Page 36 - swp0000.dvi
P. 36

1.9.4     Solution for Rogue Wave


                        The standard form of NLS equation is (1.11); namely


                                                            2
                                                          
                                                                       2
                                                     +        +  ||  =0                          (1.16)
                                                          2
                   where  is a temporal variable, and  is the spatial variable in the moving

                   frame.
                       Equation (1.16) has a general rational solution in the form [107],


                                       s
                                             µ                                   ¶
                                         2              ( )+   ( )
                                                   
                           ( )=            (−1) +                               (2)         (1.17)
                            
                                                                ( )
                   where   ( ),   ( ) and   ( ) are polynomials in  and ,and

                     ( ) has no real zeros. Index  is the order of solution. In lowest-order

                   solution  =1, one gets [88-108]

                                                  s
                                                        ∙                         ¸
                                                    2       4(1 + 4)
                                      ( )=                                − 1  2              (1.18)
                                       1
                                                         1+16  +4       2
                                                                  2
                                                                    2
                    In second-order solution with  =2, the polynomials  2 ( ),  2 ( )
                   and  2 ( ) in this solution are given by [107]


                                                                                 √
                                                                           ¢
                                                                      2
                                                              ¡
                                                   4
                           2 ( )= 36 − 48 − 144        2  4() +1 − 24 2 1  − 960 ()      4
                                                    2
                                         −864 () +48 2 ()                                        (1.19)


                                      h    ³                      √      ´            ¡        ¢
                                                                                          2
                                                     4
                                                             2
                    2 ( )= 24  15 − 4 +12 +2 2 1  − 8()                 3  2 +1 − 16 ()     5
                                       µ                   ¶¸
                                                          1
                                                2     2
                                   + 2 2() −  −                                                   (1.20)
                                                          2




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