Handout19

# Handout19 - Course 18.327 and 1.130 Wavelets and Filter...

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Unformatted text preview: Course 18.327 and 1.130 Wavelets and Filter Banks Numerical solution of PDEs: Galerkin approximation; wavelet integrals (projection coefficients, moments and connection coefficients); convergence Numerical Solution of Differential Equations Main idea: look for an approximate solution that lies in Vj. Approximate solution should converge to true solution as j → ∞. Consider the Poisson equation · § leave boundary ∂2µ = f(x) --------------- © conditions till later ¹ ∂x2 Approximate solution: uapprox(x) = ¦ c[k]2j/2 φ(2j x – k) ----------k φj,k(x) trial functions 2 1 Method of weighted residuals: Choose a set of test functions, gn(x), and form a system of equations (one for each n). ³ ∂2uapprox ∂x2 gn(x)dx = ³ f(x)gn(x) dx One possibility: choose test functions to be Dirac delta functions. This is the collocation method. gn(x) = δ(x – n/2j) ¦ c[k]φj,k(n/2j) = f(n/2j) ″  n integer ---------------------------- k 3 Second possibility: choose test functions to be scaling functions. • Galerkin method if synthesis functions are used (test functions = trial functions) • Petrov-Galerkin method if analysis functions are used e.g. Petrov-Galerkin ~ gn(x) = φj,n(x)  ∞ ∂2 2 -∞ ∂x ¦ c[k] ³ k ∈ ~ Vj ∞ ~ ~ φj,k(x) . φj,n(x) dx = ³ f(x)φj,n(x) dx -------- Note: Petrov-Galerkin -∞ ≡ Galerkin in orthogonal case 4 2 Two types of integrals are needed: (a) Connection Coefficients ∞ ∞ ~ 2 ~ ³ ∂∂x2 φj,k(x) . φj,n(x)dx = 22j ³ 2j/2φ″(2jx - k)2j/2φ(2jx - n)dx -∞ -∞ ∞ ~ = 22j ³ φ″(τ)φ(τ + k – n) dτ φ″ -∞ = 22jh∂2/∂x2 [n – k] where h∂2/∂x2 [n] is defined by ∞ ~ h∂2/∂x2 [n] = ³ φ″(t)φ(t – n)dt -----------------∞ connection coefficients 5 (b) Expansion coefficients ∞ ~ The integrals ³ f(x)φj,n(x)dx are the coefficents for -∞ the expansion of f(x) in Vj. fj(x) = ¦ rj[k] φj,k(x) -------------------------- k with ∞ rj[k] = ³ f(x) ~j,k(x) dx φ -∞ -------------------------- So we can write the system of Galerkin equations as a convolution: 22j ¦ c[k]h∂2/∂x2[n – k] = rj[n] ---------------k 6 3 Solve a deconvolution problem to find c[k] and then find uapprox using equation . Note: we must allow for the fact that the solution may be non-unique, i.e. H∂2/∂x2(ω) may have zeros. Familiar example: 3-point finite difference operator h∂2/∂x2[n] = {1, -2, 1} H∂2/∂x2(z) = 1 –2z–1 + z–2 = (1 – z–1)2  H∂2/∂x2(ω) has a 2nd order zero at ω = 0. Suppose u0(x) is a solution. Then u0(x) + Ax + B is also a solution. Need boundary conditions to fix uapprox(x). 7 Determination of Connection Coefficients ∞ ~ h∂2/∂x2[n] = ³ 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 ¦ f0[k]φ(2t – k) k φ″(t) = 8 ¦ f0[k]φ″(2t – k) k ~ ~ φ(t – n) = 2 ¦ h0[φ(2t – 2n - ) Multiply and Integrate  So h∂2/∂x2[n] = 8 ¦ f0[k] ¦ h0[h∂2/∂x2[2n +  - k] k  8 4 Daubechies 6 scaling function First derivative of Daubechies 6 scaling function 9 Reorganize as h∂2/∂x2[n] = 8¦ h0[m – 2n](¦ f0[m – k]h∂2/∂x2[k]) m k Matrix form h∂2/∂x2 = 8 A B h∂2/∂x2 m = 2n + eigenvalue problem Need a normalization condition use the moments of the scaling function: If h0[n] has at least 3 zeros at π, we can write ∞ ~ ¦ µ2[k]φ(t – k) = t2 ; µ2[k] = ³ t2φ(t – k)dt k -∞ ~ Differentiate twice, multiply by φ(t) and integrate: ¦ µ2[k]h∂2/∂x2[- k] = 2! k Normalizing condition 10 5 Formula for the moments of the scaling function ∞  µ k _ ³τφ(τ - k)dτ -∞ ∞ Recursive formula ∞ µ0 = ³ φ(τ)dτ = 1 0 µr = 0 -∞  µk = r-1 N 1 r r – i)µ i 0 2r - 1 ¦ ( i ) (¦ h0[k]k k=0 i=0  r  ¦ ( r )k-r µ 0 r=0 11 How to enforce boundary conditions? One idea – extrapolate a polyp-1 nomial: u(x) = ¦ c[k]φj,k(x) = ¦ a[x k =0 Relate c[k] to a[ through moments. Extend c[k] by extending underlying polynomial. Extrapolated polynomial should satisfy boundary constraints: Dirichlet: p-1  u(x0) = α  ¦ a[x0 = α =0 Neumann: p-1 u'(x0) = β  ¦ a[x0-1 = β =0 Constraint on a[ 12 6 Convergence Synthesis scaling function: φ(x) = 2 ¦ f0[k]φ(2x – k) k We used the shifted and scaled versions, φj,k(x), to synthesize the solution. If F0(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 like u(x) = ¦ c[k]φj,k(x) + O(hp) k where h= 1 2j = spacing of scaling functions 13 Reduction in error as a function of h 14 7 Multiscale Representation e.g. ∂2u/∂x2 = f Expand as J u = ¦ ckφ(x – k) + ¦ ¦ dj,kw(2j x-k) j=0 k k Galerkin gives a system Ku = f with typical entries ∞ ∂2 Km,n = 22j ³ ∂∂x2 w(x – n)w(x-m)dx -∞ 15 Effect of Preconditioner • • Multiscale equations: (WKWT)(Wu) = Wf Preconditioned matrix: Kprec = DWKWTD Simple diagonal preconditioner ª1 «1 « « « D=« « « « « ¬ 1 2 1 2 1 4 1 4 º » » » » » » » » 1» 4¼ 16 8 Matlab Example Numerical solution of Partial Differential Equations 17 The Problem 1. Helmholtz equation: uxx + a u = f • p=6; % Order of wavelet scheme (pmin=3) • a=0 • L = 3; % Period. • nmin = 2; % Minimum resolution • nmax = 7; % Maximum resolution 18 9 Solution at Resolution 2 19 Solution at Resolution 3 20 10 Solution at Resolution 4 21 Solution at Resolution 5 22 11 Solution at Resolution 6 23 Solution at Resolution 7 24 12 Convergence Results >> helmholtz slope = 5.9936 25 13 ...
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