Page 71 - Mathematical Techniques of Fractional Order Systems
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60  Mathematical Techniques of Fractional Order Systems


            methods. The purpose of this section is to solve and to show how one can
            find out existence of dual solutions for the problem (2.87) in the fractional
            order domain. The one-dimensional fractional order Bratu equation is
            given by
                                     α
                                    D uðtÞ 1 λe uðtÞ  5 0;            ð2:90Þ
                                     t
            where 1 , α # 2. To apply the Controlled Picard’s method, suppose that
            u (0) 5 E, so the boundary conditions (2.88) become:
             0
                                             0
                                    uð0Þ 5 0; u ð0Þ 5 E;              ð2:91Þ
            with additional forcing condition
                                        uð1Þ 5 0:                     ð2:92Þ

               By applying the iterative formula for Eq. (2.90) with initial conditions
            (2.91) and by taking H(t) 5 1, then
                                    h  ð t
                                                   α
             u m11 ðt; h; EÞ 5 u m ðt; h; EÞ 2  ðt2τÞ α21  D ½u m ðτ; h; Eފ 1 λe u m ðτ;h;EÞ  dτ
                                                   τ
                                   ΓðαÞ  0
                                                                      ð2:93Þ
            and using the property (2.4) then the Eq. (2.93) becomes
                                               h  ð t    α21
              u m11 ðt; h; EÞ 5 ð1 2 hÞu m ðt; h; EÞ 1 hEt 2  ðt2τÞ  λe u m ðτ;h;EÞ dτ ð2:94Þ
                                              ΓðαÞ  0
                        Ð  t  α21
               The term  ðt2τÞ   λe u m ðτ;h;EÞ dτ in Eq. (2.94) is difficult to obtain the
                        0
            exact value. Therefore, the value of this term is approximated by using com-
            posite Simpson method (Atkinson, 1989). This method is as follows:
                                   (                           )
                                                         l21
                                               l
                        ð t       t           X          X
                  IðtÞ 5  zðτÞdτ     y 0 1 y 2l 1  4y 2i21 1  2y 2i   ð2:95Þ
                         0        6l          i21        i51

                                                                       it
            where 2l is the number of subintervals, z 0 5 z(0), z 2l 5 z(t), and z i 5 z  i:1
                                                                      2l
            (1)2l 2 1. Then the iterative formula (2.94) to solve the problem (2.90) and
            (2.91) becomes:
                                             (                          )
                                                         l        l21
                                            th         X          X
             u m11 ðt; hÞ 5 ð1 2 hÞu m ðt; h; EÞ 1 hEt 2  z 0 1 z 2l 1  4z 2i11 1  2z 2i ;
                                            6l
                                                        i21       i51
                                                                      ð2:96Þ

                     α21
            where  ðt2τÞ  λe  u m ðt;h;EÞ    it ; ’i:0ð1Þ2l:
                   ΓðαÞ           τ5
                                   2l
               By    selecting  l 5 5   and   using   the   initial  solution
                           α      α
            u 0 ðt; EÞ 5 Et 2  λt  2  λEt 1 1  which at least satisfies the initial conditions
                        Γðα 1 1Þ  Γðα 1 2Þ
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