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Differential models for grid generation  139

                        5.7.2 Numerical aspects


                        Weight function equation
                        Here we present a standard finite-difference scheme for solving eqn (5.83). The interval
                        0   ξ   1 is divided into m equal intervals, and we can label (m + 1) discrete points
                        ξ i = i ξ, i = 0, 1,. ..,m, with  ξ = 1/m. It is useful to be able to evaluate certain
                        quantities at intermediate points, so that we have the finite-difference approximations
                        at the point corresponding to i:

                                     d   dx/dξ       1     dx/dξ         dx/dξ
                                        !      "
                                                                     −              ,      (5.95)
                                     dξ    ϕ          ξ      ϕ      1      ϕ      1
                                                 i                i+            i−
                                                                    2             2
                        with
                                         dx        x i+1 − x i  dx       x i − x i−1

                                                           ,                               (5.96)
                                         dξ    1      ξ         dξ   1       ξ
                                             i+                    i−
                                               2                     2
                        Then this finite-difference version of eqn (5.83) becomes

                                     1      1           1      1          1
                                               x i−1 −     +        x i +   x i+1  = 0,
                                   ( ξ) 2  ϕ  1       ϕ   1   ϕ  1      ϕ  1
                                           i−  2        i−  2  i+  2     i+  2
                        or
                                     −a i x i−1 + b i x i − c i x i+1 = 0,  i = 1, 2,..., (m − 1) ,  (5.97)

                                   1           1     1           1
                        with a i =    , b i =     +      , c i =   , i = 1, 2,...,(m − 1). Note that
                                  ϕ  1        ϕ  1  ϕ  1       ϕ  1
                                   i−  2      i−  2  i+  2      i+  2
                        b i = a i + c i .
                          Incorporating the end-conditions x 0 = a, x m = b, we obtain the tridiagonal matrix
                        equation
                                                                                        
                             b 1  −c 1  0    0     −       –       0         x 1        a 1 a
                          −a 2   b 2  −c 2  0     –       –       –     x 2         0   
                          0     −a 3   b 3  −c 3  –       –       –                 0   
                         
                                                                                             
                                                                         − 
                                                                                     
                          0      0    –     –     –       –       –                 −    .
                                                                                             
                         
                                                                         −  = 
                          −      –    –     –     –       –       0     −           −   
                                                                                        
                                                                                        
                            −    −−    –     0   −a m−2    b m−2  −c m−2     −           0
                            0     –    –     0     0     −a m−1   b m−1     x m−1      c m−1 b
                                                                                           (5.98)
                        The matrix of coefficients is also symmetric, since c i = a i+1 . Solutions may be obtained
                        efficiently using the Thomas Algorithm or SOR. On the accompanying disk the subdi-
                        rectory Book/one.d.gds contains the file line.SOR.f, which applies SOR to the problem.
                        Control function equation
                        A numerical scheme for solving eqn (5.80) is shown here. With the same uniform grid
                        as above along the computational ξ-axis, we have
                                    d x      x i+1 − 2x i + x i−1  and  dx     x i+1 − x i−1  ,
                                      2
                                    dξ 2  i       ( ξ) 2            dξ  i    2( ξ)
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