Page 47 - swp0000.dvi
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2
                                (22)      0  ¡  2         2   ¢
                               1  =            3   − 2∆   
                                          4
                                        2  2
                                             
                                (22)       ¡  2          2   ¢
                                2  =            − 2∆   
                                          3
                                       2   2
                                             
                                                             ⎡⎛                                  ⎞⎤
                                                                      2
                                                                                            2
                                                                                2
                                               2
                                                 2           3    0 (    0 +   0 )
                               (22)  =             0       ⎣⎝                                  ⎠⎦  
                                1        2            2    3
                                      2(    0 − 3  0 )       −2∆ (    0 − 3  0 )  2
                                                                            2
                                                                                         2
                                                             ⎛                                 ⎞
                                                                             2
                                                                                          2
                                                                   2
                                                                     0 (    0 +9  0 )
                                (22)           0
                                  =                         ⎝                                 ⎠  
                                2    2(    0 − 3  0 )  3  −2∆ (    0 − 3  0 )  2
                                                       2
                                          2
                                                                                       2
                                                                          2
                   where
                                                                          3
                                                             3
                                                                       3
                                                                                             2
                                                                             4
                                                                                 2
                                −2  3  ∙   0   0  3 4  µ    0     (    0 +   0 )  ¶¸
                          ∆ =                −     +              +     0                         
                                                                           2
                                                                                        2
                                  3 2     2    2   4    2        (    0 − 3  0 )  3
                                              3
                       For third order; ( ), the nonlinear self-interaction of the carrier wave
                   also leads to the creation of a zeroth order harmonic. Its strength is
                   analytically determined by taking  =0 component of the third-order
                   reduced equations which can be expressed as;
                                                                    ¯   ¯ 2
                                                      (2)       (20) ¯ (1)¯
                                                         =        ¯ ¯ 
                                                      0         3     1
                                                                    ¯   ¯ 2
                                                      (2)  =  (20) ¯ (1)¯
                                                                    ¯ ¯ 
                                                       0       1    1
                                                                    ¯   ¯ 2
                                                      (2)  =  (20) ¯ (1)¯
                                                                    ¯ ¯ 
                                                       0       2    1
                                                                    ¯   ¯ 2
                                                                    ¯ ¯ 
                                                      (2)  =  (20) ¯ (1)¯
                                                       0       1    1
                                                                    ¯   ¯ 2
                                                      (2)       (20) ¯ (1)¯
                                                         =        ¯ ¯                              (2.15)
                                                       0       2    1
                   where    (20)  ( =1 and 2)and     (20)
                                                    3
                                                               35
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