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Nonlinear Fractional Order Boundary-Value Problems Chapter | 2  61


             (2.91) which can be used to obtain the mth-successive approximations (2.96).
             The first approximation solution u 1 (t, h, E) is given by:
                                          α
                                   α
                             0:033ht λ  ht λ
             u 1 ðt; h; EÞ  5 tE 2    2      f 1 1 f 2 1 f 3 1 f 4 1 f 5 1 f 6 1 f 7 1 f 8 1 f 9
                               ýአ    Γ½αŠ
                                      α
                            α
                           t λ       ht λ     t 11α Eλ  ht 1:1α Eλ
                       2         1         2         1
                         Γ½1 1 አ Γ½1 1 አ Γ½2 1 አ Γ½2 1 αŠ
                                                                       ð2:97Þ
             where,
                       20:105α10:1tE2 e 22:30α t α λ 2 0:1e 22:30α t 1:1α Eλ
             f 1 5 0:1481e        ý11አ ý11አ ;-
             f 2 5 0:083e 20:223α10:2tE2 e 21:60α t α λ 2 0:2e 21:609α t 1:1α Eλ ;
                                         Γ½21αŠ
                                 Γ½11αŠ
                      20:356α10:3tE2 e 21:20α t α λ 2 0:3e 21:203α t 1:1α Eλ
             f 3 5 0:190e        ý11አ ý11አ ;-
                      20:510α10:4tE2 e 20:916α t α λ 2 0:4e 20:916α t 1:1α Eλ
             f 4 5 0:111e        ý1:1አ ý2:1አ ;
             f 5 5 0:266e 20:693α10:5tE2 e 20:693α t α λ 2 0:5e 20:69α t 1:1α Eλ ;
                                 Γ½11αŠ
                                          Γ½21αŠ
             f 6 5 0:166e 20:916α10:6tE2 e 20:510α t α λ 2 0:6e 20:510α t 1:1α Eλ ;
                                          Γ½21αŠ
                                 Γ½1:1αŠ
                      21:203α10:7tE2 e 20:35α t α λ 2 0:7e 20:356α t 1:1α Eλ
             f 7 5 0:444e        ý11አ ý21አ ;
                     21:609α10:8tE2 e 20:223α t α λ 2 0:8e 20:223α t 1:1α Eλ
             f 8 5 0:33e        ý1:1አ ý2:1አ ;
             f 9 5 1:333e 22:302α10:9tE2 e 20:1053α t α λ 2 0:899e 20:1053α t 1:1α Eλ ; and so on. With the help of
                                           Γ½2:1αŠ
                                  Γ½1:1αŠ
             forcing condition u(1) 5 0, then
                                   uð1Þ 5 u m11 ð1; E; hÞ 5 0:         ð2:98Þ
                According to the above equation and after six iterations calculated
             (m 5 5), in Fig. 2.12 the E as a function of auxiliary parameter h, has been
             plotted in the h range [0, 2] implicitly, for different values of α and λ. Two





















             FIGURE 2.12 The h 2 E curves of Eq. (2.98) for different value of λ and α.
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