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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
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