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Slides19 - Course 18.327 and 1.130 Course Wavelets and...

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Unformatted text preview: Course 18.327 and 1.130 Course Wavelets and Filter Banks Numerical solution of PDEs: Galerkin Numerical approximation; wavelet integrals (projection coefficients, moments and connection coefficients); convergence Numerical Solution of Differential Equations Numerical Main idea: look for an approximate solution that lies in Vj. Main Approximate solution should converge to true solution as j → ∞. Consider the Poisson equation leave boundary 2µ ∂ = f(x) ---------------� conditions till later ∂x2 Approximate solution: Approximate uapprox(x) = ∑ c[k]2j/2 φ(2j x – k) -----------� c[k]2 k 14243 14243 φj,k(x) trial functions 2 Method of weighted residuals: Choose a set of test Method functions, gn(x), and form a system of equations (one for each n). ∫ ∂2uapprox ∂x2 gn(x)dx = ∫ f(x)gn(x) dx f(x)g One possibility: choose test functions to be Dirac delta functions. This is the collocation method. gn(x) = δ(x – n/2j) ⇒ n integer ∑ c[k]φj,k(n/2j) = f(n/2j) c[k] ″ ----------------------------� k 3 Second possibility: choose test functions to be Second scaling functions. • Galerkin method if synthesis functions are used Galerkin (test functions = trial functions) • Petrov-Galerkin method if analysis functions are used Petrov e.g. Petrov-Galerkin ~ gn(x) = φj,n(x) ⇒ ∈ ~ Vj ∞ ~ ~ -------∑ c[k] ∫ ∂x2 φj,k(x) . φj,n(x) dx = ∫ f(x)φj,n(x) dx --------� c[k] f(x) ∞ k ∂2 -∞ Note: Petrov-Galerkin -∞ ≡ Galerkin in orthogonal case 4 Two types of integrals are needed: Two (a) Connection Coefficients ∞ ~ 2 ~ (x)dx = 22j ∞ 2j/2φ″(2jx - k)2j/2φ(2jx - n)dx ∂ ∫ ∫ φ (x) . φ 2 -∞ ∂x j,k j,n = -∞ ∞ ~ 22j ∫ φ″(τ)φ(τ -∞ + k – n) dτ = 22jh∂2/∂x2 [n – k] where h∂2/∂x2 [n] is defined by ~ h∂2/∂x2 [n] = ∫ φ″(t)φ(t – n)dt -----------------� ----------------∞ -∞ connection coefficients connection 5 (b) Expansion coefficients (b) ∞ ~ The integrals ∫ f(x)φj,n(x)dx are the coefficents for f(x) -∞ the expansion of f(x) in Vj. fj(x) = ∑ rj[k] φj,k(x) --------------------------� k with with rj[k] = ∫ f(x) ~j,k(x) dx f(x) φ ∞ -∞ --------------------------� So we can write the system of Galerkin equations as So a convolution: 22j ∑ c[k]h∂2/∂x2[n – k] = rj[n] ----------------� c[k]h k 6 ⇒Solve a deconvolution problem to find c[k] and then find uapprox using equation �. using 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 (1 ⇒ H∂2/∂x2(ω) has a 2nd order zero at ω = 0. order 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 Determination ∞ ~ 2 2[n] = ∫ φ ″(t) φ(t – n)dt h∂ /∂x (t) -∞ Simple numerical quadrature will not converge if φ ″ (t) behaves badly. (t) Instead, use the refinement equation to formulate an eigenvalue problem. φ″(t) = 8 ∑ f0[k]φ″(2t – k) k ~ ~ φ(t – n) = 2 ∑ h0[l]φ(2t – 2n - l) k l 678 678 φ(t) = 2 ∑ f0[k]φ(2t – k) Multiply and Multiply Integrate So h∂2/∂x2[n] = 8 ∑ f0[k] ∑ h0[l]h∂2/∂x2[2n + l - k] k l 8 Daubechies 6 scaling function First derivative of Daubechies 6 scaling function 9 Reorganize as Reorganize 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 +l 2n eigenvalue problem eigenvalue Need a normalization condition use the moments of the scaling function: If h0[n] has at least 3 zeros at π, we can write ∞ 2 ; µ [k] = ∫ t2~(t – k)dt ∑ µ2[k]φ(t – k) = t φ 2 -∞ k ~ Differentiate twice, multiply by φ(t) and integrate: ∑ µ2[k]h∂2/∂x2[- k] = 2! k Normalizing condition 10 10 Formula for the moments of the scaling function Formula ∞ l� µ k _- ∫τlφ(τ ∞ - k)dτ Recursive formula Recursive µ0 = ∫ φ(τ)dτ = 1 0 ∞ µr 0 N r-1 1 r r – i) µ i 0 2r - 1 ∑ ( i ) (∑ h0[k]k k=0 i=0 l r l ∑ ( r )kl-r µ 0 )k r=0 r=0 -∞ = l µk = 11 11 How to enforce boundary conditions? How One idea – extrapolate a polynomial: p-1 u(x) = ∑ c[k]φj,k(x) = ∑ a[l]xl a[ c[k] k l=0 Relate c[k] to a[l] through moments. Extend c[k] Relate by extending underlying polynomial. Extrapolated polynomial should satisfy boundary constraints: Dirichlet: p-1 l u(x0) = α ⇒ ∑ a[l]x0 = α a[ l=0 Neumann: Neumann: p-1 u'(x0) = β ⇒ ∑ a[l]lx0l-1 = β a[ l=0 Constraint Constraint on a[l] 12 12 Convergence 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) c[k] k where h= 1 2j = spacing of scaling functions 13 13 Reduction in error as a function of h Reduction 14 14 Multiscale Representation Multiscale e.g. ∂2u/∂x2 = f e.g. Expand as u = ∑ ckφ(x – k) + ∑ ∑ dj,kw(2j x-k) J j=0 k k Galerkin gives a system Ku = f with typical entries 2 2j ∫ ∂ w(x 2 ∂x2 w(x -∞ ∞ Km,n = – n)w(x-m)dx 15 15 Effect of Preconditioner Effect • • Multiscale equations: (WKWT)(Wu) = Wf Multiscale Preconditioned matrix: Kprec = DWKWTD DWKW Preconditioned Simple diagonal preconditioner 1 1 D= 1 2 1 2 1 4 1 4 1 4 16 16 Matlab Example Matlab Numerical solution of Partial Numerical Differential Equations 17 17 The Problem The 1. Helmholtz equation: uxx + a u = f 1. • p=6; % Order of wavelet scheme (pmin=3) =3 p=6; Order • a=0 • L = 3; % Period. • nmin = 2; % Minimum resolution nmin • nmax = 7; % Maximum resolution nmax 18 18 Solution at Resolution 2 Solution 19 19 Solution at Resolution 3 Solution 20 20 Solution at Resolution 4 Solution 21 21 Solution at Resolution 5 Solution 22 22 Solution at Resolution 6 Solution 23 23 Solution at Resolution 7 Solution 24 24 Convergence Results Convergence >> helmholtz slope = >> 5.9936 25 25 ...
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This note was uploaded on 12/04/2011 for the course ESD 18.327 taught by Professor Gilbertstrang during the Spring '03 term at MIT.

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