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Exact Solution of Linear Fractional Distributed Order Systems Chapter | 4  119


             exponential weight function. However, it is possible to obtain a simpler
             expression for the solution of (4.82) which is done in the next Theorem.
             This system describes anomalous nonexponential distributed order relaxa-
             tion processes (Kochubei, 2009). Thetypicalmethodusedto treat (4.82) in
             the literature is through the Laplace transform. In this method, the solution
             is first derived in the Laplace domain. Then the Mellin’s inverse formula
             is utilized to return the obtained solution back to time domain, which
             involves solving a complex integral (Naber, 2004). This approach leads to
             a solution expressed by a Laplace-type integral eventually (Naber, 2004;
             Mainardi and Pagnini, 2007). For instance, in case of a unitary weight func-
             tion, substituting w αðÞ 5 1; αA 0; 1Š in the solution obtained by (Naber,
                                        ½
             2004), gives
                                             e  λ r 1 1Þ
                                    ð 1N      2rt
                                                 ð
                            xtðÞ 5 x 0ðÞ                   dr          ð4:83Þ
                                          2 2             2
                                     0  rλ π 1 rr112λlnrð  Þ
                In the following Theorem we present an alternative representation of this
             solution expressed in terms of Gamma functions, which may be more coher-
             ent with the concept of distributed order differential equations as the general-
             ization of the fractional order counterparts.

                           Exact  solution  of  (4.82)  under  the  assumption
             Theorem   2:
             w αðÞ 5 1; αA 0; 1Š is given by
                         ½
                            t
                           ð        λτ21        λτ
                               ð t2τÞ      ð t2τÞ    τ
                   xtðÞ 5 x 0ðÞ         2           e dτ;    t . 0     ð4:84Þ
                                           ð
                                  ð
                            0   ΓλτÞ      Γλτ 1 1Þ
                Proof: Let us define the auxiliary function
                                        ð t     λτ
                                                    τ
                                  vtðÞ 5   ð t2τÞ  e dτ                ð4:85Þ
                                           ð
                                         0  Γλτ 1 1Þ
                In terms of (4.85) we will derive the actual solution of (4.82) as follows.
             At first, note that it is possible to write (4.85) in the form
                                   ð 1N      λτ
                                                 τ
                              vtðÞ 5     ð t2τÞ  e Ht 2 τð  Þdτ        ð4:86Þ
                                         ð
                                    0   Γλτ 1 1Þ
             in which H :ðÞ is the Heaviside unit step function. Using the variable change
             τ 52 lnα in the integral (4.86) gives

                                ð  1     2λlnα
                                   ð t1lnαÞ    22lnα
                           vtðÞ 5              e   Ht 1 lnαÞdα         ð4:87Þ
                                                     ð
                                    ð
                                 0  Γ 2λlnα 1 1Þ
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