设$X$是一个赋范空间, $I$是一个开区间.对任何$x\in I$和某个$\varepsilon\geq 0$,如果对任何满足微分不等式

$\begin{eqnarray*}\|a_n(x)y^{(n)}(x) + a_{n-1}(x)y^{(n-1)}(x) + \ldots + a_1(x)y'(x) + a_0(x)y(x) + h(x)\| \leq \varepsilon\end{eqnarray*}$

的映射$f : I \rightarrow X$,都存在微分方程

$\begin{eqnarray*}a_n(x)y^{(n)}(x) + a_{n-1}(x)y^{(n-1)}(x) + \ldots + a_1(x)y'(x) + a_0(x)y(x) + h(x)=0\end{eqnarray*}$

的一个解$f_0 : I \rightarrow X $使得对任何的$x\in I$不等式$\|f(x)-f_0(x)\|\leq K(\varepsilon)$都成立,这里$K(\varepsilon)$仅仅是$\varepsilon$的表达式,那么称该微分方程具有Hyers-Ulam稳定性.

在上面的定义中,如果用$\phi(x)$和$\Phi(x)$分别代替$\varepsilon$和$K(\varepsilon)$时,其结论还是成立的,那么我们称该微分方程有Hyers-Ulam-Rassias稳定性,这里$\phi, \Phi: I\rightarrow[0, \infty)$是不明确依赖于$f$和$f_0$的非负函数.该定义也可以参考文献[4].

Obloza是第一个研究线性微分方程Hyers-Ulam稳定性的学者(参考文献[12-13]).从此开始,很多数学家和学者把注意力放在了研究微分方程的Hyers-Ulam稳定性上,得到了一系列丰富的研究结果(可参考文献[1, 3-7, 11, 15, 17-18, 21, 23-24]和这些文献中的相关参考文献).此外,关于函数空间上的算子及具有不同结构的空间上函数方程的Hyers-Ulam-Rassias稳定性研究读者可以参考文献[20, 22, 25-27].

分数阶微分方程在模拟科学和工程各个领域的很多现象中已经变成最有价值和最有用的工具之一,譬如在粘弹性理论、电化学、电磁学、经济、最优控制等学科领域的应用.分数阶导数最主要的优点是能够描述很多材料的记忆性质和遗传性质.因此,近些年来,分数阶微分方程研究得到了较大的关注,这方面可参考专著[9, 14],以及专著内所列的相关参考文献.

近些年来,一些学者研究了某些分数阶系统的解和Hyers-Ulam-Rassias稳定性.在文献[8]中,应用矩阵方法, Jung讨论了常系数一阶线性微分系统的Hyers-Ulam稳定性.在文献[2]中, Gejji和Babakhani给出了一类分数阶系统初始值问题的存在唯一性定理.通过应用Laplace变换方法, Rezaei, Jung和Rassias讨论了线性微分方程的Hyers-Ulam稳定性(参考文献[16]). Wang和Li^[19]应用Laplace变换方法研究了一类线性分数阶微分方程的Hyers-Ulam稳定性. Shen和Chen^[17]也通过应用Laplace变换方法研究了一类带有常系数的线性分数阶微分方程的Ulam稳定性,得到了一些新颖的结果.

在本文中,我们研究了以下类型的分数阶微分系统的Hyers-Ulam-Rassias稳定性.该类系统包括分数阶系统

(1.1) $ \begin{equation}\label{e1.1} \left\{ \begin{array}{l} ^C_0\!D_t^\alpha\vec{x}(t)=A\vec{x}(t)+\vec{f}(t), 0<\alpha<1, t\geq0, \\ \vec{x}(0)=\vec{\eta} \end{array} \right. \end{equation}$

和分数阶系统

(1.2) $ \begin{equation}\label{e1.2} \left\{ \begin{array}{l} ^C_0\!D_t^\alpha\vec{x}(t)=A\ ^C_0\!D_t^\beta\vec{x}(t)+\vec{g}(t), 0<\beta<\alpha<1, t\geq0, \\ \vec{x}(0)=\vec{\eta}, \end{array} \right. \end{equation}$

这里$^C_0\!D_t^\alpha\vec{x}(\cdot)$和$^C_0\!D_t^\beta\vec{x}(\cdot)$都是Caputo分数阶微分算子, $A$是$n\times n$常数矩阵, $\vec{f}(t)$和$\vec{g}(t)$都是$n$ -维连续的向量值函数,通常它们也被称为系统的强迫项.本文中我们给出了使得系统(1.1)和(1.2)是Hyers-Ulam-Rassias稳定的一些充分条件.

在这一部分,我们给出一些定义和引理,在证明主要结论时要用到它们.

定义2.1  一个向量值函数$\vec{h}:(0, \infty)\rightarrow {\Bbb R}^n$的$\alpha (0<\alpha<1)$阶Riemann-Liouville分数阶积分定义为

(2.1) $\begin{equation}\label{e2.1}_0I_t^\alpha\vec{h}(t)=\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}\vec{h}(s)\text{d}\mathsf{s}.\end{equation}$

定义2.2  一个向量值函数$\vec{h}:(0, \infty)\rightarrow {\Bbb R}^n$的$\alpha (0<\alpha<1)$阶Riemann-Liouville分数阶导数定义为

(2.2) $\begin{equation}_0D_t^\alpha\vec{h}(t)=\frac{1}{\Gamma(1-\alpha)}\frac{{{\rm{d}}}}{{{\rm{d}}}t}\int_0^t(t-s)^{-\alpha}\vec{h}(s){{\rm{d}}}s.\end{equation}$

定义2.3  一个向量值函数$\vec{h}:(0, \infty)\rightarrow {\Bbb R}^n$的$\alpha (0<\alpha<1)$阶Caputo分数阶导数定义为

(2.3) $\begin{equation}\label{e2.3}_0^CD_t^\alpha\vec{h}(t)= {}_0D_t^\alpha[\vec{h}(t)-\vec{h}(0)].\end{equation}$

注2.1  当$\vec{h}\in AC[0, T]$,我们有

(2.4) $\begin{equation}_0^CD_t^\alpha\vec{h}(t)=\frac{1}{\Gamma(1-\alpha)}\int_0^t(t-s)^{-\alpha}\vec{h}'(s){{\rm{d}}}s.\end{equation}$

定义2.4   Mittag-Leffler函数$E_{\alpha, \beta}(z)$定义为

(2.5) $\begin{eqnarray} E_{\alpha, \beta}(z)=\sum\limits_{k=0}^{\infty}\frac{z^k}{\Gamma(\alpha k+\beta)} (z, \beta \in {\Bbb C}; \Re(\alpha)>0). \end{eqnarray} $

当$\alpha=\beta=1$时,我们得到$E_{1, 1}(z)=e^z$.关于该函数的更多性质可以参考文献[9].此外, $\alpha$ -指数函数定义为

$ \begin{eqnarray*} e_\alpha^{\lambda z}:=z^{\alpha-1}E_{\alpha, \alpha}(\lambda z^\alpha), \end{eqnarray*} $

这里$z\in{\Bbb C}\backslash \{0\}, {\mathfrak R}(\alpha)>0, $$\lambda\in{\Bbb C}, $${\Bbb C}$是复平面.

定义2.5  设$\vec{f}(t)$是一个$n$ -维向量值函数. $\vec{f}(t)$的Laplace变换定义为

(2.6) $\begin{equation} \vec{F}(s)={\cal L}\{\vec{f}(t)\}(s):=\int_0^\infty e^{-st}\vec{f}(t){{\rm{d}}}t. \end{equation}$

定义2.6  设$\vec{f}(t)$是定义在区间$[0, \infty)$上的向量值函数,如果存在两个实常数$M>0$和$c$,对充分大的$t$都有

(2.7) $\begin{eqnarray} \|\vec{f}(t)\|\leq Me^{ct} \end{eqnarray} $

成立,我们称向量值函数$\vec{f}(t)$是指数有界的.

引理2.1^[10]  设${\Bbb C}$是复平面, ${\mathfrak R}(s)$表示复数$s$的实部,对任何的$\alpha>0, \beta>0$及$A\in {\Bbb C}^{n\times n}$,若${\mathfrak R}(s)>\|A\|^{\frac{1}{\alpha}}$,则

(2.8) $\begin{equation} {\cal L}\{t^{\beta-1}E_{\alpha, \beta}(A t^\alpha)\}=s^{\alpha-\beta}(s^\alpha I-A)^{-1} \end{equation}$

成立.

在文献[10]中,作者得到了下面的引理2.2,并通过应用引理2.2和逆Laplace变换得到了一类微分系统的解.

引理2.2  设系统(1.1)有惟一的连续解$\vec{x}(t)$,如果$\vec{f}(t)$在$[0, \infty)$上是连续的并且是指数有界的,那么$\vec{x}(t)$和它的Caputo导数$^C_0\!D_t^\alpha\vec{x}(t)$都是指数有界的,于是它们的Laplace变换都存在.

在下文中,我们假设在系统(1.1)和(1.2)中的$\vec{f}(t)$和$\vec{g}(t)$都是指数有界的.

在这一部分,我们将要应用Laplace变换方法证明分数阶系统(1.1)是Hyers-Ulam-Rassias稳定的.

定理3.1  设$A\in M_{n\times n}({\Bbb R})$, $0<\alpha<1$,向量值函数$\vec{f}(t)$是指数有界的.如果向量值函数$\vec{x}(t): [0, +\infty)\rightarrow{\Bbb R}^n$对所有的$t\in[0, +\infty)$和某个$\varepsilon>0$满足不等式

(3.1) $ \begin{equation}\label{e3.1} \left\|^C_0\!D_t^\alpha\vec{x}(t)-A\vec{x}(t)-\vec{f}(t)\right\|\leq\varepsilon, \end{equation} $

那么存在分数阶系统(1.1)的一个解$\vec{x}^\ast(t)$使得

(3.2) $ \begin{equation}\label{e3.2} \left\|\vec{x}(t)-\vec{x}^\ast(t)\right\|\leq\varepsilon t^\alpha E_{\alpha, \alpha+1}\left(\|A\|t^\alpha\right). \end{equation} $

证  设$\vec{Z}(t)= \ ^C_0\!D_t^\alpha\vec{x}(t)-A\vec{x}(t)-\vec{f}(t)$,我们有

(3.3) $ \begin{eqnarray}\label{e3.3} {\cal L}\left\{\vec{Z}(t)\right\} &=& {\cal L}\left\{\ ^C_0\!D_t^\alpha\vec{x}(t)-A\vec{x}(t)-\vec{f}(t)\right\} \nonumber\\&=&s^\alpha I{\cal L}\left\{\vec{x}(t)\right\}-s^{\alpha-1}\vec{\eta}-A{\cal L}\left\{\vec{x}(t)\right\}-{\cal L}\left\{\vec{f}(t)\right\} \nonumber\\ &=&\left(s^\alpha I-A\right){\cal L}\left\{\vec{x}(t)\right\}-s^{\alpha-1}\vec{\eta}-{\cal L}\left\{\vec{f}(t)\right\}, \end{eqnarray} $

这里$I$是单位矩阵.由(3.3)式,我们得到

(3.4) $ \begin{equation}\label{e3.4} {\cal L}\left\{\vec{x}(t)\right\}=\left(s^\alpha I-A\right)^{-1}\left\{{\cal L}\left\{\vec{Z}(t)\right\}+s^{\alpha-1}\vec{\eta}+{\cal L}\left\{\vec{f}(t)\right\}\right\}. \end{equation}$

取

(3.5) $\begin{equation}\label{e3.5} \vec{x}^\ast(t)=\int_0^te_\alpha^{A(t-\xi)}\vec{f}(\xi){{\rm{d}}}\xi+\int_0^te_\alpha^{A(t-\xi)}A\vec{\eta}{{\rm{d}}}\xi+\vec{\eta}, \end{equation}$

通过应用Laplace变换的卷积性质,可证明

(3.6) $ \begin{eqnarray}\label{e3.6} {\cal L}\left\{\vec{x}^\ast(t)\right\} &=& {\cal L}\left\{\int_0^te_\alpha^{A(t-\xi)}\vec{f}(\xi){{\rm{d}}}\xi\right\}+ {\cal L}\left\{\int_0^te_\alpha^{A(t-\xi)}A\vec{\eta}{{\rm{d}}}\xi\right\}+{\cal L}\left\{\vec{\eta}\right\}\nonumber\\&=& {\cal L}\left\{e_\alpha^{At}\right\}{\cal L}\left\{\vec{f}(t)\right\}+{\cal L}\left\{e_\alpha^{At}\right\}{\cal L}\left\{A\vec{\eta}\right\}+\frac{1}{s}\vec{\eta} \nonumber\\&=&\left(s^\alpha I-A\right)^{-1}{\cal L}\left\{\vec{f}(t)\right\}+\left(s^\alpha I-A\right)^{-1}\frac{1}{s}A\vec{\eta}+\frac{1}{s}\vec{\eta}. \end{eqnarray} $

应用引理2.2和(3.5)式,并进行简单计算,可得

(3.7) $ \begin{eqnarray}\label{e3.7} {\cal L}\left\{^C_0\!D_t^\alpha\vec{x}^\ast(t)-A\vec{x}^\ast(t)\right\} &=& s^\alpha I{\cal L}\left\{\vec{x}^\ast(t)\right\}(s)-s^{\alpha-1}\vec{\eta}-A{\cal L}\left\{\vec{x}^\ast(t)\right\}(s) \nonumber\\ &=& (s^\alpha I-A){\cal L}\left\{\vec{x}^\ast(t)\right\}(s)-s^{\alpha-1}\vec{\eta} \nonumber\\ &=& (s^\alpha I-A)\left[\left(s^\alpha I-A\right)^{-1}{\cal L}\left\{\vec{f}(t)\right\}\right. \nonumber\\ &&\mbox{}+\left.\left(s^\alpha I-A\right)^{-1}\frac{1}{s}A\vec{\eta}+\frac{1}{s}\vec{\eta}\right]-s^{\alpha-1}\vec{\eta} \nonumber\\ &=& {\cal L}\left\{\vec{f}(t)\right\}+\frac{1}{s}A\vec{\eta}+\left(s^\alpha I-A\right)\frac{1}{s}\vec{\eta}-s^{\alpha-1}\vec{\eta}\nonumber\\ &=& {\cal L}\left\{\vec{f}(t)\right\}.\end{eqnarray}$

因为变换${\cal L}$是一对一的,这就证明了$\vec{x}^\ast(t)$是系统(1.1)的一个解.由(3.4)式和(3.6)式,我们得到

(3.8) $ \begin{eqnarray}\label{e3.8} {\cal L}\left\{\vec{x}(t)-\vec{x}^{\ast}(t)\right\} &=& {\cal L}\left\{\vec{x}(t)\right\}-{\cal L}\left\{\vec{x}^{\ast}(t)\right\} \nonumber\\ &=& \left(s^\alpha I-A\right)^{-1}\left\{{\cal L}\left\{\vec{Z}(t)\right\}+s^{\alpha-1}\vec{\eta}+{\cal L}\left\{\vec{f}(t)\right\}\right\}\nonumber\\ &\quad&-\left(s^\alpha I-A\right)^{-1}{\cal L}\left\{\vec{f}(t)\right\}-\left(s^\alpha I-A\right)^{-1}\frac{1}{s}A\vec{\eta}-\frac{1}{s}\vec{\eta} \nonumber\\ &=& \left(s^\alpha I-A\right)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}.\end{eqnarray}$

应用卷积性质和引理2.1,我们得到

(3.9) $\begin{eqnarray}\label{e3.9} {\cal L}\left\{e_\alpha^{Ax}\ast\vec{Z}(t)\right\} &=& {\cal L}\left\{e_\alpha^{Ax}\right\}{\cal L}\left\{\vec{Z}(t)\right\}\nonumber\\ &=& \left(s^\alpha I-A\right)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}.\end{eqnarray}$

由(3.8)和(3.9)式,我们可得

(3.10) $\begin{equation}\label{e3.10}\vec{x}(t)-\vec{x}^{\ast}(t)=e_\alpha^{Ax}\ast\vec{Z}(t), \end{equation}$

结合(3.1)式,可证明

(3.11) $\begin{eqnarray} \left\|\vec{x}(t)-\vec{x}^{\ast}(t)\right\| &=& \left\|e_\alpha^{Ax}\ast\vec{Z}(t)\right\| \nonumber\\ &=&\left\|\int_0^te_\alpha^{A(t-\xi)}\vec{Z}(\xi){{\rm{d}}}\xi\right\|\nonumber\\ &=&\left\|\int_0^t(t-\xi)^{\alpha-1}\sum\limits_{k=0}^\infty A^k\frac{(t-\xi)^{\alpha k}}{\Gamma(\alpha k+\alpha)}\vec{Z}(\xi){{\rm{d}}}\xi\right\|\nonumber\\ &=&\left\|\int_0^t\sum\limits_{k=0}^\infty \frac{A^k(t-\xi)^{\alpha k+\alpha-1}}{\Gamma(\alpha k+\alpha)}\vec{Z}(\xi){{\rm{d}}}\xi\right\|\nonumber\\ &\leq&\sum\limits_{k=0}^\infty\int_0^t\left\|\frac{A^k(t-\xi)^{\alpha k+\alpha-1}}{\Gamma(\alpha k+\alpha)}\right\|\left\|\vec{Z}(\xi)\right\|{{\rm{d}}}\xi\nonumber\\ &=&\varepsilon\sum\limits_{k=0}^\infty\frac{\|A\|^k}{\Gamma(\alpha k+\alpha)}\int_0^t(t-\xi)^{\alpha k+\alpha-1}{{\rm{d}}}\xi\nonumber\\ &=&\varepsilon t^\alpha\sum\limits_{k=0}^\infty\frac{\|A\|^kt^{\alpha k}}{\Gamma(\alpha k+\alpha+1)}\nonumber\\ &=&\varepsilon t^\alpha E_{\alpha, \alpha+1}\left(\|A\|t^\alpha\right).\end{eqnarray}$

这就完成了定理的证明.

推论3.1  设$A\in M_{n\times n}({\Bbb R})$, $0<\alpha<1$,向量值函数$\vec{f}(t)$是指数有界的.如果向量值函数$\vec{x}(t): [0, +\infty)\rightarrow{\Bbb R}^n$对任何的$t\in [0, +\infty)$都满足不等式

(3.12) $\begin{equation}\label{e3.12} \left\|^C_0\!D_t^\alpha\vec{x}(t)-A\vec{x}(t)-\vec{f}(t)\right\|\leq F(t), \end{equation}$

这里数量值函数$F(t)\geq 0$,那么存在分数阶系统(1.1)的一个解$\vec{x}^\ast(t)$使得

(3.13) $\begin{equation}\label{e3.13} \left\|\vec{x}(t)-\vec{x}^\ast(t)\right\|\leq\sum\limits_{k=0}^\infty\frac{\|A\|^k}{\Gamma(\alpha k+\alpha)}\int_0^t(t-\xi)^{\alpha k+\alpha-1}F(\xi){{\rm{d}}}\xi. \end{equation} $

证  应用定理3.1,容易证明该结论.

例3.1  考虑分数阶系统

(3.14) $\begin{equation}\label{3.14}^C_0\!D_t^{\frac{1}{2}}\vec{x}(t)=A\vec{x}(t)+\vec{f}(t), \end{equation}$

初始值为$\vec{x}(0)=\left[\begin{array}{cc}0\\0\end{array}\right]=\vec{\eta}$,这里$\vec{x}(t)=\left[\begin{array}{cc}x_1(t)\\x_2(t)\end{array}\right]$, $A=\left[\begin{array}{cc}1&2\\3&4\end{array}\right]$,对任何的$\varepsilon>0$,

$\vec{f}(t)=\left[\begin{array}{cc}-2t^3-t^2+\frac{8}{3\sqrt{\pi}}t^{\frac{3}{2}}+\frac{\varepsilon}{2}\\ -4t^3-3t^2+\frac{16}{5\sqrt{\pi}}t^{\frac{5}{2}}+\frac{\varepsilon}{3}\end{array}\right].$

对任何的$\varepsilon>0$,容易验证向量值函数$\vec{x}_1(t)=\left[\begin{array}{cc}t^2\\t^3\end{array}\right]$满足

(3.15) $ \begin{equation}\label{e3.15} \left\|^C_0\!D_t^{\frac{1}{2}}\vec{x}_1(t)-A\vec{x}_1(t)-\vec{f}(t)\right\|<\varepsilon, \end{equation}$

并且初始值为$\vec{x}_1(0)=\left[\begin{array}{cc}0\\0\end{array}\right]=\vec{\eta}$.应用(3.5)式和系统(3.14)的初始值,我们得到系统的一个精确解,也就是

(3.16) $ \begin{equation}\label{e3.16} \vec{x}^\ast(t)=\int_0^te_{\frac{1}{2}}^{A(t-\xi)}\vec{f}(\xi){{\rm{d}}}\xi. \end{equation}$

应用定理3.1, $\vec{x}_1(t)$的控制函数取为$\varepsilon t^{\frac{1}{2}}E_{\frac{1}{2}, \frac{3}{2}}\left(10 t^{\frac{1}{2}}\right)$,于是有

(3.17) $\begin{equation}\label{e3.17} \|\vec{x}_1(t)-\vec{x}^\ast(t)\|\leq\varepsilon t^{\frac{1}{2}}E_{\frac{1}{2}, \frac{3}{2}}\left(10 t^{\frac{1}{2}}\right). \end{equation}$

因此,我们能够估计逼近解$\vec{x}_1(t)$的误差.

在这一部分,我们将要应用Laplace变换方法研究分数阶系统(1.2)的Hyers-Ulam-Rassias稳定性.

定理4.1  设$A\in M_{n\times n}({\Bbb R})$, $0<\beta<\alpha<1$,向量值函数$\vec{g}(t)$是指数有界的.如果向量值函数$\vec{x}(t): [0, +\infty)\rightarrow{\Bbb R}^n$对任意的$t\in[0, +\infty)$和某个$\varepsilon>0$满足不等式

(4.1) $\begin{equation}\label{e4.1} \left\|\ ^C_0\!D_t^\alpha\vec{x}(t)-A \ ^C_0\!D_t^\beta\vec{x}(t)-\vec{g}(t)\right\|\leq\varepsilon, \end{equation}$

那么存在分数阶系统(1.2)的一个解$\vec{x}^\ast(t)$使得

(4.2) $\begin{equation}\label{e4.2} \left\|\vec{x}(t)-\vec{x}^\ast(t)\right\|\leq\varepsilon t^\alpha E_{\alpha-\beta, \alpha+1}\left(\|A\|t^{\alpha-\beta}\right). \end{equation}$

证  对任何的$t>0$,设$\vec{Z}(t)= \ ^C_0\!D_t^\alpha\vec{x}(t)-A\ ^C_0\!D_t^\beta\vec{x}(t)-\vec{g}(t)$,我们得到

(4.3) $\begin{eqnarray}\label{e4.3} {\cal L}\left\{\vec{Z}(t)\right\} &=& {\cal L}\left\{^C_0\!D_t^\alpha\vec{x}(t)-A \ ^C_0\!D_t^\beta\vec{x}(t)-\vec{g}(t)\right\} \nonumber\\ &=& {\cal L}\left\{\ ^C_0\!D_t^\alpha\vec{x}(t)\right\}-A{\cal L}\left\{\ ^C_0\!D_t^\beta\vec{x}(t)\right\}-{\cal L}\left\{\vec{g}(t)\right\} \nonumber\\ &=&s^\alpha I{\cal L}\left\{\vec{x}(t)\right\}-s^{\alpha-1}\vec{\eta}-A\left\{s^\beta I{\cal L}\left\{\vec{x}(t)\right\}-s^{\beta-1}\vec{\eta}\right\}-{\cal L}\left\{\vec{g}(t)\right\} \nonumber\\ &=&\left(s^\alpha I-As^\beta \right){\cal L}\left\{\vec{x}(t)\right\}-s^{\alpha-1}\vec{\eta}+As^{\beta-1}\vec{\eta}-{\cal L}\left\{\vec{g}(t)\right\}, \end{eqnarray}$

这里$I$是单位矩阵.由(4.3)式,得到

(4.4) $\begin{equation} {\cal L}\left\{\vec{x}(t)\right\}=\left(s^\alpha I-As^\beta \right)^{-1}\left[s^{\alpha-1}\vec{\eta}-As^{\beta-1}\vec{\eta}+{\cal L}\left\{\vec{g}(t)\right\}\right]+\left(s^\alpha I-As^\beta \right)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}. \end{equation}$

取

(4.5) $\begin{equation}\label{e4.5} \vec{x}^\ast(t)=x_0(t)\vec{\eta}+\int_0^t(t-\xi)^{\alpha-1}E_{\alpha-\beta, \alpha}\left[A(t-\xi)^{\alpha-\beta}\right]\vec{g}(\xi){{\rm{d}}}\xi, \end{equation}$

这里

(4.6) $ \begin{equation}\label{e4.6} x_0(t)=E_{\alpha-\beta, 1}(At^{\alpha-\beta})-At^{\alpha-\beta}E_{\alpha-\beta, \alpha-\beta+1}(At^{\alpha-\beta}). \end{equation}$

应用引理2.1,我们得到

(4.7) $ \begin{eqnarray}\label{e4.7} {\cal L}\{\vec{x}^\ast(t)\} &=&{\cal L}\Bigg{\{}x_0(t)\vec{\eta}+\int_0^t(t-\xi)^{\alpha-1}E_{\alpha-\beta, \alpha}\left[A(t-\xi)^{\alpha-\beta}\right]\vec{g}(\xi){{\rm{d}}}\xi\Bigg{\}}\nonumber\\ &=&{\cal L}\Big{\{}E_{\alpha-\beta, 1}(At^{\alpha-\beta})\Big{\}}\vec{\eta}-{\cal L}\Big{\{}At^{\alpha-\beta}E_{\alpha-\beta, \alpha-\beta+1}(At^{\alpha-\beta})\Big{\}}\vec{\eta}\nonumber\\ &&\mbox{}+{\cal L}\left\{\int_0^t(t-\xi)^{\alpha-1}E_{\alpha-\beta, \alpha}\left[A(t-\xi)^{\alpha-\beta}\right]\vec{g}(\xi){{\rm{d}}}\xi\right\}\nonumber\\ &=&{\cal L}\Big{\{}E_{\alpha-\beta, 1}(At^{\alpha-\beta})\Big{\}}\vec{\eta}-A{\cal L}\Big{\{}t^{\alpha-\beta}E_{\alpha-\beta, \alpha-\beta+1}(At^{\alpha-\beta})\Big{\}}\vec{\eta}\nonumber\\ &&\mbox{}+{\cal L}\left\{t^{\alpha-1}E_{\alpha-\beta, \alpha}\left[At^{\alpha-\beta}\right]\right\}{\cal L}\left\{\vec{g}(t)\right\}\nonumber\\ &=&s^{\alpha-\beta-1}\left(s^{\alpha-\beta}I-A\right)^{-1}\vec{\eta}-As^{\alpha-\beta-(\alpha-\beta+1)}\left(s^{\alpha-\beta}I-A\right)^{-1}\vec{\eta} \nonumber\\ &&\mbox{}+s^{-\beta}\left(s^{\alpha-\beta}I-A\right)^{-1}{\cal L}\left\{\vec{g}(t)\right\} \nonumber\\ &=&\left(s^{\alpha-\beta}I-A\right)^{-1}\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]. \end{eqnarray}$

由(4.7)式,可以得到

(4.8) $\begin{eqnarray}\label{e4.8} &&{\cal L}\left\{\ ^C_0\!D_t^\alpha\vec{x}^\ast(t)-A \ ^C_0\!D_t^\beta\vec{x}^\ast(t)\right\}\\ &=&{\cal L}\left\{^C_0\!D_t^\alpha\vec{x}^\ast(t)\right\}-A{\cal L}\left\{^C_0\!D_t^\beta\vec{x}^\ast(t)\right\}\\ &=&s^\alpha I{\cal L}\left\{\vec{x}^\ast(t)\right\}-s^{\alpha-1}\vec{\eta}-A\left\{s^\beta I{\cal L}\left\{\vec{x}^\ast(t)\right\}-s^{\beta-1}\vec{\eta}\right\}\\ &=&s^\alpha\left(s^{\alpha-\beta}I-A\right)^{-1}\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]-s^{\alpha-1}\vec{\eta}\\ &&-A\left\{s^\beta\left(s^{\alpha-\beta}I-A\right)^{-1}\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]-s^{\beta-1}\vec{\eta}\right\}\\ &=&s^\alpha\left(s^{\alpha-\beta}I-A\right)^{-1}\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]-s^{\alpha-1}\vec{\eta}\\ &&-As^\beta\left(s^{\alpha-\beta}I-A\right)^{-1}\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]+As^{\beta-1}\vec{\eta}\\ &=&\left(s^\alpha I-As^\beta\right)\left(s^{\alpha-\beta}I-A\right)^{-1}\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]\\ &&-\left(s^{\alpha-1}I-As^{\beta-1}\right)\vec{\eta}\\ &=&s^\beta\left(s^{\alpha-\beta}I-A\right)\left(s^{\alpha-\beta}I-A\right)^{-1}\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]\\ &&-\left(s^{\alpha-1}I-As^{\beta-1}\right)\vec{\eta}\\ &=&s^\beta\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]-\left(s^{\alpha-1}I-As^{\beta-1}\right)\vec{\eta}\\ &=&{\cal L}\left\{\vec{g}(t)\right\}, \end{eqnarray}$

因此, $\vec{x}^\ast(t)$是系统(1.2)的一个解.应用(4.4)和(4.7)式,我们得到

(4.9) $ \begin{eqnarray}\label{e4.9} {\cal L}\left\{\vec{x}(t)-\vec{x}^\ast(t)\right\} &=&{\cal L}\left\{\vec{x}(t)\right\}-{\cal L}\left\{\vec{x}^\ast(t)\right\} \nonumber\\ &=&\left(s^\alpha I-As^\beta \right)^{-1}\left[s^{\alpha-1}\vec{\eta}-As^{\beta-1}\vec{\eta}+{\cal L}\left\{\vec{g}(t)\right\}\right]+\left(s^\alpha I-As^\beta\right)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}\nonumber\\ &&\mbox{}-\left(s^{\alpha-\beta}I-A\right)^{-1}\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]\nonumber\\ &=&\left(s^\alpha I-As^\beta \right)^{-1}\left[s^{\alpha-1}\vec{\eta}-As^{\beta-1}\vec{\eta}+{\cal L}\left\{\vec{g}(t)\right\}\right]+\left(s^\alpha I-As^\beta \right)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}\nonumber\\ &&\mbox{}-\left(s^\alpha I-As^\beta\right)^{-1}\cdot s^\beta\left[s^{\alpha-\beta-1}\vec{\eta}-A\cdot s^{-1}\vec{\eta}+s^{-\beta}{\cal L}\left\{\vec{g}(t)\right\}\right]\nonumber\\ &=&\left(s^\alpha I-As^\beta \right)^{-1}\left[s^{\alpha-1}\vec{\eta}-As^{\beta-1}\vec{\eta}+{\cal L}\left\{\vec{g}(t)\right\}\right]+\left(s^\alpha I-As^\beta \right)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}\nonumber\\ &&\mbox{}-\left(s^\alpha I-As^\beta\right)^{-1}\left[s^{\alpha-1}\vec{\eta}-A\cdot s^{\beta-1}\vec{\eta}+{\cal L}\left\{\vec{g}(t)\right\}\right]\nonumber\\ &=&\left(s^\alpha I-As^\beta \right)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}. \end{eqnarray}$

另一方面,我们有

(4.10) $ \begin{eqnarray}\label{e4.10} {\cal L}\left\{\left[t^{\alpha-1}E_{\alpha-\beta, \alpha}\left(At^{\alpha-\beta}\right)\right]\ast \vec{Z}(t)\right\} &=&{\cal L}\left[t^{\alpha-1}E_{\alpha-\beta, \alpha}\left(At^{\alpha-\beta}\right)\right]{\cal L}\left\{\vec{Z}(t)\right\}\nonumber\\ &=&s^{-\beta}(s^{\alpha-\beta}I-A)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}\nonumber\\ &=&(s^\alpha I-As^\beta)^{-1}{\cal L}\left\{\vec{Z}(t)\right\}. \end{eqnarray}$

由(4.9)和(4.10)式,可得

(4.11) $ \begin{equation}\label{e4.11} \vec{x}(t)-\vec{x}^\ast(t)=\left[t^{\alpha-1}E_{\alpha-\beta, \alpha}\left(At^{\alpha-\beta}\right)\right]\ast \vec{Z}(t). \end{equation} $

因此,应用(4.11)式,得到

(4.12) $\begin{eqnarray}\label{e4.12} \left\|\vec{x}(t)-\vec{x}^\ast(t)\right\| &=& \left\|\left[t^{\alpha-1}E_{\alpha-\beta, \alpha}\left(At^{\alpha-\beta}\right)\right]\ast \vec{Z}(t)\right\| \nonumber\\ &=&\left\|\int_0^t(t-\xi)^{\alpha-1}E_{\alpha-\beta, \alpha}\left[A(t-\xi)^{\alpha-\beta}\right]\vec{Z}(\xi){{\rm{d}}}\xi\right\|\nonumber\\ &=&\left\|\int_0^t\sum\limits_{k=0}^\infty \frac{A^k(t-\xi)^{\alpha k-\beta k+\alpha-1}}{\Gamma[(\alpha-\beta) k+\alpha]}\vec{Z}(\xi){{\rm{d}}}\xi\right\|\nonumber\\ &=&\left\|\sum\limits_{k=0}^\infty \int_0^t\frac{A^k(t-\xi)^{\alpha k-\beta k+\alpha-1}}{\Gamma[(\alpha-\beta) k+\alpha]}\vec{Z}(\xi){{\rm{d}}}\xi\right\|\nonumber\\ &\leq&\sum\limits_{k=0}^\infty\left\|\int_0^t\frac{A^k(t-\xi)^{\alpha k-\beta k+\alpha-1}}{\Gamma[(\alpha-\beta) k+\alpha]}\vec{Z}(\xi){{\rm{d}}}\xi\right\|\nonumber\\ &\leq&\sum\limits_{k=0}^\infty\int_0^t\left\|\frac{A^k(t-\xi)^{\alpha k-\beta k+\alpha-1}}{\Gamma[(\alpha-\beta) k+\alpha]}\right\|\left\|\vec{Z}(\xi)\right\|{{\rm{d}}}\xi\nonumber\\ &\leq&\varepsilon\sum\limits_{k=0}^\infty\frac{\|A\|^k}{\Gamma[(\alpha-\beta) k+\alpha]}\int_0^t(t-\xi)^{\alpha k-\beta k+\alpha-1}{{\rm{d}}}\xi\nonumber\\ &=&\varepsilon t^\alpha\sum\limits_{k=0}^\infty\frac{\left(\|A\|t^{\alpha-\beta}\right)^k}{\Gamma[(\alpha-\beta) k+\alpha+1]}\nonumber\\ &=&\varepsilon t^\alpha E_{\alpha-\beta, \alpha+1}\left(\|A\|t^{\alpha-\beta}\right).\end{eqnarray}$

这就完成了定理的证明.

推论4.1  设$A\in M_{n\times n}({\Bbb R})$, $0<\beta<\alpha<1$,向量值函数$\vec{g}(t)$是指数有界的.如果向量值函数$\vec{x}(t): [0, +\infty)\rightarrow{\Bbb R}^n$对任意的$t>0$和某个函数$G(t)\geq 0$满足不等式

(4.13) $\begin{equation}\label{e4.13} \left\|\ ^C_0\!D_t^\alpha\vec{x}(t)-A \ ^C_0\!D_t^\beta\vec{x}(t)-\vec{g}(t)\right\|\leq G(t), \end{equation} $

那么存在分数阶系统(1.2)的一个解$\vec{x}^\ast(t): [0, +\infty)\rightarrow{\Bbb R}^n$使得

(4.14) $ \begin{equation}\label{e4.14} \left\|\vec{x}(t)-\vec{x}^\ast(t)\right\|\leq\sum\limits_{k=0}^\infty\frac{\|A\|^k}{\Gamma\left[(\alpha-\beta) k+\alpha\right]}\int_0^t(t-\xi)^{\alpha k-\beta k+\alpha-1}G(\xi){{\rm{d}}}\xi. \end{equation}$

证  应用定理4.1,容易证明该结论成立.

例4.1  考虑分数阶系统

(4.15) $\begin{equation}\label{4.15}{}^C_0\!D_t^{\frac{1}{2}}\vec{x}(t)=A \ ^C_0\!D_t^{\frac{1}{3}}\vec{x}(t)+\vec{g}(t), \end{equation}$

初始值为$\vec{x}(0)=\left[\begin{array}{cc}0\\0\end{array}\right]$, $\vec{x}(t)=\left[\begin{array}{cc}x_1(t)\\x_2(t)\end{array}\right]$, $A=\left[\begin{array}{cc}0&1\\-1&0\end{array}\right]$,这里的扰动项为

$\vec{g}(t)=\left[\begin{array}{cc}\frac{\sqrt{\pi}}{2}-\frac{3}{2\Gamma(\frac{2}{3})}t^{\frac{2}{3}}-\frac{1}{200}\sin t\\ 2\sqrt{\frac{t}{\pi}}+\frac{3\sqrt{\pi}}{\Gamma(\frac{1}{6})}t^{\frac{1}{6}}-\frac{1}{300}\cos t\end{array}\right].$

对于$\varepsilon=\frac{1}{100}$,向量值函数$\vec{x}_1(t)=\left[\begin{array}{cc}\sqrt{t}\\t\end{array}\right]$满足

(4.16) $ \begin{equation}\label{e4.16} \left\|\ ^C_0\!D_t^{\frac{1}{2}}\vec{x}_1(t)-A \ ^C_0\!D_t^{\frac{1}{3}}\vec{x}_1(t)-\vec{g}(t)\right\|<\frac{1}{100}, \end{equation} $

初始值满足$\vec{x}_1(0)=\left[\begin{array}{cc}0\\0\end{array}\right]$.应用公式(4.5)和系统(4.15)的初始值,我们得到该系统的一个精确解,也就是

(4.17) $ \begin{equation}\label{e4.17} \vec{x}^\ast(t)=\int_0^t(t-\xi)^{-\frac{1}{2}}E_{\frac{1}{6}, \frac{1}{2}}\left[A(t-\xi)^{\frac{1}{6}}\right]\vec{g}(\xi){{\rm{d}}}\xi. \end{equation}$

应用定理4.1, $\vec{x}_1(t)$的控制函数取为$\frac{1}{100}t^{\frac{1}{2}}E_{\frac{1}{6}, \frac{3}{2}}\left(2t^{\frac{1}{6}}\right)$,于是有

(4.18) $ \begin{equation}\label{e4.18} \|\vec{x}_1(t)-\vec{x}^\ast(t)\|\leq\frac{1}{100}t^{\frac{1}{2}}E_{\frac{1}{6}, \frac{3}{2}}\left(2t^{\frac{1}{6}}\right). \end{equation}$

这样,我们就能够估计逼近解$\vec{x}_1(t)$的误差.