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    Applications of Wavelets inNumerical Mathematics

    Kees Verhoeven

    1. Brief summary

    2. Data compression

    3. Denoising

    4. Preconditioning

    5. Adaptive grids

    6. Integral equations

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    1. Brief Summary

    (t): scaling function.For the following 2-scale relation holds

    (t) =

    k=

    pk(2t k), t IR.

    (t): mother wavelet.For the following 2-scale relation holds

    (t) =

    k=

    qk(2t k), t IR.

    The decomposition for reads

    (2tk) =

    m=

    h2mk(tm)+g2mk(tm), t IR.

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    2. Data Compression

    We consider a function f

    f : [0, 1] IR.

    We want to approximate this function by a function v de-fined by

    v =

    k

    ckk,

    where {k|k = 1, . . . , N } is a basis for the linear functionspace V.

    The quality of the approximation can be expressed in termsof a norm

    f v.

    An alternative is to expand f periodically. We thereforelook at the Fourier series of f

    f(x) =

    m=

    cme2imx

    and approximate this by

    v(x) =M

    m=M

    cme2imx.

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    So

    V =

    e2mix |m = M , . . . , M

    ,

    with dimension N = 2M + 1. The basis functions form anorthonormal system. Therefore

    ck = (k, f) =

    10

    f(x)e2imxdx.

    Error estimates

    Given f, g : [0, 1] IR with the Fourier expansions

    f =

    m=

    cme2imx, g =

    m=

    dme2imx.

    Then10

    f(x)g(x)dx =

    m=

    cmdm.

    So 10

    |f(x)|2dx =

    m=

    |cm|2.

    The error then reads

    2M := f v2 =

    |m|>M

    |cm|2.

    In many cases properties of f lead to an error estimate ofthe type

    M CM, C, > 0.

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    Example

    Given

    f(x) = x(x 1

    2)(x 1), x [0, 1].

    We can derive

    cm =3

    4i3m3.

    The error M can therefore be estimated via

    2M =9

    86

    m=M+1

    m6 9

    86

    M

    y6dy =9

    406M5.

    M L2-error

    9406 M

    5

    10 0.426 104 0.484 104

    20 0.803 105 0.855 105

    40 0.147 105 0.151 105

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    How about localized functions?

    It seems sensible to approximate a localized function withbasis functions which also have compact support.

    Stepfunctions:

    k =

    h1

    2 , x [(k 1)h,kh),

    0, else.

    Note that:

    h = N1 and k = 1

    more efficient for function evaluations

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    Comparison

    We consider the following function.

    The error of the approximation using the Fourier series atM = 64, is approximately 0.001. The approximation using

    first order splines is 0.01 withN

    = 2M

    + 1 = 129.We would like to use localized basis functions only wherethe function to be approximated behaves like a localizedfunction.

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    Therefore, we would like basis functions which all have the

    same shape but different scales. Then, if we have

    v =N

    k=1

    ckk

    and |cj| , we also have

    f

    k

    ckk

    f

    k=j

    ckk

    + cjj.

    This leads to data compression.

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    Example

    We denote by VJ the space of piecewise constant basis func-tions on [0,1] with width h = 2J and dimension N = 2J.

    The space V0 has one basis function: the constant function1 = 1.For V1 the usual basis functions are depicted here:

    The coefficients ck behave like

    ck = h1

    2 a+h/2

    ah/2

    f(x)dx h1

    2f(a).

    Can we chose a more appropriate basis?

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    First approach: construct basis for VJ by expanding the

    basis of VJ1.

    Figure 1: An alternative basis for V1.

    Figure 2: Together with the functions above an alternative basis for V2.

    But: the new basis is no longer orthogonal.For the test function f(x) = x, the drop in the coefficientsseems to be like h3/2 (23/2 3).

    generation |ck|

    0 0.3541 0.1252 0.0443 0.016

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    A better alternative basis for VJ

    Construct an orthogonal basis {i} for VJ. This leads to

    Figure 3: A better alternative basis for V1.

    Figure 4: Together with the two functions above a better alternative basisfor V2.

    Note that these are the Haar wavelets!

    (x) = 0,0(x) = 2(x), 0,1(x) = 3(x), 1,1(x) = 4(x).

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    Now:

    ck = h1

    2

    aah/2 f(x)dx

    a+h/2a f(x)dx

    = h

    1

    2

    aah/2(f(a) + (x a)f

    (a) + . . .)dx

    a+h/2

    a (f(a) + (x a)f(a) + . . .)dx

    1

    4f(a)h3/2.

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    Comparison of this basis and homogeneous one:

    Figure 5: Approximation of f (top left) with 32, 10, 9 basis functions,respectively.

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    Figure 6: Approximation of f with 22, 10 homogeneous basis functions,respectively.

    If wavelets used as basis functions have several momentsequal to zero, then reduction will be better.

    Because, if

    v =lZZ

    cl,Il,I +J1k=I

    lZZ

    dl,kl,k,

    then

    |dk,J| Chd+3/2,

    where d is the number of moments equal to zero.

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    Example

    M L2-error for spline d = 1 L2-error for spline d = 3

    256 1.149 103 2.829 104

    128 1.149 103 2.829 104

    64 1.356 103 2.829 104

    32 2.623 103 1.143 103

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    3. Denoising

    Suppose we have a signal with some noise. We can make awavelet decomposition up to a certain depth.If the coefficients of the wavelets remain relatively large(say |dj, k| > ), then we have some localized noise.So cancel these contributions by forcing dj,k = 0 at L levelsdeep.

    Then, we can reconstruct the filtered signal.

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    The role of L can be seen as follows:

    Figure 7: Filtering of f with L = 1, 3, 5 levels ( = 0.1).

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    The role of is shown here:

    Figure 8: Filtering of f (top) with L = 1, 2 levels ( = 0.01).

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    But: if the original signal is not periodic, we encounter

    problems at the boundaries.

    Figure 9: Filtering of non-periodic f with 3 levels ( = 0.01).

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    4. Preconditioning

    Consider

    D(u) = f

    on a domain , with the differential operator D elliptic. IfD is linear, this would lead to a linear system

    Dc = r,

    with

    Dj,k = (j, Dk), rj = (j, f).

    The matrix D is called the stiffness matrix. If we use aniterative method to solve this system, the speed of con-vergence strongly depends on the condition number of thematrix D, (D) = DD1, with

    D = supc=1

    cTDc, D11 = infc=1

    cTDc.

    For symmetric matrices, we can express the condition num-ber in terms of the eigenvalues:

    (D) =max

    min.

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    Example

    We consider

    D(u) = d2u

    dx2+ u = f, x (0, 1),

    with periodical boundary conditions.We assume that the numerical solution v can be writtenas a linear combination of certain scaling functions whichspan VJ.

    Galerkins method and integration by parts gives us

    Dc = r,

    with

    Di,j =

    di,J

    dx,

    dj,J

    dx

    + (i,J, j,J),

    and, as before

    rj = (j, f).

    For linear B-splines:di,J

    dx,

    dj,J

    dx

    =

    2J/2

    d

    dx(2Jxi)2J/2

    d

    dx(2Jxj)dx.

    The derivatives are piecewise constant, and therefore wederive

    Di,j =

    2N2 + 23

    , i = j,

    N2 + 16 , i = (j 1) mod N,0 else.

    This is a circulant matrix, that is Di,j = d(i j).

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    We define the symbol of a circulant matrix as

    D(z) =

    j

    Di,jzij.

    For D now follows

    max = maxzN=1

    |D(z)|, min = minzN=1

    |D(z)|.

    For the differential equation we have

    D(z) =

    j

    di,Jdx ,

    dj,J

    dx

    + (i,J, j,J)

    zij.

    We write

    D(z) = D1(z) + D2(z),

    with

    D1(z) = j

    di,J

    dx,

    dj,J

    dx

    zij, D2(z) =

    j

    (i,J, j,J)zij.

    The second term can be recognized as D2(z) = R(z). Inthe same manner we can derive D1(z) = 2

    2JR(z).Using this and the 2-scale relations, we can derive

    D(z) = N2(2 z z1) +1

    6(4 + z+ z1).

    Calculating the biggest and smallest eigenvalue, we see that

    1 = 4N2 + 13

    .

    If N , 1 .

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    We now use wavelets

    v =k,j

    dk,jk,j.

    The stiffness matrix D then looks like

    Dl,m,j,k =

    dm,l

    dx,

    dk,j

    dx

    + (m,l, k,j).

    After some algebra using Riesz functions and 2-scale rela-tions, we can show that

    2(D) C, for all J.

    Comparison

    2J 1 216 1024.3 45.432 4096.3 49.7

    64 16384.3 52.9128 65536.3 55.4256 262144.3 57.3

    Again: this strongly depends on the periodicity of the bound-ary conditions.

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    5. Adaptive grids

    We consider a hyperbolic PDE

    u(x, t)

    t= F(u,

    u

    x, . . .),

    together with initial condition and periodic boundary con-ditions.

    The approximation of the initial condition u(x, 0) is done

    by

    v(x, 0) =N1i=0

    ci,I(0)i,I(x) +J1j=I

    iIj

    di,j(0)i,j(x).

    Here N = 2J is the amount of intervals on the coarsestgrid I, the set Ij is a subset of all possible wavelets on thegrids j = I , . . . , J 1. These sets Ij are found by making awavelet decomposition and leave out all wavelet coefficients

    below a certain threshold .

    But: this would mean that we have to approximate thefunction on the finest grid first!

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    We ignore all contributions below a wavelet for which

    |dk,l| . (filled circles mean |dk,l| > , open circles mean:|dk,l| )

    Adaptivity means that wavelets left out in previous timesteps can occur again.

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    Example: Burgers equation

    Figure 10: Approximation on t = 0, 112

    , 212

    , 312

    , 412

    , 512

    , for Burgers equation.

    The number of basis functions with coefficient above threshold is 32, 56,122, 114, 114 and 114, respectively.

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    Example: wave equation

    Figure 11: Approximation on t = 0, 0.3 and 0.5, respectively, for the wave

    equation. The number of basis functions with coefficient above threshold isapproximately 60.

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    6. Integral Equations

    We consider

    u(x) =

    K(x; t)u(t)dt + f(x).

    We take

    v(x) =

    j

    cjj(x).

    Galerkins method would leave us with

    Ac = r,

    with

    Aj,k = (j, k)

    j(x)K(x; t)k(t)dxdt, rj = (j, f).

    Often this A is well conditioned, but full.

    Using wavelets reduces the number of nonzero elements.

    We represent

    v(x) =lZZ

    cl,Il,I +J1k=I

    lZZ

    dl,kl,k.

    Look at the second term of A

    Kl,m,j,k =

    m,l(x)K(x; t)k,j(t)dxdt.

    We make the following assumption on K(x; t) d

    xdK(x; t)

    +

    d

    tdK(x; t)

    Cd 1|x t|d+1 ,for a certain d > 0.

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    Making use of the Taylor series of K(x; t) and taking a

    wavelet with d zero moments, we can derive

    |Kl,m,j,k| C1

    |x0 t0|d+1.

    Using this we can bring down the number of nonzero ele-ments (or better: the number of elements with value abovea certain threshold ).

    Example

    N = 106 = 109 = 1012

    24 74% 92% 92%48 19% 85% 96%96 5.1% 54% 78%

    192 1.1% 16% 50%384 0.34% 3.5% 25%

    Table 1: The number of elements above threshold , with Daubechieswavelets with K = 2.

    N = 106 = 109 = 1012

    24 66% 92% 92%48 12% 93% 96%96 3.1% 47% 90%

    192 0.85% 12% 56%384 0.32% 2.4% 21%

    Table 2: The number of elements above threshold , with Daubechieswavelets with K = 5.

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