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¦¦¦ φφφ
∂∂∂ μμμ ∂∂∂
φφφ
Course 18.327 and 1.
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
∂^2 μ leave boundary ∂x^2 = f(x) --------------- ^
© (^) conditions till later
Approximate solution:
uapprox(x) = ¦ c[k]2 j/2^ φ(2 j^ x – k) -----------
k
φj,k(x) trial functions
2
δδδ
∂ ∂∂ ∂∂∂
φφφ ∈∈∈
Method of weighted residuals: Choose a set of test functions, gn (x), and form a system of equations (one for each n).
³
∂^2 u (^) approx gn (x)dx = ³ f(x)gn (x) dx ∂x^2 One possibility: choose test functions to be Dirac delta functions. This is the collocation method.
gn (x) = δ(x – n/2 j) n integer
¦¦¦¦ c[k]φφφφj,k″″″″(n/2 j) = f(n/2 j) ---------------------------- 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) ∈ V (^) j
¦¦¦¦ c[k] ³³³³ φφφφj,k(x). φφφφj,n (x) dx = ³³³³ f(x)φφφφj,n (x) dx
k - ∞∞∞∞^ - ∞∞∞∞
∂∂∂∂^2 ∂∂ ∂∂x^2
Note: Petrov-Galerkin ≡ Galerkin in orthogonal case
4
∂∂∂ ∂∂∂ ωωω
∂∂∂ ∂∂∂ ∂∂∂ ∂∂∂ (^) ∂∂∂ ∂∂∂ ωωω ωωω
∂∂∂ ∂∂∂ ³³³^ φφφ^ φφφ
φφφ
φφφ ¦¦¦ φφφ φ″φ″φ″ ¦¦¦ φ″φ″φ″ φφφ ¦¦¦ """ φφφ """
∂∂∂ ∂∂∂ ¦¦¦^ ¦¦¦^ """^ ∂∂∂ ∂∂∂ """
∞ ∞∞ ∞∞∞
"""
"""
"
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 /∂x^2 (ω) may have zeros.
Familiar example: 3-point 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 uapprox(x).
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Determination of Connection Coefficients ∞ (^) ~ h∂^2 /∂x^2 [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 ¦ f 0 [k]φ(2t – k) k φ″(t) = 8 ¦ f 0 [k]φ″(2t – k) (^) Multiply and ~ k ~ φ(t – n) = 2 ¦ h 0 ["]φ(2t – 2n - ") Integrate So h∂^2 /∂x^2 [n] = 8 ¦ f 0 [k] ¦ h 0 ["]h∂^2 /∂x^2 [2n + " - k] k (^) " 8
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Daubechies 6 scaling function
First derivative of Daubechies 6 scaling function
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Reorganize as
h∂∂∂∂^2 /∂∂∂∂x^2 [n] = 8¦¦¦¦ h 0 [m – 2n](¦¦¦¦ f 0 [m – k]h∂∂∂∂^2 /∂∂∂∂x^2 [k])
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 ππππ, we can write
¦¦¦¦ μμμμ 2 [k]φφφφ(t – k) = t^2 ; μμμμ 2 [k] = ³³³³ t^2 φφφφ(t – k)dt
Differentiate twice, multiply by φφφφ(t) and integrate:
¦¦¦¦ μμμμ 2 [k]h∂∂∂∂^2 /∂∂∂∂x^2 [- k] = 2! Normalizing condition
m k
k - ∞∞∞∞ ~
k
m = 2n +""""
φφφ ¦¦¦ φφφ
φφφ ωωω πππ
¦¦¦ φφφ
Convergence
Synthesis scaling function:
φ(x) = 2 ¦ f 0 [k]φ(2x – k) k
We used the shifted and scaled versions, φj,k(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 like
u(x) = ¦ c[k]φj,k(x) + O(h p^ ) k where
h = = spacing of scaling functions 1 2 j 13
Reduction in error as a function of h
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Multiscale Representation
e.g. ∂∂∂∂^2 u/∂∂∂∂x 2 = f Expand as
u = ¦¦¦¦ c (^) kφφφφ(x – k) + ¦¦¦¦ ¦¦¦¦ dj,kw(2 j^ x-k)
Galerkin gives a system
Ku = f
with typical entries
Km,n = 2 2j^ ³³³³ w(x – n)w(x-m)dx
k (^) j=0 k
J
∞∞∞∞
∂∂∂∂^2 ∂∂∂∂x^2
16
Effect of Preconditioner
- Multiscale equations: (WKW T)(Wu) = Wf
- Preconditioned matrix: Kprec = DWKW TD
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(^14)
(^14)
(^14)
(^12)
(^12)
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D
Simple diagonal preconditioner
Solution at Resolution 2
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Solution at Resolution 3
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Solution at Resolution 4
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Solution at Resolution 5
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Convergence Results
>> helmholtz slope = 5.
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