tailieunhanh - Sabatier Agrawal Machado Advances in Fractional Calculus Episode 4

Tham khảo tài liệu 'sabatier agrawal machado advances in fractional calculus episode 4', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | RIESZ POTENTIALS AS CENTRED DERIVATIVES 109 D - 2r -a sin an 2 7 t-l - -1 d 69 -X On the existence of a inverse Riesz potential In current literature 7 10 the Riesz potentials are only defined for negative orders verifying -1 a 0. However our formulation is valid for every a -1. This means that we can define those potentials even for positive orders. However we cannot guaranty that there is always an inverse for a given potential. The theory presented in section allows us to state that The inverse of a given potential when existing is of the same type the inverse of the type k k 1 2 potential is a type-k potential. The inverse of a given potential exists iff its order a verifies a 1. The order of the inverse of an a order potential is a -a order potential. The inverse can be computed both by 33 respectively 34 and by 43 respectively 44 . This is in contradiction with the results stated in 10 about this subject and will have implications in the solution of differential equations involving centred derivatives. An analytic derivative An interesting result can be obtained by combining 53 with 65 to give a complex function Hd w Hd1 ÍHd2 70 We obtain a function that is null for 0. This means that the operator defined by 44 is the Hilbert transform of that defined in 43 . The inverse Fourier transform of 70 is an analytic signal and the corresponding analytic derivative is given by the convolution of the function at hand with the operator r-1 sg HD t 2r -a cos arn 2 1 2r -a sin an 2 This leads to a convolution integral formally similar to the Riesz-Feller potentials 10 . We can give this formula another aspect by noting that 110 Ortigueira 1 r a 1 .sin are r a 1 sin are 2 72 2r -a cos are 2 2recos are 2 n and 1 r a 1 .sin are r a 1 cos are 2 73 2r -a sin are 2 2resin are 2 n We obtain easily HD t - F a 1 t 1sin are 2 -i t 1sgn t cos are 2 74 that can be rewritten as HD t ơ 7 t ----sgn t eiW2sgn t 75 This impulse response leads to the following potential DDf t

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