（四）分数阶微积分

我们重点考察$R-L$型分数阶微积分的性质，简记${}_{0}^{RL}D_{t}^{\beta}=D_{t}^{\beta}$，若无特殊说明。
a). 线性性
$$D_{t}^{\beta}[f(t)+g(t)]=D_{t}^{\beta}f(t)+D_{t}^{\beta}g(t)$$
$$D_{t}^{\beta}\lambda f(t)=\lambda D_{t}^{\beta}f(t) $$
证明:直接带入定义验算即可.设$m=[\beta]+1$
$${}_{0}^{RL}D_{t}^{\beta}f(t)=\frac{1}{\Gamma(m-\beta)}\frac{d^{m}}{dt^{m}}\int_{0}^{t}(t-\tau)^{m-\beta-1}f(\tau)d\tau$$
b). 积分的叠加性
$$D_{t}^{-\alpha}D_{t}^{-\beta}f(t)=D_{t}^{-\alpha-\beta}f(t)\ \ \ \ \ (\alpha,\beta>0)$$
证明:对整数阶积分结论是显然的,对于分数阶R-L积分仍然具有叠加性。
由定义知
$${}_{0}D_{t}^{-\beta}f(t)=\frac{1}{\Gamma(\beta)}\int_{0}^{t}(t-x)^{\beta-1}f(x):=g(t)$$
那么
\begin{eqnarray*}
{}_{0}^{}D_{t}^{-\alpha}g(t)&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-\tau)^{\alpha-1}g(\tau) d\tau\\
&=&\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{t}(t-\tau)^{\alpha-1}d\tau \int_{0}^{\tau}(\tau-x)^{\beta-1}f(x)dx\\
&=&\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{t}f(x)dx \int_{x}^{t}(t-\tau)^{\alpha-1}(\tau-x)^{\beta-1}d\tau(交换积分次序)\\
&=&\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{t}f(x)dx\int_{0}^{1}(t-x)^{\alpha+\beta-1}(1-\xi)^{\alpha-1}\xi ^{\beta-1}d\xi \ \ \ \ (Let\ \xi=\frac{\tau-x}{t-x})\\
&=&\frac{B(\alpha,\beta)}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{t}(t-x)^{\alpha+\beta-1}f(x)dx\\
&=&\frac{1}{\Gamma(\alpha+\beta)}\int_{0}^{t}(t-x)^{\alpha+\beta-1}f(x)dx\\
&=&{}_{0}D_{t}^{-\alpha-\beta}f(t)
\end{eqnarray*}
由此我们也得到了积分满足交换性，即
$$D_{t}^{-\alpha}D_{t}^{-\beta}f(t)=D_{t}^{-\alpha-\beta}f(t)=D_{t}^{-\beta}D_{t}^{-\alpha}f(t)\ \ \ \ \ (\alpha,\beta>0)$$
c). 上式考虑了积分叠加的情形，对于连续函数$f(t)$考虑混合运算“先积分再微分”.（还记得R-L定义思路$D^{\beta}=D^{m}D^{-(m-\beta)}$）
$${}_{0}^{}D_{t}^{\alpha}{}_{0}D_{t}^{-\beta}f(t)={}_{0}^{}D_{t}^{\alpha-\beta}f(t)\ \ \ \ \ \ (\alpha>0,\beta>0)$$
证明:先探讨一种特殊的情形
$$D^{\lambda}D^{-\lambda}f(t)=f(t)\ \ \ \ \ (\lambda>0)$$
当$\lambda$为整数时结论显然成立。不妨设$k-1 \leq \lambda<k$,即$k=[\lambda]+1$.故由定义有
$$D^{\lambda}=D^{k}D^{-(k-\lambda)}$$
代入下式
$$D^{\lambda}D^{-\lambda}=D^{k}D^{-(k-\lambda)}D^{-\lambda}=D^{k}D^{-k}=I\ \ \ (use \ b).)$$
$\bullet$若$\alpha<\beta$,记$m=[\alpha]+1,n=[\alpha-\beta]+1$，则
\begin{eqnarray*}
D^{\alpha}D^{-\beta}f(t)&=&D^{m}D^{-m-\alpha}D^{-\beta}f(t)
&=&D^{m}D^{-(m-\alpha+\beta)}f(t)\\
&=&D^{m}D^{-m}D^{-(\beta-\alpha)}f(t)\\
&=&D^{\alpha-\beta}f(t)\\
\end{eqnarray*}
$\bullet$若$\alpha>\beta$,则
\begin{eqnarray*}
D^{\alpha}D^{-\beta}f(t)&=&D^{m}D^{-m-\alpha}D^{-\beta}f(t)\\
&=&D^{m}D^{-(m-\alpha+\beta)}f(t)\\
&=&D^{n}D^{m-n}D^{-(\beta-\alpha)}f(t)\\
&=&D^{n}D^{-[n-(\alpha-\beta)]}\\
&=&D^{\alpha-\beta}f(t)\\
\end{eqnarray*}
综上有
$${}_{0}^{}D_{t}^{\alpha}{}_{0}D_{t}^{-\beta}f(t)={}_{0}^{}D_{t}^{\alpha-\beta}f(t)\ \ \ \ \ \ (\alpha>0,\beta>0)$$
d).\ 我们前面讲了一般$D^{\alpha}D^{-\beta}f(t)\neq D^{-\beta}D^{-\alpha}f(t)$,也就是说积分算子与微分算子的交换性并不总是成立的，在经典的微积分里同样要满足一些特定的条件。
例如$f(x)\equiv C$
先微分再积分：
$$D^{-1}Df(x)=0$$
先积分再微分：
$$DD^{-1}f(x)=C$$
下面我们研究混合运算中的“先微分再积分”情形，即$D^{-\beta}D^{\alpha},(\beta>0,\alpha>0)$形式.
首先,我们有结论:
$$D^{-\lambda}D^{\lambda}f(t)=f(t)-\sum_{j=1}^{m}\frac{D^{\lambda-j}f(0)}{\Gamma(\lambda-j+1)}t^{\lambda-j}$$
证明:回忆当$f(x,y)$满足连续性条件且可微时
$$\frac{d}{dx}\int_{a}^{b}f(x,y)dy=\int_{a}^{b}\frac{\partial f(x,y)}{\partial x}dy$$
$$\frac{d}{dx}\int_{\alpha (x)}^{\beta(x)}f(x,y)dy=f(x,\beta(x))\beta'(x)-f(x,\alpha(x))\alpha'(x)+\int_{\alpha(x)}^{\beta(x)}\frac{\partial f(x,y)}{\partial x}dy$$
特殊地，
$$\frac{d}{dx}\int_{0}^{x}f(x,y)dy=f(x,x)+\int_{0}^{x}\frac{\partial f(x,y)}{\partial x}dy$$
设$g(t,\tau)=(t-\tau)^{\lambda}D^{\lambda}f(\tau)$, 则$g(\tau,\tau)=0$
$$\frac{\partial g}{\partial t}=\lambda (t-\tau)^{\lambda -1}D^{\lambda}f(\tau)$$
所以
\begin{eqnarray*}
D^{-\lambda}D^{\lambda}f(t)&=&\frac{1}{\Gamma(\lambda)}\int_{0}^{t}(t-\tau)^{\lambda-1}D^{\lambda}f(\tau)d\tau\\
&=&\frac{1}{\Gamma(\lambda+1)}\frac{d}{dt}\int_{0}^{t}(t-\tau)^{\lambda}D^{\lambda}f(\tau)d\tau\\
\end{eqnarray*}
注意上式中不能轻易使用分部积分法，因为尚未验证分数阶微分是否具有叠加性。
不妨设$m-1 \leq \lambda <m$,即$m=[\lambda]+1$,由定义和$\ c).$知
\begin{eqnarray*}
\frac{1}{\Gamma(\lambda+1)}\int_{0}^{t}(t-\tau)^{\lambda}D^{\lambda}f(\tau)d\tau&=&
\frac{1}{\Gamma(\lambda+1)}\int_{0}^{t}(t-\tau)^{\lambda}D^{m}D^{-(m-\lambda)}f(\tau)d\tau\\
&=&\frac{1}{\Gamma(\lambda+1)}\int_{0}^{t}(t-\tau)^{\lambda}dD^{m-1}D^{-(m-\lambda)}f(\tau)\\
&=&\frac{D^{\lambda-1}f(\tau)}{\Gamma(\lambda+1)}(t-\tau)^{\lambda}|_{0}^{t}
+\frac{1}{\Gamma(\lambda)}\int_{0}^{t}(t-\tau)^{\lambda-1}D^{\lambda-1}f(\tau)d\tau\\
&=&\frac{1}{\Gamma(\lambda)}\int_{0}^{t}(t-\tau)^{\lambda-1}D^{\lambda-1}f(\tau)d\tau-\frac{D^{\lambda-1 }f(0)}{\Gamma(\lambda+1)}t^{\lambda}\\
&=&\frac{1}{\Gamma(\lambda-1)}\int_{0}^{t}(t-\tau)^{\lambda-2}D^{\lambda-2}f(\tau)d\tau-\frac{D^{\lambda -2}f(0)}{\Gamma(\lambda)}t^{\lambda-1}-\frac{D^{\lambda-1}f(0)}{\Gamma(\lambda+1)}t^{\lambda}\\
&=&\cdots \ \ \ \cdots \ \ \ \cdots \\
&=&\frac{1}{\Gamma(\lambda-m+1)}\int_{0}^{t}(t-\tau)^{\lambda-m}D^{-m-\lambda}f(\tau)d\tau-\sum_{j=1}^{m}\frac{D^{\lambda-j}f(0)}{\Gamma(\lambda-j+2)}t^{\lambda-j+1}\\
&=&D^{-(\lambda-m+1)}D^{-(m-\lambda)}f(t)-\sum_{j=1}^{m}\frac{D^{\lambda-j}f(0)}{\Gamma(\lambda-j+2)}t^{\lambda-j+1}\\
&=&D_{t}^{-1}f(t)-\sum_{j=1}^{m}\frac{D^{\lambda-j}f(0)}{\Gamma(\lambda-j+2)}t^{\lambda-j+1}\\
\end{eqnarray*}
故
\begin{eqnarray*}
D^{-\lambda}D^{\lambda}f(t)&=&\frac{d}{dt}[D_{t}^{-1}f(t)-\sum_{j=1}^{m}\frac{D^{\lambda-j}f(0)}{\Gamma(\lambda-j+2)}t^{\lambda-j+1}]\\
&=&f(t)-\sum_{j=1}^{m}\frac{D^{\lambda-j}f(0)}{\Gamma(\lambda-j+1)}t^{\lambda-j}\\
\end{eqnarray*}
下面我们考虑$D^{-\beta}D^{\alpha},(\beta>0,\alpha>0)$形式.
$\bullet$若$\beta \leq \alpha ,$则
\begin{eqnarray*}
D^{-\beta}D^{\alpha}f(t)&=&D^{\alpha-\beta}D^{-(\alpha-\beta)}D^{-\beta}D^{\alpha}f(t)\\
&=&D^{\alpha-\beta}D^{-\alpha}D^{\alpha}f(t)\ \ \ \ (use \ \ b).\ \ )\\
&=&D^{\alpha-\beta}[f(t)-\sum_{j=1}^{m}\frac{D^{\alpha-j}f(0)}{\Gamma(\alpha-j+1)}t^{\alpha-j}]\\
&=&D^{\alpha-\beta}f(t)-\sum_{j=1}^{m}\frac{D^{\alpha-j}f(0)}{\Gamma(\alpha-j+1)}D^{\alpha-\beta}t^{\alpha-j}\\
&=&D^{\alpha-\beta}f(t)-\sum_{j=1}^{m}\frac{D^{\lambda-j}f(0)}{\Gamma(\beta-j+1)}t^{\beta-j}\\
\end{eqnarray*}
$\bullet$若$\beta>\alpha$,则
\begin{eqnarray*}
D^{-\beta}D^{\alpha}f(t)&=&D^{-(\beta-\alpha)}D^{-\alpha}D^{\alpha}f(t)\\
&=&D^{-(\beta-\alpha)}[f(t)-\sum_{j=1}^{m}\frac{D^{\alpha-j}f(0)}{\Gamma(\alpha-j+1)}t^{\alpha-j}]\\
&=&D^{\alpha-\beta}f(t)-\sum_{j=1}^{m}\frac{D^{\alpha-j}f(0)}{\Gamma(\alpha-j+1)}D^{-(\beta-\alpha)}t^{\alpha-j}\\
&=&D^{\alpha-\beta}f(t)-\sum_{j=1}^{m}\frac{D^{\lambda-j}f(0)}{\Gamma(\beta-j+1)}t^{\beta-j}\\
\end{eqnarray*}
总而言之
$$D^{-\beta}D^{\alpha}f(t)=D^{\alpha-\beta}f(t)-\sum_{j=1}^{m}\frac{D^{\alpha-j}f(0)}{\Gamma(\beta-j+1)}t^{\beta-j}$$
e). 前面我们讨论了积分叠加的情形和“先积分再求导”和“先求导再积分”的情形，现在我们考虑微分是否具有叠加性，即
$$D^{\beta}D^{\alpha}f(t)=D^{\beta+\alpha}f(t)\ \ (\alpha>0,\beta>0) \ \ \ \ \ \ \ \ \ \ (\textbf{?})$$
一般结论并不是总成立的，例如$f(t)=1$,则有
$$D^{\beta}Df(t)=0$$
$$DD^{\beta}f(t)=D\frac{t^{-\beta}}{\Gamma(1-\beta)}=-\beta \frac{t^{-\beta-1}}{\Gamma(1-\beta)}\neq 0$$
设$n-1\leq \beta <n$即$n=[\beta]+1$利用$ d).\ $ 可得
$$D^{\beta}D^{\alpha}=D^{n}D^{-(n-\beta)}D^{a}=D^{n}[D^{\alpha+\beta-n}f(t)-\sum_{j=1}^{m}\frac{D^{\alpha-j}f(0)}{\Gamma(n-\beta-j+1)}t^{n-\beta-j}]$$
f ). 回忆
二项式定理
$$(f(x)+g(x))^{n}=C_{n}^{k}f^{n-k}(x)g^{k}(x)$$
Leibnitz公式
$$(f(x)g(x))^{(n)}=C_{n}^{k}f^{(n-k)}(x)g^{(k)}(x)$$
分数阶R-L积分的Leibniz公式为
$$D^{-\nu}[f(x)+g(x)]=\sum_{k=0}^{\infty}C_{-\nu}^{k}D^{k}g(x)D^{-\nu-k}f(x)$$
证明：根据定义有
$$D^{-\nu}f(x)g(x)=\frac{1}{\Gamma(\nu)}\int_{0}^{x}(x-\tau)^{\nu-1}f(\tau)g(\tau)d\tau$$
将$g(\tau)$在$\tau=x$处展成$Taloy$公式
$$g(\tau)=g(x)+\sum_{k=1}^{\infty}\frac{D^{k}g(x)}{k!}(\tau-x)^k$$
代入定义式得
\begin{eqnarray*}
D^{-\nu}f(x)g(x)&=&\frac{1}{\Gamma(\nu)}g(x)\int_{0}^{x}(x-\tau)^{\nu-1}f(\tau)d\tau+
\sum_{k=1}^{\infty}\frac{D^{k}g(x)}{k!}\frac{(-1)^{k}}{\Gamma(\nu)}\int_{0}^{x}(x-\tau)^{k+\nu-1}f(\tau)d\tau \\
&=&g(x)D^{-\nu}f(x)+\sum_{k=1}^{\infty}C_{-\nu}^{k}D^{k}g(x)D^{-\nu-k}f(x)\\
&=&\sum_{k=0}^{\infty}C_{-\nu}^{k}D^{k}g(x)D^{-\nu-k}f(x)
\end{eqnarray*}
其中$$C_{-\nu}^{k}=\frac{(-1)^k\Gamma(k+\nu)}{k!\Gamma(\nu)}$$.
再根据分数阶导数定义可得到分数阶导数的Leibniz公式
\begin{eqnarray*}
D^{\nu}f(x)g(x)&=&D^{m}D^{-(m-\nu)}f(x)g(x)\\
&=&D^{m}[\sum_{k=0}^{\infty}C_{\nu-m}^{k}D^{k}g(x)D^{\nu-m-k}f(x)]
\end{eqnarray*}
g ). R-L型分数阶导数与Caputo型导数的关系
一般两者并不相同,但在下面定理的条件下两者等价。
定理：若$f(t)$具有$m$阶连续导（$m=[\beta]+1$），且$f^{k}(a)=0\ (k=0,1,2,\cdots,m-1)$,则有
$$_{a}^{RL}D_{t}^{\beta}f(t)=_{a}^{C}D_{t}^{\beta}f(t)$$
证明：反复利用分部积分可得
\begin{eqnarray*}
{}_{a}^{C}D_{t}^{\beta}f(t)&=&\frac{1}{\Gamma(m-\beta)}\int_{a}^{t}(t-\tau)^{m-\beta-1}f^{(m)}(\tau)d\tau\\
&=&\frac{(m-\beta-1)(m-\beta-2)\cdots (-\beta)}{\Gamma(m-\beta)}\int_{a}^{t}(t-\tau)^{-\beta-1}f(\tau)d\tau\\
\end{eqnarray*}
另一方面反复利用
$$\frac{d}{dt}\int_{a}^{t}f(t,\tau)d\tau=f(t,t)+\int_{a}^{t}\frac{\partial f}{\partial x}$$
\begin{eqnarray*}
{}_{a}^{RL}D_{t}^{\beta}f(t)&=&\frac{1}{\Gamma(m-\beta)}\frac{d^{m}}{dt^{m}}\int_{a}^{t}(t-\tau)^{m-\beta-1}f(\tau)d\tau\\
&=&\frac{1}{\Gamma(m-\beta)}\frac{d^{m-1}}{dt^{m-1}}\frac{d}{dt}\int_{a}^{t}(t-\tau)^{m-\beta-1}f(\tau)d\tau\\
&=&\frac{m-\beta-1}{\Gamma(m-\beta)}\frac{d^{m-1}}{dt^{m-1}}\int_{a}^{t}(t-\tau)^{m-\beta-2}f(\tau)d\tau\\
&=&\cdots \ \ \ \ \cdots\ \ \ \ \cdots \ \ \ \ \\
&=&\frac{(m-\beta-1)(m-\beta-2)\cdots (-\beta)}{\Gamma(m-\beta)}\int_{a}^{t}(t-\tau)^{-\beta-1}f(\tau)d\tau\\
\end{eqnarray*}
所以此时
$${}_{a}^{RL}D_{t}^{\beta}f(t)={}_{a}^{C}D_{t}^{\beta}f(t)$$

h ).分数阶导数与整数阶导数的关系
评注：分数阶导数是整数阶导数的推广，然而这种推广并不是唯一的，历史上出现了多种版本的分数阶导数，例如$R_L$ 型、$Caputo$型、$Resiz$ 型等各个版本计算结果都不尽相同，都有独立的应用，$Mill$ 和$Ross$ 在此基础上提出了序列微积分的概念在形式上统一了分数阶微积分。
i). 序列分数阶导数实际上是基于观察所得出的定义，对整数阶的导数有
$$D^{n}f(t)=DDD \cdots Df(t)$$
推广到更一般的形式
$$D^{a}f(t)=D^{\alpha_{1}}D^{\alpha_{2}}\cdots D^{\alpha_{n}}f(t)$$
其中$\alpha=\alpha_{1}+\alpha_{2}+ \cdots + \alpha_{n}$.
从上面的形式可以看到分解并不是唯一的
当分解形式是
$$ D^{\alpha}=D^{n}D^{-(n-\alpha)}$$
为$Riemann-Liouville$型分数阶导数.
当分解形式是
$$ D^{\alpha}=D^{-(n-p)}D^{n}$$
为$Caputo$型分数阶导数.

j ). Remark: 利用卷积工具可推广到广义函数的分数阶导数,经典分析理论到近代分析跨越.