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By Boling Guo, Xueke Pu, Fenghui Huang
This e-book goals to introduce a few new tendencies and effects at the research of the fractional differential equations, and to supply a great figuring out of this box to novices who're drawn to this box, that's the authors' appealing desire. This publication describes theoretical and numerical features of the fractional partial differential equations, together with the authors' researches during this box, akin to the fractional Nonlinear Schrödinger equations, fractional Landau Lifshitz equations and fractional Ginzburg Landau equations. It additionally covers sufficient primary wisdom at the fractional derivatives and fractional integrals, and sufficient history of the fractional PDEs.
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Additional info for Fractional Partial Differential Equations and Their Numerical Solutions
Sample text
When μ and ν are integers, it reduces to the classical situation, and is consistent to the multiple iteration integrals above. Now, we give several discussions on this definition. 1. The class C includes functions which behave asymptotically like ln t or tμ near t = 0 for −1 < μ < 0, as well as functions like f (τ ) = |τ − a|μ for μ > −1 and 0 < a < t. 2. 2) in the Stieltjes integral, we have −ν 0 Dt f (t) = 1 Γ(ν + 1) t f (τ )dg(ξ), 0 where g(τ ) = −(t − τ )ν is a monotone increasing function on the closed 1 f (ξ)tν by interval [0, t].
For readers’ convenience, some basics of Fourier transform, Laplace transform and Mittag-Leffler function are given at the end of the chapter. 1 Fractional integrals and derivatives Riemann-Liouville fractional integrals To introduce R-L fractional integral, consider first the following iteration integrals t D−1 [f ](t) = f (τ )dτ, 0 t D −2 [f ](t) = τ1 dτ1 0 f (τ )dτ, 0 ··· t D−n [f ](t) = τ1 dτ1 0 τn−1 dτ2 · · · 0 f (τ )dτ. 0 ··· These multiple iteration integrals can all be expressed as t Kn (t, τ )f (τ )dτ, 0 for a certain kernel function Kn(t, τ ).
Let m be the smallest integer greater than μ, then by definition of fractional derivative Dμ [tp f (t)] = Dm [D −m+μ tn f (t)]. 2, we have n D −(m−μ) [tn f (t)] = k Cν−m [D k tn ][Dμ−m−k f (t)]. 15) k=0 We can show that if f ∈ C , then for arbitrary l = 0, 1, 2, · · · , there holds D l [D μ−m−k f (t)] = Dl+μ−m−k f (t). Hence, if f ∈ C , n k Cμ−m Dm [Dk tn ][D −m+μ−k f (t)] μ n D [t f (t)] = k=0 n m k Cμ−m = k=0 j Cm [D j+k tn ][D μ−j−k f (t)]. Eqn. & Their Numerical Solu. 16) Cμr [D r tn ][Dμ−r f (t)], = μ > 0.