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Exact Solution of Linear Fractional Distributed Order Systems Chapter | 4  121


             Theorem 2 reveals an identity between two expressions (4.83) and (4.84) as
             stated in the following Remark.

             Remark 2: : The following identity holds true for all λ; t . 0

              ð t    λτ21                 ð 1N
                                                     ð
                ð t2τÞ   ð ð  λ 1 1Þτ 2 tÞ  τ       λ r 1 1Þ      2rt
                                    e dτ 5                      2  e  dr ð4:94Þ
                                                2 2
                       ð
               0     Γλτ 1 1Þ              0  rλ π 1 rr112λlnrð  Þ
             4.4  NUMERICAL EXAMPLES

             In this section, four examples are presented to evaluate the obtained solutions
             of distributed order differential Eqs. (4.17) and (4.82) for different coeffi-
             cients and weight functions.

             Example 1: Assume x 0ðÞ 5 1. Solutions of (4.82) with
                                     λA 12; 10; 8; 6; 4g               ð4:95Þ
                                        f
             and the unitary weight function w αðÞ 5 1  ðαA 0; 1ŠÞ are plotted in Fig. 4.1.
                                                    ½
             As it is expected, decay in solutions is more intense for greater values of λ.

             Example 2: The first elements of the solutions of (4.17) with utðÞ 5 HtðÞ,a
             zero initial condition and different matrix pairs ðA; BÞ given in Table 4.2 are
             depicted in Fig. 4.2.


                    1
                                                                λ = 12
                   0.9
                   0.8                                          λ = 10
                   0.7                                          λ = 8
                   0.6
                                                                λ = 6
                   0.5
                                                                λ = 4
                   0.4
                   0.3
                   0.2
                   0.1
                    0
                     0     0.1  0.2   0.3   0.4   0.5   0.6   0.7   0.8
                                             t
             FIGURE 4.1 Solution of (4.82) with x 0ðÞ 5 1 and the weight function w αðÞ 5 1;αA 0; 1Š for
                                                                      ½
             different values of λ given by (4.95).
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