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248                                               8.  Stability  and  Transition


         where  now
                              C3            ci  =  £1,  c 4  =  £ 2         (8.2.6)
                                  6  + 6'
         The  difference  equations  (8.2.3)  and  (8.2.5)  have trivial solutions  fj  =  Sj  =  (f)j  —
         gj  =  0 for  all  j  and  we shall  use the  iteration  procedure  described  below  to  find
         the  special  parameter  values  for  which  nontrivial  solutions  exist.
            Since  the  Orr-Sommerfeld  equation  and  the  boundary  conditions  are  ho-
         mogenous, the trivial solution  (f)(y) =  0 is valid  for  all values of a,  /?, uo and  R.  For
         this  reason  to  compute  the  eigenvalues  and  the  eigenfunctions  we  first  replace
         the  boundary  condition  0'(O)  =  0  (that  is  f 0  =  0)  of  Eq.  (8.2.5)  by  0"(O)  =  1,
         that  is  so  =  1.  Now  the  difference  equations  have  a  non-trivial  solution  since
         ^"(O)  7^ 0  and  we  seek  to  adjust  or  to  determine  parameter  values  so that  the
         original  boundary  condition  is satisfied.  This  is achieved  by  an  iteration  scheme
         based  on  Newton's  method.  Specifically,  we  first  write  the  wall  boundary  con-
         ditions
                              0o  =  (ri)o  =  O,  so  =  (r 2 )o  =  l     (8.2.7)
         and  the  edge  boundary  conditions  in  Eq.  (8.2.5b)  and  Eq.  (8.2.3)  in  matrix-
        vector  form  as  in  Eq.  (4.4.29),  with  Sj  and  fj  defined  by

                                       4>j         (nh
                                       S J
                                 *;  =    .  f J  =                        (8.2.
                                       h
                                       9j
        and  the  Aj,  Bj^  Cj  denote  4  4 matrices  given  by

                 1    0   0  0           1      0    -{c 3)j  0
                 0    1 0  0            0       1      0     -(c 3)j
            =                 ,  Aj  —
        A 0                                                          1  <  j  <  J
               (ci)i  (c 3 )i  1  0   (Cl)j+1  (C3)i+1  1     0
               (C2)i  (c 4 )i  0  1   (C2)j+1  (C4)j + 1  0   1
                                     1    0    -{c z)j  0
                                     0    1    0     "(C3)j
                             Aj  =
                                    (ci)  (cs)  1     0
                                     0  (c 4)  0       1
                                - 1  0  -(c 3 )j  0
                                                c
                                0  - 1   0    "( s)i
                         Bj  =                         1  <  3  <  J
                                0   0    0      0
                                0   0    0      0
                                0      0     0  0
                                0      0     0  0
                          =                           0  <  j  <  J  -  1  (8.2.9)
                       c 3
                             (ci)j+i  (c 3 )i+i  - 1  0
                             (c 2 )i+i  (c 4 ) J + i  0  - 1
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