By Changpin Li, Fanhai Zeng

Numerical equipment for Fractional Calculus provides numerical tools for fractional integrals and fractional derivatives, finite distinction equipment for fractional usual differential equations (FODEs) and fractional partial differential equations (FPDEs), and finite aspect equipment for FPDEs.

The ebook introduces the fundamental definitions and homes of fractional integrals and derivatives earlier than protecting numerical equipment for fractional integrals and derivatives. It then discusses finite distinction equipment for either FODEs and FPDEs, together with the Euler and linear multistep tools. the ultimate bankruptcy indicates how you can remedy FPDEs through the use of the finite point method.

This booklet presents effective and trustworthy numerical tools for fixing fractional calculus difficulties. It bargains a primer for readers to additional increase state-of-the-art examine in numerical fractional calculus. MATLAB® features can be found at the book’s CRC Press internet page.

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The remedy is to use the following technique α RL D0,t f (t) t=t n = ≈ α RL D0,t ( f (t) − 1 Δtα f (0)) t=tn + f (0)tn−α Γ(1 − α) n+p ω(α) f (tn− j+p ) − f (0) + j j=0 f (0)tn−α . 55) is exact when f (t) is a constant. 56) and α 1 Δtα n+1 ω(α) j f (tn− j+1 ) =α j=0 α RL D0,t f (t) t=t n Γ(μ + 1)tμ−1−α Δt + O(Δt2 ). 57) yields α RL D0,t f (t) t=t n ⎡ 1 ⎢⎢⎢⎢⎢ (2 − α) = α ⎢⎢⎣ Δt 2 n ω(α) j f (tn− j ) + j=0 α 2 n+1 j=0 ⎤ ⎥⎥ ⎥⎥⎥ 2 ω(α) j f (tn− j+1 )⎥⎥⎦ + O(Δt ). 58) is obtained to approximate RL Dα0,t f (t).

78) It is easy to calculate that G(s) = L{tα−1 ; s} = Γ(α)s−α . 79) Hence L{D−α 0,t f (t); s} = 1 L{tα−1 ∗ f (t); s} = s−α L{ f (t); s} = s−α F(s). 80) Now let us turn to the Laplace transform of the Riemann–Liouville derivative operator with order α, m − 1 ≤ α < m. Let Then g(t) = D−(m−α) f (t). 82) = g(m) (t). 77) gives L{RL Dα0,t f (t); s} = L{g(m) (t); s} = sm L{g(t); s} − m−1 sk g(m−k−1) (0). 80) one has f (t); s} = s−(m−α) L{ f (t); s}. 84) gives the Laplace transform of the Riemann–Liouville derivative as L{RL Dα0,t f (t); s} = sα L{ f (t); s} − m−1 sk k=0 α−k−1 f (t) , RL D0,t t=0 m − 1 ≤ α < m.

Then the function Φ(s) = f (t − s)s−1−α possesses the Hadamard property at the point s = 0 for each t and α < m + λ. f. Γ(−α) ∞ 0 f (t − s) 1 ds = Γ(−α) s1+α 1 + Γ(−α) where α < m + λ, α ∞ 0 1 m k sk (k) k=0 (−1) k! f (t) s1+α f (t − s) − 0 m f (k) (t) f (t − s) (−1)k , ds + k! 115) 0, 1, 2, · · · . 2 with m − 1 < α ≤ m. 115) for any α > 0, α 1, 2, · · ·. 3 Directional Integrals and Derivatives in R2 Definition 7 Let α > 0, θ ∈ [0, 2π) be given. The α-th order fractional integral in the direction of θ is given by D−α θ u(x, y) = 1 Γ(α) ∞ ξα−1 u(x − ξ cosθ, y − ξ sin θ) dξ.