertible, since it corresponds to the vector Laplace operator

with Dirichlet boundary conditions. The rst equation reads

Aw +Bp = f () w = A

1

(f Bp): (4:2)

Substituting this to the second equation we obtain

B

T

w = g () B

T

A

1

(f Bp) = g () (B

T

A

1

B)p = g +B

T

A

1

f: (4:3)

The second assertion follows from the Theorem (3.1) because this discretization also satises

the Babuska-Brezzi-Ladyzhenskaya condition.

To solve the system (4.1) we propose the following conjugate gradient algorithm:

1. Initial guess p

h

= 0

Solve Aw = f ()

Set r = B

T

w

Set = (r; r)

Set q = r

2. Solve Av = Bq ()

Set z = B

T

v

Set = =(q; z)

Set p

h

= p

h

q

Set w = w + v

Set r = r z

Set

old

= and = (r; r)

3. If > acc

2

then

Set = =

old

Set q = r + q

Goto 2.

endif

In steps marked with () we must solve systems with the matrix block A: The matrix

block A is a discrete counterpart of the vector Laplace operator with Dirichlet boundary

conditions

A =

"

^

A 0

0

^

A

#

"

0

0

#

: (4:4)

The disadvantage of this approach is that we must solve these systems with very high

accuracy in order to reach convergence. To solve the systems we can apply ctitious domain

method. In the next chapter we will consider the ctitious domain method for solving the

Poisson equation with Dirichlet boundary conditions.

10

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4.2 Preconditioner for the Laplace operator based on the ctitious domain

method

In the Uzawa's algorithm we need to solve linear systems, where the coecient matrix A

corresponds to the vector Laplace operator. Since the matrix A is a block diagonal matrix,

the solution can be obtained by solving two separate systems. These systems are a discrete

counterpart of Poisson equations for the two velocity components. In this chapter we intro-

duce a fast algorithm which is based on the ctitious domain approach [1, 2, 13, 30, 32] for

solving the Poisson equations with Dirichlet boundary conditions.

Fictitious domain method is a preconditioned iterative method which utilizes so called

fast direct methods [5, 6, 7, 18, 19, 34, 35, 36]. The restriction of these methods is that the

mesh should be a rectangular one and the domain should be a rectangle.

4.2.1 Fast direct Poisson solver in rectangle

In this chapter we present a brief overview on one fast direct Poisson solver in a rectan-

gle. This method is based on the Fast Fourier Transform (FFT) [11, 18]. More detailed

descriptions on this method and some others can be found from [32].

The domain R

2

is supposed to be the rectangle [a

1

; b

1

] [a

2

; b

2

]: For the fast direct

Poisson solver presented we need a rectangular mesh for the domain : First we dene a

division for the intervals [a

1

; b

1

] and [a

2

; b

2

] such that

a

k

= x

k

0

; x

k

1

; : : : ; x

k

N

k

+1

= b

k

; k = 1; 2:

In order to get a triangulation we must divide every rectangular cell [x

1

i

; x

2

j

] [x

1

i+1

; x

2

j+1

];

i = 0; : : : ; N

1

; j = 0; : : : ; N

2

; in to two triangles. The inner nodes are numbered from one to

N

1

N

2

from left to right and from bottom to up.

Remark 4.1 The velocity mesh of a given rectangular domain is rectangular one, if we use

P

1

isoP

2

=P

1

elements to discretize the Stokes equation in a rectangle. This holds true, if we

construct the pressure triangulation on a rectangular mesh.

With the mesh presented previously the nite element discretization with the linear trian-

gle elements and the nite dierence discretization with the ve point scheme gives the same

stiness matrix A. The matrix A is a symmetric block tridiagonal matrix whose codiagonal

blocks are diagonal matrices and diagonal blocks are tridiagonal matrices. The diagonal and

codiagonal blocks are commutative.

The matrix A can be expressed in the tensor product form

A = K

2

A

1

+ A

2

K

1

: (4:5)

Here the tensor product for the matrices B 2 R

NN

and C 2 R

MM

is dened by

B C = (B

ij

C)

N

i;j=1

2 R

NMNM

:

In the tensor product (4.5) the matrices A

i

2 R

N

k

N

k

are tridiagonal matrices and the

matrices K

i

2 R

N

k

N

k

are diagonal matrices. The matrices A

i

and K

i

are depending only

on the mesh division in the x

i

-direction.

11

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Remark 4.2 The blocks

^

A of the matrix A have the structure (4.5) if we use rectangular

mesh on a rectangular domain.

The method which we consider is so called method of separation of variables which is

based on an observation that the matrix A in (4.5) can be expressed in a matrix product

form. First we dene the generalized eigenvalue problem

A

2

w

i

=

i

K

2

w

i

; kw

i

k = 1; i = 1; : : : ; N

2

(4:6)

and then we can dene the matrices

W = [w

1

; w

2

; : : : ; w

N

2

] and = diagf

1

;

2

; : : : ;

N

2

g:

Now the matrix product form for the matrix A is

A = (K

1

2

2

W I

1

)(I

2

A

1

+ K

1

)(W

T

K

1

2

2

I

1

);

where the matrices I

i

2 R

N

i

N

i

; i = 1; 2; are the identity matrices. By inverting the matrix

A we obtain

A

1

= (K

1

2

2

W I

1

)(I

2

A

1

+ K

1

)

1

(W

T

K

1

2

2

I

1

): (4:7)

Now we consider the arithmethic complexity of computing matrix A

1

times vector f:

We divide the multiplication to four steps in the following way.

1. Solve the generalized eigenvalue problem (4.6).

2. Compute g = (W

T

K

1

2

2

I

1

)f:

3. Solve (I

2

A

1

+ K

1

)v = g:

4. Compute u = (K

1

2

2

W I

1

)v:

The complexity for solving the generalized eigenvalue problem (4.6) is O(N

1

N

2

) opera-

tions, because the matrix A

2

is a tridiagonal matrix and the matrix K

2

is a diagonal matrix.

If the mesh is uniform in x

2

-direction then the analytical solution is known.

The steps 2. and 4. are similar regarding the complexity. Both matrices contains N

1

N

2

2

non zero matrix entries and therefore O(N

1

N

2

2

) operations are required. The multiplications

can be considered as a Fourier transformation and an inverse Fourier transformation. In

the case that the mesh is uniform in x

2

-direction the multiplications can be carried out in

the FFT manner and then the complexity is only O(N

1

N

2

logN

2

): The most ecient FFT

algorithms require that N

2

= 2

i

1 for some integer i.

The matrix in the step 3. is a symmetric block diagonal matrix and all N

2

diagonal blocks

areN

1

N

1

tridiagonal matrices. It requires O(N

1

) operations to solve one tridiagonal system.

Therefore the total complexity for this step is O(N

1

N

2

) operations.

As the previous analysis shows the total complexity is O(N

1

N

2

2

) for a nonuniform mesh in

x

2

-direction and O(N

1

N

2

logN

2

) for an uniform mesh in x

2

-direction. The Fourier transfor-

mation direction can be changed to be x

1

-direction and if the mesh is uniform in x

1

-direction

then the total complexity is O(N

1

N

2

logN

1

): The method of separation of variables is usually

used only when mesh is uniform at least in one direction.

12

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4.2.2 Partial solution

The partial solution is a method for solving a part of unknowns in linear systems where the

coecient matrix is of the form (4.5) with sparse right hand side vector. The methodology

was introduced independently in [25] and [3]. It is useful in the ctitious domain method,

because the solution is needed only near the boundary in the ctitious domain iteration and

the right hand side vector is also sparse.

We suppose that the force vector is sparse and the wanted solution is sparse. In here by

sparse we mean O(

p

N), where N is the total number of unknowns. Also we suppose that

p

N N

1

N

2

: Now we apply the method of separation of variables to this problem. The

complexity of step 2. and 4. of multiplication is reduced in this case. Each row and column

in the multiplying matrix contains N

2

non zero entries which have to be multiplied O(

p

N)

times. Therefore the complexity of partial solution is only O(N):

4.2.3 The ctitious domain method for the Dirichlet boundary value problem

Let us consider the following partial dierential equation

(

u = f; x 2

1

u = 0; x 2 @ =: :

(4:8)

where

1

2 R

2

is, for the sake of simplicity, a bounded polygonal domain and f 2 L

2

():

Furthermore, also for the sake of simplicity, we assume that the boundary is compatible

to triangulated rectangular (not necessary uniform) mesh.

Remark 4.3 The ctitious domain method can be applied to problems with more general

kind of domains. See, for example [32].

Instead of considering only the equation (4.8), we do the following: rst, we embed the

domain

1

into a suciently large rectangle : After that we solve together two equations,

namely

(

u

1

= f; x 2

1

;

u

1

= u

2

; x 2 ;

(4:9)

and

8

>

<

>

:

u

2

= 0; x 2 n

1

=:

2

;

u

2

= 0; x 2 @;

@u

2

@n

= 0; x 2 :

(4:10)

It follows, that u

2

= 0 is the unique solution of (4.10). Thus, the equation (4.9) is

equivalent to the original equation (4.8).

The discretization is carried out as follows. First we dene a rectangular mesh T

h

on

rectangle ; which after suitable triangulation coincides the boundary : As an example, the

gure 4.2. illustrates one such mesh for the domain of the gure 4.1. This approach gives us

the triangulations of

1

(denoted further on by T

1

h

) and

2

:= n (denoted by T

2

h

). Next

we dene the weak formulations for the problems (4.9) and (4.10). The weak formulation for

the problem (4.10) reads: Find u

2

2 H

1

(

2

) V

2

such that

Z

2

ru

2

rv dx = 0; 8v 2 V

2

; (4:11)

13

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Π

Ω

∂∂u

n=0

u1=u2

2

u2=0

−∆u1=f

−∆u2=0

Fig. 4.1. A polygonal domain embedded to a rectangle

and the equivalent extended problem.

Fig. 4.2. A compatible triangulated rectangular mesh.

14

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and for the problem (4.9): Find u

1

2

~

V

1;u

2

; where V

1;u

2

= fv 2 H

1

(

1

); vj

= u

2

j

; vj

@

= 0g

such that:

Z

1

ru

1

rv dx =

Z

1

fv dx; 8v 2 V

1

H

1

0

(

1

): (4:12)

We dene the corresponding spaces of nite element functions by setting

V

h

2

= fv 2 V

2

j v is linear in each triangle of T

2

h

g;

V

h

1

= fv 2 V

1

j v is linear in each triangle of T

1

h

g;

~

V

h

1;u

2

= fv 2

~

V

1;u

2

j v is linear in each triangle of T

1

h

g;

and

V

h

= fv 2 H

1

() j vj

1

2

~

V

h

1;u

2

; vj

2

2 V

h

2

g:

By using these triangulations and nite element spaces we obtain an algebraic system for

problems (4.9) and (4.10). This system is

Au =

2

6

6

6

6

6

4

A

11

A

1

0 0 0

A

1

A

A

0 0

0 0 A

(2)

A

+

0

0 0 A

+

A

+ +

A

+2

0 0 0 A

2 +

A

22

3

7

7

7

7

7

5

2

6

6

6

6

6

4

u

1

u

u

u

+

u

2

3

7

7

7

7

7

5

=

2

6

6

6

6

6

4

f

1

f

0

0

0

3

7

7

7

7

7

5

: (4:13)

The block

"

A

11

A

1

0

A

1

A

A

#

corresponds to the Dirichlet boundary value problem (4.9) and the block

2

6

4

A

(2)

A

+

0

A

+

A

+ +

A

+2

0 A

2 +

A

22

3

7

5

to the extended Neumann boundary value problem (4.10). The system (4.13) is called non-

symmetric ctitious domain algebraic system. Here denotes the grid boundary between

1

and

2

; denotes the grid line next to in

1

; and + the grid line next to in

2

: For

example, the block A

is computed by

A

(i; j) =

Z

1

r

i

r

j

dx;

where

i

is the basis function corresponding to the node i in and

j

is the basis function

corresponding to the node j in :

We wish to solve (4.13) by using preconditioned conjugate gradiend method. As a pre-

conditioner we use the matrix B arising from the nite element discretization of the Poisson

equation with Dirichlet boundary conditions on the rectangle by using the triangulated

rectangular mesh T

h

: It follows that the preconditioner B is of the form

B =

2

6

6

6

6

6

4

A

11

A

1

0 0 0

A

1

A

A

0 0

0 A

A

A

+

0

0 0 A

+

A

+ +

A

+2

0 0 0 A

2 +

A

22

3

7

7

7

7

7

5

:

15

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It should be noted that the system to be solved is a nonsymmetric one and thus, the use of

the preconditioned conjugate gradient method must be justied. It appears that if we solve

for the initial guess the system

Bu

0

= f; (4:14)

where

f =

h

f

t

1

f

t

0

t

0

t

0

t

i

t

;

the residual

0

= Au

0

f will have a special form. Since the matrices A and B have the

same rst two and last two block rows, it follows that blocks

0

1

;

0

;

+

and

0

2

will be zero

blocks. So, the rst residual

0

2 im(C) U

A

; where C = B A and U

A

is a subspace of

R

N

consisting of vectors x 2 R

N

; which have zero blocks x

1

; x

; x

+

and x

2

:

U

A

R

N

; U

A

= fx 2 R

N

j x = [0

t

0

t

x

t

0

t

0

t

]

t

g: (4:15)

The error vector z

0

= u

0

u

; where u

= A

1

f belongs to the subspace U = A

1

U

A

R

N

:

This subspace has the representation

U = f 2 R

N

jA

11

1

+ A

1

= 0; A

1

1

+ A

+ A

= 0g (4:16)

Now it is easy to show that the matrix A is symmetric and positive denite with respect to

the subspace U i. e., (Ax; y) = (x;Ay); x; y 2 U

A

and x

t

Ax > 0; x 2 U n f0g: Also, the

matrix A

1

is s.p.d. in the subspace U

A

: Then, it follows that B

1

A is A s.p.d. in U and

AB

1

is A

1

s.p.d. in U

A

:

It is easy to see, that the subspaces U

A

and U are invariant i. e., AB

1

U

A

U

A

and

B

1

AU U: Thus, it follows that all the residuals

k

= Au

k

f belong to the subspace U

A

and the error vectors z

k

= u

k

u

belong to the subspace U during the conjugate gradient

iteration, if the rst residual belongs to U

A

: Therefore, we can apply the preconditioned

conjugate gradient method to system (4.13) if the initial guess is chosen by (4.14).

The following Theorem can be proved:

Theorem 4.1 The condition number Condj

U

A

;A

(B

1

A) is bounded from above with a positive

constant which is independend of the mesh step size h:

This Theorem states that the number of iterations needed to reach a prescribed accuracy is

bounded.

The next preconditioned conjugate gradient algorithm solves the Dirichlet boundary value

problem with the nonsymmetric ctitious domain method. In the following algorithm the

subscript [ means both the and blocks of a vector.

1. Entrance to the subspace. Solve Bu = f:

2. Preconditioned conjugate gradient iteration.

(a) First iteration.

= (Au f)

[

= (B

1

[0 0

T

0 0]

T

)

[

w

[

=

[

=

T

v

=

^

A

+ (A

(2)

^

A

)

16

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(b) Next iteration step.

=

w

T

v

u

= u

w

=

v

If j

j > " then

[

= (B

1

[0 0

T

0 0]

T

)

[

old

=

=

T

=

old

w

[

=

[

+ w

[

v

=

^

A

+ (A

(2)

^

A

)

+ v

Goto 2.(b)

Endif

3. Exit from the subspace. Solve Bu = g; where

g =

0

B

B

B

B

B

B

@

f

1

f

+

^

A

u

0

0

1

C

C

C

C

C

C

A

:

Theorem 4.2 It takes O(N logN) arithmetical operations to solve the system (4.13) by using

the previous algorithm.

Proof: In steps 1 and 3 we have to solve a linear system with coecient matrix B: Since

the mesh T

h

is a rectangular mesh it follows that the matrix B has a separable structure

and we can apply the method of separation of variables. Thus, the arithmetical complexity

of the steps 1 and 3 is of order O(N logN): Since the condition number of the matrix B

1

A

is bounded from above with a constant independent of the mesh size h (and simultaniously

of N , h

1

p

N

), the number of preconditioned conjugate gradient iterations needed to reach

a prescribed accuracy " is bounded (it depends only on the domain

1

). In each iteration

we need to solve systems of the form B = : However, the residual vector is sparse (it is

nonzero only on ) and we need the solution only on the grid curves and : Thus, we

can apply the partial solution technique to solve these systems. Therefore, the arithmetical

complexity of each iteration step and the whole iteration 2 is of order O(N log

2

"

): So, the

asympthotical complexity is O(N logN):

5 The Lanczos method of minimized iterations with a block-

diagonal preconditioner

As is known, the preconditioned Lanczos method of minimized iterations with a symmetric

positive denite preconditioner can be used for solving algebraic systems with symmetric

indenite matrices. In this section we propose a method of constructing such a preconditioner

for solving problem (1.1).

17

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5.1 Block-diagonal preconditioner

The idea of a block-diagonal preconditioner for the Lanczos method is based on the following

simple statement which can be easily veried. The eigenvalue problem

"

A B

B

T

0

# "

w

p

#

=

"

A 0

0 B

T

A

1

B

# "

w

p

#

; p ? kerB; p 6= 0

has only two dierent eigenvalues:

1

p

5

2

and

1+

p

5

2

; besides, if p 2 kerB, there will be an

eigenvalue = 1. This means that if we apply the generalized Lanczos method of minimized

iterations with the preconditioner

"

A 0

0 B

T

A

1

B

#

to problem (1.1) the method will converge

to the exact solution in three iterations. However, the solution of system with the matrix

"

A 0

0 B

T

A

1

B

#

presents considerable diculties. Therefore, instead of this matrix, we

suggest to choose, as a preconditioner, a certain \perturbed" matrix

~

G =

"

~

A 0

0

~

S

#

, where

~

A =

~

A

T

> 0;

~

S =

~

S

T

> 0. Thus the matrix

~

G is symmetric and positive denite.

The fact that the matrix

~

G is an ecient preconditioner for the matrixG in the generalized

Lanczos method of minimized iterations is justied in the following theorem:

Theorem 5.1 Suppose that A = A

T

> 0;

~

A =

~

A

T

> 0;

~

S =

~

S

T

> 0 and there exist

such positive constants

1

;

2

;

1

;

2

independent of the grid step h that the following two

conditions are satised:

1

(Aw; w)

(

~

Aw; w)

2

; 8w 2 R

2N

v

; w 6= 0; (5.1)

1

(Sp; p)

(

~

Sp; p)

2

; 8p 2 R

N

p

; p ? kerB; (5.2)

where S = B

T

A

1

B.

Then the spectrum of the generalized eigenvalue problem

G

"

w

p

#

=

~

G

"

w

p

#

; p ? kerB (5:3)

belongs to the set

2

4

1

+

q

2

1

+ 4

1

1

2

;

2

+

q

2

2

+ 4

2

2

2

3

5

S

2

4

2

q

2

2

+ 4

2

2

2

;

1

q

2

1

+ 4

1

1

2

3

5

S

[

1

;

2

] ;

where the bounds of each of these segments do not depend on h.

Proof. Let us rewrite problem (5.3) as

(

Aw + Bp =

~

Aw

B

T

p =

~

Sp

18

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From the rst equation we obtain (

~

A A)w = Bp.

(I) First let us examine the case >

2

.

Here the matrix

~

A A is positive denite, and consequently we may write w = (

~

A

A)

1

Bp. Therefore, from the second equation we have

B

T

(

~

A A)

1

Bp =

~

Sp: (5:4)

Taking the scalar product of (5.4) by p, we obtain

(

~

A A)

1

Bp; Bp

=

~

Sp; p

:

From condition (5.2) we nd

2

A

1

Bp; Bp

(

~

A A)

1

Bp; Bp

1

A

1

Bp; Bp

:

Let us denote Bp by w

1

. Then we have

2

(

~

A A)

1

w

1

; w

1

(A

1

w

1

; w

1

)

1

:

Designating

(

(

~

AA)

1

w

1

; w

1)

(A

1

w

1

; w

1

)

= we get

2

1

: (5:5)

On the other hand, considering the generalized eigenvalue problem

(

~

A A)

1

w

1

= A

1

w

1

and using condition (5.1) it is easy to obtain inequalities:

min

max

; (5:6)

where

min

=

1

1

;

max

=

2

2

.

From (5.5) and (5.6) it follows that should satisfy the following system of inequalities:

8

>

>

<

>

>

:

1

1

2

2

2

1

:

(5:7)

But system (5.7) has a solution if and only if

1

1

2

2

2

; (5:8)

or

1

1

1

2

2

: (5:9)

19

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Taking into account that >

2

, from inequalities (5.8) we obtain

2

2

4

max

0

@

1

+

q

2

1

+ 4

1

2

2

;

2

1

A

;

2

+

q

2

2

+ 4

2

2

2

3

5

:

Similarly, inequalities (5.9) lead to

2

2

4

max

0

@

1

+

q

2

1

+ 4

1

1

2

;

2

1

A

;

2

+

q

2

2

+ 4

2

1

2

3

5

:

Finally we have:

2

2

4

max

0

@

1

+

q

2

1

+ 4

1

1

2

;

2

1

A

;

2

+

q

2

2

+ 4

2

2

2

3

5

:

(II) Now consider the case <

1

.

Here the matrix

~

A A is negative denite, and similar to case (I) we have

B

T

(

~

A A)

1

Bp =

~

Sp:

Further,

(

~

A A)

1

Bp; Bp

=

~

Sp; p

:

Thus, it is easy to see that < 0. From condition (5.2) we obtain

2

(A

~

A)

1

w

1

; w

1

(A

1

w

1

; w

1

)

1

:

Using the same considerations as in the case >

2

, we nally obtain:

2

2

4

2

q

2

2

+ 4

2

2

2

;

1

q

2

1

+ 4

1

1

2

3

5

:

Adding the segment [

1

;

2

] to the set of possible values of we arrive at the required

conclusion.

Now a question arises: how to choose matrices

~

S and

~

A satisfying conditions (5.1), (5.2)?

The choice of

~

S is obvious:

~

S = I; (5:10)

then it follows from Theorem 3.1 that condition (5.2) is satised with the constants

1

= c

1

,

2

= 1, where c

1

is that of (3.1). Here we suggest to choose a matrix

~

A as follows:

~

A = A

2

4

I

r

Y

j=1

I

j

~

K

1

A

3

5

1

; (5:11)

20

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where

j

are the Chebyshev iterative parameters, and

~

K is some matrix. It can be easily

shown that if

~

K satises the inequality

cond (

~

K

1

A) (5:12)

with independent of h then condition (5.1) will be satised with

1

= 1 ;

2

= 1 + ;

where

=

2q

r

1 + q

2r

; q =

p

1

p

+ 1

:

In the next section we shall consider the construction of matrix

~

K based on the MGDD

method.

5.2 Multigrid domain decomposition method

In this section the domain is supposed to be a union of M

0

> 1 squares, whose sides are

parallel to the coordinate axes, i. e.,

=

M

0

S

i=1

!

(i)

; here !

(i)

are open squares which do not

intersect, and @!

(i)

denotes their boundaries, i = 1; : : : ; M

0

. We dene the basic grid at the

0-th level as

0

=

M

0

S

i=1

@!

(i)

. The grid

0

is rectangular and uniform; its step h

0

= 1. We

shall choose a positive integer t 1 and, for l = 1; : : : ; t, construct a hierarchical sequence

of rectangular uniform basic grids

1

; : : : ;

t

such that a step of the grid

l1

contains three

steps of

l

: h

l

=

1

3

h

l1

= 3

l

; l = 1; : : : ; t. For each basic grid

l

we construct three

staggered grids

u

l

;

v

l

;

p

l

, as in Section 3.1. Thus, we obtain a hierarchical sequence of

staggered grids in the domain (see Fig. 5.1).

We shall assume that problem (1.1) is a discretization of problem (2.1), (2.2) on the grids

u

t

;

v

t

;

p

t

. The matrix A in (1.1) can be represented as

A =

1

h

2

t

"

A

u

0

0 A

v

#

; (5:13)

where

1

h

2

t

A

u

is the dierence analogue of the operator on the grid

u

t

, and

1

h

2

t

A

v

is the

dierence analogue of on the grid

v

t

. Accordingly, the matrix

~

K will be constructed in

the form

~

K =

1

h

2

t

"

~

K

u

0

0

~

K

v

#

: (5:14)

Further we shall consider construction of the matrix

~

K

u

. The matrix

~

K

v

can be constructed

similarly.

Let us now consider the grid

u

l

; 1 l t, in more detail. We partition the nodes of

the grid

u

l

into three groups (Fig. 5.2). Those nodes of the grid

u

l

which are also nodes of

u

l1

will form the rst group. The second group will consist of the nodes of

u

l

which belong

to the edges of

u

l1

and do not belong to the rst group. The third group will contain the

nodes of

u

l

which are inside the cells of

u

l1

.

21

Page 22: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

g g

g g

g g

w w

w w

s

c

s

c

s

c

s

c

s

c

s

c

c

s

c

s

c

s

c

s

c

s

c

s

c

c

s

c

s

c

s

c

s

c

s

c

s

c

c

s

c

s

c

s

c

s

c

s

c

s

c

c

c

c

c

c

c

c

c

c

s

s

s

s

s

s

s

s

Fig. 5.1. The hierarchical sequence of staggered grids.

nodes of the grid

u

l

nodes of the grid

v

l

nodes of the grid

p

l

nodes of the grids

u

l1

;

v

l1

;

p

l1

We also dene a sequence of grids

u

l1=2

; l = 1; : : : ; t, which are obtained by eliminating

the nodes of the third group (with the corresponding edges) from grids

u

l

; l = 1; : : : ; t.

For the grids

u

t

and

u

t1=2

we also dene the superelements

Z

(j)

t

= !

(j)

u;t1

\

u

t

; Z

(j)

t1=2

= !

(j)

u;t1

\

u

t1=2

;

here !

(j)

u;t1

is a cell of the grid

u

t1

.

We dene a space V

(l)

h

as a set of functions which are dened on

u

l

, linear on each edge

of

u

l

and vanishing on @. Similarly, V

(l1=2)

h

is dened as a set of functions on

u

l1=2

which

are linear on each edge of

u

l1=2

and vanish on @.

According to the above node groups for

u

t

, we represent the matrix A

u

in the block-

tridiagonal form:

A

u

=

2

6

4

A

u;11

A

u;12

0

A

u;21

A

u;22

A

u;23

0 A

u;32

A

u;33

3

7

5

: (5:15)

The following statement can be directly checked.

Lemma 5.1 The matrix A

u

in (5.15) can be dened by the relation

(A

u

u; w) =

h

t

2

X

i

Z

@!

(i)

u;t

du

h

ds

dw

h

ds

ds; (5:16)

which is supposed to be true for all vectors u; v.

22

Page 23: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

Fig. 5.2. Grid

u

1

.

nodes of the 1st group

nodes of the 2nd group

nodes of the 3rd group

The sum in (5.16) is taken for all cells !

(i)

u;t

of the grid

u

t

; u

h

and w

h

denote the extensions

of vectors u and w onto the space V

(t)

h

.

Now we dene a matrix

^

K

u;11

by the equality

(

^

K

u;11

u; w) =

h

t

2

X

i

Z

@!

(i)

u;t1

du

h

ds

dw

h

ds

ds; (5:17)

which is supposed to be true for all u, v. Here the sum is taken for all cells of the grid

u

t1

;

u

h

and w

h

are the extensions of vectors u and w onto the space V

(t1=2)

h

. It can be shown

that the matrix

^

K

u;11

admits the following block representation:

^

K

u;11

=

"

A

u;11

A

u;12

A

u;21

K

u;22

#

: (5:18)

We also introduce a symmetric matrix

K

u

=

2

6

4

A

u;11

A

u;12

0

A

u;21

K

u;22

+ A

u;23

A

1

u;33

A

u;32

A

u;23

0 A

u;32

A

u;33

3

7

5

: (5:19)

It can be readily veried that the following statement is valid.

Lemma 5.2

(K

u;22

w; w)

(A

u;22

A

u;32

A

1

u;33

A

u;32

)w; w

; 8w:

Let us now examine the generalized eigenvalue problem

K

u

u = A

u

u: (5:20)

Using Lemma 5.2, we can easily prove the following fact.

23

Page 24: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

Lemma 5.3 The eigenvalues of problem (5.20) belong to the segment [1; c], where c =

(K

1

u

A

u

).

It was shown in [23] that the constant c in Lemma 5.3 is not greater than the largest

eigenvalue of the spectral problem

(j)

K

(j)

u

= A

(j)

u

u

(j)

: (5:21)

Here the matrix A

(j)

u

is a restriction of A

u

on an arbitrary internal superelement Z

(j)

t

; it is

dened by the following relation:

(A

(j)

u

u

j

; w

j

) =

h

t

2

Z

Z

(j)

t

du

h

j

ds

dw

h

j

ds

ds; 8u

h

; w

h

2 V

(t)

h

; (5:22)

where u

h

j

and w

h

j

are the restrictions of functions u

h

and w

h

, on Z

(j)

t

, u

j

and w

j

are vectors

corresponding to the functions u

h

j

and w

h

j

, and

K

(j)

u

=

"

^

K

(j)

u;11

0

0 0

#

;

^

K

(j)

u;11

is dened by the relation

(

^

K

(j)

u

u

j

; w

j

) =

h

t

2

Z

Z

(j)

t1=2

du

h

j

ds

dw

h

j

ds

ds; 8u

h

; w

h

2 V

(t1=2)

h

; (5:23)

where u

h

j

and w

h

j

are the restrictions of functions u

h

and w

h

, on a superelement Z

(j)

t1=2

, u

j

and w

j

are vectors corresponding to the functions u

h

j

and w

h

j

.

For an internal superelement Z

(j)

t

we write down 1616 matrices A

(j)

u

and K

(j)

u

using the

enumeration of the nodes given in Fig. 5.3.

2 7 8 3

1 9 10 4

12

11

6

5

15

13

16

14

Fig. 5.3. The enumeration of grid nodes for the superelement Z

(j)

t

.

24

Page 25: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

A

(j)

u

=

1

2

2

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

4

2 0 0 0 0 -1 0 0 -1 0 0 0 0 0 0 0

0 2 0 0 -1 0 -1 0 0 0 0 0 0 0 0 0

0 0 2 0 0 0 0 -1 0 0 -1 0 0 0 0 0

0 0 0 2 0 0 0 0 0 -1 0 -1 0 0 0 0

0 -1 0 0 4 -1 0 0 0 0 0 0 -2 0 0 0

-1 0 0 0 -1 4 0 0 0 0 0 0 0 0 -2 0

0 -1 0 0 0 0 4 -1 0 0 0 0 -2 0 0 0

0 0 -1 0 0 0 -1 4 0 0 0 0 0 -2 0 0

-1 0 0 0 0 0 0 0 4 -1 0 0 0 0 -2 0

0 0 0 -1 0 0 0 0 -1 4 0 0 0 0 0 -2

0 0 -1 0 0 0 0 0 0 0 4 -1 0 -2 0 0

0 0 0 -1 0 0 0 0 0 0 -1 4 0 0 0 -2

0 0 0 0 -2 0 -2 0 0 0 0 0 8 -2 -2 0

0 0 0 0 0 0 0 -2 0 0 -2 0 -2 8 0 -2

0 0 0 0 0 -2 0 0 -2 0 0 0 -2 0 8 -2

0 0 0 0 0 0 0 0 0 -2 0 -2 0 -2 -2 8

3

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

5

(5.24)

K

(j)

u

=

1

2

2

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

6

4

2 0 0 0 0 -1 0 0 -1 0 0 0 0 0 0 0

0 2 0 0 -1 0 -1 0 0 0 0 0 0 0 0 0

0 0 2 0 0 0 0 -1 0 0 -1 0 0 0 0 0

0 0 0 2 0 0 0 0 0 -1 0 -1 0 0 0 0

0 -1 0 0 2 -1 0 0 0 0 0 0 0 0 0 0

-1 0 0 0 -1 2 0 0 0 0 0 0 0 0 0 0

0 -1 0 0 0 0 2 -1 0 0 0 0 0 0 0 0

0 0 -1 0 0 0 -1 2 0 0 0 0 0 0 0 0

-1 0 0 0 0 0 0 0 2 -1 0 0 0 0 0 0

0 0 0 -1 0 0 0 0 -1 2 0 0 0 0 0 0

0 0 -1 0 0 0 0 0 0 0 2 -1 0 0 0 0

0 0 0 -1 0 0 0 0 0 0 -1 2 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

3

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

7

5

(5.25)

Direct calculation of all the eigenvalues in problem (5.21) with the matrices (5.24) and

(5.25) gives the following result.

Lemma 5.4 In problem (5.21)

(j)

max

= 3 +

p

2, and, consequently, the eigenvalues of the

matrix K

1

u

A

u

belong to the segment [1; 3 +

p

2].

Introduce the matrix

K

u;11

= A

u;11

A

u;12

K

1

u;22

A

u;21

(5:26)

and denote K

u;33

= A

u;33

. Then the matrix K

u

in (5.19) may be rewritten as

K

u

=

2

6

6

4

K

u;11

+ A

u;12

K

1

u;22

A

u;21

A

u;12

0

A

u;21

K

u;22

+ A

u;23

K

1

u;33

A

u;32

A

u;23

0 A

u;32

K

u;33

3

7

7

5

=

F

T

u

2

6

6

4

K

u;11

0 0

0 K

u;22

0

0 0 K

u;33

3

7

7

5

F

u

; (5.27)

where

F

u

=

2

6

6

4

I

11

0 0

K

1

u;22

A

u;21

I

22

0

0 K

1

u;33

A

u;32

I

33

3

7

7

5

: (5:28)

25

Page 26: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

Note that K

u;22

is a block-diagonal matrix with blocks 2 2, and K

u;33

is a block-diagonal

matrix with blocks 4 4.

The following fundamental result is of major importance for further considerations. It

may be checked directly.

Lemma 5.5

K

u;11

=

1

3

A

u;t1

:

Here the matrix

1

h

2

t1

A

u;t1

is the dierence analogue of the operator on the grid

u

t1

.

In view of this, the matrix K

u

may be called the two-grid preconditioner for the matrix A

u

.

Then, similar to [23], we shall construct a multigrid preconditioner for the matrix A

u

.

For l = 1; : : : ; t we introduce the matrices

A

u;l

=

2

6

6

4

A

(l)

u;11

A

(l)

u;12

0

A

(l)

u;21

A

(l)

u;22

A

(l)

u;23

0 A

(l)

u;32

A

(l)

u;33

3

7

7

5

; (5:29)

so that matrices

1

h

2

l

A

u;l

are the dierence analogues of the operator on the grids

u

l

.

Using (5.17), (5.18), (5.26), (5.27) and (5.28) we construct the corresponding two-grid pre-

conditioners

K

u;l

= F

T

u;l

2

6

6

4

K

(l)

u;11

0 0

0 K

(l)

u;22

0

0 0 K

(l)

u;33

3

7

7

5

F

u;l

; (5:30)

where K

(l)

u;11

=

1

3

A

u;l1

and K

(l)

u;33

= A

(l)

u;33

. Thus A

u;t

A

u

and K

u;t

K

u

. Then we choose

a positive integer s 1 and let H

(2)

u;11

=

~

K

1

u;1

K

1

u;1

. For l = 2; : : : ; t we construct the

sequence of matrices

R

(l)

u;11

=

2

4

I

(l)

11

s

Y

j=1

I

(l)

11

(l)

j

H

(l)

u;11

A

u;l1

3

5

A

1

u;l1

; (5.31)

~

K

(l)

u;11

=

1

3

h

R

(l)

u;11

i

1

; (5.32)

~

K

u;l

= F

T

u;l

2

6

6

4

~

K

(l)

u;11

0 0

0 K

(l)

u;22

0

0 0 K

(l)

u;33

3

7

7

5

F

u;l

; H

(l+1)

u;11

=

~

K

1

u;l

: (5.33)

From Lemma 5.4 it follows that eigenvalues of the matrices K

1

u;l

A

u;l

belong to the segment

[1; 3 +

p

2]. In particular, the eigenvalues of the matrix H

(2)

u;11

A

u;1

K

1

u;1

A

u;1

belong to this

segment. Let us take an integer l 2 and assume that eigenvalues of the matrix H

(l)

u;11

A

u;l1

belong to the segment [

l1

;

l1

]; 0 <

l1

<

l1

, where

1

= 1 and

1

= 3 +

p

2. Then,

if we take the parameters

(l)

j

; j = 1; : : : ; s in (5.31) to be reciprocal to the roots of the

Chebyshev polynomial

C

(l)

s

(z) = C

s

l1

+

l1

2z

l1

l1

!

;

26

Page 27: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

we shall nd that the eigenvalues of spectral problem

~

K

(l)

u;11

w = K

(l)

u;11

w

belong to the segment [1

l

; 1 +

l

], where

l

=

2q

s

1 + q

2s

; q

l

=

p

l1

1

p

l1

+ 1

;

l1

=

l1

l1

:

Consequently, the eigenvalues of the matrix

~

K

1

u;l

A

u;l

belong to the segment [

l

;

l

], where

l

= 1

l

and

l

= (3 +

p

2)(1 +

l

). The matrix

~

K

u

~

K

u;t

is the MGDD-preconditioner

for the matrix A

u

. The above facts lead to the following conclusion.

Lemma 5.6

cond

~

K

1

u

A

u

t

; (5:34)

where

1

= 3 +

p

2 and

l

= (3 +

p

2)

"

(

p

l1

+ 1)

s

+ (

p

l1

1)

s

(

p

l1

+ 1)

s

(

p

l1

1)

s

#

2

; l = 2; : : : ; t: (5:35)

We shall consider the case s = 3. From (5.35) we obtain

l

= (3 +

p

2)

l1

"

l1

+ 3

3

l1

+ 1

#

2

; l = 2; : : : ; t: (5:36)

Analysis of the function '(x) = (3+

p

2)x

x + 3

3x+ 1

2

shows that f

l

g in (5.36) is a monotone

non-decreasing sequence for all

1

> 1. Besides, for all

1

2 (1;

max

], the quantities

l

in

(5.36) also belong to the interval (1;

max

]. Here

max

=

1

17

4(3 +

p

2)

1=2

(6 +

p

2) + 12

p

2 + 21

5:9

is the largest root of the equation

(3 +

p

2)

"

+ 3

3 + 1

#

2

= 1:

Thus we have proved the following statement.

Lemma 5.7 The estimate

cond

~

K

1

u

A

u

< 5:9

is valid for s = 3.

Other values of s may be considered similarly.

Absolutely the same considerations can be applied to the construction of the matrix

~

K

v

.

Therefore, we arrive at the following result.

Theorem 5.2

cond

~

K

1

A

t

;

where

t

is taken as in Lemma 5.6. The value of

t

is independent of the shape of the domain

.

Remark 5.1 For s 3 the condition number of the matrix

~

K

1

A has an upper bound which

is independent of the grid step h (number of levels t) and the shape of the domain . Thus,

the above matrix

~

K

1

can be used for constructing the matrix

~

A in (5.11).

27

Page 28: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

5.3 The Lanczos method of minimized iterations

As noted above, in problem (1.1) the matrix G is symmetric and the matrix

~

G is symmetric

and positive denite. This fact allows us to use the Lanczos method of minimized iterations

[12, 27, 28, 29]:

q

i

=

8

>

>

<

>

>

:

~

G

1

0

; i = 1

~

G

1

Gq

1

2

q

1

; i = 2

~

G

1

Gq

i1

i

q

i1

i

q

i2

; i 2

i

=

kGq

i1

k

2

~

G

1

G

~

G

1

kGq

i1

k

2

~

G

1

;

i

=

kGq

i1

k

2

~

G

1

kGq

i2

k

2

~

G

1

; (5.37)

i

=

(

i1

; Gq

i

)

~

G

1

kGq

i

k

2

~

G

1

;

"

w

i

p

i

#

=

"

w

i1

p

i1

#

i

q

i

;

i

= G

"

w

i

p

i

#

"

f

g

#

:

The convergence rate of this method is estimated in the following lemma [37].

Lemma 5.8 The estimate

k

k

k

~

G

1

C

[k=2]

~

M

2

+ ~m

2

~

M

2

~m

2

!!

1

k

0

k

~

G

1

"k

0

k

~

G

1

; (5:38)

is valid for method (5.37). Here C

[k=2]

is the Chebyshev polynomial of degree [k=2],

~

M =

(

~

G

1

G), and 1= ~m = (

~

GG

+

).

Note that according to Theorem 5.1 the numbers

~

M and ~m are independent of h. There-

fore, we arrive at the following important result.

Theorem 5.3 Iterative method (5.37) with block-diagonal preconditioner (1.2) constructed

by using the MGDD technique and (5.10) is optimal in the sense that the computational

cost of solving problem (1.1) by this method with the accuracy " according to (5.38) can be

estimated from above by cN

un

ln "

1

, where c is a constant independent of N

un

(the number

of unknowns).

6 Numerical experiments

The aim of the numerical experiments was to investigate the rate of convergence of Lanczos

method of minimized iterations with block-diagonal preconditioner and Uzawa's method as

a function of grid step and the domain shape. Besides, it was interesting to compare the

CPU times used by these two methods in identical domains on grids with equal number of

nodes. All the computations were made in double precision using computer HP 9000/720

with performance of 17 MFlops.

We have considered a model problem in \inverse stair" domain. It is shown on Fig. 6.1.

28

Page 29: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

-

6

- -

6

?

6

?

u = 1

d

D

U < 1

y

x

0

L

2

L

1

h

Fig. 6.1. Model region.

In all these experiments the boundary conditions were taken as

x = L

1

:

u = 1

y (h+ d=2)

d=2

!

2

;

v = 0;

x = L

2

;

u =

0

@

1

y (D=2)

D=2

!

2

1

A

U; U = d=D;

v = 0:

(the Poiseuille ow). On solid borders we take boundary condition of zero velocity:

u = 0; v = 0:

In order to describe the domain shape, hereafter in this section we use the designations of

Fig. 6.1.

The rst group of experiments was carried out using Lanczos method of minimized it-

erations with block-diagonal preconditioner. As a model domains, we used the L-shaped

domains with dierent parameters:

1

= [4; 12] [0; 4] n [4; 0] [0; 2] (d = 2; D = 4; L

1

= 4; L

2

= 12)

2

= [6; 4] [0; 3] n [6; 0] [0; 1] (d = 2; D = 3; L

1

= 6; L

2

= 4)

First, we experimentally determined the optimal parameters s (number of Chebyshev

iterations in MGDD, see (5.31)) and r (number of Chebyshev iterations for constructing the

matrix

~

A in (5.11). Tables 1 and 2 shows the results of these experiments.

s = 3 s = 4

r N

it

T ime N

it

T ime

1 81 1620 81 1800

2 52 1695 51 1886

3 43 1988 42 2219

s = 3 s = 4

r N

it

T ime N

it

T ime

1 103 574 103 638

2 68 608 67 664

3 58 734 55 766

Table 1:

1

; n

x

= 1296; n

y

= 324 Table 2:

2

; n

x

= 810; n

y

= 243

29

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Hereafter n

x

denotes the maximal number of steps of basic grid in the x-direction, and n

y

in the y-direction; N

it

is the number of outer iterations of generalized Lanczos method of

minimized iterations (the stopping criterion was a reduction in the residual by a factor of

10

6

: k

N

it

k 10

6

k

0

k), T ime is the total CPU time (in seconds). As is seen from Tables 1

and 2 s = 3 and r = 1 are optimal parameters.

Second, we made the computations for various dimensions of the algebraic problem (num-

ber of levels in MGDD method). The results are given in Tables 3, 4.

l n

x

n

y

N

it

T ime

2 48 12 90 1.5

3 144 36 87 15.6

4 432 108 83 157.0

5 1296 324 81 1619.7

l n

x

n

y

N

it

T ime

2 30 9 98 0.5

3 90 27 106 3.6

4 270 81 105 46.3

5 810 243 103 573.8

Table 3:

1

; s = 3; r = 1 Table 4:

2

; s = 3; r = 1

Here l denotes the number of levels. We used optimal iterative parameters. Tables 3, 4

shows that the number of outer iterations is practically independent of the dimension of the

problem.

The second group of experiments was carried out using Usawa's method. The experimen-

tations with the Uzawa's algorithm were done on the model region

1

: The stopping criterion

for the Uzawa's iteration was kresidualk

1

10

6

: The stopping criteria for the ctitious do-

main iteration was kresidualk

1

10

12

: The results of the numerical experiments are listed

in the table 5. The rst column gives the number of velocity nodes in x and ydirection.

The number of Uzawa iterations is listed in the second column. The CPU-time for solving

the equation is given in the third column.

n

x

n

y

N

it

T ime

33 9 36 0.7

65 17 36 2.1

129 33 35 7.4

257 65 34 28.0

513 129 32 110.7

1025 257 30 464.9

2049 513 24 1397.0

Table 5. Uzawa's algorithm

with ctitious domain

Notice, that the iteration count is bounded as could be expected by Lemma 4.1. The

CPU-time usage is plotted in gure 6.2 versus the number of velocity nodes. It appears, that

the Uzawa's algorithm is about two times faster that the Lanczos method in this example.

30

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Uzawa

Lanczos

CPU seconds

nodes0.5

1

2

5

10

20

50

100

200

500

1000

2000

300 1000 3000 10000 30000 100000 300000 1000000

Fig. 6.2. The CPU-time usage versus the number of velocity nodes

31

Page 32: Iliash Rossi Jari anen oiv T - Jyväskylän yliopistousers.jyu.fi/~tene/papers/report93-2.pdf · Rossi Jari anen oiv T. e Iterativ Metho ds for Solving the es Stok Problem uri Y Iliash

7 Acknowledgements

The authors want to acknowledge Prof. Yuri Kuznetsov, Prof. Pekka Neittaanmaki and

Prof. Jacques Periaux for their many valuable advice and comments concerning this work.

Yuri Iliash is also grateful to Alexei Kourbatov for his help in translating his part of the

report from Russian.

This work has been nancially supported by the Council of Technology of the Academy

of Finland and Dassault Aviation, France.

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