1 问题的提出

在工程技术和科学研究中，有许多现象是由微分方程来描述的。随着科学技术的发展，人们对分数阶微分方程的研究越来越多，取得了很多成果^[1-13]。

微分方程组通常用来描述涉及到多个状态变量的运动系统，文献[14-15]研究了具有Caputo导数的分数阶微分方程组边值问题解的存在性。

本文研究一类高阶Riemann-Liouville分数阶微分方程组边值问题。

$\left\{ \begin{split} & {{\rm {D}}^\alpha }{ {U}}(t) + { \varLambda} {{\rm {D}}^{\alpha - 1}}{ {U}}(t) = { {F}}(t,{ {U}}(t)),\;\;0 < t < 1\\ & {u_1}({\eta _1}) = {u_2}({\eta _2}) = {u_3}({\eta _3}) = \cdots = {u_n}({\eta _n}) = 0 \\ & {u_1}(1) = {u_2}(1) = {u_3}(1) = \cdots = {u_n}(1) = 0 \end{split} \right.$

(1)

其中，

${ {U}}(t) = \left( {\begin{array}{*{20}{c}} {{u_1}(t)} \\ {{u_2}(t)} \\ \vdots \\ {{u_n}(t)} \end{array}} \right),\;\;{{\rm {D}}^\alpha }{ {U}}(t) = \left( \begin{gathered} {{\rm {D}}^\alpha }{u_1}(t) \\ {{\rm {D}}^\alpha }{u_2}(t) \\ {\rm{ }} \vdots \\ {{\rm {D}}^\alpha }{u_n}(t) \\ \end{gathered} \right)\\{ {F}} = \left( {\begin{array}{*{20}{c}} {{f_1}} \\ {{f_2}} \\ \vdots \\ {{f_n}} \end{array}} \right), \;{ {\varLambda}} = \left( {\begin{array}{*{20}{c}} {{\lambda _1}}&0&0& \cdots &0 \\ 0&{{\lambda _2}}&0& \cdots &0 \\ 0&0&{{\lambda _3}}& \cdots &0 \\ \vdots & \vdots & \vdots & \; & \vdots \\ 0&0&0& \cdots &{{\lambda _n}} \end{array}} \right)$

${{\rm {D}}^\alpha }$ 表示Riemann-Liouville导数， $1 < \alpha < 2,0 < {\lambda _i} <$ $ 1,0 < {\eta _i} < 1,{f_i} \in C(I \times {\mathbb{R}^n},{\rm{ }}\mathbb{R})$ ， $t \in [0,1],\;\;$ $i = 1,2, \cdots ,n$ 。

2 预备知识与引理

为了后面证明的需要，现给出一些定义和引理。

定义1^[16] 　设函数 $f(t)$ 当 $t \geqslant 0$ 时有定义，且积分 $\displaystyle\int_0^{ + \infty } {f(t){{\rm e}^{ - st}}\operatorname{d} } t$ （ $s$ 是一个复变量）在s的某一个域内收敛，则由此积分所确定的函数 $F(s) = \displaystyle\int_0^{ + \infty } {f(t){{\rm e}^{ - st}}{\rm d} t} $ 称为函数 $f(t)$ 的Laplace变换，记 $F(s) = {\cal{L}}\left( {f(t)} \right),$ F(s)称为 $f(t)$ 的像函数. 在相同条件下，称

$f(t) = {{\cal{L}}^{ - 1}}\left( {F(s)} \right) = \frac{1}{{2\pi {\rm {i}}}}\int_{{{s}} - {\rm i}\infty }^{{ {s}} + {\rm i}\infty } {F(p)} {{\rm e}^{pt}}{\rm d} p$

为 $F(s)$ 的Laplace逆变换。

引理1^[16] 　若 ${\cal{L}}\left( {{f_1}(t)} \right) = {F_1}(s),{\rm{ }}{\cal{L}}\left( {{f_2}(t)} \right) = {F_2}(s),$ 则

$\begin{split} &{\cal{L}}\left( {{f_1}(t)*{f_2}(t)} \right) = {F_1}(s){F_2}(s) \\ &{{\cal{L}}^{ - 1}}\left( {{F_1}(s)*{F_2}(s)} \right) = {f_1}(t){f_2}(t) \end{split} $

定义2^[16] 　函数 $u:{\mathbb{R}_ + } \to \mathbb{R}$ 的 $\alpha $ 阶Riemann-Liouville分数阶积分定义为

${{\rm I}^\alpha }u(t) = \frac{1}{{\Gamma (\alpha )}}\int_0^t {{{(t - \tau )}^{\alpha - 1}}} u(\tau ){\rm d} \tau$

对任意的 $\alpha > 0$ ，右端积分在 ${\mathbb{R}_ + }$ 上逐点可积， $\Gamma ( \cdot )$ 是Gamma函数。

定义3^[17] 　函数 $u:{\mathbb{R}_ + } \to \mathbb{R}$ 的 $\alpha $ 阶Riemann-Liouville分数阶导数定义为

${{\rm {D}}^\alpha }u(t) = {{\rm {D}}^n}{{\rm I}^{n - \alpha }}u(t) = \frac{1}{{\Gamma (n - \alpha )}}{\left(\frac{{\rm {d}}}{{{\rm {d}}{t^n}}}\right)^n}\int_0^t {\frac{{u(\tau )}}{{{{(t - \tau )}^{\alpha + 1 - n}}}}} {\rm d} \tau$

对任意的 $\alpha > 0$ ，右端积分在 ${\mathbb{R}_ + }$ 上逐点可积，n为大于或等于 $\alpha $ 的最小整数。

引理2^[17] 　对任意的 $\alpha > 0,$ 函数 $u:{\mathbb{R}_ + } \to \mathbb{R}$ 的 $\alpha $ 阶Laplace变换公式为

${\cal{L}}({{\rm {D}} ^ \alpha} u)(s) = {s ^ \alpha}{\cal{L}}(u)(s) - \sum\limits_{j = 1}^l {{d_j}} {s^{j - 1}},\;\;\;l - 1 \leqslant \alpha \leqslant l,{\rm{ }}l \in \mathbb{N}$

${d_j} = ({{\rm {D}}^{\alpha - j}}u)(0^+ ),\;\;j = 1,2, \cdots ,l$

定义4^[17] 　令 $\alpha ,\beta > 0$ ，函数 ${E_{\alpha ,\beta }}$ 的定义为

${E_{\alpha ,\beta }}(z) = \sum\limits_{k = 0}^\infty {\frac{{{z^k}}}{{\Gamma (\alpha k + \beta )}}}$

当级数收敛时，称级数是关于参数 $\alpha ,\beta $ 的二元Mittag-Leffler函数。

引理3^[17] 　Mittag-Leffler函数的Laplace变换公式为

$\begin{split}&{\cal{L}}\left( {{t^{\alpha k + \beta - 1}}E_{\alpha ,\beta }^{(k)}( \pm a{t^\alpha })} \right) = \int_0^\infty {{{\rm e}^{ - st}}{t^{\alpha k + \beta - 1}}} E_{\alpha ,\beta }^{(k)}( \pm a{t^\alpha }){\rm d} t = \\ &\qquad\qquad\frac{{k!{s^{\alpha - \beta }}}}{{{{({s^\alpha } { m}a)}^{k + 1}}}},\;\; {\rm {Re}}(s) > |a{|^{\frac{1}{\alpha }}}\end{split}$

引理4^[17] 　广义Mittag-Leffler导数定义为

$\begin{split}&E_{\alpha ,\beta }^{(n)}(z) = {\left(\frac{{\rm d}}{{{\rm d}z}}\right)^n}\left( {E_{\alpha ,\beta }^{n + 1}(z)} \right) = n!E_{\alpha ,\beta + \alpha n}^{n + 1}(z),\\ &\qquad z \in \mathbb{C},\;\alpha ,\beta ,\rho \in \mathbb{C},\;R (\alpha ) > 0\end{split}$

其中，

$\begin{split} & E_{\alpha ,\beta }^\rho (z) = \sum\limits_{k = 0}^\infty {\frac{{{{(\rho )}_k}}}{{\Gamma (\alpha k + \beta )}}} \frac{{{z^k}}}{{k!}} \\ & {(\rho )_k} = \rho (\rho + 1) \cdots (\rho + k - 1) \end{split} $

引理5 　设 ${\rm{0}} < {\lambda _1} < 1$ , 则级数

$\begin{split} &\sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}}}{{n!}}} {t^{n + \alpha - 1}}E_{1,\alpha }^{(n)}( - t) = \\ & \quad\sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}}}{{n!}}} {t^{n + \alpha - 1}}\sum\limits_{k = 0}^\infty {\frac{{\Gamma (n + k + 1)}}{{\Gamma (n + k + \alpha )}}} \frac{{{{( - t)}^k}}}{{k!}} \buildrel \displaystyle\Delta \over = {g_{11}}(t) \\ &\sum\limits_{n = 0}^\infty {\frac{{(1 + {\lambda _1}){{(1 - {\lambda _1})}^n}}}{{n!}}} {t^{n + \alpha - 1}}E_{1,\alpha }^{(n)}( - t) =\\ & \quad\sum\limits_{n = 0}^\infty {\frac{{(1 + {\lambda _1}){{(1 - {\lambda _1})}^n}}}{{n!}}} {t^{n + \alpha - 1}}\sum\limits_{k = 0}^\infty {\frac{{\Gamma (n + k + 1)}}{{\Gamma (n + k + \alpha )}}} \frac{{{{( - t)}^k}}}{{k!}} \buildrel \displaystyle\Delta \over = {g_{12}}(t)\\ &{t^{2 - \alpha }}\sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}}}{{n!}}} {t^{n + \alpha - 2}}E_{1,\alpha - 1}^{(n)}( - t) = \\ & \quad \sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}}}{{n!}}} {t^n}\sum\limits_{k = 0}^\infty {\frac{{\Gamma (n + k + 1)}}{{\Gamma (n + k + \alpha - 1)}}} \frac{{{{( - t)}^k}}}{{k!}} \buildrel \displaystyle\Delta \over = {t^{2 - \alpha }}{g_{13}}(t) \end{split} $

在 $[0,1]$ 上一致收敛。证明从略。

因为，

$ \begin{split}& \mathop {\lim }\limits_{t \to {0^ + }}\; {t^{2 - \alpha }}{g_{13}}(t) =\\ &\qquad \mathop {\lim }\limits_{t \to {0^ + }} \left(\sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}}}{{n!}}{t^n}\sum\limits_{k = 0}^\infty {\frac{{\Gamma (n + k + 1)}}{{\Gamma (n + k + \alpha - 1)}} \frac{{{{( - t)}^k}}}{{k!}}} } \right) =\\ &\qquad\frac{1}{{\Gamma (\alpha - 1)}} \end{split}$

所以， ${t^{2 - \alpha }}{g_{13}}(t){|_{t = 0}} = \dfrac{1}{{\Gamma (\alpha - 1)}}$ ，即函数 ${t^{2 - \alpha }}{g_{13}}(t)$ 在 $t $ = 0时连续。

引理6 　函数 ${g_{11}}(t),{g_{12}}(t),{g_{13}}(t)$ 的性质：

a. $ 0 < {g_{11}}(t) < \dfrac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _1})}};$

b. 函数 ${g_{11}}(t),{g_{12}}(t),{t^{2 - \alpha }}{g_{13}}(t)$ 在[0, 1]上连续。

证明 　 a. 由引理5可知， ${g_{11}}(t) > 0$ 显然成立。

$\begin{split} {g_{11}}(t) =& \sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}}}{{n!}}} {t^{n + \alpha - 1}}\sum\limits_{k = 0}^\infty {\frac{{\Gamma (n + k + 1)}}{{\Gamma (n + k + \alpha )}}} \frac{{{{( - t)}^k}}}{{k!}} = \\ & \sum\limits_{k = 0}^\infty {{{( - 1)}^k} \frac{{{t^{k + \alpha - 1}}}}{{\Gamma (k + \alpha )}}} + \\ &\sum\limits_{n = 1}^\infty {\frac{{{{(1 - {\lambda _1})}^n}}}{{n!}}} {t^{n + \alpha - 1}}\sum\limits_{k = 0}^\infty {\frac{{\Gamma (n + k + 1)}}{{\Gamma (n + k + \alpha )}}} \frac{{{{( - t)}^k}}}{{k!}} \end{split} $

由引理5可知，级数 $\displaystyle\sum\limits_{k = 0}^\infty {{{( - 1)}^k} \dfrac{{{t^{k + \alpha - 1}}}}{{\Gamma (k + \alpha )}}} $ 关于t在 $[0,1]$ 上一致收敛，

因为，

$\frac{{{t^{k + \alpha }}}}{{\Gamma (k + \alpha + 1)}} \frac{{\Gamma (k + \alpha )}}{{{t^{k + \alpha - 1}}}} = \frac{t}{{k + \alpha }} < 1$

所以，函数列 $\left\{ \dfrac{{{t^{k + \alpha - 1}}}}{{\Gamma (k + \alpha )}}\right\} $ 关于k单调递减。

结合Leibniz判别法可知，

$0 < \sum\limits_{k = 0}^\infty {{{( - 1)}^k}} \frac{{{t^{k + \alpha - 1}}}}{{\Gamma (k + \alpha )}} < \frac{{{t^{\alpha - 1}}}}{{\Gamma (\alpha )}} < \frac{1}{{\Gamma (\alpha )}}$

由引理5可知，级数

$\begin{split}&\sum\limits_{n = 1}^\infty {\frac{{(1 - {\lambda _1})}}{{n!}}} {t^{n + \alpha - 1}}\sum\limits_{k = 0}^\infty {\frac{{\Gamma (n + k + 1)}}{{\Gamma (n + k + \alpha )}}} \frac{{{{( - t)}^k}}}{{k!}} < \\ &\qquad\sum\limits_{n = 0}^\infty {\frac{{{{\left( {t(1 - {\lambda _1})} \right)}^n}}}{{n!}}} = {{\rm e}^{t(1 - {\lambda _1})}} < {{\rm e}^{(1 - {\lambda _1})}}\end{split}$

即证 $ 0 < {g_{11}}(t) < \dfrac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _1})}}$ 。

b. 函数 ${g_{11}}(t),{g_{12}}(t),{t^{2 - \alpha }}{g_{13}}(t)$ 在[0, 1]上连续，显然成立。

证毕。

引理7 　设 $1 \!\!<\!\! \alpha \!\! <\!\! 2,$ ${\varDelta _1}\!\!=\!\! {g_{12}}({\eta _1}){g_{13}}(1) \!-\! $ ${g_{12}}(1){g_{13}}({\eta _1})\!\! \ne $ 0，则对任意 ${h_1} \in C(I,\mathbb{R})$ ，边值问题

$\left\{ \begin{split} &{{\rm D}^\alpha }{u_1}(t) + {\lambda _1}{{\rm D}^{\alpha - 1}}{u_1}(t) = {h_1}(t),\;\;\;{\rm{ }}0 < t < 1 \\ &{u_1}({\eta _1}) = 0,\;\;\;{\rm{ }}{u_1}(1) = 0 \end{split} \right.$

(2)

存在唯一解

$\begin{split} &{u_1}(t) = \int_0^t {{g_{11}}(t - \tau )} {h_1}(\tau ){\rm d} \tau + \hspace{80pt}\\ &\quad\frac{{{g_{12}}(1){g_{13}}(t) - {g_{13}}(1){g_{12}}(t)}}{{{\varDelta _1}}}\!\! \int_0^{{\eta _1}} {{g_{11}}({\eta _1} - \tau )} {h_1}(\tau ){\rm d} \tau + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad \frac{{{g_{13}}({\eta _1}){g_{12}}(t) \!- \!{g_{12}}({\eta _1}){g_{13}}(t)}}{{{\varDelta _1}}}\!\! \int_0^1 {{g_{11}}(1 \!-\! \tau )} {h_1}(\tau ){\rm d} \tau \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \end{split} $

(3)

证明 　由引理2可知，

$\begin{split} &{\cal{L}}({{\rm D}^\alpha }{u_1}(t))(s) = {s^\alpha }{\cal{L}}(u(t))(s) - {d_{11}} - {d_{12}}s\\ &{\cal{L}}({{\rm D}^{\alpha - 1}}{u_1}(t))(s) = {s^{\alpha - 1}}{\cal{L}}(u(t))(s) - {d_{11}} \end{split} $

所以，对 ${{\rm D}^\alpha }{u_1}(t) + {\lambda _1}{{\rm D}^{\alpha - 1}}{u_1}(t) = {h_1}(t)$ 进行Laplace变换，可得

$({s^\alpha } + {\lambda _1}{s^{\alpha - 1}}){\cal{L}}({u_1}(t))(s) - (1 + {\lambda _1}){d_{11}} - {d_{12}}s = {\cal{L}}({h_1}(t))(s)$

整理得到

$\begin{split}{\cal{L}}({u_1}(t))(s) =& \frac{1}{{{s^\alpha } + {\lambda _1}{s^{\alpha - 1}}}}{\cal{L}}({h_1}(t))(s) + \\ &\frac{{1 + {\lambda _1}}}{{{s^\alpha } + {\lambda _1}{s^{\alpha - 1}}}}{d_{11}} +\frac{s}{{{s^\alpha } + {\lambda _1}{s^{\alpha - 1}}}}{d_1}_2\end{split}$

又因为，

$\begin{split}&\dfrac{1}{{{s^\alpha } + {\lambda _1}{s^{\alpha - 1}}}} = \dfrac{{{s^{1 - \alpha }}}}{{s + {\lambda _1} + 1 - 1}} = \dfrac{{{s^{1 - \alpha }}}}{{1 + s}} \dfrac{1}{{1 + \dfrac{{{\lambda _1} - 1}}{{1 + s}}}} =\hspace{40pt} \\ &\qquad\quad\dfrac{{{s^{1 - \alpha }}}}{{1 + s}}\sum\limits_{n = 0}^\infty {\dfrac{{{{(1 - {\lambda _1})}^n}}}{{{{(1 + s)}^n}}}} = \sum\limits_{n = 0}^\infty {\dfrac{{{{(1 - {\lambda _1})}^n}{s^{1 - \alpha }}}}{{{{(1 + s)}^{n + 1}}}}}\end{split} $

$\begin{split}&\dfrac{s}{{{s^\alpha } + {\lambda _1}{s^{\alpha - 1}}}} = \dfrac{{{s^{2 - \alpha }}}}{{s + {\lambda _1} + 1 - 1}} = \dfrac{{{s^{2 - \alpha }}}}{{1 + s}} \dfrac{1}{{1 + \dfrac{{{\lambda _1} - 1}}{{1 + s}}}} = \hspace{40pt}\\ &\qquad\quad\dfrac{{{s^{2 - \alpha }}}}{{1 + s}}\sum\limits_{n = 0}^\infty {\dfrac{{{{(1 - {\lambda _1})}^n}}}{{{{(1 + s)}^n}}}} = \sum\limits_{n = 0}^\infty {\dfrac{{{{(1 - {\lambda _1})}^n}{s^{2 - \alpha }}}}{{{{(1 + s)}^{n + 1}}}}}\end{split}$

所以，

$\begin{split} & {\cal{L}}({u_1}(t))(s) = \sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}{s^{1 - \alpha }}}}{{{{(1 + s)}^{n + 1}}}}} {\cal{L}}({h_1}(t))(s) +\\ &\qquad\quad\sum\limits_{n = 0}^\infty {\frac{{(1 + {\lambda _1}){{(1 - {\lambda _1})}^n}{s^{1 - \alpha }}}}{{{{(1 + s)}^{n + 1}}}}} {d_{11}} +\\ &\qquad\quad\sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}{s^{2 - \alpha }}}}{{{{(1 + s)}^{n + 1}}}}} {d_{12}} \end{split} $

(4)

由引理3可知，

$\begin{split} &\frac{{{s^{1 - \alpha }}}}{{{{(1 + s)}^{n + 1}}}} = \frac{1}{{n!}}{\cal{L}}\left( {{t^{n + \alpha - 1}}E_{1,\alpha }^{(n)}( - t)} \right)(s) \\ &\frac{{{s^{2 - \alpha }}}}{{{{(1 + s)}^{n + 1}}}} = \frac{1}{{n!}}{\cal{L}}\left( {{t^{n + \alpha - 2}}E_{1,\alpha - 1}^{(n)}( - t)} \right)(s) \end{split} $

(5)

将式（5）代入式（4）并整理，得到

$\!\!\!\!\!\!\begin{split} \!\!\!\!\!\!\!\!&{\cal{L}}({u_1}(t))(s) \!\!=\!\! {\cal{L}}\left( {\sum\limits_{n = 0}^\infty {\frac{{{{(1 \!\!-\!\! {\lambda _1})}^n}}}{{n!}}{t^{n + \alpha - 1}}E_{1,\alpha }^{(n)}( - t)} } \right)(s) {\cal{L}}({h_1}(t))(s) +\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \\ &\qquad\quad {d_{11}}{\cal{L}}\left( {\sum\limits_{n = 0}^\infty {\frac{{(1 + {\lambda _1}){{(1 - {\lambda _1})}^n}}}{{n!}}{t^{n + \alpha - 1}}E_{1,\alpha }^{(n)}( - t)} } \right)(s) +\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ & \qquad\quad {d_{12}}{\cal{L}}\left( {\sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _1})}^n}}}{{n!}}{t^{n + \alpha - 2}}E_{1,\alpha - 1}^{(n)}( - t)} } \right)(s) =\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \\ &\qquad \quad {\cal{L}}({g_{11}}(t))(s) {\cal{L}}({h_1}(t))(s) + {d_{11}}{\cal{L}}({g_{12}}(t))(s) + \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \\ &\qquad\quad{d_{12}}{\cal{L}}({g_{13}}(t))(s)= {\cal{L}}\left( {{g_{11}}(t)*{h_1}(t)} \right)(s) +\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \\ & \qquad\quad{d_{11}}{\cal{L}}({g_{12}}(t))(s) + {d_{12}}{\cal{L}}({g_{13}}(t))(s) = \\ &\qquad\quad {\cal{L}}\left( {\int_0^t {{g_{11}}(t - \tau )} {h_1}(\tau ){\rm d} \tau } \right)(s) + \\ &\qquad\quad{d_{11}}{\cal{L}}({g_{12}}(t))(s) + {d_{12}}{\cal{L}}({g_{13}}(t))(s)\end{split}$

(6)

对式（6）两边进行Laplace逆变换，得到

${{ u}_1}(t) = \int_0^t {{g_{11}}(t - \tau )} {h_1}(\tau ){\rm d} \tau + {d_{11}}{g_{12}}(t) + {d_{12}}{g_{13}}(t)$

(7)

将边界条件 ${u_1}(1) = 0,{\rm{ }}{u_1}({\eta _1}) = 0$ 代入式（7），得到

$\begin{split} &\int_0^{{\eta _1}} {{g_{11}}({\eta _1} - \tau )} {h_1}(\tau ){\rm d} \tau + {d_{11}}{g_{12}}({\eta _1}) + {d_{12}}{g_{13}}({\eta _1}) = 0\\ & \quad\int_0^1 {{g_{11}}(1 - \tau )} {h_1}(\tau ){\rm d} \tau + {d_{11}}{g_{12}}(1) + {d_{12}}{g_{13}}(1) = 0 \end{split} $

设 $ {\varDelta _1} \ne 0$ 时，解得

$\begin{split}{d_{11}} =& \frac{1}{{{\varDelta _1}}}\left( {{g_{13}}({\eta _1})\int_0^1 {{g_{11}}(1 - \tau )} {h_1}(\tau ){\rm d} \tau -}\right.\\ &\left.{ {g_{13}}(1)\int_0^{{\eta _1}} {{g_{11}}({\eta _1} - \tau )} {h_1}(\tau ){\rm d} \tau } \right)\end{split}$

$\begin{split}{d_{12}} =& \frac{1}{{{\varDelta _1}}}\left( {{g_{12}}(1)\int_0^{{\eta _1}} {{g_{11}}({\eta _1} - \tau )} {h_1}(\tau ){\rm d} \tau -}\right.\\ &\left.{{g_{12}}({\eta _1})\int_0^1 {{g_{11}}(1 - \tau )} {h_1}(\tau ){\rm d} \tau } \right)\end{split}$

将d₁，d₂带入式（7），可得式（3）成立。

证毕。

由引理7即可得到引理8。

引理8 　边值问题（1）等价于积分方程

$\begin{split} {u_i}(t) =& \int_0^t {{g_{i1}}(t - \tau )} {f_i}(\tau ,{u_1}(\tau ), \cdots ,{u_n}(\tau )){\rm d} \tau +\\ &\quad\frac{{{g_{i2}}(1){g_{i3}}(t) - {g_{i3}}(1){g_{i2}}(t)}}{{{\varDelta _i}}}{\text{·}}\\ &\quad\int_0^{{\eta _i}} {{g_{i1}}({\eta _i} - \tau )} {f_i}(\tau ,{u_1}(\tau ), \cdots ,{u_n}(\tau )){\rm d} \tau + \\ &\quad\frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}}{\text{·}}\\ &\quad\int_0^1 {{g_{i1}}(1 - \tau )} {f_i}(\tau ,{u_1}(\tau ), \cdots ,{u_n}(\tau )){\rm d} \tau \end{split} $

其中，

$i = 1,2, \cdots ,n,\;\;{\varDelta _i} = {g_{i2}}({\eta _i}){g_{i3}}(1) - {g_{i2}}(1){g_{i3}}({\eta _i}) \ne 0$

$\begin{split} &{g_{i1}}(t) = \sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _i})}^n}}}{{n!}}} {t^{n + \alpha - 1}}E_{1,\alpha }^{(n)}( - t) \\ & {g_{i2}}(t) = \sum\limits_{n = 0}^\infty {\frac{{(1 + {\lambda _i}){{(1 - {\lambda _i})}^n}}}{{n!}}} {t^{n + \alpha - 1}}E_{1,\alpha }^{(n)}( - t) \\ & {g_{i3}}(t) = \sum\limits_{n = 0}^\infty {\frac{{{{(1 - {\lambda _i})}^n}}}{{n!}}} {t^{n + \alpha - 2}}E_{1,\alpha - 1}^{(n)}( - t) \end{split} $

引理9^[18]（Leray-Shauder抉择）　令 $E$ 是Banach空间，假设 $T:E \to E$ 是全连续算子，令 $V(T) = \{ x \in E:$ 存在 $\mu \in (0,1),$ 使得 $x = \mu T(x)\} $ ，则集合 $V(T)$ 是无界集或者算子T至少存在1个不动点。

3 解的存在性与唯一性

空间 ${E_1} = C_{2 - \alpha }^0[0,1] = \{ u \in C(0,1]:\mathop {\lim }\limits_{t \to {0^ + }}\; {t^{2 - \alpha }}u(t)$ 存在 $\} $ ，以范数 $||u|| = \mathop {\sup }\limits_{t \in (0,1]}\; {t^{2 - \alpha }}|u(t)|$ 构成了Banach空间。所以， $E = \underbrace {{E_1} \times {E_1} \times \cdots \times {E_1}}_n$ 以范数 $||{ U}|{|_E}{\rm{ = ||}}{u_1}|| + $ $ {\rm{||}}{u_2}|| + \cdots + {\rm{||}}{u_n}||$ 构成了Banach空间。

设 ${\varDelta _i} \ne 0,{\rm{ }}i = 1,2, \cdots ,n$ ，为了方便，给出记号：

${l_{i1}} = \mathop {\sup }\limits_{t \in (0,1]} \;\left|{t^{2 - \alpha }} \frac{{{g_{i2}}(1){g_{i3}}(t) - {g_{i3}}(1){g_{i2}}(t)}}{{{\varDelta _i}}}\right|$

${l_{i2}} = \mathop {\sup }\limits_{t \in (0,1]} \;\left|{t^{2 - \alpha }} \frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}}\right|$

由引理8可知，对任意的 $ { U}(t) = {({u_1}(t),{u_2}(t), \cdots ,}$ ${{u_n}(t))^{\rm T}} \in E,$ 定义算子

$T({ U})(t) = {\left( {{T_1}({ U})(t), \cdots ,{T_n}({ U})(t)} \right)^{\rm {\rm T}}}$

即

$\begin{split} &{T_i}({ U}(t)) = \int_0^t {{g_{i1}}(t - \tau )} {f_i}(\tau ,{ U}(\tau ))\operatorname{d} \tau + \\ &\quad \frac{{{g_{i2}}(1){g_{i3}}(t) - {g_{i3}}(1){g_{i2}}(t)}}{{{\varDelta _i}}}{\text{·}}\\ &\quad \int_0^{{\eta _i}} {{g_{i1}}({\eta _i} - \tau )} {f_i}(\tau ,{ U}(\tau )){\rm d} \tau+ \\ &\quad\frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}}{\text{·}}\\ &\quad \int_0^1 {{g_{i1}}(1 - \tau )} {f_i}(\tau ,{ U}(\tau )){\rm d} \tau \end{split} $

(8)

其中， $i = 1,2,\cdots, n$ 。所以，算子 $T:E \to E$ 。

为了证明的方便，给出记号：

${\gamma _i} = \frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}},{\rm{ }}{\theta _i} = 1 + {\eta _i}{l_{i1}} + {l_{i2}}$

(9)

$\quad{\varPhi _0} = {\gamma _1}{\theta _1}{k_{10}} + {\gamma _2}{\theta _2}{k_{20}} + \cdots + {\gamma _n}{\theta _n}{k_{n0}}\quad\quad\quad$

$ {\varPhi _i} = \frac{1}{{\alpha - 1}}({\gamma _1}{\theta _1}{k_{1i}} + {\gamma _2}{\theta _2}{k_{2i}} + \cdots + {\gamma _n}{\theta _n}{k_{ni}}) $

(10)

定理1 　设 ${\varDelta _i} \ne 0$ , 存在实数 ${k_{ij}}(i,j = 1,2, \cdots ,n)$ 且 ${k_{i0}} > 0$ ，使得

$|{f_i}(t,{u_1}, \cdots ,{u_n})| \leqslant {k_{i0}} + {k_{i1}}|{u_1}| + \cdots + {k_{in}}|{u_n}|$

并且设 $\max\; \{ {\varPhi _1},{\varPhi _2}, \cdots ,{\varPhi _n}\} < 1$ , 其中， ${\varPhi _i}$ 是由式（10）给出的，则边值问题（1）在[0, 1]上至少存在1个解。

证明 　由假设条件可知，任意

$\begin{split} &{ U}(t) = {({u_1}(t),{u_2}(t), \cdots ,{u_n}(t))^{\rm T}} \in E, \\ &|{f_i}(t,{u_1}(t),{u_2}(t), \cdots ,{u_n}(t))| \leqslant{k_{i0}} + {k_{i1}}|{u_1}(t)| + \\ & \quad {k_{i2}}|{u_2}(t)| + \cdots + {k_{in}}|{u_n}(t)|= {k_{i0}} + {t^{\alpha - 2}}({k_{i1}}{t^{2 - \alpha }}|{u_1}(t)| +\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ & \quad{k_{i2}}{t^{2 - \alpha }}|{u_2}(t)| \!+\! \cdots \!+\! {k_{in}}{t^{2 - \alpha }}|{u_n}(t)|) \!= \\ &\quad{k_{i0}} + {t^{\alpha - 2}}({k_{i1}}||{u_1}|| + {k_{i2}}||{u_2}|| + \cdots + {k_{in}}||{u_n}||) \end{split} $

(11)

a. 证明算子T是全连续。

（a） $T$ 是连续算子。

设 $\{ {{ U}_m}\} \subset E,{\rm{ }}{ U} \in E$ ，当 $m \to \infty $ 时， $ ||{{ U}_m} - { U}|{|_E} \to$ $ 0$ ，则存在一个常数 $\sigma > 0$ ，使得 $||{{ U}_m}|{|_E} \leqslant \sigma ,{\rm{ ||}}{ U}{\rm{|}}{{\rm{|}}_E} \leqslant $ $ \sigma $ ，即对任意的 $t \in [0,1]$ ，有 $|{t^{2 - \alpha }}{u_i}_m| \leqslant \sigma $ , ${\rm{|}}{t^{2 - \alpha }}{u_i}| \leqslant $ $ \sigma $ 。因此，对几乎处处的 $t \in [0,1]$ ，有

$\begin{split}&\mathop {\lim }\limits_{m \to \infty } \;{f_i}(t,{{ U}_m}(t)) = \mathop {\lim }\limits_{m \to \infty }\;{f_i}(t,{t^{\alpha - 2}}{t^{2 - \alpha }}{{ U}_m}(t)) = \\ &\qquad\quad{f_i}(t,{t^{\alpha - 2}}{t^{2 - \alpha }}{ U}(t)) = {f_i}(t,{ U}(t))\end{split}$

由引理6和式（11）可知，

$ \begin{split}&|{g_{i1}}(t - \tau )\left( {{f_i}(t,{{ U}_m}(t)) - {f_i}(t,{ U}(t))} \right)| \leqslant\\ &\quad 2\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm {\rm e}}^{(1 - {\lambda _i})}}\right)\left( {{k_{i0}} + {t^{\alpha - 2}}({k_{i1}}||{u_1}|| + \cdots + {k_{in}}||{u_n}||)} \right)\end{split} $

由Lebesgue控制收敛定理可得

$ \begin{split} &\mathop {\lim }\limits_{m \to \infty } \;||{T_i}{{ U}_m} - {T_i}{ U}|| = \mathop {\lim }\limits_{m \to \infty } \mathop {\sup }\limits_{t \in (0,1]} \;{t^{2 - \alpha }}|{T_i}{{ U}_m} - {T_i}{ U}| =\\ &\quad \mathop {\lim }\limits_{m \to \infty } \mathop {\sup }\limits_{t \in (0,1]} {t^{2 - \alpha }}\Bigg( {\int_0^t {{g_{i1}}(t \!-\! \tau )|{f_i}(\tau ,{{ U}_m}(\tau ))} \!-\! {f_i}(\tau ,{ U}(\tau ))|{\rm d} \tau } + \\ &\quad \frac{{{g_{i2}}(1){g_{i3}}(t) \!-\! {g_{i3}}(1){g_{i2}}(t)}}{{{\varDelta _i}}}\int_0^{{\eta _i}} {{g_{i1}}({\eta _i} \!-\! \tau )|{f_i}(\tau ,{{ U}_m}(\tau ))} - \\ &\quad{f_i}(\tau ,{ U}(\tau ))|{\rm d} \tau + \frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}} {\text{·}}\\ &\quad{\int_0^1 {{g_{i1}}(1 - \tau )|{f_i}(\tau ,{{ U}_m}(\tau ))} - {f_i}(\tau ,{ U}(\tau ))|{\rm d} \tau } \Bigg) = 0 \end{split} $

故 $\mathop {\lim }\limits_{m \to \infty }\; ||T{{ U}_m} - T{ U}|{|_E} \to 0$ ，则算子 $T$ 是连续算子。

（b） $T$ 是相对列紧。

首先证明算子 $T$ 在E上一致有界。令 $\varOmega \subset E$ 有界，对任意 ${ U} \in \varOmega $ ，存在 $M > 0$ ，使得 $||{ U}|{|_\varOmega } \leqslant M$ 。由引理6可知，

$|{g_{11}}(t)| < \frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _1})}}$

由式（11）可得

$\begin{split} &|{f_i}(t,{ U}(t))|\leqslant \\ &\quad{k_{i0}} + {t^{\alpha - 2}}({k_{i1}}||{u_1}|| + {k_{i2}}||{u_2}|| + \cdots + {k_{in}}||{u_n}||) \leqslant \\ & \quad{k_{i0}} + M{t^{\alpha - 2}}\max \;\;\{ {k_{i1}},{k_{i2}}, \cdots ,{k_{in}}\} \end{split} $

(12)

对任意 ${ U} \in \varOmega ,{\rm{ }}t \in [0,1]$ ，结合式（12），有

$\begin{split} & {\rm{ |}}{t^{2 - \alpha }}{T_i}({ U}(t))| \leqslant \int_0^t {|{g_{i1}}(t - \tau )} {f_i}(\tau ,{ U}(\tau ))|{\rm d} \tau + \left|{t^{2 - \alpha }} \frac{{{g_{i2}}(1){g_{i3}}(t) - {g_{i3}}(1){g_{i2}}(t)}}{{{\varDelta _i}}}\right|\int_0^{{\eta _i}} {|{g_{i1}}({\eta _i} - \tau ){f_i}(\tau ,{ U}(\tau )} )|{\rm d} \tau + \\ &\quad \left|{t^{2 - \alpha }} \frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}}\right|\int_0^1 {|{g_{i1}}(1 - \tau )} {f_i}(\tau ,{ U}(\tau ))|{\rm d} \tau \leqslant \\ & \quad \left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left( {\int_0^1 {({k_{i0}} + M{\tau ^{\alpha - 2}}\max\; \{ {k_{i1}},{k_{i2}}, \cdots ,{k_{in}}\} )} {\rm d} \tau } \right. + {l_{i1}}\int_0^{{\eta _i}} {({k_{i0}} + M{\tau ^{\alpha - 2}}\max\; \{ {k_{i1}},{k_{i2}}, \cdots ,{k_{in}}\} )} {\rm d} \tau+ \\ &\quad {l_{i2}}\left. {\int_0^1 {({k_{i0}} + M{\tau ^{\alpha - 2}}\max\; \{ {k_{i1}},{k_{i2}}, \cdots ,{k_{in}}\} )} {\rm d} \tau } \right) \leqslant (1 + {\eta _i}{l_{i1}} + {l_{i2}})\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left({k_{i0}} + \frac{M}{{\alpha - 1}}\max\; \{ {k_{i1}},{k_{i2}}, \cdots ,{k_{in}}\} \right) \end{split} $

所以，对任意 ${ U} \in \varOmega ,{\rm{ }}t \in [0,1]$ ，有

$\begin{split}&||{T_i}({ U})|| \leqslant (1 + {\eta _i}{l_{i1}} + {l_{i2}})\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right){\text{·}}\\ &\qquad\left({k_{i0}} + \frac{M}{{\alpha - 1}}\max\;\; \{ {k_{i1}},{k_{i2}}, \cdots ,{k_{in}}\} \right)\end{split}$

(13)

由不等式（13）可得

$\begin{split} ||T({ U}) &|{|_E} = \sum\limits_{i = 1}^n {||{T_i}({ U})||} \leqslant \hspace{90pt}\\ &\sum\limits_{i = 1}^n {(1 + {\eta _i}{l_{i1}} + {l_{i2}})\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right){\text{·}}}\\ &{\left({k_{i0}} + \frac{M}{{\alpha - 1}}\max\;\; \{ {k_{i1}},{k_{i2}}, \cdots ,{k_{in}}\} \right)} \end{split} $

因此, 算子 $T$ 一致有界。

然后再证算子 $T$ 等度连续。只需证明算子 ${T_i}$ 等度连续。

函数 ${g_{i1}}(t),\;{t^{2 - \alpha }}{g_{i2}}(t),\;{t^{2 - \alpha }}{g_{i3}}(t)$ 在[0,1]上连续，所以，函数 ${g_{i1}}(t),\;{t^{2 - \alpha }}{g_{i2}}(t),\;{t^{2 - \alpha }}{g_{i3}}(t)$ 在[0,1]上一致连续，即对任意 $\varepsilon > 0$ ，存在常数 $\delta > 0$ ，当 ${t_1},{t_2} \in $ $ [0,1]$ ， ${t_1} < {t_2}$ ， ${\rm{|}}{t_2} - {t_1}{\rm{| < }}\delta $ 时，都有

$ \quad\begin{split} &|{g_{i1}}({t_2} - \tau ) - {g_{i1}}({t_1} - \tau )| < \frac{\varepsilon }{{5({k_{i0}} + M{{(\alpha - 1)}^{ - 1}}\max\; \{ {k_{i1}}, \cdots ,{k_{in}}\} )}} \\ &\quad |t_2^{\alpha - 1} -t_1^{\alpha - 1}| < \frac{{(\alpha - 1)\varepsilon }}{{5({\Gamma ^{ - 1}}(\alpha ) + {{\rm e}^{(1 - {\lambda _i})}})M\max\; \{ {k_{i1}}, \cdots ,{k_{in}}\} }}\\ & \quad|t_2^{2 - \alpha }{g_{i2}}({t_2}) - t_1^{2 - \alpha }{g_{i2}}({t_1})| <\frac{{|{\varDelta _i}|{{({\eta _i}|{g_{i3}}(1)| + |{g_{i3}}({\eta _i})|)}^{ - 1}}\varepsilon }}{{5({\Gamma ^{ - 1}}(\alpha ) + {{\rm e}^{(1 - {\lambda _i})}})({k_{i0}} + M{{(\alpha - 1)}^{ - 1}}\max\; \{ {k_{i1}}, \cdots ,{k_{in}}\} )}} \\ &\quad|t_2^{2 - \alpha }{g_{i3}}({t_2}) - t_1^{2 - \alpha }{g_{i3}}({t_1})| < \frac{{|{\varDelta _i}|{{({\eta _i}|{g_{i2}}(1)| + |{g_{i2}}({\eta _i})|)}^{ - 1}}\varepsilon }}{{5({\Gamma ^{ - 1}}(\alpha ) + {{\rm e}^{(1 - {\lambda _i})}})({k_{i0}} + M{{(\alpha - 1)}^{ - 1}}\max\; \{ {k_{i1}}, \cdots ,{k_{in}}\} )}} \end{split} $

所以，对上述的 $\varepsilon > 0$ ，存在常数 $\delta \in ( {0,\varepsilon {{(5{k_{i0}}({\Gamma ^{ - 1}}(\alpha ) +}}}$ $ {{{{{\rm e}^{(1 - {\lambda _i})}}))}^{ - 1}}} ) $ , 当 ${t_1},{t_2} \in [0,1]$ , ${t_1} < {t_2}$ , ${\rm{|}}{t_2} - {t_1}{\rm{| < }}\delta $ ，则有

$\begin{split} |t_2^{2 - \alpha }&{T_i}({ U})({t_2}) - t_1^{2 - \alpha }{T_i}({ U})({t_1})|\leqslant \int_0^{{t_1}} {|\left( {{g_{i1}}({t_2} - \tau ) - {g_{i1}}({t_1} - \tau )} \right)} {f_i}(\tau ,{ U}(\tau ))|{\rm d}\tau + \\ &\int_{{t_1}}^{{t_2}} {|{g_{i1}}({t_2} - \tau )} {f_i}(\tau ,{ U}(\tau ))|{\rm d}\tau + |{g_{i2}}(1)( {t_2^{2 - \alpha }{g_{i3}}({t_2}) -}{ t_1^{2 - \alpha }{g_{i3}}({t_1})} ) - {g_{i3}}(1)( {t_2^{2 - \alpha }{g_{i2}}({t_2}) -t_1^{2 - \alpha }{g_{i2}}({t_1})} )| {\text{·}} \\ &\frac{1}{{|{\varDelta _i}|}}\int_0^{{\eta _i}} {|{g_{i1}}({\eta _i} - \tau )} {f_i}(\tau ,{ U}(\tau ))|{\rm d}\tau + |{g_{i3}}({\eta _i})(t_2^{2 - \alpha }{g_{i2}}({t_2}) - t_1^{2 - \alpha }{g_{i2}}({t_1})) - {g_{i2}}({\eta _i})(t_2^{2 - \alpha }{g_{i3}}({t_2}) - t_1^{2 - \alpha }{g_{i3}}({t_1}))| {\text{·}}\\ &\frac{1}{{|{\varDelta _i}|}}\int_0^1 {|{g_{i1}}(1 - \tau )} {f_i}(\tau ,{ U}(\tau ))|{\rm d}\tau\leqslant \int_0^1 {|{g_{i1}}({t_2} - \tau ) - {g_{i1}}({t_1} - \tau )|({k_{i0}} + M{\tau ^{\alpha - 1}}\max\; \{ {k_{i1}}, \cdots ,{k_{in}}\} )} {\mathop{\rm d}\nolimits} \tau+ \\ & \int_{{t_1}}^{{t_2}} {|{g_{i1}}({t_2} - \tau )|({k_{i0}} + M{\tau ^{\alpha - 1}}\max\; \{ {k_{i1}}, \cdots ,{k_{in}}\} )} {\mathop{\rm d}\nolimits} \tau +\\ &\left( {|{g_{i2}}(1)(t_2^{2 - \alpha }{g_{i3}}({t_2}) - t_1^{2 - \alpha }{g_{i3}}({t_1}))| + |{g_{i3}}(1)(t_2^{2 - \alpha }{g_{i2}}({t_2}) - t_1^{2 - \alpha }{g_{i2}}({t_1}))|} \right) {\text{·}}\\ &\frac{1}{{|{\varDelta _i}|}}\int_0^{{\eta _i}} {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)({k_{i0}} + M{\tau ^{\alpha - 1}}\max\; \{ {k_{i1}}, \cdots ,{k_{in}}\} )} {\mathop{\rm d}\nolimits} \tau+ \\ &\left( {|{g_{i3}}({\eta _i})(t_2^{2 - \alpha }{g_{i2}}({t_2}) - t_1^{2 - \alpha }{g_{i2}}({t_1}))| + |{g_{i2}}({\eta _i})(t_2^{2 - \alpha }{g_{i3}}({t_2}) - t_1^{2 - \alpha }{g_{i3}}({t_1}))|} \right) {\text{·}}\\ &\frac{1}{{|{\varDelta _i}|}}\int_0^1 {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)({k_{i0}} + M{\tau ^{\alpha - 1}}\max\; \{ {k_{i1}}, \cdots ,{k_{in}}\} )} {\mathop{\rm d}\nolimits} \tau \leqslant \frac{1}{5}\varepsilon + \frac{1}{5}\varepsilon + \frac{1}{5}\varepsilon + \frac{1}{5}\varepsilon + \frac{1}{5}\varepsilon = \varepsilon \end{split}$

因此，算子 $T$ 等度连续。由Arzela-Ascoli定理可知，算子 $T$ 相对列紧。又因为算子 $T$ 是连续算子，所以，算子 $T$ 是全连续的。

b. 记 $V(T) \!\!=\!\! \{ { U} \!\in\! E:{ U} \!=\! \mu T{ U},$ $0 \!<\! \mu \!<\! 1\} $ ，令 ${ U} \!\in \!\!$ $ V(T)$ ，则 ${ U} = \mu T{ U}$ 。对任意 $t \in [0,1]$ ，有 $ {u_i}(t) =$ $ \mu {T_i}({ U}(t))$ 。

由式（8）和式（11）可知，

$\begin{split} &{\rm{|}}{t^{2 - \alpha }}{u_i}{\rm{(}}t{\rm{)| = |}}{t^{2 - \alpha }}\mu {T_i}({ U}(t))|\leqslant \int_0^t {|{g_{i1}}(t - \tau )} {f_i}(\tau ,{ U}(\tau ))|\operatorname{d} \tau + \\ &\quad\left|{t^{2 - \alpha }} \frac{{{g_{i2}}(1){g_{i3}}(t) - {g_{i3}}(1){g_{12}}(t)}}{{{\varDelta _i}}}\right|\int_0^{{\eta _i}} {|{g_{i1}}({\eta _i} - \tau ){f_i}(\tau ,{ U}(\tau )} )|\operatorname{d} \tau + \\ &\quad \left|{t^{2 - \alpha }} \frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}}\right|\int_0^1 |{{g_{i1}}(1 - \tau )} {f_i}(\tau ,{ U}(\tau ))|\operatorname{d} \tau \leqslant\\ &\quad\int_0^1 {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left( {{k_{i0}} + {\tau ^{\alpha - 2}}({k_{i1}}||{u_1}|| + {k_{i2}}||{u_2}|| + \cdots + {k_{in}}||{u_n}||)}\right)\operatorname{d} \tau }+ \\ &\quad {l_{i1}}\int_0^{{\eta _i}} {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left( {{k_{i0}} + {\tau ^{\alpha - 2}}({k_{i1}}||{u_1}|| + {k_{i2}}||{u_2}|| + \cdots + {k_{in}}||{u_n}||)} \right){\rm d} \tau } + \\ &\quad {l_{i2}}\int_0^1 {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left( {{k_{i0}} + {\tau ^{\alpha - 2}}({k_{i1}}||{u_1}|| + {k_{i2}}||{u_2}|| + \cdots + {k_{in}}||{u_n}||)} \right){\rm d} \tau } = \\ &\quad \left (\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left( {{k_{i0}}(1 + {\eta _i}{l_{i1}} + {l_{i2}}) + \frac{{{k_{i1}}}}{{\alpha - 1}}(1 + {\eta _i}{l_{i1}} + {l_{i2}})||{u_1}|| + \cdots } +{\frac{{{k_{in}}}}{{\alpha - 1}}(1 + {\eta _i}{l_{i1}} + {l_{i2}})||{u_n}||} \right)=\\ &\quad {\gamma _i}{\theta _i}\left({k_{i0}} + \frac{{{k_{i1}}}}{{\alpha - 1}}||{u_1}|| + \cdots + \frac{{{k_{in}}}}{{\alpha - 1}}||{u_n}||\right) \end{split} $

$\begin{gathered} \end{gathered} $

因此

$\qquad||{u_i}|| \leqslant {\gamma _i}{\theta _i}\left({k_{i0}} + \frac{{{k_{i1}}}}{{\alpha - 1}}||{u_1}|| + \cdots + \frac{{{k_{in}}}}{{\alpha - 1}}||{u_n}||\right)\!\!\!\!\!\!\!\!\!\!$

(14)

由式（14）可得

$\begin{split} & ||{u_1}|| + ||{u_2}|| + \cdots + ||{u_n}|| \leqslant\\ &\quad({\gamma _1}{\theta _1}{k_{10}} + {\gamma _2}{\theta _2}{k_{20}} + \cdots + {\gamma _n}{\theta _n}{k_{n0}}) + \\ &\quad\frac{1}{{\alpha - 1}}({\gamma _1}{\theta _1}{k_{11}} + {\gamma _2}{\theta _2}{k_{21}} + \cdots + {\gamma _n}{\theta _n}{k_{n1}})||{u_1}|| + \cdots + \\ &\quad\frac{1}{{\alpha - 1}}({\gamma _1}{\theta _1}{k_{1n}} + {\gamma _2}{\theta _2}{k_{2n}} + \cdots + {\gamma _n}{\theta _n}{k_{nn}})||{u_n}|| =\\ &\quad {\varPhi _0} + \max\; \{ {\varPhi _1},{\varPhi _2}, \cdots ,{\varPhi _n}\} (||{u_1}|| + \cdots + ||{u_n}||) \end{split} $

所以，

$||{ U}|{|_E} = ||{u_1}|| + \cdots + ||{u_n}|| \leqslant \frac{{{\varPhi _0}}}{{1 - \max \;\;\{ {\varPhi _1},{\varPhi _2}, \cdots ,{\varPhi _n}\} }}$

即证得集合 $V$ 是有界集。因此，由引理9可知，算子 $T$ 至少存在1个不动点。所以，边值问题（1）至少存在1个解。

证毕。

接下来证明边值问题（1）解的唯一性，采用Banach压缩映射原理。

为了证明的方便，引入一些记号：

${\omega _i} = \frac{1}{{\alpha - 1}}({\gamma _1}{\theta _1}{N_{1i}} + {\gamma _2}{\theta _2}{N_{2i}} + \cdots + {\gamma _n}{\theta _n}{N_{ni}})$

(15)

$ \qquad\qquad\quad{\xi _i} = \frac{{{\gamma _i}{\theta _i}}}{{\alpha - 1}}({N_{i1}} + {N_{i2}} + \cdots + {N_{in}}) $

(16)

定理2 　令 ${f_i}:[0,1] \times {\mathbb{R}^n} \to \mathbb{R}$ 是连续函数，设 ${\varDelta _i} \ne 0$ ，存在实数 ${N_{ij}} \geqslant 0$ ， $i,j = 1,2, \cdots ,n$ , 对任意 $t \in [0,1],{\rm{ }}{u_i}(t),{\nu _i}(t) \in E$ ，使得

$|{f_i}(t,{u_1}, \cdots ,{u_n}) - {f_i}(t,{\nu _1}, \cdots ,{\nu _n})| \leqslant \sum\limits_{j = 1}^n {{N_{ij}}|{u_j} - {v_j}|}$

设 ${\omega _1} + {\omega _2} + \cdots + {\omega _n} < 1$ ，其中， ${\omega _i}$ 是由式（15）定义的，则有：

a. 边值问题（1）在[0, 1]上有唯一解 ${{ U}^*}(t)$ ；

b. 对任意初始点 ${{ U}_0} \in E$ ，令

$\begin{split}&{{ U}_1} = T{{ U}_0},{\rm{ }}{{ U}_2} = T{{ U}_1} = \\ &\quad {T^2}{{ U}_0}, \cdots ,{{ U}_m} = T{{ U}_{m - 1}} = {T^m}{{ U}_0}\end{split}$

(17)

边值问题（1）的解 ${{ U}^*}(t) = \mathop {\lim }\limits_{m \to \infty }\; {{ U}_m}(t)$ , 有误差估计

$||{{ U}^*} - {{ U}_m}|{|_E} \leqslant \frac{{{{({\omega _1} + {\omega _2} + \cdots + {\omega _n})}^m}}}{{1 - ({\omega _1} + {\omega _2} + \cdots + {\omega _n})}}||{{ U}_1} - {{ U}_0}|{|_E}$

证明 　设 $\mathop {\sup }\limits_{t \in (0,1]}\; |{f_i}(t,0,0, \cdots ,0)| = {\rho _i}$ ，使得

$r \geqslant \max\; \left\{ \frac{{n{\gamma _1}{\theta _1}{\rho _1}}}{{1 - n{\xi _1}}},\frac{{n{\gamma _2}{\theta _2}{\rho _2}}}{{1 - n{\xi _2}}}, \cdots ,\frac{{n{\gamma _n}{\theta _n}{\rho _n}}}{{1 - n{\xi _n}}}\right\}$

先证 $T{B_r} \subset {B_r}$ ，其中， ${B_r} = \{ U \in E:||U|{|_E} \leqslant r\} $ 。对任意 ${ U}(t) \in {B_r} $ ，则有

$\begin{split} &|{t^{2 - \alpha }}{T_i}({u_1}(t), \cdots ,{u_n}(t))| \leqslant \int_0^t {|{g_{i1}}(t - \tau )|} {\rm{ }}\left( {|{f_i}(\tau ,{u_1}(\tau ), \cdots ,{u_n}(\tau )) - {f_i}(\tau ,0, \cdots ,0)| + |{f_i}(\tau ,0, \cdots ,0)|} \right){\rm d} \tau + \\ &\quad \left|{t^{2 - \alpha }} \frac{{{g_{i2}}(1){g_{i3}}(t) - {g_{i3}}(1){g_{i2}}(t)}}{{{\varDelta _i}}}\right|\int_0^{{\eta _i}} {|{g_{i1}}({\eta _i} - \tau )\left( {{f_i}(\tau ,{u_1}(\tau ), \cdots ,{u_n}(\tau )) - {f_i}(\tau ,0, \cdots ,0)} \right)|} {\rm d} \tau + \\ & \quad \left|{t^{2 - \alpha }} \frac{{{g_{i2}}(1){g_{i3}}(t) - {g_{i3}}(1){g_{i2}}(t)}}{{{\varDelta _i}}}\right|\int_0^{{\eta _i}} {|{g_{i1}}({\eta _i} - \tau ){f_i}(\tau ,0, \cdots ,0)|} {\rm d} \tau + \\ & \quad \left|{t^{2 - \alpha }}\frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}}\right| \int_0^1 {|{g_{i1}}(1 - \tau )} \left( {{f_i}(\tau ,{u_1}(\tau ), \cdots ,{u_n}(\tau )) - {f_i}(\tau ,0, \cdots ,0)} \right)|{\rm d} \tau + \\ &\quad \left|{t^{2 - \alpha }} \frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}}\right|\int_0^1 {|{g_{i1}}(1 - \tau ){f_i}(\tau ,0, \cdots ,0)|} {\rm d} \tau \leqslant {\rho _i}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)(1 + {\eta _i}{l_{i1}} + {l_{i2}}) +\\ &\quad \int_0^t {\left(\frac{1}{{\Gamma (\alpha )}} \!\!+\! {{\rm e}^{(1 - {\lambda _i})}}\right)\left(\sum\limits_{j \!=\! 1}^n {{N_{ij}}\!|\!{u_j}(\tau )\!|\!} \right)} {\rm d} \tau \!+\! {l_{i1}}\int_0^{{\eta _i}} {\left(\frac{1}{{\Gamma (\alpha )}} \!+\! {{\rm e}^{(1 \!-\! {\lambda _i})}}\right)\left(\sum\limits_{j = 1}^n {{N_{ij}}|{u_j}(\tau )|} \right)} {\rm d} \tau \!+\! {l_{i2}}\int_0^1 {\left(\frac{1}{{\Gamma (\alpha )}} \!+\! {{\rm e}^{(1 \!-\! {\lambda _i})}}\right)\left(\sum\limits_{j = 1}^n {{N_{ij}}|{u_j}(\tau )|} \right)} {\rm d} \tau \leqslant \\ & \quad {\rho _i}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)(1 + {\eta _i}{l_{i1}} + {l_{i2}}) + \int_0^1 {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)} {\tau ^{\alpha - 2}}\left(\sum\limits_{j = 1}^n {{N_{ij}}{\tau ^{2 - \alpha }}|{u_j}(\tau )|} \right){\rm d} \tau + \\ & \quad {l_{i1}}\int_0^{{\eta _i}} {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right){\tau ^{\alpha - 2}}\left(\sum\limits_{j = 1}^n {{N_{ij}}{\tau ^{2 - \alpha }}|{u_j}(\tau )|} \right)} {\rm d} \tau + {l_{i2}}\int_0^1 {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right){\tau ^{\alpha - 2}}\left(\sum\limits_{j = 1}^n {{N_{ij}}{\tau ^{2 - \alpha }}|{u_j}(\tau )|} \right)} {\rm d} \tau\leqslant \\ & \quad {\rho _i}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)(1 + {\eta _i}{l_{i1}} + {l_{i2}}) + \frac{1}{{\alpha - 1}}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left(\sum\limits_{j = 1}^n {{N_{ij}}||{u_j}||} \right) + \\ & \quad \frac{{{\eta _i}{l_{i1}}}}{{\alpha - 1}}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left(\sum\limits_{j = 1}^n {{N_{ij}}||{u_j}||} \right) + \frac{{{l_{i2}}}}{{\alpha - 1}}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)\left(\sum\limits_{j = 1}^n {{N_{ij}}||{u_j}||} \right) \leqslant \\ &\quad \left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)(1 + {\eta _i}{l_{i1}} + {l_{i2}})\left(\frac{1}{{\alpha - 1}}({N_{i1}} + {N_{i2}} + \cdots + {N_{in}})r + {\rho _i}\right)= {\gamma _i}{\theta _i}\left( {\frac{1}{{\alpha - 1}}({N_{i1}} + {N_{i2}} + \cdots + {N_{in}})r + {\rho _i}} \right) \end{split} $

因此,

$||{T_i}({ U})|| \leqslant {\gamma _i}{\theta _i}\left( {\frac{1}{{\alpha - 1}}({N_{i1}} + {N_{i2}} + \cdots + {N_{in}})r + {\rho _i}} \right) \leqslant \frac{r}{n}$

即

$ ||T({ U})|{|_E}=\left|\sum\limits_{i = 1}^n {{T_i}({ U})}\right| \leqslant \sum\limits_{i = 1}^n {{\gamma _i}{\theta _i}\left( {\frac{1}{{\alpha - 1}}({N_{i1}} + {N_{i2}} + \cdots + {N_{in}})r + {\rho _i}} \right)}\leqslant \frac{r}{n} + \cdots + \frac{r}{n} = r $

所以， $T{B_r} \subset {B_r} $ 。

接下来证明算子 $T$ 是压缩的。对任意 $t \in [0,1], $ ${{ U}}(t) = {({u_1}(t), \cdots ,{u_n}(t))^{\rm T}} \in E,{ V}(t) = {({\nu _1}(t), \cdots ,{\nu _n}(t))^{\rm T}} \!\in\! E$ ，则有

$\begin{split} &|{t^{2 - \alpha }}{T_i}({ U})(t) - {t^{2 - \alpha }}{T_i}({ V})(t)|\leqslant \int_0^t {|{g_{i1}}(t - \tau )} \left( {{f_i}(\tau ,{ U}(\tau )) - {f_i}(\tau ,{ V}(\tau ))} \right)|{\rm d}\tau+ \\ &\quad \left|{t^{2 - \alpha }} \frac{{{g_{i2}}(1){g_{i3}}(t) - {g_{i3}}(1){g_{i2}}(t)}}{{{\varDelta _i}}}\right|\int_0^{{\eta _i}} {|{g_{i1}}({\eta _i} - \tau )} \left( {{f_i}(\tau ,{ U}(\tau )) - {f_i}(\tau ,{ V}(\tau ))} \right)|{\rm d}\tau +\\ &\quad \left|{t^{2 - \alpha }} \frac{{{g_{i3}}({\eta _i}){g_{i2}}(t) - {g_{i2}}({\eta _i}){g_{i3}}(t)}}{{{\varDelta _i}}}\right|\int_0^1 {|{g_{i1}}(1 - \tau )} \left( {{f_i}(\tau ,{ U}(\tau )) - {f_i}(\tau ,{ V}(\tau ))} \right)|{\rm d}\tau\leqslant \\ &\quad \int_0^1 {\left(\frac{1}{{\Gamma (\alpha )}} \!+\! {{\rm e}^{(1 - {\lambda _i})}}\right){\tau ^{\alpha - 2}}}\left(\sum\limits_{j = 1}^n {{N_{ij}}{\tau ^{2 - \alpha }}|{u_j} \!-\! {v_j}|}\right){\rm d}\tau+ {l_{i1}}\int_0^{{\eta _i}} {\left(\frac{1}{{\Gamma (\alpha )}} \!+\! {{\rm e}^{(1 - {\lambda _i})}}\right)}{\tau ^{\alpha - 2}}\left(\sum\limits_{j = 1}^n {{N_{ij}}{\tau ^{2 - \alpha }}|{u_j} \!-\! {v_j}|} \right){\rm d}\tau +\\ &\quad {l_{i2}}\int_0^1 {\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)}{\tau ^{\alpha - 2}}\left(\sum\limits_{j = 1}^n {{N_{ij}}{\tau ^{2 - \alpha }}|{u_j} - {v_j}|} \right){\rm d}\tau\leqslant \\ &\quad\frac{1}{{\alpha - 1}}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)({N_{i1}}||{u_1} - {\nu _1}|| + {N_{i2}}||{u_2} - {\nu _2}|| + \cdots + {N_{in}}||{u_n} - {\nu _n}||) + \\ &\quad \frac{{{\eta _i}{l_{i1}}}}{{\alpha - 1}}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)({N_{i1}}||{u_1} - {\nu _1}|| + {N_{i2}}||{u_2} - {\nu _2}|| + \cdots + {N_{in}}||{u_n} - {\nu _n}||) + \\ &\quad \frac{{{l_{i2}}}}{{\alpha - 1}}\left(\frac{1}{{\Gamma (\alpha )}} + {{\rm e}^{(1 - {\lambda _i})}}\right)({N_{i1}}||{u_1} - {\nu _1}|| + {N_{i2}}||{u_2} - {\nu _2}|| + \cdots + {N_{in}}||{u_n} - {\nu _n}||) = \\ &\quad \frac{{{\gamma _i}{\theta _i}}}{{\alpha - 1}}({N_{i1}}||{u_1} - {\nu _1}|| + {N_{i2}}||{u_2} - {\nu _2}|| + \cdots + {N_{in}}||{u_n} - {\nu _n}||) \end{split} $

因此，对任意 $t \in [0,1],{{ U}}(t) = {({u_1}(t), \cdots ,{u_n}(t))^{\rm T}},$ ${ V}(t) = {({\nu _1}(t), \cdots ,{\nu _n}(t))^{\rm T}} \in E$ ，则有

$ |{t^{2 - \alpha }}{T_i}({ U})(t) - {t^{2 - \alpha }}{T_i}({ V})(t)| \leqslant \frac{{{\gamma _i}{\theta _i}}}{{\alpha - 1}}({N_{i1}}||{u_1} - {\nu _1}|| + {N_{i2}}||{u_2} - {\nu _2}|| + \cdots + {N_{in}}||{u_n} - {\nu _n}||) $

(18)

由式（18）可知，

$ \begin{split}&{\rm{||}}T({ U}) - T({ V})|{|_E} \leqslant \frac{1}{{\alpha - 1}}({\gamma _1}{\theta _1}{N_{11}} + {\gamma _2}{\theta _2}{N_{21}} + \cdots + {\gamma _n}{\theta _n}{N_{n1}})||{u_1} - {u_1}|| + \cdots + \hspace{100pt}\\ &\qquad\frac{1}{{\alpha - 1}}({\gamma _1}{\theta _1}{N_{1n}} + {\gamma _2}{\theta _2}{N_{2n}} + \cdots + {\gamma _n}{\theta _n}{N_{nn}})||{u_n} - {\nu _n}|| \leqslant ({\omega _1} + \cdots + {\omega _n})(||{u_1} - {\nu _1}|| + \cdots + ||{u_n} - {\nu _n}||)\end{split} $

(19)

因为， ${\omega _1} \!+\! {\omega _2} \!+\! \cdots \!+\! {\omega _n} \!<\! 1$ ，所以，算子 $T$ 是压缩算子。由Banach压缩映射定理可知，算子 $T$ 有唯一的不动点，所以，边值问题（1）在 $t \in [0,1]$ 上有唯一解 ${{ U}^*}(t)$ 。

对任意的 $k > m$ ，有

$\begin{split} &{\rm{ ||}}{{ U}_m} - {{ U}_k}|{|_E} \leqslant ||{{ U}_m} - {{ U}_{m + 1}}|{|_E} + ||{{ U}_{m + 1}} - {{ U}_{m + 2}}|{|_E} + \cdots + ||{{ U}_{k - 1}} - {{ U}_k}|{|_E} = \\ &\quad ||T{{ U}_{m - 1}} - T{{ U}_m}|{|_E} + ||T{{ U}_m} - T{{ U}_{m + 1}}|{|_E} + \cdots + ||T{{ U}_{k - 2}} - T{{ U}_{k - 1}}|{|_E} \leqslant \\ &\quad \left( {{{({\omega _1} + {\omega _2} + \cdots + {\omega _n})}^m} + {{({\omega _1} + {\omega _2} + \cdots + {\omega _n})}^{m + 1}} + \cdots + {{({\omega _1} + {\omega _2} + \cdots + {\omega _n})}^{k - 1}}} \right)||{{ U}_1} - {{ U}_0}|{|_E} = \\ &\quad {({\omega _1} + {\omega _2} + \cdots + {\omega _n})^m}\left( {1 + ({\omega _1} + {\omega _2} + \cdots + {\omega _n}) + \cdots + {{({\omega _1} + {\omega _2} + \cdots + {\omega _n})}^{k - m - 1}}} \right)||{{ U}_1} - {{ U}_0}|{|_E} = \\ & \quad {({\omega _1} + {\omega _2} + \cdots + {\omega _n})^m}\frac{{1 - {{({\omega _1} + {\omega _2} + \cdots + {\omega _n})}^{k - m}}}}{{1 - ({\omega _1} + {\omega _2} + \cdots + {\omega _n})}}||{{ U}_1} - {{ U}_0}|{|_E} \leqslant \frac{{{{({\omega _1} + {\omega _2} + \cdots + {\omega _n})}^m}}}{{1 - ({\omega _1} + {\omega _2} + \cdots + {\omega _n})}}||{{ U}_1} - {{ U}_0}|{|_E} \end{split} $

又因为， ${\omega _1} + {\omega _2} + \cdots + {\omega _n} < 1$ ，得到迭代序列 $\{ {{ U}_m}\} $ 收敛，所以，存在极限 $\mathop { U}\limits^ - (t) = \mathop {\lim }\limits_{m \to \infty }\; {{ U}_m}(t)$ 。

由三角不等式和式（19）可得

$\begin{split} &||\mathop { U}\limits^ - - T\mathop { U}\limits^ - |{|_E} \leqslant ||\mathop { U}\limits^ - - {{ U}_m}|{|_E} + {\rm{||}}{{ U}_m} - T\mathop { U}\limits^ - |{|_E} \leqslant \\ & \quad ||\mathop { U}\limits^ - - {{ U}_m}|{|_E} + ({\omega _1} + {\omega _2} + \cdots + {\omega _n})||{{ U}_{m - 1}} - \mathop { U}\limits^ - |{|_E} \end{split} $

因为， $\mathop { U}\limits^ - (t) = \mathop {\lim }\limits_{m \to \infty } \;{{ U}_m}$ ，所以，当 $m \to \infty $ 时，

$||\mathop { U}\limits^ - - {{ U}_m}|{|_E} + ({\omega _1} + {\omega _2} + \cdots + {\omega _n})||{{ U}_{m - 1}} - \mathop { U}\limits^ - |{|_E} \to 0$

得到 $||\mathop { U}\limits^ - - T\mathop { U}\limits^ - |{|_E}{\rm{ = }}0$ ，即 $\mathop { U}\limits^ - {\rm{ = }}T\mathop { U}\limits^ - $ 。由定理2可知，边值问题（1）存在唯一解 ${{ U}^*}$ ，所以，

${{ U}^*}{\rm{ = }}\mathop { U}\limits^ - $

迭代序列式（17）对任意的 ${{ U}_0} \in E$ 都收敛到T的唯一不动点 ${{ U}^*}$ ，且有误差估计

$||{{ U}^*} - {{ U}_m}|{|_E} \leqslant \frac{{{{({\omega _1} + {\omega _2} + \cdots + {\omega _n})}^m}}}{{1 - ({\omega _1} + {\omega _2} + \cdots + {\omega _n})}}||{{ U}_1} - {{ U}_0}|{|_E}$

证毕。

4 例　题

为了说明所得结论具有较好的适用性，考虑几个具体的问题。

例1 　考虑分数阶微分方程组边值问题

$\left\{ \begin{split} &{{\rm D}^\alpha }{u_1}(t) + 0.99{{\rm D}^{\alpha - 1}}{u_1}(t) = {f_1}(t,{u_1}(t),{v_1}(t)),\\ &\quad 0 < t < 1 \\ & {{\rm D}^\alpha }{u_2}(t) + 0.98{{\rm D}^{\alpha - 1}}{u_2}(t) = {f_2}(t,{u_1}(t),{v_1}(t)),\\ &\quad 0 < t < 1 \\ & {u_1}({\eta _1}) = {u_1}(1) = 0\\ & {u_2}({\eta _2}) = {u_2}(1) = 0 \end{split} \right.$

(20)

其中， $\alpha {\rm{ = }}\dfrac{3}{2}$ , ${\lambda _1}{\rm{ = }}0.99,{\rm{ }}{\lambda _2} = 0.98$ 是（0, 1）上的2个正实数，取 ${\eta _1}{\rm{ = }}\dfrac{1}{2},{\rm{ }}{\eta _2}{\rm{ = }}\dfrac{1}{4}$ 。

通过计算。得到

$\left\{\begin{align} & {\varDelta _1} \approx - 0.321\;394 \ne 0,{\rm{ }}\;\;{\varDelta _2} \approx - 0.235\;721 \ne 0\\ &{\gamma _1} \approx 2.138\;429,{\gamma _2} \approx 2.148\;581 \\ &{l_{11}} \approx 2.133\;16,{\rm{ }}{l_{12}} \approx 0.572\;013 \\ &{l_{21}} \approx 2.026\;437,{\rm{ }}{l_{22}} \approx 2.276\;976 \\ &{\theta _1} \approx 4.093\;02,{\rm{ }}{\theta _2} \approx 3.419\;979 \end{align} \right.$

(4.2)

令

$\begin{split} & {f_1}(t,{u_1},{u_2}) = \frac{{{t^2}}}{{900}}\left(1 + \frac{1}{{789}}{u_1} + \frac{1}{{596}}\sin\; {u_2}\right),\\ &\qquad\quad t \in [0,1],{\rm{ }}{u_1},{u_2} \in \mathbb{R} \\ & {f_2}(t,{u_1},{u_2}) = \frac{t}{{999}}\left(2 + \frac{1}{{900}}\sin\; {u_1} + \frac{1}{{127}}{\cos ^2}\;{u_2}\right),\\ &\qquad\quad t \in [0,1],{\rm{ }}{u_1},{u_2} \in \mathbb{R} \end{split} $

即

$\begin{split} & |{f_1}(t,{u_1},{u_2})| = \left|\frac{{{t^2}}}{{900}}\left(1 + \frac{1}{{789}}{u_1} + \frac{1}{{596}}\sin \;{u_2}\right)\right| \leqslant \\ &\qquad\quad 1 + \frac{1}{{789}}|{u_1}| + \frac{1}{{1\;200}}|{u_2}| \\ &|{f_2}(t,{u_1},{u_2})| = \left|\frac{t}{{999}}\left(2 + \frac{1}{{900}}\sin \;{u_1} + \frac{1}{{127}}{\cos ^2}\;{u_2}\right)\right| \leqslant\\ &\quad\qquad2 + \frac{1}{{900}}|{u_1}| + \frac{1}{{137}}|{u_2}| \end{split} $

并且, ${k_{11}} = \dfrac{1}{{789}},\;\;{k_{12}} = \dfrac{1}{{1\;200}},\;\;{k_{21}} = \dfrac{1}{{900}},\;\;{k_{22}} = \dfrac{1}{{137}} $ 。

所以，

$\begin{gathered} \frac{1}{{\alpha - 1}}({\gamma _1}{\theta _1}{k_{11}} + {\gamma _1}{\theta _1}{k_{21}}) \approx 0.041\;636\;9 < 1 \\ \frac{1}{{\alpha - 1}}({\gamma _2}{\theta _2}{k_{12}} + {\gamma _2}{\theta _2}{k_{22}}) \approx 0.140\;376 < 1 \end{gathered} $

因此，所有的条件都满足定理1，所以，由定理1可知，微分方程组（20）至少存在1个解。

例2 　在例1的基础上，只改变函数 ${f_1},{\rm{ }}{f_2}$ ，其他的条件保持不变， ${\varDelta _i},{\rm{ }}{l_{1i}},{\rm{ }}{l_{2i}},$ ${\gamma _i},{\rm{ }}{\theta _i}(i = 1,2)$ 是式（20）中的，其中， $\alpha {\rm{ = }}\dfrac{3}{2}$ 。令

$\begin{split} &{f_1}(t,{u_1},{u_2}) = 1 + \frac{{{t^2}}}{{790}}{u_1} + \frac{1}{{960}}\arctan\; {u_2},\\ &\quad\qquad t \in [0,1],{\rm{ }}{u_1},{u_2} \in \mathbb{R}\\ &{f_2}(t,{u_1},{u_2}) = 2 + \frac{t}{{460}}\sin\; {u_1} + \frac{5}{{239}}\cos\; {u_2},{\rm{ }}t \in [0,1],\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\\ &\quad\qquad {u_1},{u_2} \in \mathbb{R} \end{split} $

(22)

即

$\begin{gathered} |{f_1}(t,{u_1},{u_2}) - {f_1}(t,{v_1},{v_2})| \leqslant \frac{1}{{790}}{\rm{|}}{u_1} - {v_1}| + \frac{1}{{960}}{\rm{|}}{u_2} - {\nu _2}|\\ |{f_2}(t,{u_1},{u_2}) - {f_2}(t,{v_1},{v_2})| \leqslant \frac{1}{{460}}{\rm{|}}{u_1} - {v_1}| + \frac{5}{{239}}{\rm{|}}{u_2} - {\nu _2}| \end{gathered} $

并且， $ {N_{11}} = \dfrac{1}{{790}},{N_{12}} = \dfrac{1}{{960}},{N_{21}} = \dfrac{1}{{460}},{N_{22}} = \dfrac{5}{{239}}$ 。

所以，

$\begin{split}&\frac{1}{{\alpha - 1}}[({\gamma _1}{\theta _1}{N_{11}} + {\gamma _2}{\theta _2}{N_{21}}) +\\ &\qquad ({\gamma _1}{\theta _1}{N_{12}} + {\gamma _2}{\theta _2}{N_{22}})] \approx 0.379\;794\end{split}$

所有的条件满足定理2，因此，微分方程组（20）有唯一解( ${f_1},{f_2}$ 是式（21）中给定的)。

对任意 ${{ U}_0} \!\!=\!\! {(0,{\rm{ }}0)^{\rm T}}$ ，则 ${{ U}_1}\! \!=\!\! {(0.041\;565,} $ ${\;0.043\;789)^{\rm T}}$ ，且有误差估计

$||{{ U}^*} - {{ U}_{200}}|| \leqslant 1.272\;61 \times {10^{ - 84}}||{{ U}_1} - {{ U}_0}||$