Page 137 - Basic Structured Grid Generation
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126  Basic Structured Grid Generation

                        involving a tridiagonal matrix (one in which the only non-zero entries appear along
                        the main diagonal and the two adjacent parallel diagonals on either side). These can
                        usually be solved efficiently using the Thomas Algorithm (otherwise known as the
                        Tridiagonal Matrix Algorithm), which is based on Gaussian Elimination. A typical set
                        of equations for the (N − 1) unknowns u 1 ,u 2 ,. ..,u N−1 ,is
                                     −a i u i−1 + b i u i − c i u i+1 = d i ,  i = 1, 2,. ..,(N − 1),  (5.41)

                        where the values of u 0 and u N are supposed known due to the given boundary
                        conditions.
                          The equivalent matrix equation is
                                                         Au = d                            (5.42)

                        where
                                                                                    
                                         b 1  −c 1  0    0      −       –       0
                                                        0      –       –       –    
                                        −a 2  b 2  −c 2
                                                                                    
                                         0   −a 3   b 3  −c 3   –       –       –
                                                                                    
                                                                                    
                                         0     0    –    –      –       –       –     ,
                                                                                    
                                 A = 
                                                                                    
                                      −       –    –    –      –       –       0    
                                                                                    
                                      −      −−    –    0    −a N−2  b N−2   −c N−2  
                                         0     –    –    0      0     −a N−1  b N−1
                                                           
                                         u 1              d 1
                                         u 2              d 2
                                                           
                                                           
                                         −                d 3
                                                           
                                                           
                                                           
                                         −    ,          −    .
                                 u =              d = 
                                                           
                                         −                −
                                                           
                                                           
                                         −
                                                     d N−2  
                                        u N−1            d N−1
                        Here we have transferred the terms −a 1 u 0 and −c N−1 u N to the right-hand side, where
                        they are assumed to have been incorporated into the d 1 and d N−1 terms, respectively.
                        So now we can assume a 1 = c N−1 = 0 in eqn (5.41).
                          A is diagonally dominant if
                                      b i > 0and b i > |a i |+ |c i |,  i = 1, 2,..., (N − 1)  (5.43)
                          The Thomas Algorithm replaces eqn (5.41) with a simpler recurrence relation from
                        which u i−1 has been eliminated, thereby reducing A to an upper triangular matrix. The
                        equations can then be solved by simple back substitution.
                          Suppose that we seek a recurrence relation of the form
                                          u i = P i u i+1 + Q i ,  i = 1, 2,...,(N − 1)    (5.44)
                        This is consistent with eqn (5.41) if
                                −a i (P i−1 u i + Q i−1 ) + b i u i − c i u i+1 = d i ,  i = 2, 3,...,(N − 1)
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