Wavelets As Galerkin Basis

Aleksandr YakovlevDepartment of Computational Physics

St Petersburg State University

Contents

1. Introduction

2. Galerkin Method

3. Multiresolution analysis

4. Multiresolution analysis and Galerkin Method

5. Solution for an example equations

5.1 Finite differences method

5.2 Wavelet Galerkin solution

5.3 Incorporation of boundary conditions

5.4 Offsetting boundary conditions to control error

5.5 Comparison of results

Introduction

1. Differential equations ODE PDE constant coefficients variable coefficients

2. Domain and boundary conditions Dirichlet Neuman Cyclic

3. Method Galerkin method

4. Improvements Use Wavelet basis Increase resolution Increase order

The Galerkin Method

1

01

0 1

Equation on ( , ), ( ) 0 on

Basis ( , )

Decomposition ( , ) ( , ) ( , ) ( , )

Define [ ( , ) ] [ ( , )] ( , ..., , , )

Re

, 0

u

q

N

i i

N

a i ii

i i Ni

D x y S u D

g x y

u x y u x y u x y a g x y

L a g x y L g x y R a

L u x

a x

y

y

21

11

1 1 1

1

ire

( , ..., , , ), ( , ) 0, 1, 2, ...,

Substitution

( , ..., , , ), [ ( , )], ( , ) [ ( , )], ( , ) 0

Matrix equation

[ ], [ ],

[ ], [ ],

N i L

N

N i j j i a ij

N

N N N

R a a x y g x y i N

R a a x y g a L g x y g x y L u x y g x y

L g g L g g

L g g L g g

1 1[ ],

[ ],

a

N a N

a L u g

a L u g

Multiresolution Analysis

/ 2,

1

2

1

0

1.

2.

3. 0

4. if ( ) (2 )

5. if ( ) ( )

6. scaling function to define basis

if we define as an ortho

( ) 2 (2 )

j j

jj

j j

j j

j

j

j

j

jj k

V V

V L

V

f t V f t V

f t V f t k V

k

k

V

P

t t

Z

Z

2

2,

2

gonal projector on then

lim for every ( )

this means that if then { } becomes complete in ( )

Examle: Haar's multiresolution analysis

( ); :

j

jj

j k k

j

V

P f f f L

j L

V f L k f

2 ,2 ( 1)j jk kconst

Daubechies D6 scaling function

Multiresolution Analysis and Galerkin method

1/ 21, , ,

0 , ,

, ,

Having Multiresolution analysis Scaling relation 2

Select resolution level: let it be

Decomposition ( , ) ( , ) ( , ) ( ) ( )

where ( ) ( , ), ( )

With

j l k l j kk

n n k n kk

n k n k

p

n

u x y u x y u x y a y x

a y u x y x

0 ,

, , , , 0 ,

the boundary conditions ( , ) ( , ) , ( ) 0

Matrix equation

where , , ( ) -solution vector, [ ],

In oder to simplify calculation one can use quadrature

n n k

i j n i n j i n i n j

L u x y L u x y x

B L A a y R

BA R

L u

,

formula for example

( ), ( ) ( / 2 ), ( )nn kf x x mf k m x dx

Finite difference solution to DE

Consider the equation

let , , (0) ( ), (0) ( )

consider point discretizati on [0,

periodic in with peri

,

] so ( ), ( ),

where 0,1,2..

( ), ( )

1 and /

finit

od

i i

xx

u f u u d f f d

n d u u i x f f i x

i n

x

u u f u

x d n

d

u x f f x

1 1 1 12

21 1

21e differences

( )

suibstituting obtain ( 2 ) ( ) , define 2

1 0 0 1

1 1 0 0

0 1 0 0in matrix form

0 0 0 1

1 0 0 1

i i i i i i ixx i

i i i i

u u u u u u uu

x x x x

u x u u f x r x

r

r

r

r

r

0 0

1 1

2 2

2 2

1 1

n n

n n

u f

u f

u f

u f

u f

Finite difference solution to DE (1)

1,1 2,1 3,1 ,1

0 0

matrix equation

Define kernel vector ( , , ... ) ( ,1,0...0,1)

Convolution *

ˆ ˆ ˆ ˆ ˆ ˆDiscrete Fourier transform . or /

Particular case =0 then det 0

ˆ ˆ0 and is unde

n

CU F

K c c c c r

K U F

K U F U F K

C

K U

0 0

0

fined but we initially concider

ˆ ˆmean over the period is zero this maens 0. We can set 0

ˆand 0

U

U K

U

Wavelet Galerkin Method

/ 2

/ 2

Concider the equation

Scaling equation (2 )

Galerkin approximation ( ) 2 (2 )

substitution 2 and 2 gives

( ) ( ) ( ), ( ) is periodic in with

xx

kk

m mk

k

m mk k

kk

u u f

h x k

u x c x k

y x c c

U y u x c y k U y y

period 2

Discretize ( ) letting take only integer values.Thus ( ) ( )

then obtain matrix equationi i

i k i k i k ik k

md

U y y U U i y U y

U c c

0 2 2 1 1

1 1 3 2 2

2 2 1 4 3 3

2 3 4

2 3

1 3 1 2

0 0 0

0 0 0

0 0

0 0

0 0 0

0 0 0 0

N

N N N

N N

n N n

U c

U c

U c

U c

Wavelet Galerkin Method (1)

Wavelet Galerkin Method (2)

/ 2

/ 2

2

2

as before

wavelet expancion for is ( ) 2 (2 )

Define ( ) ( ) ( ) where 2 obtain

substitute expancions for ( ) and ( ) into original equatio

*

n

*

(

m m

k

mk k k

k

kk

f f x g x k

F y f x g y k g g

u x f x

c yx

U K c

F K g

2

) ( ) ( ) or

( ) ( ) ( ) taking inner product

2 ( ) ( ) ( ) ( )

( ) ( )

k kk k

k k kk k k

mk k

k k

kk

k c y k g y k

c y k c y k g y k

c y k y j dy c y k y j dy

g y k y j dy

Wavelet Galerkin Method (3)

,

1 2 2 1

1 3 2 2

2 1 1

2

2

2

Remembering ( ) ( ) we may write

where ( ) ( )

get matrix equation where

0

0

0

0

2

2

k j

j k

N N

N N

N N

Nm

mk j k j j

k

c c g

Tc

y k y j dy

y k y j dy

T

g

1

1 2

2 1 3

1 2 2 3

20

0

0 0

0

0 0

where 2

N

N N

N N

m

Wavelet Galerkin Method (4)

we obtain convolution *

Taking Fourier transforms gives

ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ. , . , .

ˆ ˆ ˆ ˆ ˆfrom which .(( / ) / )

ˆ/r ˆ ˆo

K c g

U K c F K g K c g

U

U F K

K F K K

Incorporation of boundary conditions

consider the problem in [ , ]

with boundary conditions ( ) , ( )

suppose that and are periodic withperiod , where 0

let be periodic solution on [0, ] let then

a b

xx

xx

a b

u a u u b u

u f d a

u v

b d

v

u u f

d

v v

w

f

in [ , ]

0 in [ , ]

but in [0, ] where ( ) ( ) ( )

keeping in mind Green Function ( ) from this

( ) * ( - ) ( - ) wich on bounds

( ) (0) ( - )

xx

xx a b

xx

a b

a b a

a b

w w a b

w w X d X X x X x a X x b

G G x

w x G X X G x a X G x b

w a X G X G a b u

( )

( ) ( - ) (0) ( ) get matrix equation

( )(0) ( - )

( )( - ) (0)

a b b

a a

b b

v a

w b X G b a X G u v b

X u v aG G a b

X u v bG b a G

Incorporation of boundary conditions

Offsetting boundary conditions to control error

kk

1 1

wavelet decomposition of delta function

( )= g ( ) where 2 ( )

It supply congruent with supply of basis functions on current resolution level

suppose ,

( - ) ( - ) wich ar

mk

a b

x y k g k

a a s b b s

X X x a X x b

1 1

1 1

e given by

( )( ) ( )

( )( ) ( )

The offset must be at least supply of basis functions on the level

Let us concider as an example the following equation

a a

b b

xx

X u v aG a a G a b

X u v bG b a G b b

u u

2 2

4

4 sin 1 sin9 6 4 2

with (1) 3, (2) 2 3, D12 wavelet, level 4, period 3

number of coefficients is 2 48 the offset 0.5

error decay 10 times

m

x x

u u d

d s

D12 wavelet coefficients of the delta function

Error in wavelet solution to boundary value problem

Offsetting of boundary sources to control error

Error in wavelet solution to boundary value problem with offset sources

Comparsion of results

22 2

Let us concider as an example the following equation

16 4 16 4 16256 sin sin 128 cos cos

9 3 3 3 3

with (0) 1, (1) 6.25

exact solution

4 169sin sin c

3 3

xxu u x x x x

u u

u x x

4os

3x

Decay in error of wavelet and finite difference solutions withincreasing sample size

Variation of computation time with increasing sample size

Acknowlegements

• Prof. S.Yu. Slavyanov, St Petersburg State Univercity

• Prof. A.V. Tsiganov, St Petersburg State Univercity

• Prof. S.L. Yakovlev, St Petersburg State Univercity

• And other colleagues of mine from the Department of Computational Physics