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410 V. Mishra and Sabina

)2()1()(1

21 k xa x

N k k

k .

2. Wavelet Galerkin Technique

Galerkin Method. It was Russian engineer V.I. Galerkin who proposed a projectionmethod based on weak form. In it a set of test functions are selected such thatresidual of differential equation becomes orthogonal to test functions [8].

Consider one-dimensional differential equation [12, pp. 451-479]10),()( x x f x Lu (2.1)with Dirichlet boundary conditions

.)1(,)0( buau f is real valued and continuous functions of x on [0,1]. L is a uniformly elliptic differential operator .

L2([0, 1]) is a Hilbert space with inner product

.)()(,1

0

dt xg x f g f

Suppose that jv is a complete orthonormal system for L2 ([0, 1]) and that every jv

is 2C on [0, 1] such that.)1(,)0( bvav j j

Select a finite set of indices j and consider the subspace jvspanS j :

Let the approximate solution su of the given equation be

.S v xu k k

k s (2.2)We would like to determine k x in a way that su behaves as if is a true solution on S ,i.e.

jv f vu L j js ,, (2.3)such that the boundary conditions 0)1()0( ss uu are satisfied. Substituting su in(2.3),

.,, jv f xvv L jk jk k

(2.4)

Let X and Y denote the vectors k k x and ,, k k k v f y and A the matrixk jk j

a A,,

, where .,, jk k j vv La (2.4) reduces to the system of linear equations

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Wavelet Galerkin solutions 411

jk k

k j y xa , or equivalently AX = Y . (2.5)

Thus in Galerkin method, for each subset , we find an approximation su in S to u by solving (2.5) for X and then substituting its components in (2.2). It is expectedthat as we increase in some systematic way, su converges to u , the actual solution.

Condition Number of a Matrix. We know that a linear system Y AX has aunique solution X for every Y if a square matrix A is invertible. It is often observedthat for two close values of Y and 'Y , X and ' X are far apart. Such a linear system

is called badly conditioned . Thus data Y is expected to be fairly accurate. Conditionnumber of A is given by ,)( 1# A A AC 1)(# AC .

Thus ),(# AC is the measure of stability of the linear system under perturbation of thedata .Y Small condition number near 1 is desirable. In case it is high, replace thesystem by equivalent system , BY BAX B is a preconditioning matrix such that

)()( ## AC BAC .To facilitate easy calculation, A is considered to be sparse , i.e., A should have high

proportion of entries 0. The best one is when A is in diagonal form.

Wavelet Galerkin methodLet )(, xk j 2 )2(

2/ k x j j be a wavelet basis for ])1,0([2 L with boundaryconditions

.0)1()0( ,, k jk j For each k jk j ,,),( is

2C .

The scale of approximates j2 and is centralized near point k j2 and equatesto zero outside the interval centred at k j2 of length proportional to .2 j In Wavelet Galerkin method equations (2.2) and (2.3) may thus be replaced by

k jk j

k js xu ,,

,

and.),(,, ,,

,,, ml f x L mlk j

k jmlk j

So that .Y AX (2.6)Where A= ,

),(),,(,;, k jmlk jmla ),(, )( k jk j x X , ),(, )( mlml yY .

In it mlk jk jml La ,,,;, , , ., ,, mlml f y The pairs (l,m) and (j,k) represent respectively row and column of A. Consider A to be sparse. Represent Y AX by equivalent

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V MZ (2.7)in which case M has relatively low condition number, if A has not. This system isnow well conditioned. Again M to be sparse is desirable.The matrix M in the preconditioned system (4.6) has condition number boundedindependently of . So as we increase to approximate solution with moreaccuracy, the condition number maintains it boundedness, which is much better thanthe finite difference method in which case condition number grows as 2 N . Thus thedata errors, may be due to rounding off, has no effect in wavelet Galerkin solutionover [0, 1] as we approach for better and better accuracy.

3. Connection Coefficients

In order to find the solution of differential equation by using wavelet Galerkinmethod there is need to find the connection coefficients as explored in Latto et al. [1]as

.)()( 22

1

1

21

21 dx x x d ld ld d ll (3.1)Taking derivatives of the scaling function d times. we get

.)2(2)(

1

0

N

k

d k k

d d

k xa x (3.2)

After simplification and considering it for all 2121

d d ll , gives a system of linear

equations with 21d d

as unknown vector:

,2

12121

1d d

d d d T

where 21 d d d and i

ilqi aaT 2 .

The moments k i M of i are defined as

dx x x M ik k i )(

with 100 M .Latto et al. [1] derives a formula to compute the moments by induction on k .

where the ia s are the Daubechies wavelet coefficients.Finally the system will be

,)12(2

1 1

0

1

00

N

i

lt i

t

l

t j j

t j

ji ial

t i

t

j M

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Matlab software is used to compute the connection coefficients and moments atdifferent scales. We have computed the connection coefficients by substituting thevalues of Daubechies coefficients in the matrix T and by evaluating moments byusing the programme that is given in [8].Latto et al. [1] computed the connection coefficients at j= 0 and N= 6 only. We havecomputed the connection coefficients at all values of j and N . The values ofconnection coefficients, for example, at j= 0, 4, 7 and N= 6 and at j= 5 and N= 12 areshown in Table 3.1.

Table 3.1: 2- term connection coefficients

Connection Coefficients at N= 6 , j= 0 , d1= 2 , d2= 0 ]4[ 5.357142857144194e-003]3[ 1.142857142857108e-001

]2[ -8.761904761904359e-001

]1[ 3.390476190476079e+000]0[ -5.267857142857051e+000]1[ 3.390476190476190e+000]2[ -8.761904761904867e-001]3[ 1.142857142857139e-001]4[ 5.357142857141956e-003

Connection Coefficients at N= 6 , j= 7 , d1= 2 , d2= 0 ]4[ 8.777142857143141e+001

]3[ 1.872457142857136e+003]2[ -1.435550476190484e+004]1[ 5.554956190476204e+004

]0[ -8.630857142857112e+004]1[ 5.554956190476144e+004]2[ -1.435550476190458e+004]3[ 1.872457142857130e+003]4[ 8.777142857143379e+001

.!

02

1211

d M

I T d d d

d

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Connection Coefficients at N= 12 , j= 5 , d= 2 ]10[ -1.294452960894988e-008

]9[ 2.693100643685327e-005]8[ -3.549272110888916e-003]7[ -5.566810017417083e-002]6[ -6.727300176318085e-002]5[ 6.633534506945016e+000]4[ -5.054628929484702e+001]3[ 2.098232006991124e+002]2[ -6.458708407940719e+002]1[ 2.367351361198240e+003]0[ -3.774529005718791e+003

]1[ 2.367351361198260e+003]2[ -6.458708407940832e+002]3[ 2.098232006991146e+002

]4[ -5.054628929484773e+001]5[ 6.633534506945424e+000

]6[ -6.727300176361031e-002]7[ -5.566810017416311e-002

]8[ -3.549272110869743e-003]9[ 2.693100643242748e-005

]10[ -1.294445771987477e-008

4. Wavelet Methods for ODE

Consider the equation [Amaratunga et al.]

,22

f u x

u (4.1)

where u , f are periodic in x of period Z d .The Wavelet Galerikin solution of periodic problem is slightly more complicatedthan the finite difference approach as the former involves solving a set of

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simultaneous equations in wavelet space and not in physical space. The solution inwavelet space is then transformed back to physical space by FTT.Let the wavelet expansion )( xu at scale j be

,,)2(2)( 2/ Z k k xc xuk

j jk (4.2)

where k c s are periodic wavelet coefficients of u, periodic in x.

Put x y j2 so that

.2),()()( 2/ k j

k k k cC k yC xu yU

If d is the period of u, then )( yU and so also k C is periodic in y with period .2 d j

Let us discretize )( yU at all dyadic points Z y y x j ,2

.1,....,2,1,0, niC C U k k

k ik ik

k i

The matrix representation is C k U * , where k is the convolution kernal, i.e. thefirst column of the scaling function matrix.Similarly the wavelet expansion for ),( x f

.,)2(2)( 2/ Z k k xd x f k

j jk (4.3)

k j

k k k d Dk y D x f yF 2/2),()()( .

And the matrix representation is.* Dk F

Substitute the expansions of )( xu and )( x f in (4.1) and then take inner product on both sides with Z j j y ),( .Use dy j yk ynk j )()(

)( and dx j yk y jk )()( , we obtain.. gC k Now take Fourier Transforms

.. C k U

.. Dk F

.. gC k Subsequently, k F U / . Inverse FT will give U.Wherein . and / denote component by component multiplication and divisionrespectively.

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.)(&)( ba ubuuau Let u and f are periodic with period d , d ba 0 .If f is not periodic, it can be make periodic making it zero or extending smoothlyoutside ].,[ ba Let )( xv be the solution in ],[ ba with periodic boundary conditions.The solution )( xu with Dirichlet boundary conditions is by adding another functions

)( xw such that.wvu (4.8)

From (18) f wv x

w

x

v 2

2

2

2

, i.e.

022

w x

w in .,ba

However on or outside xxwba ],,[ may take such values as to make u satisfy thegiven boundary conditions. The desired effect may be achieved by placing sources(or delta functions) along the boundary ],0[ d which encompasses ],[ ba . In otherwords, we need the solution w to

X w x

w 2

2

in ],,0[ d

where ).()()( b x X a x X x X X ba ba X X , are constants, and stands for delta function.

Now Greens function )( xG (or impulse response) of the differential equation isgiven by

).(22

xG x

G

So the solution ).()()(*)( b xG X a xG X x X xGw ba (4.9)From (4.8) and (4.9)

).()()()( avubaG X oG X aw aba ).()()()( bvuoG X abG X bw bba

Equivalently,

.)(

)(

)()()()(

bvu

avu

X

X

oGabG

baGoG

b

a

b

a

Solving for ,, ba X X the solution w is obtained.

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Offsetting the boundary sources. Placing the sources at the boundary in the waveletmethod amounts to large error due to finite support of ; i.e, number of non zerowavelet coefficients. Actually, support of in wavelet space is equal to support of that is used to define it.

k

k k yg x ),()(

where wavelet coefficients)(2 k g mk .

Number of wavelet coefficients .2 d m

The magnitude s of the offset is so chosen that number of discritization pointinvolved is at least equal to the support of scaling functionSuppose

sbbsaa 11 , .Then )()( 11 b x X a x X X ba so that ba X X , are solutions of

.)(

)(

)()(

)()(

11

11

bvu

avu

X

X

bbGabG

baGaaG

b

a

b

a

5. Test Problem

We apply fictitious boundary approach to analyse harmonic wave (differential)equation:

]1,0[,022

xu x

u (5.1)

with Dirichlet boundary conditions 1)0( u and 0)1( u . = ( 9.5 ) = 891. In [8], this problem has been treated with boundary 1)0( u & 0)1( u .Exact solution is: = .

Let the solution be

j

N k

j jk k xc xu

2

1

2/

)2(2)( .

Fictitious boundary approach is applied in which original boundary [0,1] is extendedon both ends by margin j N 2

1 . Proceeding as in Latto et al. [1], (5.1) reduces to

0,, k

k nk k

k nk C C (5.2)

where

dxk xn x j jk n )2()2(, and

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dxk xn x j jk n )2()2(, .

We compute the solution of equation (5.2) by replacing the first and last equations by the equations obtained by using the right and left boundaries which also representthe relations of the coefficients k c as explained in Section 4. After replacing the rows,we get the following matrix form with N= 6:

0TC

where C =

Now by applying Gauss elimination method and using the programming of Matlabthis matrix can be easily solved. Thus solution is obtained directly by substituting thevalues of .

)62*(12

1

2

3

4

5

::

j jC

C

C

C

C

C

)62)*(42(

034

4101

00...001......000...00::::::::::::::

0......00..................::::::::::::::

00...000...00...00...000...100...000

.

j j

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For N= 8 , j= 8, the graph of solution of exact solution and wavelet solution

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-2

-1

0

1

2

x

D8, 256 points.

Exact solution

WG Solution

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.005

0.01

0.015Error.

x

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For N= 6 , j= 8, the graph of solution of exact solution and wavelet solution

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -2

-1

0

1

2

x

D6, 256 points.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0

0.005

0.01

0.015Error.

x

Exact solution

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For N= 6 , j= 9 , graph shows that error decreases between exact and wavelet solution.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -2

-1

0

1

2

x

D6, 512 points.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0

1

2

3

4x 10

-3 Error.

x

Exact solution

WG solution

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For N= 12 , j= 10 , graph shows that error decreases between exact and waveletsolution.

6. Conclusion

The wavelet method has been shown to be a powerful numerical tool for the fast andaccurate solution of differential equations. Solutions obtained using the Daubechies6, 8 and 12 coefficients wavelets have been compared with the exact solutions. Insolving harmonic equation is chosen to be 891. Matching solutions are obtainedfor N= 6 , j= 9 and N= 12 , j= 10. Condition numbers for D12 are constantly lower thanfor D6, less errors are shown for former. Dianfeng et al. [11] is silent for highervalues of resolution near resonance point but here good solution is shown to exist for

D12.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-2

-1

0

1

2

x

D12, 1024 points.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

2

4

6

8x 10

-4 Error.

x

Exact solution

WG Solution

)2(2)()(1

0

2/1 nt nht N

n

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References

[1] A. Latto, H.L. Resnikoff and E. Tenenbaum, The Evaluation of ConnectionCoefficients of Compactly Supported Wavelets, in: Proceedings of the French-USAWorkshop on Wavelets and Turbulence , Princeton, New York, 1991, Springer-Verlag, 1992.[2] Bjorn Jawerth and Wim Sweldens, Wavelets Multiresolution Analysis Adaptedfor Fast Solution of Boundary Value Ordinary Differential Equations, Proc. 6 th Cop.

Mount Multi. Conf., April 1993, NASA Conference Pub., 259--273.

[3] G. Belkin, R. Coifman and V. Rokhlin, Fast Wavelet Transforms and NumericalAlgorithms, Comm. Pure Appl. Math. 44 (1997), 141-183.[4] J.-C. Xu and W.-C. Shann, Wavelet-Galerkin Methods for Two-point BoundaryValue Problems, Num. Math. Eng. 37(1994), 2703-2716.[5] J. R. Williams and K. Amaratunga, Wavelet Based Greens Function Approachto 2D PDEs, Engg. Comput. 10 (1993), 349-367.[6] J. R. Williams and K. Amaratunga, High Order wavelet Extrapolation Schemesfor Initial Problems and Boundary Value Problems, July 1994, IESL Tech. Rep., No.94-07, Intelligent Engineering Systems Laboratory, MIT.[7] J.R. Williams and Kelvin Amaratunga, Simulation Based Design using Wavelets,Intelligent Engineering Systems Laboratory, MIT (USA).

[8] Jordi Besora, Galerkin Wavelet Method for Global Waves in 1D , Master Thesis,Royal Inst. of Tech. Sweden, 2004.[9] K. Amaratunga, J.R. Williams, S. Qian and J. Weiss , Wavelet-Galerkin solutionsfor one Dimensional Partial Differential Equations , IESL Technical Report No. 92-05, Intelligent Engineering Systems Laboratory, M. I. T ., 1992.[10] K. Amaratunga and J.R. William, Wavelet-Galerkin Solutions for One-dimensional Partial Differential Equations, Inter. J. Num. Meth. Eng. 37(1994),2703-2716.[11] L.U. Dianfeng, Tadashi Ohyoshi and Lin ZHU, Treatment of BoundaryCondition in the Application of WaveletGalerkin Method to a SH Wave Problem,1996, Akita Univ. (Japan).

[12] M.W. Frazier, An Introduction to Wavelets through Linear Algebra, Springer, New York, 1999.[13] Stephan Dhalke and Ilona Weinreich, Wavelet-Galerkin Methods: An AdaptedBiorthogonal Wavelet Basis, Constructive Approximation 9 (1993), 237-262.[14] Ole Christensen, Frames, Riesz Basdes, and Discrete Gabor/WaveletExpansions, Bulletin (New Series) of the American Mathematical Socie ty 38 (2001),273-291.

Received: September, 2010

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