Course and Wavelets and Filter Banks

Transcription

1 Course and Wavelets and Filter Bans Numerical solution of PDEs: Galerin approximation; wavelet integrals (projection coefficients, moments and connection coefficients); convergence

2 Numerical Solution of Differential Equations Main idea: loo for an approximate solution that lies in V j. Approximate solution should converge to true solution as j. Consider the Poisson equation leave boundary = f(x) conditions till later 2 µ x 2 Approximate solution: u approx (x (x)) = c[]2 j/2 j/2 φ(2 (2 j x ) φ j, (x) trial functions 2

3 Method of weighted residuals: Choose a set of test functions, g n (x), and form a system of equations (one for each n). 2 u approx x 2 g n (x)dx = f(x)g n (x) dx One possibility: choose test functions to be Dirac delta functions. This is the collocation method. g n (x) ) = δ(x n/2 j ) n integer c[]φ j, (n/2 j ) = f(n/2 j )

4 Second possibility: choose test functions to be scaling functions. Galerin method if synthesis functions are used (test functions = trial functions) Petrov-Galerin method if analysis functions are used e.g. Petrov-Galerin ~ ~ g n (x) ) = φ j,n (x) V j 2 φ j, (x). φ j,n (x) dx - x 2 - c[] ~ ~ dx = f(x)φ j,n (x) dx Note: Petrov-Galerin Galerin in orthogonal case 4

5 Two types of integrals are needed: (a) Connection Coefficients φ j, (x). φ j,n (x)dx = 2 2j ~ 2 j/2 φ (2 j x - )2 j/2 ~ 2 φ(2 j x - n)dx - x 2 - = 2 2j ~ φ (τ)φ(τ + n) dτ d - = 2 2j h [n ] 2 / x 2 where h 2 / x 2 [n] is defined by ~ h [n] = φ (t) (t)φ(t n)dt 2 / x connection coefficients 5

6 (b) Expansion coefficients ~ The integrals f(x)φ j,n (x)dx are the coefficents for the expansion of f(x) in V j. with f j (x) ) = r j [] φ j, (x) r j [] ] = f(x) ~ φ j, (x) dx So we can write the system of Galerin equations as a convolution: 2j c[]h 2 / x 2 [n ] = r j [n] j 6

7 Solve a deconvolution problem to find c[] and then find u approx using equation. Note: we must allow for the fact that the solution may be non-unique, i.e. H 2 / x 2 (ω)) may have zeros. Familiar example: 3-point 3 finite difference operator h 2 / x 2 [n] = {1, -2, 1} H 2 / x 2 (z) = 1 2z 1 + z 2 = (1 z 1 ) 2 H 2 / x 2 (ω)) has a 2 nd order zero at ω = 0. Suppose u 0 (x) is a solution. Then u 0 (x) + Ax + B is also a solution. Need boundary conditions to fix (x). u approx (x 7

8 Determination of Connection Coefficients ~ h 2 / x 2 [n] = φ (t) φ(t n)dt - Simple numerical quadrature will not converge if φ (t) behaves badly. Instead, use the refinement equation to formulate an eigenvalue problem. φ(t) = 2 f 0 []φ(2t ) φ (t) = 8 f 0 []φ φ (2t ) ~ ~ φ(t n) = 2 h 0 [l]φ(2t 2n - l) So l Multiply and Integrate h 2 / x 2 [n] = 8 f 0 [] h 0 [l]h 2 / x 2 [2n + l - ] l 8

9 Daubechies 6 scaling function First derivative of Daubechies 6 scaling function 9

10 Reorganize as h 2 / x 2 [n] = 8 8 h 0 [m 2n]( f 0 [m ]h 2 / x 2 []) m m = 2n +l + Matrix form h 2 / x 2 = 8 A B h 2 / x 2 eigenvalue problem Need a normalization condition use the moments of the scaling function: If h 0 [n] has at least 3 zeros at π,, we can write µ 2 []φ(t ) = t 2 ~ ; µ 2 [] = t 2 φ(t )dt - ~ Differentiate twice, multiply by φ(t) and integrate: µ 2 []h 2 / x 2 [- ] = 2! Normalizing condition 10

11 Formula for the moments of the scaling function l - µ _ τ l φ(τ - )dτ Recursive formula µ 0 = φ(τ)dτ = 1 0 µ r = ( ) ( ( h 0 [] r i )µ 0 l r - 1 l r-1 i=0 r i l µ = ( ) l-r µ r=0 r N =0 r 0 i 0 11

12 How to enforce boundary conditions? One idea extrapolate a polynomial: p-1 u(x) ) = c[]φ j, (x) ) = a[l]x l l=0 Relate c[] ] to a[l] ] through moments. Extend c[] by extending underlying polynomial. Extrapolated polynomial should satisfy boundary constraints: Dirichlet: p-1 u(x 0 ) = α a[l]x 0 = α l=0 Neumann: p-1 u'(x 0 ) = β a[l]lx l-1 0 = β l=0 l Constraint on a[l] 12

13 Convergence Synthesis scaling function: φ(x) = 2 f 0 []φ(2x ) We used the shifted and scaled versions, φ j, (x), to synthesize the solution. If F 0 (ω)) has p zeros at π,, then we can exactly represent solutions which are degree p 1 polynomials. In general, we hope to achieve an approximate solution that behaves lie u(x) ) = c[]φ j, (x) ) + O(h p ) where h = 1 2 j = spacing of scaling functions 13

14 Reduction in error as a function of h 14

15 Multiscale Representation e.g. 2 u/ x 2 = f Expand as u = c φ(x ) + d j, w(2 j x-) Galerin gives a system Ku = f with typical entries m,n = 2 2j K m,n j=0 2j - 2 x 2 J w(x n)w(x-m)dx 15

16 Effect of Preconditioner Multiscale equations: (WKW T )(Wu) = Wf Preconditioned matrix: K prec = DWKW T D Simple diagonal preconditioner D =

17 Matlab Example Numerical solution of Partial Differential Equations 17

18 The Problem 1. Helmholtz equation: u xx + a u = f p=6; % Order of wavelet scheme (p( min =3) a = 0 L = 3; % Period. nmin = 2; % Minimum resolution nmax = 7; % Maximum resolution 18

19 Solution at Resolution 2 19

20 Solution at Resolution 3 20

21 Solution at Resolution 4 21

22 Solution at Resolution 5 22

23 Solution at Resolution 6 23

24 Solution at Resolution 7 24

25 Convergence Results >> helmholtz slope =