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4
                                      4
                                   3
                                        2
                      4 =           0           5  
                                  2
                                                 2
                              2(    0 − 3 0  )
                                                    ¢
                                 ¡  2              2 2            2    2
                      5 =6∆     0 − 3 0        −    0  (5    0
                                            3 0  2
                                     2                        2   2
                             +21 0  ) −           [3   0  (    0 +
                                           2
                                              0
                                         ¡
                                                            ¢
                                 2
                                            2
                              0  ) − 2∆     0 − 3 0  2 2 ]
                                     µ          ¶      2 2  µ                   ¶
                                 (20)   0   0          2      (20)    (20)
                                                       
                      6 =              −       −              0    +       
                                 3      2      2       2             2      1
                                                       
                                         2
                                            2
                                          2     ∙            (20)     1    ¡                  ¢  (20) ¸
                                                                                  2
                             +             0        −2 0      −               0 +3 0  2     
                                                 2 2
                                  2
                               (    0 − 3 0  )             2       0                      1
                                                         µ                    ¶
                                                       2
                                                           0             0
                                                 7 =          +(1+3)          
                                                      2      3             3
                   A negative sign for the product of the coefficients  is required for
                   amplitude modulation stability. For positive  the amplitude grows
                   and large wave is created. Many analytical methods have been used to
                   find rouge wave solutions from NLS equation [88, 106, 109]. The rational
                   solution is the important one that is given by [105]
                                           s
                                                  ∙                       ¸
                                              2      4(1 + 4)
                                        =                             − 1 exp(2)                  (2.19)
                                                           2
                                                  1+16  +4       2
                                                              2
                                                         (1)
                       For simplicity, we have set         ≡  The solution, equation (2.19), refers
                                                         1
                   to an amount of wave energy concentrated in a relatively very small area
                   in space. The rogue wave is usually an envelope of a carrier wave with a

                   wavelength smaller than the central region of the envelope.




                   2.4      Results and Discussion


                    In this chapter, the nonlinear DAWs in an unmagnetized dusty plasma

                   having Boltzmann electrons, nonthermal ions, hot and cold dust grains

                   have been investigated. The application of the derivative expansion per-


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