Penurunan persamaan 2.18 dan 2.19 Penyelesaian masalah gelombang dispersi taklinear dengan menggunakan metode homotopi

b. Penurunan persamaan 2.18 dan 2.19

Tinjau persamaan 2.17 berikut: cosh sinh , b a b        2.17a 2 1 1 2 2 cosh sinh cosh sinh sinh , B A B A B             2.17b dengan bergantung pada  dan memenuhi sinh .       Jika persamaan 2.17a diturunkan terhadap ,  maka diperoleh sinh sinh cosh sinh a b           atau 2 cosh sinh sinh . b a          2.17c Turunan kedua dari persamaan 2.17a terhadap  adalah   2 2 cosh cosh sinh sinh sinh sinh 2 sinh cosh sinh b a                 atau 2 2 3 2 2 cosh sinh sinh 2 cosh sinh b b a             Karena 2 2 cosh sinh 1,     maka   2 2 3 2 2 1 sinh sinh sinh 2 cosh sinh b b a              atau 2 2 3 2 sinh 2 cosh sinh 2 sinh . b a b            2.17d Turunan ketiga dari persamaan 2.17a terhadap  adalah   3 3 2 cosh sinh 2 cosh 2sinh cosh sinh sinh sinh sinh 6 sinh cosh sinh b a b                     atau 3 2 2 4 3 3 cosh sinh 4 cosh sinh 2 sinh 6 cosh sinh b a a b                atau   3 2 2 4 3 3 cosh sinh 4 1 sinh sinh 2 sinh 6 cosh sinh b a a b                 atau 3 2 4 3 3 cosh sinh 4 sinh 6 sinh 6 cosh sinh . b a a b               2.17e Turunan pertama dari persamaan 2.17b terhadap  adalah   2 2 1 1 2 2 sinh sinh cosh sinh cosh sinh sinh 2 sinh cosh sinh A B A B                    atau     2 2 3 1 1 2 2 2 sinh cosh sinh 1 sinh sinh sinh 2 cosh sinh A B A B                   atau 2 2 2 1 1 2 3 2 sinh cosh sinh sinh 2 cosh sinh 2 sinh . A B A B A                 2.17f Turunan kedua dari persamaan 2.17b terhadap  adalah     2 2 3 2 2 1 1 2 2 2 2 2 2 cosh sinh cosh sinh sinh 2 sinh cosh 2 sinh sinh 2 cosh sinh cosh sinh 6 sinh cosh sinh A B A B A                            atau     2 2 3 2 2 1 1 2 4 2 2 3 2 2 cosh sinh 1 sinh sinh sinh 2 sinh cosh 2 sinh 21 sinh sinh 6 sinh cosh A B A B A                          atau 2 2 2 1 2 2 1 2 3 3 4 1 2 2 sinh cosh sinh 4 sinh 2 cosh sinh 2 sinh 6 cosh sinh 6 sinh B A B A B A B                      2.17g Selanjutnya, tinjau persamaan 2.16 Berikut: 2 2 0, k                         2.1a   3 2 2 3 2 0. k k                         2.16b  Penurunan persamaan 2.18 Sebstitusikan persamaan 2.17c , 2.17d, dan 2.17f ke dalam persamaan 2.16a, maka diperoleh           2 2 2 2 1 1 2 3 2 2 2 3 cosh sinh sinh cosh sinh cosh sinh sinh sinh cosh sinh sinh 2 cosh sinh 2 sinh sinh 2 cosh sinh 2 sinh b a b a b b a A B A B A k b a b                                       atau   2 2 2 2 2 2 2 3 2 2 1 1 2 3 2 2 2 3 cosh sinh sinh cosh sinh sinh cosh sinh cosh sinh cosh sinh sinh sinh sinh cosh sinh sinh 2 cosh sinh 2 sinh sinh 2 cosh sinh 2 sinh b a b b b a ab a b ab b A B A B A b k a k b k                                                atau     2 2 2 2 2 2 2 3 2 2 1 1 2 3 2 2 2 3 cosh sinh sinh cosh sinh sinh 1 sinh sinh cosh sinh cosh sinh sinh sinh sinh cosh sinh sinh 2 cosh sinh 2 sinh sinh 2 cosh sinh 2 sinh b a b b b a ab a b ab b A B A B A b k a k b k                                                 atau           2 1 2 2 2 2 1 2 3 2 sinh cosh sinh sinh 2 2 cosh sinh 2 2 2 sinh A ab b k b b B b ab A a a b B a k A ab b k                              Penurunan persamaan 2.19 Substitusikan persamaan 2. 17c, 2.17e, 2.17f, dan 2.17g ke dalam persamaan 2.16b, maka diperoleh          2 2 2 1 1 2 3 2 2 1 2 2 3 2 1 2 2 2 1 1 2 2 2 sinh cosh sinh sinh 2 cosh sinh 2 sinh cosh sinh sinh cosh sinh sinh 2 cosh sinh 2 sinh cosh sinh sinh cosh sinh cosh sinh sinh cosh si A B A B A b a b A B A B A b a B A B A B k b                                                    2 4 3 2 2 1 2 2 1 3 3 4 1 2 2 nh 4 sinh 6 sinh 6 cosh sinh sinh cosh sinh 4 sinh 2 cosh sinh 2 sinh 6 cosh sinh 6 sinh a a b k B A B A B A B                            atau 2 2 2 1 1 2 3 2 2 2 1 1 2 3 2 2 2 2 2 2 2 1 1 2 2 sinh cosh sinh sinh 2 cosh sinh 2 sinh sinh cosh sinh sinh 2 cosh sinh 2 sinh cosh sinh cosh sinh cosh sinh 2 cosh sinh 2 cosh sinh A B A B A b A b B b A b B b A aA aB aA aB aA                                            3 2 2 3 2 1 1 3 4 2 2 2 2 2 2 1 1 2 3 2 2 3 2 1 1 3 4 2 2 sinh cosh sinh sinh 2 cosh sinh 2 sinh cosh sinh cosh sinh cosh sinh cosh sinh cosh sinh sinh cosh sinh sinh cosh sinh sinh bA bB bA bB bA bB bA bB bA bB aB aA aB aA aB                                          2 2 2 2 3 2 4 1 2 2 2 3 2 1 1 3 4 2 2 cosh sinh 4 sinh 6 cosh sinh 6 sinh sinh cosh sinh 4 sinh 2 cosh sinh 2 sinh 6 cosh sinh 6 sinh b k a k b k a k B k A k B k A k B k A k B k                                       atau             2 1 1 1 2 2 2 1 2 1 2 2 1 2 2 2 1 2 1 2 1 1 2 3 2 1 1 1 2 2 2 2 2 sinh cosh sinh 2 2 4 4 sinh 2 2 2 2 2 cosh sinh 2 2 2 2 2 sinh 3 3 6 6 cos b A bA aB B k A aA bB b B b k A k B b A aB bA aB a k B k A aA b B bB A k B b A aB bA B k A aA bB b k A k                                                       3 2 4 2 2 2 2 2 h sinh 2 2 6 6 sinh bA aB bA aB a k B k            

c. Penyelesaian masalah nilai awal 2.46