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Optimal Finite Difference Grids
                      Oleksiy Varfolomiyev
                  Advisor Prof. Michael S. Siegel
                          Prof. Michael R. Booty




NJIT, June 2011
Motivation question
Motivation question


     Finite Differences???


  But they are only 2nd order
     accurate, aren’t they?
Motivation answer
Motivation answer



  How about exponential
  super-convergence and
 spectrally accurate grids?!
Outline


✤   Model Problem
✤   Discretization techniques
✤   Optimal discretization grids
✤   Numerical results
What are we solving?

1   Model problem
                 Laplace equation on a semi-infinite strip
Consider the following boundary value problem: Laplace equation on a semi-
 nfinite strip with the given Neumann data on the left boundary and Dirichlet
data elsewhere
           ∂ 2 w(x, y) ∂ 2 w(x, y)
         −        2
                      −       2
                                   = 0,       (x, y) ∈ [0, ∞) × [0, 1],    (1)
               ∂y          ∂x
                                    ∂w
                                        (0, y) = −ϕ(y), y ∈ [0, 1],        (2)
                                    ∂x
                                                          w|x=∞ = 0,       (3)
                         w(x, 0) = 0,         w(x, 1) = 0,   x ∈ [0, ∞).   (4)

Assume
                                        m
                              ϕ(y) =          ai sin(iπy)                  (5)
                                        i=1
The∂w (0, y) = −ϕ(y), ybe [0, 1], as a problem
   is defined on span{sin (πy), . . . , sin (mπy)}, spA =
                                                           above problem can ∈ studied
 e can obtain Dirichlet data on the left boundary using
                                             2 w(x)
                                                          ∂x                        w|x=∞ = 0, (2)                  (3)
t map. Equation (6) gives Aw(x) = d dx2 , therefore
                                                    w(x, 0) = 0,
nd now we can use given in (7) Neumann data to get at                            2
                                                                  w(x, 1) = 0, =d0,w(x) ∞). (3)
                                                                        w|x=∞ x ∈ [0,                    (4)
                                                                       Aw(x) −          = 0,     x ∈ [0, ∞)

      Model Problem
           w(0) = f (A)ϕ,                  w(x, 0) = 0,    w(x, 1) = 0, x ∈ [0, ∞).
                                                                                  dx 2       (4)
      Assume
mpedance function.
                                                                                          dw
                                                              m                              (x = 0) = −ϕ
  Assume                                                                                  dx
                                                     ϕ(y) =
                                                        m        ai sin(iπy)                             (5)
                                               ϕ(y) =     ai sin(iπy)
                                                             i=1                             (5) x=∞ = 0,
                                                                                               w|
                                                              i=1
    The above problem can bewhere Aas = problem is defined on span{sin (πy), . . . , s
                                 studied        a − ∂ 22
  The above problem can be studied as a problem ∂y
                                {π 2d2 w(x) 2 }. We can obtain Dirichlet data on the
                                    , . . . , (mπ)
                              2 w(x)
                             d −
                        Aw(x)Neumann-to-Dirichlet ∈ [0, ∞)Equation (6) gives Aw(x)
                    Aw(x) −            = 0, = x ∈ [0, ∞)map.
                                            2
                                                   0,    x                   (6)  (6)
                               dx2 dx
                                  dw(x)          1
                                − dx dw A 2 dw and now we can use given in (7) Neum
                                             =     w(x)
                                x = 0 dx (x = 0) = −ϕ
                                                      (x = 0) = −ϕ           (7)  (7)
          2
                                                   dx
                                                        w|x=∞ = 0, w(0) = f (A)ϕ,
                                                 w|x=∞ = 0,                  (8)  (8)
                                                                        1
                                  ∂2                                   −2
  where A A = − ∂ 2 defined on (λ) onλ span{sin.(πy), . . . , sin (mπy)}, spA =
    where         = −         is     here f span{sin (πy), . . , sin (mπy)}, spA =
                                               =       is impedance function.
                                  isNeumann-to-Dirichlet Map
                                  ∂y 2defined
                             ∂y 2
  {π 2 , .2 . , (mπ)2 }. 2 We can obtain Dirichlet data on the left boundary using
          .
    {π , . . . , (mπ) }. We can obtain Dirichlet data on d2 w(x)left boundary using
                                                                       the
  Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = dx2 , therefore        d2 w(x)
   Neumann-to-Dirichlet map. Equation
   dw(x)   1                                                                (6) gives Aw(x) =        dx2
                                                                                                            , therefore
  − dx = A 2 w(x) and now we can use given
     dw(x)     1                                                            in (7) Neumann data to   get   at
    −
  x = 0 dx           = A 2 w(x) and now we can use given in (7) Neumann data to get at
      x=0                                                  w(0) = f (A)ϕ,
                             1
                            −2                                  w(0) = f (A)ϕ,
  here f (λ) = λ                 is impedance function.
                                  1
                                 −2
      here f (λ) = λ                     is impedance function.
Idea


 ✦   Continuum NtD maps are Stieltjes-Markov functions of A

 ✦   Finite Difference NtD maps are rational functions of A

 ✦   Grids can be computed by minimizing the rational
     approximation error of NtD maps on the spectrum of A
Method outline


๏   Write down the Finite Difference Scheme

๏   Find the Neumann-to-Dirichlet map as f(A)

๏   Find the Rational Approximation of f(A) on the spectrum of A,
    obtain the grid steps

๏   Solve the discrete system
2.1 Explicit scheme xi . Denote hi = xi+1 − xi , hi = xi − xi−1 Then after
 tives be defined at nodes ˆ                           ˆ    ˆ
 discretization in x we have the following semi-discrete problem
Let the finite difference solution be defined at nodes xi , the finite difference deriva-
                  1 wi+1 x .        wi − w =
tives be defined at nodes −iwiDenote hi i−1 xi+1 0, xi , i == x. .−kxi−1 Then after
                                                         ˆ
Approximate NtD Map
          Awi −              ˆ   −             =  −      hi 2, ˆi , ˆ ,
                                                               .        x       (9)
                  ˆ       hi           hi−1
discretization in hi we have the following semi-discrete problem
                   x
                                  1 w2 − w1        1
                         Aw −
                 1 wi+1 −1wi ˆ wi − wi−1 = ˆ ϕ,             wk+1 = 0,         (10)
         Awi −                   −1
                                 h       h1    = 0, 1
                                                  h      i = 2, . . . , kx ,     (9)
                 ˆ
                 hi       hi           hi−1
 where wi = w(xi , y).           1 w2 − w1         1
                           Neumann-to-Dirichlet Map w
                         Aw −
 The crucial fact is that we1 canˆobtain hNeumann-to-Dirichlet k+1 = 0,
                                                =     ϕ,        map             (10)
                                 h1        1       ˆ
                                                   h1
                                               m
where wi = w(xi , y).1 (y) = fk (A)ϕ(y) =
                   w                          ai fk (i2 π 2 ) sin(iπy),                         (11)
The crucial fact is that we can obtain Neumann-to-Dirichlet map
                                          i=1
                                            m
 where approximate impedance function fk (λ) can 2 2written as a continued frac-
                                                       be
 tion             w1 (y) = fk (A)ϕ(y) =       ai fk (i π ) sin(iπy),         (11)
          Approximate Impedance function can be written as a
                                          i=1 1
                      fk (λ) =                                              (12)
                               hcontinued fraction
                               ˆ 1λ +              1
where approximate impedance function fkλ+···+can be written as a continued frac-
                                      h1 + ˆ (λ)
                                           h2
                                                     1
                                                         1
                                                  hk−1 +    1
tion                                                     ˆ λ+ 1
                                                         hk
                                                              hk
                                               1
                      fk (λ) =                                               (12)
 The L2 error of the semidiscreteλsolution at x = 0 can be estimated by
                               ˆ
                               h1 + h +            1
                                                     1
                                           1   ˆ                    1
                                               h2 λ+···+            1                  1
                                                           hk−1 +                     −2
           ek = ||w(0, y) − w1 (y)||L2 [0,1] ≤ ||ϕ||         maxhk λ+ 1
                                                                ˆ
                                                                      h |fk (λ) − λ        |.   (13)
                                                       λ∈[π 2 ,(mπ)2 ] k
The L2 error of the semidiscrete solution at x = 0 can be estimated by
 Thus the problem of grid optimization with respect to the Neumann-to-Dirichlet
ˆ
                  hi        h
                        fk (λ)i =     hi−1                                                         (12)
                              ˆ 1λ +
                              h                   1
                                                    1
                                     h1 + ˆ− w
                                 1 w2 h2 λ+···+         1 1
                                                         1
                                               1 h
                          Aw1 −                    k−1 + ˆ ϕ,1
                                                    = hk λ+                  wk+1 = 0,                    (10)
                                 ˆ
                                 h1       h1            ˆ
                                                        h 1 hk
Approximation error
 The L2 error of the semidiscrete solution at x = 0 can be estimated by
where w = w(x , y).
        i          i
                                                                                    1
                                                                                   −2
The crucial kfact is that− w1can L2 [0,1] ≤ Neumann-to-Dirichlet λ
          e = ||w(0, y) we (y)|| obtain ||ϕ||     max    |fk (λ) −                map
                                                                                    |.             (13)
                                                  2    2
                                                  λ∈[π ,(mπ) ]
                                              m
 Thus the problemw (y) = f (A)ϕ(y) = respect(i2 π 2 ) sin(iπy),
                   of grid optimization with a f to the Neumann-to-Dirichlet (11)
                    1        k                  i k
 map error can be reduced to the uniformi=1 rational approximation of the inverse
           Impedance function rational approximation error
 square root.
wherelarge m the optimal approximation onfa (λ) can be written as λmax ] yields frac-
 For approximate impedance function k spectral interval [λmin , a continued
tion
                                         1
                                        −2
                                                  1   π2k
                      maxk (λ) = − λ | = O exp 1
                            f      |fk (λ)                                    ,                    (14) (12)
                   λ∈[π 2 ,(mπ)2 ]      ˆ
                                        h1 λ + h + 1
                                                        λmin
                                                    log λmax
                                             ˆ  1                 1
                                             h2 λ+···+                1
                                                         hk−1 +
 Now if we perform standard finite difference discretization in y, our problem
                                                                  hk λ+ 1
                                                                  ˆ
                                                                        hk
 becomes
     2         Boundary condition approximation error
The L error of the semidiscrete solution at x = 0 can be estimated by
               j−1      j    j+1                                                          1
                   − 2wi + wi
              wi ||w(0, y) − w1 (y)||1 2 [0,1]i+1 − wi wmax i−1 |fk (λ)                  −2
            ek =
            −                    −   L w ≤ ||ϕ|| − i2− w 2                         −λ         |.          (13)
                       2            ˆi                λ∈[π ,(mπ) ] = 0,                            (15)
                      hy            h           hi        hi−1
Thus the problem of grid optimization with respect jto the .Neumann-to-Dirichlet
                                       i = 2, . . . , kx , = 2, . . , ky
                     j−1      j   j+1             j       j
map error can be reduced to the wi
                   wi − 2wi + uniform rationalwapproximation of the inverse
                                         1 w2 − 1                1 j
                 −                    −                      =      ϕ ,  (16)
square root.               hy2           ˆ
                                         h1          h1          ˆ
                                                                h1
Grids with Geometric Sequences
of Steps
 For large condition number, the optimal
 grids are close to geometric progressions
                                                                                                      Grid for Nx = 10, Ny = 10
                                                                                 1


                                                      √                         0.9

                                         α = eπ/            k
                                                                                0.8
                                                √
                                               − kπ
                                        h1 = e                                  0.7

                                                 √                              0.6
                            i−1         π(i−k−1)/ k
            hi = h1 α             =e
                                                    √                           0.5
                           ˆ
                           h1 = h1 /(1 + α)                                     0.4


                     ˆ i = h1 αi−1 =
                           ˆ            hi                                      0.3
                     h                   √
                                      1+ α                                      0.2

                                                                                0.1

                                                                                 0
                                                                                      0   0.1   0.2        0.3        0.4         0.5   0.6   0.7
                             1
                                                            √
                                    1+o(1)
  max
   √            |fk (λ) − λ− 2 | ≤ h1        = e[−π+o(1)]       k
                                                                    , h1 → +0
λ∈[e   2π ,∞]
1
                             fk (λ) =                                                                       (12)
                                        ˆ
                                        h1 λ +                     1
                                                                       1
                                                 h1 + ˆ                     1
                                                      h2 λ+···+                1
                                                                  hk−1 +
                                                                            ˆ λ+ 1
                                                                            hk
                                                                                 hk


Discretization method
  2 L2Discretization techniques= 0 can be estimated by
 The  error of the semidiscrete solution at x
  2.1    Explicit scheme
          ek = ||w(0, y) − w1 (y)||                  ≤ ||ϕ||           max          |fk (λ) − λ
                                                                                                   1
                                                                                                  −2
                                                                                                       |.   (13)
                                          L2 [0,1]
                                                                  λ∈[π 2 ,(mπ)2 ]
  Let the finite difference solution be defined at nodes xi , the finite difference deriva-
                             ˆ                              ˆ
 Thus the problem of grid optimization withi respect to the Neumann-to-Dirichlet after
  tives be defined at nodes xi . Denote h = xi+1 − xi , hi = xi − xi−1 Then
                                                                 ˆ    ˆ
  discretization in x we have Discretization in x
 map error can be reduced to the uniform rational approximation of the inverse
                               the following semi-discrete problem
 square root.
 For large m the optimalwi+1 − wi
                   1                 w − a spectral interval [λmin , λmax ] yields
                         approximationi on wi−1
             Awi −                           −                             = 0,         i = 2, . . . , k,           (9)
                        ˆ
                        hi         hi            hi−1
                                             1              π2k
                   max          |fk (λ) − λ− 2w2 − w1
                                         1     | = O exp 1 λm in                             ,              (14)
                λ∈[π 2 ,(mπ)2 Aw1 −
                              ]                         = logϕ, ax                         wk+1 = 0,               (10)
                                        ˆ1
                                        h         h1      ˆ
                                                          h 1 λm
 Now if we perform standard finite difference discretization in y, our problem
  where w = w(xi , y).
 becomes i                  Discretization in y
  The crucial fact is that we can obtain Neumann-to-Dirichlet map
              j−1          j     j+1            m
             wi     −   2wi + wi= f (A)ϕ(y) =− wi a f i(i2 π 2i−1
                          w1 (y)       1 wi+1        w − w ) sin(iπy),                                             (11)
         −                           −
                                     k             − k
                                                   i              = 0,                                      (15)
                          2
                         hy            ˆi
                                       h    hi i=1      hi−1
                                                          i = 2, . . . , kx , j = 2, . . . , ky
  where approximate impedance function fk (λ) jcan be written as a continued frac-
                    j−1      j     j+1              j
  tion            wi − 2wi + wi          1 w2 − w1           1 j
                −                      −                =         ϕ ,   (16)
                            2
                           hy
                     fk (λ) =            h 1 1 h1
                                         ˆ                   ˆ
                                                             h1               (12)
                               ˆ                 1
                               h1 λ + h + w1 = 0, 1i = 2, . . . , kx ,  (17)
                                                          1   ˆ i
                                                              h λ+···+              1
ˆ j−1       ˆ        j      ˆ j+1                  j
−h1 h1 w1 + 2h1 h1 + h2 w1 − h1 h1 w1 − h1 h2 ϕj = h2 w2 , j = 2, . . . , ky (22)
                      y                     y       y
                                                                      1
                                                                     wi = 0,              i = 2, . . . , kx , (23)
Numerical Scheme                                               j
                                                                     ky
                                                                    wi     = 0,           i = 2, . . . , kx , (24)
                                                              w1 = w(yj ),                 j = 1, . . . , ky (25)
                                                                 j
                                                   k
                                                                wkx +1 = 0,                j = 1, . . . , ky (26)
                                                  wi y   = 0,       i = 2, . . . , kx ,              (18)
                                          j
So, starting from w1 (y) we consequentially obtain),the solutionkon the lines parallel
                                         w1 = w(yj    j = 1, . . . , y    (19)
to y axis w2 (y), w3 (y), . . . wkx (y).    j
                                          wk +1 = 0, j = 1, . . . , ky    (20)
                                                   x


     After regrouping terms in the first two discrete equations, we have
                                     ˆ
                                     hi h2                     h i h2 j
            ˆ j−1          ˆ     2       y     j      ˆ j+1         y            j
        −hi hi wi +    2hi hi + hy +         w i − hi hi w i −        wi−1 = h2 wi+1 , (21)
                                                                              y
                                     hi−1                      hi−1
                                                         i = 2, . . . , kx ,     j = 2, . . . , ky
          ˆ j−1       ˆ        j      ˆ j+1                  j
      −h1 h1 w1 + 2h1 h1 + h2 w1 − h1 h1 w1 − h1 h2 ϕj = h2 w2 , j = 2, . . . , ky (22)
                            y                     y       y
                                                                 1
                                                                wi = 0,        i = 2, . . . , kx , (23)
                                                                k
                                                              wi y = 0,        i = 2, . . . , kx , (24)
                                                           j
                                                          w1 = w(yj ),           j = 1, . . . , ky (25)
                                                             j
                                                            wkx +1 = 0,          j = 1, . . . , ky (26)

     So, starting from w1 (y) we consequentially obtain the solution on the lines parallel
     to y axis w2 (y), w3 (y), . . . wkx (y).
Numerical scheme not using the
NtD map

                                                                                                                Sparse discrete system matrix
                                                                                                   0
          Numerical scheme with the
2.2 Implicit scheme                                                                                5
          sparse block-banded matrix
The discrete system (15) above can also be solved using only the knowledge of                     10
given Neumann data on the left boundary. I.e. we proceed not using Neumann-
to-Dirichlet map described above.                                                                 15
We rewrite the system in the form
                                                                                                  20
                            ˆ
                            h h2                      h h2
    ˆ j−1
−hi hi wi +       ˆ i + h2 + i y
              2hi h                  j      ˆ i wj+1 − i y wj − h2 wj = 0, (27)
                                   w i − hi h i
                         y                                       y i+1
                            hi−1                      hi−1 i−1                                    25
                                                i = 2, . . . , kx ,    j = 2, . . . , ky
                                                                                                  30
     ˆ j−1       ˆ        j      ˆ j+1       j
 −h1 h1 w1 + 2h1 h1 + h2 w1 − h1 h1 w1 − h2 w2 = h1 h2 ϕj , j = 2, . . . , ky (28)
                       y                  y          y
                                                       1                                          35
                                                      wi = 0,         i = 2, . . . , kx , (29)
                                                       k
                                                     wi y = 0,        i = 2, . . . , kx , (30)    40
                                                   j
                                                  w1 = w(yj ),         j = 1, . . . , ky (31)
                                                    j                                             45
                                                   wkx +1 = 0,         j = 1, . . . , ky (32)
                                                                                           (33)        0   10          20              30       40
                                                                                                                            nz = 212
We can see that the system matrix is sparse and block-banded.
                Sparse discrete system matrix
 0
Numerical Results

                                                              Analytic Solution

    4      Numerical results
    Boundary Condition         0.35

    We apply our method for the problem (1) with Neumann b
                                0.3
                       Analytic Solution
                               0.25                           −πx s
    ϕ(y) = sin(πy). Then the analytic solution is w(x, y) = e
                                0.2

                               0.15
    Analytic solution           0.1
 ) with Neumann boundary condition
                             0.05

n is w(x, y) = e−πx sin(πy).    0
                                1
                                  Analytic Solution
                                      0.8                                                   1.5
                                            0.6
                                                                                        1
                                                  0.4
                                                        0.2                       0.5
                                                              0   0

    0.35
Error for the solution on the
boundary
                 −11            Abs. error in approximation of Dir BC
             x 10
       1.4


       1.2


        1


       0.8


       0.6


       0.4


       0.2


        0
             0      0.1   0.2    0.3    0.4     0.5     0.6     0.7     0.8   0.9   1
Error for the solution on the
boundary
                 −11            Abs. error in approximation of Dir BC
             x 10
       1.4


       1.2


        1


       0.8


       0.6


       0.4


       0.2


        0
             0      0.1   0.2    0.3    0.4     0.5     0.6     0.7     0.8   0.9   1
Error for the solution on the
boundary
                 −11            Abs. error in approximation of Dir BC
             x 10
       1.4


       1.2
                                                                              ...with only 14
                                                                               discretization
        1                                                                         points!
       0.8


       0.6


       0.4


       0.2


        0
             0      0.1   0.2    0.3    0.4     0.5     0.6     0.7     0.8    0.9   1
Errors for the standard methods

Method with matrix inversion                                                                         Equally spaced
     and optimal grid                                                                           Finite Difference method
and the one obtained with the implicit method.
           −4               Abs. error in approximation of Dir BC                                                  Abs. error for node x2
       x 10
 0.5                                                                                0.035

  0
                                                                                     0.03

−0.5
                                                                                    0.025
 −1


−1.5                                                                                 0.02


 −2
                                                                                    0.015

−2.5
                                                                                     0.01
 −3

                                                                                    0.005
−3.5


 −4                                                                                    0
       0        0.1   0.2    0.3    0.4     0.5     0.6     0.7     0.8   0.9   1           0    0.1   0.2   0.3    0.4     0.5     0.6     0.7   0.8   0.9   1
Error depending on the # of pts

  Table 1: Max and l2 error in approximation of the solution on the boundary given
  by Neumann-to-Dirichlet map

             of steps         Max error                  l2 error
                4       4.435465319653220e-04      0.001507654161160
                6       2.516176484440580e-05    5.419957700473768e-05
                8       2.071832056138589e-06    4.462819749004914e-06
               10       8.695114162016182e-08    1.872966830920883e-07
               12       2.251853870038900e-09    4.850594880461687e-09
               14       1.253025461167567e-11    2.699071837316973e-11
               16       2.361721929133864e-12    5.087230550790546e-12
Conclusion
                                                                 1
                            ∃! p(λ) : p(λ) = min                √ − q(λ)
                                                                    max      (41)
                                           q∈Pk λ∈[λmin ,λmax ]   λ
           We used Wolfram Mathematica intrinsic function MiniMaxApproximation to
           obtain p(λ). The next figure shows the approximation error

                11
           10


                12
           10

   ➡ Rational approximant of the NtD map can be obtained.
                13
           10


                14

      Exponential convergence at prescribed surfaces
           10



                15
           10



   ➡ Technique can be generalized to a variety of PDE problems
                            10               100                    1000                104

           To obtain grid steps we rewrite obtained approximation of the impedance
           function in the form of continued fraction (12). We proceed with Euclidean type
   ➡ Method can be used in a multidomain framework
           algorithm with 2k polynomial divisions, i.e.
                                              ck−1 λk−1 + ck−2 λk−2 + · · · + c0
                                     p(λ) =                                                        (42)
                                                dk λk + dk−1 λk−1 + · · · + d0
                                                                  1
                                                    = d λk +d λk−1 +···+d
                                                                k          k−1            0
                                                            ck−1 λk−1 +ck−2 λk−2 +···+c0
                                                            1
                        =                         dk ck−2
                                                                                               ,
                                                                                 d c0
                                         dk−1 −    ck−1
                                                            λk−1 +···+ d1 − c k         λ+d0
                             dk                                                  k−1
                            ck−1 λ   +            ck−1 λk−1 +ck−2 λk−2 +···+c0

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Optimal Finite Difference Grids

  • 1. Optimal Finite Difference Grids Oleksiy Varfolomiyev Advisor Prof. Michael S. Siegel Prof. Michael R. Booty NJIT, June 2011
  • 3. Motivation question Finite Differences??? But they are only 2nd order accurate, aren’t they?
  • 5. Motivation answer How about exponential super-convergence and spectrally accurate grids?!
  • 6. Outline ✤ Model Problem ✤ Discretization techniques ✤ Optimal discretization grids ✤ Numerical results
  • 7. What are we solving? 1 Model problem Laplace equation on a semi-infinite strip Consider the following boundary value problem: Laplace equation on a semi- nfinite strip with the given Neumann data on the left boundary and Dirichlet data elsewhere ∂ 2 w(x, y) ∂ 2 w(x, y) − 2 − 2 = 0, (x, y) ∈ [0, ∞) × [0, 1], (1) ∂y ∂x ∂w (0, y) = −ϕ(y), y ∈ [0, 1], (2) ∂x w|x=∞ = 0, (3) w(x, 0) = 0, w(x, 1) = 0, x ∈ [0, ∞). (4) Assume m ϕ(y) = ai sin(iπy) (5) i=1
  • 8. The∂w (0, y) = −ϕ(y), ybe [0, 1], as a problem is defined on span{sin (πy), . . . , sin (mπy)}, spA = above problem can ∈ studied e can obtain Dirichlet data on the left boundary using 2 w(x) ∂x w|x=∞ = 0, (2) (3) t map. Equation (6) gives Aw(x) = d dx2 , therefore w(x, 0) = 0, nd now we can use given in (7) Neumann data to get at 2 w(x, 1) = 0, =d0,w(x) ∞). (3) w|x=∞ x ∈ [0, (4) Aw(x) − = 0, x ∈ [0, ∞) Model Problem w(0) = f (A)ϕ, w(x, 0) = 0, w(x, 1) = 0, x ∈ [0, ∞). dx 2 (4) Assume mpedance function. dw m (x = 0) = −ϕ Assume dx ϕ(y) = m ai sin(iπy) (5) ϕ(y) = ai sin(iπy) i=1 (5) x=∞ = 0, w| i=1 The above problem can bewhere Aas = problem is defined on span{sin (πy), . . . , s studied a − ∂ 22 The above problem can be studied as a problem ∂y {π 2d2 w(x) 2 }. We can obtain Dirichlet data on the , . . . , (mπ) 2 w(x) d − Aw(x)Neumann-to-Dirichlet ∈ [0, ∞)Equation (6) gives Aw(x) Aw(x) − = 0, = x ∈ [0, ∞)map. 2 0, x (6) (6) dx2 dx dw(x) 1 − dx dw A 2 dw and now we can use given in (7) Neum = w(x) x = 0 dx (x = 0) = −ϕ (x = 0) = −ϕ (7) (7) 2 dx w|x=∞ = 0, w(0) = f (A)ϕ, w|x=∞ = 0, (8) (8) 1 ∂2 −2 where A A = − ∂ 2 defined on (λ) onλ span{sin.(πy), . . . , sin (mπy)}, spA = where = − is here f span{sin (πy), . . , sin (mπy)}, spA = = is impedance function. isNeumann-to-Dirichlet Map ∂y 2defined ∂y 2 {π 2 , .2 . , (mπ)2 }. 2 We can obtain Dirichlet data on the left boundary using . {π , . . . , (mπ) }. We can obtain Dirichlet data on d2 w(x)left boundary using the Neumann-to-Dirichlet map. Equation (6) gives Aw(x) = dx2 , therefore d2 w(x) Neumann-to-Dirichlet map. Equation dw(x) 1 (6) gives Aw(x) = dx2 , therefore − dx = A 2 w(x) and now we can use given dw(x) 1 in (7) Neumann data to get at − x = 0 dx = A 2 w(x) and now we can use given in (7) Neumann data to get at x=0 w(0) = f (A)ϕ, 1 −2 w(0) = f (A)ϕ, here f (λ) = λ is impedance function. 1 −2 here f (λ) = λ is impedance function.
  • 9. Idea ✦ Continuum NtD maps are Stieltjes-Markov functions of A ✦ Finite Difference NtD maps are rational functions of A ✦ Grids can be computed by minimizing the rational approximation error of NtD maps on the spectrum of A
  • 10. Method outline ๏ Write down the Finite Difference Scheme ๏ Find the Neumann-to-Dirichlet map as f(A) ๏ Find the Rational Approximation of f(A) on the spectrum of A, obtain the grid steps ๏ Solve the discrete system
  • 11. 2.1 Explicit scheme xi . Denote hi = xi+1 − xi , hi = xi − xi−1 Then after tives be defined at nodes ˆ ˆ ˆ discretization in x we have the following semi-discrete problem Let the finite difference solution be defined at nodes xi , the finite difference deriva- 1 wi+1 x . wi − w = tives be defined at nodes −iwiDenote hi i−1 xi+1 0, xi , i == x. .−kxi−1 Then after ˆ Approximate NtD Map Awi − ˆ − = − hi 2, ˆi , ˆ , . x (9) ˆ hi hi−1 discretization in hi we have the following semi-discrete problem x 1 w2 − w1 1 Aw − 1 wi+1 −1wi ˆ wi − wi−1 = ˆ ϕ, wk+1 = 0, (10) Awi − −1 h h1 = 0, 1 h i = 2, . . . , kx , (9) ˆ hi hi hi−1 where wi = w(xi , y). 1 w2 − w1 1 Neumann-to-Dirichlet Map w Aw − The crucial fact is that we1 canˆobtain hNeumann-to-Dirichlet k+1 = 0, = ϕ, map (10) h1 1 ˆ h1 m where wi = w(xi , y).1 (y) = fk (A)ϕ(y) = w ai fk (i2 π 2 ) sin(iπy), (11) The crucial fact is that we can obtain Neumann-to-Dirichlet map i=1 m where approximate impedance function fk (λ) can 2 2written as a continued frac- be tion w1 (y) = fk (A)ϕ(y) = ai fk (i π ) sin(iπy), (11) Approximate Impedance function can be written as a i=1 1 fk (λ) = (12) hcontinued fraction ˆ 1λ + 1 where approximate impedance function fkλ+···+can be written as a continued frac- h1 + ˆ (λ) h2 1 1 hk−1 + 1 tion ˆ λ+ 1 hk hk 1 fk (λ) = (12) The L2 error of the semidiscreteλsolution at x = 0 can be estimated by ˆ h1 + h + 1 1 1 ˆ 1 h2 λ+···+ 1 1 hk−1 + −2 ek = ||w(0, y) − w1 (y)||L2 [0,1] ≤ ||ϕ|| maxhk λ+ 1 ˆ h |fk (λ) − λ |. (13) λ∈[π 2 ,(mπ)2 ] k The L2 error of the semidiscrete solution at x = 0 can be estimated by Thus the problem of grid optimization with respect to the Neumann-to-Dirichlet
  • 12. ˆ hi h fk (λ)i = hi−1 (12) ˆ 1λ + h 1 1 h1 + ˆ− w 1 w2 h2 λ+···+ 1 1 1 1 h Aw1 − k−1 + ˆ ϕ,1 = hk λ+ wk+1 = 0, (10) ˆ h1 h1 ˆ h 1 hk Approximation error The L2 error of the semidiscrete solution at x = 0 can be estimated by where w = w(x , y). i i 1 −2 The crucial kfact is that− w1can L2 [0,1] ≤ Neumann-to-Dirichlet λ e = ||w(0, y) we (y)|| obtain ||ϕ|| max |fk (λ) − map |. (13) 2 2 λ∈[π ,(mπ) ] m Thus the problemw (y) = f (A)ϕ(y) = respect(i2 π 2 ) sin(iπy), of grid optimization with a f to the Neumann-to-Dirichlet (11) 1 k i k map error can be reduced to the uniformi=1 rational approximation of the inverse Impedance function rational approximation error square root. wherelarge m the optimal approximation onfa (λ) can be written as λmax ] yields frac- For approximate impedance function k spectral interval [λmin , a continued tion 1 −2 1 π2k maxk (λ) = − λ | = O exp 1 f |fk (λ) , (14) (12) λ∈[π 2 ,(mπ)2 ] ˆ h1 λ + h + 1 λmin log λmax ˆ 1 1 h2 λ+···+ 1 hk−1 + Now if we perform standard finite difference discretization in y, our problem hk λ+ 1 ˆ hk becomes 2 Boundary condition approximation error The L error of the semidiscrete solution at x = 0 can be estimated by j−1 j j+1 1 − 2wi + wi wi ||w(0, y) − w1 (y)||1 2 [0,1]i+1 − wi wmax i−1 |fk (λ) −2 ek = − − L w ≤ ||ϕ|| − i2− w 2 −λ |. (13) 2 ˆi λ∈[π ,(mπ) ] = 0, (15) hy h hi hi−1 Thus the problem of grid optimization with respect jto the .Neumann-to-Dirichlet i = 2, . . . , kx , = 2, . . , ky j−1 j j+1 j j map error can be reduced to the wi wi − 2wi + uniform rationalwapproximation of the inverse 1 w2 − 1 1 j − − = ϕ , (16) square root. hy2 ˆ h1 h1 ˆ h1
  • 13. Grids with Geometric Sequences of Steps For large condition number, the optimal grids are close to geometric progressions Grid for Nx = 10, Ny = 10 1 √ 0.9 α = eπ/ k 0.8 √ − kπ h1 = e 0.7 √ 0.6 i−1 π(i−k−1)/ k hi = h1 α =e √ 0.5 ˆ h1 = h1 /(1 + α) 0.4 ˆ i = h1 αi−1 = ˆ hi 0.3 h √ 1+ α 0.2 0.1 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 1 √ 1+o(1) max √ |fk (λ) − λ− 2 | ≤ h1 = e[−π+o(1)] k , h1 → +0 λ∈[e 2π ,∞]
  • 14. 1 fk (λ) = (12) ˆ h1 λ + 1 1 h1 + ˆ 1 h2 λ+···+ 1 hk−1 + ˆ λ+ 1 hk hk Discretization method 2 L2Discretization techniques= 0 can be estimated by The error of the semidiscrete solution at x 2.1 Explicit scheme ek = ||w(0, y) − w1 (y)|| ≤ ||ϕ|| max |fk (λ) − λ 1 −2 |. (13) L2 [0,1] λ∈[π 2 ,(mπ)2 ] Let the finite difference solution be defined at nodes xi , the finite difference deriva- ˆ ˆ Thus the problem of grid optimization withi respect to the Neumann-to-Dirichlet after tives be defined at nodes xi . Denote h = xi+1 − xi , hi = xi − xi−1 Then ˆ ˆ discretization in x we have Discretization in x map error can be reduced to the uniform rational approximation of the inverse the following semi-discrete problem square root. For large m the optimalwi+1 − wi 1 w − a spectral interval [λmin , λmax ] yields approximationi on wi−1 Awi − − = 0, i = 2, . . . , k, (9) ˆ hi hi hi−1 1 π2k max |fk (λ) − λ− 2w2 − w1 1 | = O exp 1 λm in , (14) λ∈[π 2 ,(mπ)2 Aw1 − ] = logϕ, ax wk+1 = 0, (10) ˆ1 h h1 ˆ h 1 λm Now if we perform standard finite difference discretization in y, our problem where w = w(xi , y). becomes i Discretization in y The crucial fact is that we can obtain Neumann-to-Dirichlet map j−1 j j+1 m wi − 2wi + wi= f (A)ϕ(y) =− wi a f i(i2 π 2i−1 w1 (y) 1 wi+1 w − w ) sin(iπy), (11) − − k − k i = 0, (15) 2 hy ˆi h hi i=1 hi−1 i = 2, . . . , kx , j = 2, . . . , ky where approximate impedance function fk (λ) jcan be written as a continued frac- j−1 j j+1 j tion wi − 2wi + wi 1 w2 − w1 1 j − − = ϕ , (16) 2 hy fk (λ) = h 1 1 h1 ˆ ˆ h1 (12) ˆ 1 h1 λ + h + w1 = 0, 1i = 2, . . . , kx , (17) 1 ˆ i h λ+···+ 1
  • 15. ˆ j−1 ˆ j ˆ j+1 j −h1 h1 w1 + 2h1 h1 + h2 w1 − h1 h1 w1 − h1 h2 ϕj = h2 w2 , j = 2, . . . , ky (22) y y y 1 wi = 0, i = 2, . . . , kx , (23) Numerical Scheme j ky wi = 0, i = 2, . . . , kx , (24) w1 = w(yj ), j = 1, . . . , ky (25) j k wkx +1 = 0, j = 1, . . . , ky (26) wi y = 0, i = 2, . . . , kx , (18) j So, starting from w1 (y) we consequentially obtain),the solutionkon the lines parallel w1 = w(yj j = 1, . . . , y (19) to y axis w2 (y), w3 (y), . . . wkx (y). j wk +1 = 0, j = 1, . . . , ky (20) x After regrouping terms in the first two discrete equations, we have ˆ hi h2 h i h2 j ˆ j−1 ˆ 2 y j ˆ j+1 y j −hi hi wi + 2hi hi + hy + w i − hi hi w i − wi−1 = h2 wi+1 , (21) y hi−1 hi−1 i = 2, . . . , kx , j = 2, . . . , ky ˆ j−1 ˆ j ˆ j+1 j −h1 h1 w1 + 2h1 h1 + h2 w1 − h1 h1 w1 − h1 h2 ϕj = h2 w2 , j = 2, . . . , ky (22) y y y 1 wi = 0, i = 2, . . . , kx , (23) k wi y = 0, i = 2, . . . , kx , (24) j w1 = w(yj ), j = 1, . . . , ky (25) j wkx +1 = 0, j = 1, . . . , ky (26) So, starting from w1 (y) we consequentially obtain the solution on the lines parallel to y axis w2 (y), w3 (y), . . . wkx (y).
  • 16. Numerical scheme not using the NtD map Sparse discrete system matrix 0 Numerical scheme with the 2.2 Implicit scheme 5 sparse block-banded matrix The discrete system (15) above can also be solved using only the knowledge of 10 given Neumann data on the left boundary. I.e. we proceed not using Neumann- to-Dirichlet map described above. 15 We rewrite the system in the form 20 ˆ h h2 h h2 ˆ j−1 −hi hi wi + ˆ i + h2 + i y 2hi h j ˆ i wj+1 − i y wj − h2 wj = 0, (27) w i − hi h i y y i+1 hi−1 hi−1 i−1 25 i = 2, . . . , kx , j = 2, . . . , ky 30 ˆ j−1 ˆ j ˆ j+1 j −h1 h1 w1 + 2h1 h1 + h2 w1 − h1 h1 w1 − h2 w2 = h1 h2 ϕj , j = 2, . . . , ky (28) y y y 1 35 wi = 0, i = 2, . . . , kx , (29) k wi y = 0, i = 2, . . . , kx , (30) 40 j w1 = w(yj ), j = 1, . . . , ky (31) j 45 wkx +1 = 0, j = 1, . . . , ky (32) (33) 0 10 20 30 40 nz = 212 We can see that the system matrix is sparse and block-banded. Sparse discrete system matrix 0
  • 17. Numerical Results Analytic Solution 4 Numerical results Boundary Condition 0.35 We apply our method for the problem (1) with Neumann b 0.3 Analytic Solution 0.25 −πx s ϕ(y) = sin(πy). Then the analytic solution is w(x, y) = e 0.2 0.15 Analytic solution 0.1 ) with Neumann boundary condition 0.05 n is w(x, y) = e−πx sin(πy). 0 1 Analytic Solution 0.8 1.5 0.6 1 0.4 0.2 0.5 0 0 0.35
  • 18. Error for the solution on the boundary −11 Abs. error in approximation of Dir BC x 10 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
  • 19. Error for the solution on the boundary −11 Abs. error in approximation of Dir BC x 10 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
  • 20. Error for the solution on the boundary −11 Abs. error in approximation of Dir BC x 10 1.4 1.2 ...with only 14 discretization 1 points! 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
  • 21. Errors for the standard methods Method with matrix inversion Equally spaced and optimal grid Finite Difference method and the one obtained with the implicit method. −4 Abs. error in approximation of Dir BC Abs. error for node x2 x 10 0.5 0.035 0 0.03 −0.5 0.025 −1 −1.5 0.02 −2 0.015 −2.5 0.01 −3 0.005 −3.5 −4 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
  • 22. Error depending on the # of pts Table 1: Max and l2 error in approximation of the solution on the boundary given by Neumann-to-Dirichlet map of steps Max error l2 error 4 4.435465319653220e-04 0.001507654161160 6 2.516176484440580e-05 5.419957700473768e-05 8 2.071832056138589e-06 4.462819749004914e-06 10 8.695114162016182e-08 1.872966830920883e-07 12 2.251853870038900e-09 4.850594880461687e-09 14 1.253025461167567e-11 2.699071837316973e-11 16 2.361721929133864e-12 5.087230550790546e-12
  • 23. Conclusion 1 ∃! p(λ) : p(λ) = min √ − q(λ) max (41) q∈Pk λ∈[λmin ,λmax ] λ We used Wolfram Mathematica intrinsic function MiniMaxApproximation to obtain p(λ). The next figure shows the approximation error 11 10 12 10 ➡ Rational approximant of the NtD map can be obtained. 13 10 14 Exponential convergence at prescribed surfaces 10 15 10 ➡ Technique can be generalized to a variety of PDE problems 10 100 1000 104 To obtain grid steps we rewrite obtained approximation of the impedance function in the form of continued fraction (12). We proceed with Euclidean type ➡ Method can be used in a multidomain framework algorithm with 2k polynomial divisions, i.e. ck−1 λk−1 + ck−2 λk−2 + · · · + c0 p(λ) = (42) dk λk + dk−1 λk−1 + · · · + d0 1 = d λk +d λk−1 +···+d k k−1 0 ck−1 λk−1 +ck−2 λk−2 +···+c0 1 = dk ck−2 , d c0 dk−1 − ck−1 λk−1 +···+ d1 − c k λ+d0 dk k−1 ck−1 λ + ck−1 λk−1 +ck−2 λk−2 +···+c0

Editor's Notes