GL分数阶微积分

预备公式

Γ ( z + 1 ) = z Γ ( z ) (1) \Gamma(z+1)=z\Gamma(z)\tag{1} Γ(z+1)=zΓ(z)(1)

Γ ( z ) = ∫ 0 ∞ e − t t z − 1 d t = lim ⁡ n → ∞ n ! n z z ( z + 1 ) . . . ( z + n ) = lim ⁡ n → ∞ n z [ p r ] ( p + r ) (2) \Gamma(z)=\int_{0}^{\infty} e^{-t}t^{z-1}\mathrm{d}t=\lim_{n\to\infty}\frac{n!n^{z}}{z(z+1)...(z+n)}=\lim_{n\to\infty}\frac{n^{z}}{\begin{bmatrix} p\\r \end{bmatrix}(p+r)}\tag{2} Γ(z)=∫0∞​e−ttz−1dt=n→∞lim​z(z+1)...(z+n)n!nz​=n→∞lim​[pr​](p+r)nz​(2)

B ( z , w ) = ∫ 0 1 τ z − 1 ( 1 − τ ) w − 1 d τ = Γ ( z ) Γ ( w ) Γ ( z + w ) (3) B(z,w)=\int_0^1\tau^{z-1}(1-\tau)^{w-1}\mathrm{d}\tau=\frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}\tag{3} B(z,w)=∫01​τz−1(1−τ)w−1dτ=Γ(z+w)Γ(z)Γ(w)​(3)

[ p r ] = p ( p + 1 ) . . . ( p + r − 1 ) r ! = ( − 1 ) r ( − p r ) (4) \begin{bmatrix} p\\r \end{bmatrix}=\frac{p(p+1)...(p+r-1)}{r!}=(-1)^r\begin{pmatrix}-p\\r \end{pmatrix}\tag{4} [pr​]=r!p(p+1)...(p+r−1)​=(−1)r(−pr​)(4)

( p r ) = ( p − 1 r ) + ( p − 1 r − 1 ) (5) \begin{pmatrix}p\\r \end{pmatrix}=\begin{pmatrix}p-1\\r \end{pmatrix}+\begin{pmatrix}p-1\\r-1 \end{pmatrix}\tag{5} (pr​)=(p−1r​)+(p−1r−1​)(5)

f n ( t ) = d n f d t n = lim ⁡ h → 0 1 h n ∑ r = 0 n ( − 1 ) r ( n r ) f ( t − r h ) = lim ⁡ h → 0 1 h n Δ n f ( t ) (6) f^{n}(t)=\frac{d^{n}f}{dt^{n}}=\lim_{h\to 0} \frac{1}{h^n}\sum_{r=0}^{n}(-1)^r\begin{pmatrix}n\\r \end{pmatrix} f(t-rh)=\lim_{h\to 0} \frac{1}{h^n}\Delta^nf(t)\tag{6} fn(t)=dtndnf​=h→0lim​hn1​r=0∑n​(−1)r(nr​)f(t−rh)=h→0lim​hn1​Δnf(t)(6)

变限积分求导
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将积分和导数统一

我们把n阶导数的定义，即公式（6）极限号去掉​​​​​​​，并且把p的范围从正整数改为实数，从而定义了下面这个函数
f h ( p ) ( t ) = 1 h p ∑ r = 0 p ( − 1 ) r ( p r ) f ( t − r h ) (7) f^{(p)}_h(t)=\frac{1}{h^p}\sum_{r=0}^{p}(-1)^r\begin{pmatrix}p\\r \end{pmatrix}f(t-rh) \tag{7} fh(p)​(t)=hp1​r=0∑p​(−1)r(pr​)f(t−rh)(7)
再定义极限
lim ⁡ h → 0 n h = t − a f h ( p ) ( t ) = a D t p f ( t ) (8) \lim_{h\to 0 \atop nh=t-a }f^{(p)}_h(t)=_aD^{p}_tf(t)\tag{8} nh=t−ah→0​lim​fh(p)​(t)=a​Dtp​f(t)(8)

-p<0表示积分

此时是一个变上限积分，区间从a到t，证明利用定积分定义即可
当p是整数时，(8)的极限和p阶乘有关，即
a D t − p f ( t ) = 1 ( p + 1 ) ! ∫ a t ( t − τ ) p − 1 f ( τ ) d t (9) _aD^{-p}_tf(t)=\frac{1}{(p+1)!}\int_a^t(t-\tau)^{p-1}f(\tau)\mathrm{d}t\tag{9} a​Dt−p​f(t)=(p+1)!1​∫at​(t−τ)p−1f(τ)dt(9)
p是任意阶时用gamma函数代替阶乘，证明过程在书2.2。

a D t − p f ( t ) = 1 Γ ( p + 1 ) ∫ a t ( t − τ ) p − 1 f ( τ ) d t (10) _aD^{-p}_tf(t)=\frac{1}{\Gamma(p+1)}\int_a^t(t-\tau)^{p-1}f(\tau)\mathrm{d}t \tag{10} a​Dt−p​f(t)=Γ(p+1)1​∫at​(t−τ)p−1f(τ)dt(10)
如果函数m+1阶连续，由分部积分可知
a D t − p f ( t ) = ∑ k = 0 m f ( k ) ( a ) ( t − a ) p + k Γ ( p + k + 1 ) + 1 Γ ( p + k + 1 ) ∫ a t ( t − τ ) p + m f m + 1 ( τ ) d τ (11) _aD^{-p}_tf(t)=\sum_{k=0}^m\frac{f^{(k)}(a)(t-a)^{p+k}}{\Gamma(p+k+1)}+\frac{1}{\Gamma(p+k+1)}\int_a^t(t-\tau)^{p+m}f^{m+1}(\tau)\mathrm{d}\tau \tag{11} a​Dt−p​f(t)=k=0∑m​Γ(p+k+1)f(k)(a)(t−a)p+k​+Γ(p+k+1)1​∫at​(t−τ)p+mfm+1(τ)dτ(11)

p>0表示导数

当p>0时上述定义函数具有性质（利用公式（5）证明）
f h ( p ) ( t ) = ∑ k = 0 m ( − 1 ) n − k ( p − k − 1 r n − k ) h − p Δ k f ( a + k h ) + p − 1 ∑ r = 0 n − m − 1 ( − 1 ) r ( p − m − 1 r ) Δ m + 1 f ( t − r h ) f^{(p)}_h(t)=\sum_{k=0}^{m}(-1)^{n-k}\begin{pmatrix}p-k-1\\rn-k\end{pmatrix}h^{-p}\Delta^kf(a+kh)+p^{-1}\sum_{r=0}^{n-m-1}(-1)^r\begin{pmatrix}p-m-1\\r \end{pmatrix}\Delta^{m+1}f(t-rh) fh(p)​(t)=k=0∑m​(−1)n−k(p−k−1rn−k​)h−pΔkf(a+kh)+p−1r=0∑n−m−1​(−1)r(p−m−1r​)Δm+1f(t−rh)
对上式求积分估计，和积分形式是一样的。但是需要满足m<p<m+1
a D t p f ( t ) = ∑ k = 0 m f ( k ) ( a ) ( t − a ) − p + k Γ ( − p + k + 1 ) + 1 Γ ( − p + m + 1 ) ∫ a t ( t − τ ) − p + m f m + 1 ( τ ) d τ (12) _aD^{p}_tf(t)=\sum_{k=0}^m\frac{f^{(k)}(a)(t-a)^{-p+k}}{\Gamma(-p+k+1)}+\\\frac{1}{\Gamma(-p+m+1)}\int_a^t(t-\tau)^{-p+m}f^{m+1}(\tau)\mathrm{d}\tau \tag{12} a​Dtp​f(t)=k=0∑m​Γ(−p+k+1)f(k)(a)(t−a)−p+k​+Γ(−p+m+1)1​∫at​(t−τ)−p+mfm+1(τ)dτ(12)
特殊的函数 ( t − a ) ν (t-a)^{\nu} (t−a)ν有分数阶导数
a D t p ( t − a ) ν = Γ ( ν + 1 ) Γ ( − p + ν + 1 ) ( t − a ) ν − p (13) _aD^{p}_t(t-a)^{\nu}=\frac{\Gamma(\nu+1)}{\Gamma(-p+\nu+1)}(t-a)^{\nu-p}\tag{13} a​Dtp​(t−a)ν=Γ(−p+ν+1)Γ(ν+1)​(t−a)ν−p(13)

整数阶和分数阶混合运算

利用变上限积分求导
d n d t n ( a D t p f ( t ) ) = ∑ k = 0 s f ( k ) ( a ) ( t − a ) − p + n + k Γ ( − p + n + k + 1 ) + 1 Γ ( − p + n + s + 1 ) ∫ a t ( t − τ ) − p + s + n f s + 1 ( τ ) d τ = a D t p + n f ( t ) (14) \frac{d^n}{dt^n}(_aD^{p}_tf(t))=\sum_{k=0}^s\frac{f^{(k)}(a)(t-a)^{-p+n+k}}{\Gamma(-p+n+k+1)}+\frac{1}{\Gamma(-p+n+s+1)}\int_a^t(t-\tau)^{-p+s+n}f^{s+1}(\tau)\mathrm{d}\tau \\ =_aD^{p+n}_tf(t)\tag{14} dtndn​(a​Dtp​f(t))=k=0∑s​Γ(−p+n+k+1)f(k)(a)(t−a)−p+n+k​+Γ(−p+n+s+1)1​∫at​(t−τ)−p+s+nfs+1(τ)dτ=a​Dtp+n​f(t)(14)

a D t p ( d n f ( t ) d t n ) = ∑ k = 0 s f ( n + k ) ( a ) ( t − a ) − p + k Γ ( − p + k + 1 ) + 1 Γ ( − p + s + 1 ) ∫ a t ( t − τ ) − p + s f n + s + 1 ( τ ) d τ (15) _aD^{p}_t(\frac{d^nf(t)}{dt^n})=\sum_{k=0}^s\frac{f^{(n+k)}(a)(t-a)^{-p+k}}{\Gamma(-p+k+1)}+\frac{1}{\Gamma(-p+s+1)}\int_a^t(t-\tau)^{-p+s}f^{n+s+1}(\tau)\mathrm{d}\tau\\ \tag{15} a​Dtp​(dtndnf(t)​)=k=0∑s​Γ(−p+k+1)f(n+k)(a)(t−a)−p+k​+Γ(−p+s+1)1​∫at​(t−τ)−p+sfn+s+1(τ)dτ(15)
证明技巧：s取不同的值 P57 一式s=m+n-1，二式s=m-1，两式相减就有下式
d n d t n ( a D t p f ( t ) ) = a D t p ( d n f ( t ) d t n ) + ∑ k = 0 n − 1 f ( k ) ( a ) ( t − a ) − p + n + k Γ ( − p + n + k + 1 ) (16) \frac{d^n}{dt^n}(_aD^{p}_tf(t))= _aD^{p}_t(\frac{d^nf(t)}{dt^n})+\sum_{k=0}^{n-1}\frac{f^{(k)}(a)(t-a)^{-p+n+k}}{\Gamma(-p+n+k+1)}\tag{16} dtndn​(a​Dtp​f(t))=a​Dtp​(dtndnf(t)​)+k=0∑n−1​Γ(−p+n+k+1)f(k)(a)(t−a)−p+n+k​(16)
显然当t=a，a导数为0有
d n d t n ( a D t p f ( t ) ) = a D t p ( d n f ( t ) d t n ) = a D t p + n f ( t ) \frac{d^n}{dt^n}(_aD^{p}_tf(t))= _aD^{p}_t(\frac{d^nf(t)}{dt^n})=_aD^{p+n}_tf(t) dtndn​(a​Dtp​f(t))=a​Dtp​(dtndnf(t)​)=a​Dtp+n​f(t)

分数阶和分数阶混合运算

这个地方的结果和整数阶是一样的,p和q的是任意大于零的数。
a D t q ( a D t p f ( t ) ) = a D t p ( a D t q f ( t ) ) = a D t p + q f ( t ) (17) _aD^{q}_t(_aD^{p}_tf(t))=_aD^{p}_t(_aD^{q}_tf(t))=_aD^{p+q}_tf(t)\tag{17} a​Dtq​(a​Dtp​f(t))=a​Dtp​(a​Dtq​f(t))=a​Dtp+q​f(t)(17)