分数阶微积分理论的研究最早源于Euler的一些猜想, 随后在17世纪数学家Leibniz和Hospital的信件中说到: 分数阶导数相当于在两个整数阶导数之间引入某种插入法. 此后, 越来越多的学者开始对分数阶理论进行研究. 1812年, Laplace定义了分数阶导数, 1822年, Fourier给出了函数的任意$ \mu $阶导数的定义, 1832–1837年, Riemann给出了分数阶积分的定义, 1847年, Liouville给出了分数阶微分的定义, 1967年, Caputo提出了以自己名字命名的分数阶导的定义, 1974年, Oldhun出版了第一本分数阶微积分著作, 后来越来越多的分数阶微积分著作被出版, 使得分数阶微积分的发展更加完善. 起初分数阶微分方程的研究仅仅局限在数学领域, 直到1982年, Mandel-brot首次提出分数阶理论可以应用在其他学科, 如: 分数阶很适合描述各种原材料和混合物的性质、记忆和遗传性质等各种原料的形成、混沌与湍流及电磁学等诸多方面. 因此, 分数阶研究不仅有重要的理论价值, 还有广泛的实际应用价值^[1-9].

近几十年来, 越来越多的学者注重将分数阶理论与(神经脉冲的传播, 桥梁铰链)生物模型和工程模型等结合起来^[11-21]. 特别地, 2019年Mehandiratta等^[15]考虑的分数阶导数不仅依赖于未知函数, 还取决于它的分数阶导数和由$ k $条边$ k+1 $个结点组成的星图$ G=V\cup E, V=\{v_{0}, v_{1}, \cdots , v_{k}\}, E=\{e_{i}=\overrightarrow{v_{i}v_{0}}, i=1, 2, \cdots , k\}, $其中$ v_{0} $是联结点, $ e_{i}=\overrightarrow{v_{i}v_{0}} $, $ l_{i} $是连接点$ v_{i} $和$ v_{0} $的边的长度, 即$ l_{i}=|\overrightarrow{v_{i}v_{0}}|. $在每条边$ e_{i} $上作者考虑了$ x\in(0, l_{i}) $上以$ v_{i} $为原点的局部坐标系上的非线性分数阶微分方程边值问题

$ \left\{\begin{array}{ll} {}_{C}D_{0, x}^{\alpha}u_{i}(x)=f_{i}(x, u_{i}(x), \ {}_{C}D_{0, x}^{\beta}u_{i}(x)), 0<x<l_{i}, &i=1, 2, \cdots , k, \\ u_{i}(0)=u_{i}(1)=0, & i=1, 2, \cdots , k, \\ u_{i}(l_{i})=u_{j}(l_{j}), & i, j=1, 2, \cdots , k, i\neq j, \\ { } \sum\limits_{i=1}^ku_{i}'(l_{i})=0, &i=1, 2, \cdots , k \end{array} \right. $

解的存在唯一性, 其中$ 1<\alpha\leq2, 0<\beta\leq\alpha-1, _{C}D_{0, x}^{\alpha}, _{C}D_{0, x}^{\beta} $是Caputo分数阶导数, $ f_{i} $是$ [0, 1]\times {{\Bbb R}} \times{{\Bbb R}} $上的连续函数, 其中$ i=1, 2, \cdots , k $.

受上述文献的启发, 本文运用Banach压缩映射原理和Schaefer不动点定理研究星图上的非线性分数阶微分方程边值问题

$ \left\{\begin{array}{ll} {}_{C}D_{0, x}^{\alpha}u_{i}(x)=f_{i}(x, u_{i}(x), \ _{C}D_{0, x}^{\beta}u_{i}(x)), 0<x<l_{i}, & i=1, 2, \cdots , k, \\ u_{i}'(0)=u_{i}(1)=0, & i=1, 2, \cdots , k, \\ u_{i}''(l_{i})=u_{j}''(l_{j}), & i, j=1, 2, \cdots , k, i\neq j, \\ { } \sum\limits_{i=1}^ku_{i}''(l_{i})=0, &i=1, 2, \cdots , k \end{array} \right.\;\;\;\;\;\;\;\;\;\;\;\;\begin{array}{r}(1.1)\\ (1.2)\\ (1.3)\\ (1.4)\\ \end{array} $

解的存在唯一性, 其中$ 2<\alpha\leq3, 0<\beta<1, $$ _{C}D_{0, x}^{\alpha}, $$ _{C}D_{0, x}^{\beta} $是Caputo分数阶导数, $ f_{i}, $$ i=1, 2, \cdots , k $是$ [0, 1]\times {{\Bbb R}} \times{{\Bbb R}} $上关于三个变元连续可微的函数.

值得注意的是, 本文考虑的分数阶导数不仅依赖于未知函数, 还取决于它的分数阶导数和由$ k $条边$ k+1 $个结点组成的图 1

$ G=V\cup E, V=\{v_{0}, v_{1}, \cdots , v_{k}\}, E=\{e_{i}=\overrightarrow{v_{i}v_{0}}, i=1, 2, \cdots , k\}, $

其中$ v_{0} $是联结点, $ e_{i}=\overrightarrow{v_{i}v_{0}} $, $ l_{i} $是连接点$ v_{i} $和点$ v_{0} $的边的长度, 即$ l_{i}=|\overrightarrow{v_{i}v_{0}}|. $在每条边$ e_{i} $上, 我们考虑了$ x\in(0, l_{i}) $上以$ v_{i} $为原点的局部坐标系.

记空间$ X=\{u:u\in C^{2}[0, 1], \ _{C}D_{0, t}^{\beta}u\in C[0, 1]\}, $其范数为

$ ||u||_{X}=||u||_{\infty}+||u'||_{\infty}+||u''||_{\infty}+||_{C}D_{0, t}^{\beta}u||, $

其中$ ||u||_{\infty}=\sup\limits_{t\in[0, 1]}|u(t)|. $因此$ (X, ||\cdot||_{X}) $是Banach空间, 且乘积空间$ (X^{k}=X\times X\times X\times\cdots \times X) $是Banach空间, 其范数为$ ||(u_{1}, u_{2}, \cdots , u_{k})||_{X^{k}}=\sum\limits_{i=1}^k||u_{i}||_{X}, (u_{1}, u_{2}, \cdots , u_{k})\in X^{k} $.

本文假设

(H1) $ f_{i}:[0, 1]\times {{\Bbb R}} \times{{\Bbb R}} \rightarrow {{\Bbb R}} $是关于三个变元连续可微的函数, 其中$ i=1, 2, \cdots , k. $

(H2) 在$ [0, 1] $上存在非负连续函数$ a_{i}(t), i=1, 2, \cdots , k, $使得对任意的$ (x, y), (x_{1}, y_{1}) \in {{\Bbb R}} ^{2}, $$ t\in[0, 1] $有

$ |f_{i}(t, x, y)-f_{i}(t, x_{1}, y_{1})|\leq \omega_{i}(|x-x_{1}|+|y-y_{1}|), $

其中$ \omega_{i}=\sup\limits_{t\in[0, 1]}a_{i}(t), i=1, 2, \cdots , k $.

(H3) 存在$ L_{i}>0, $使得$ |f_{i}(t, x, y)|\leq L_{i}, t\in[0, 1], x, y \in R, i=1, 2, \cdots , k. $

记

(2.1) $ \begin{eqnarray} M_{i}&=&\Big[\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)(\frac{11}{\Gamma(\alpha-1)}+\frac{2}{\Gamma(\alpha+1)}+\frac{1}{\Gamma(\alpha)}+\frac{2}{\Gamma(\alpha-\beta)} +\frac{4}{\Gamma(\alpha-2)\Gamma(3-\beta)})\Big]{}\\ &&+\frac{10}{\Gamma(\alpha-2)}\sum^k_{j=1}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)+ \sum\limits_{j=1\atop j\neq i}^{k}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\Big(\frac{6}{\Gamma(\alpha-2)}+\frac{4}{\Gamma(\alpha-2)\Gamma(3-\beta)}\Big), {\quad} \end{eqnarray} $

(2.2) $ \begin{eqnarray} P_{i}&=&\Big[l_{i}^{\alpha}\Big(\frac{2}{\Gamma(\alpha)}+\frac{2}{\Gamma(\alpha-1)}+\frac{2}{\Gamma(\alpha-2)}+\frac{1}{\Gamma(\alpha-\beta)}\Big)+\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}l_{j}^{\alpha}{}\\ &&+\sum^k_{j=1}l_{j}^{\alpha}\Big(\frac{10}{\Gamma(\alpha-2)}+\frac{6}{\Gamma(\alpha-2)\Gamma(3-\beta)}\Big)\Big]. \end{eqnarray} $

下面介绍本文的主要工具定理以及一些重要定义和引理.

定义2.1^[15]  函数$ f \in C(a, b) $的$ \alpha>0 $阶分数阶积分定义为

$ D_{a, t}^{-\alpha}f(t)=\frac{1}{\Gamma(\alpha)}\Big(\int_{a}^{t}(t-s)^{\alpha-1}f(s){\rm d}s\Big). $

定义2.2^[15]  函数$ f \in C^{n}(a, b) $的$ \alpha>0 $阶Caputo分数阶导数定义为

$ _{C}D_{a, t}^{\alpha}f(t)=\frac{1}{\Gamma(n-\alpha)}\Big(\int_{a}^{t}(t-s)^{n-\alpha-1}f^{(n)}(s){\rm d}s\Big), $

其中$ n=[\alpha]+1 $.

引理2.3^[10]   (Schaefer不动点定理)  令$ X $是一个Banach空间, $ T: X\rightarrow X $是全连续算子, 若集合$ \{x\in X, x=\mu Tx, \mu\in(0, 1)\} $是有界的, 则$ T $在$ X $中至少有一个不动点.

引理2.4^[10]   (Banach压缩映射原理)  设$ X $是完备的度量空间, $ T: X\rightarrow X $是压缩映射, 则$ T $在$ X $上有唯一的不动点.

引理2.5^[22]  设$ \alpha>0 $, 则分数阶微分方程$ _{C}D_{0, t}^{\alpha}u(t)=0 $有解

$ u(t)=c_{1}+c_{1}t+c_{3}t^{2}+\cdots +c_{n}t^{n-1}, $

$ c_{i}\in {{\Bbb R}} , i=1, 2, \cdots , n, n=[\alpha]+1. $

引理2.6^[22]  设$ \alpha>0 $, 则

$ D_{0, t}^{-\alpha}{{}_{C}D_{0, t}^{\alpha}u(t)}=u(t)+c_{1}+c_{2}t+c_{3}t^{2} +\cdots +c_{n}t^{n-1}, $

$ c_{i}\in {{\Bbb R}} , i=1, 2, \cdots , n, n=[\alpha]+1. $

引理2.7^[15]  令$ u $是$ [0, l] $上的函数, 且在$ [0, l] $上存在$ {}_{C}D_{0, x}^{\alpha}u , \alpha>0 $使得$ x\in[0, l] $, $ t=\frac{x}{l}\in[0, 1] $, 则有

$ _{C}D_{0, x}^{\alpha}u(x)=l^{-\alpha}(_{C}D_{0, t}^{\alpha}u)(t). $

引用引理2.7且做变量变换$ u(x)=u(lt) $, 则边值问题(1.1)–(1.4)等价于

$ \left\{ \begin{array}{lll} {}_{C}D_{0, t}^{\alpha}u_{i}(t)=l_{i}^{\alpha}f_{i}(t, u_{i}(t), l_{i}^{-\beta}\cdot {}_{C}D_{0, t}^{\beta}u_{i}(t)), &t\in(0, 1), \alpha\in(2, 3], \\ u_{i}'(0)=u_{i}(1)=0, & i=1, 2, \cdots , k, \\ u_{i}''(1)=u_{j}''(1), & i, j=1, 2, \cdots , k, i\neq j, \\ { } \sum\limits_{i=1}^{k}l_{i}^{-2}u_{i}''(1)=0, & i=1, 2, \cdots , k. \end{array} \right. \;\;\;\;\;\;\;\;\;\begin{array}{r} (2.3)\\ (2.4)\\ (2.5)\\ (2.6)\\ \end{array} $

引理2.8  令$ h_{i}\in C^{1}[0, 1], $则分数阶微分方程

(2.7) $ \begin{equation} {}_{C}D_{0, t}^{\alpha}u_{i}(t)=h_{i}(t) \end{equation} $

的解与积分方程

(2.8) $ \begin{eqnarray} u_{i}(t)&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h_{i}(s){\rm d}s- \frac{1}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h_{i}(s){\rm d}s\\ &&+\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}} {2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(1-t^{2})h_{j}(s){\rm d}s{} \\ &&-\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}} {2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(1-t^{2})(h_{j}(s)-h_{i}(s)){\rm d}s \end{eqnarray} $

的解是等价的.

证  首先证明微分方程(2.7)的解是积分方程(2.8)的解.

由引理2.6得

$ \begin{eqnarray*} u_{i}(t)&=&D_{0, t}^{-\alpha}h_{i}(t)-C_{i}^{(1)}-C_{i}^{(2)}t-C_{i}^{(3)}t^{2}\\ &=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h_{i}(s){\rm d}s-C_{i}^{(1)}-C_{i}^{(2)}t-C_{i}^{(3)}t^{2}, i=1, 2, \cdots , k, \end{eqnarray*} $

其中$ C_{i}^{(j)}, i=1, 2, \cdots , k, j=1, 2, 3 $都是常数, 则有

$ u_{i}'(t)=\frac{1}{\Gamma(\alpha-1)}\int_{0}^{t}(t-s)^{\alpha-2}h_{i}(s){\rm d}s-C_{i}^{(2)}-2tC_{i}^{(3)}. $

$ u_{i}''(t)=\frac{1}{\Gamma(\alpha-2)}\int_{0}^{t}(t-s)^{\alpha-3}h_{i}(s){\rm d}s-2C_{i}^{(3)}. $

由$ u_{i}'(0)=0, $得$ C_{i}^{(2)}=0. $由(2.6)式知

(2.9) $ \begin{equation} \sum\limits_{i=1}^kl_{i}^{-2}\Big(\frac{1}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha-3}h_{i}(s){\rm d}s-2C_{i}^{(3)}\Big)=0. \end{equation} $

由$ u_{i}''(1)=u_{j}''(1) $得

$ \frac{1}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha-3}h_{i}(s){\rm d}s-2C_{i}^{(3)} =\frac{1}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s-2C_{j}^{(3)}. $

两边同时求和再作差得

$ \begin{eqnarray*} &&\sum\limits_{j=1}^kl_{j}^{-2}\left[\frac{1}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s-2C_{j}^{(3)}\right]\\ &&-\sum\limits_{i=1}^kl_{i}^{-2}\left[\frac{1}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha-3}h_{i}(s){\rm d}s-2C_{i}^{(3)}\right]=0. \end{eqnarray*} $

$ \sum\limits_{j=1\atop j\neq i}^{k}l_{j}^{-2}\left[\frac{1}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s\right] +2\sum\limits_{j=1\atop j\neq i}^{k}l_{j}^{-2}C_{i}^{(3)} =0. $

(2.10) $ \begin{equation} \sum\limits_{j=1\atop j\neq i}^{k}l_{j}^{-2}\left[\frac{1}{\Gamma(\alpha-2)} \int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s+2C_{i}^{(3)}\right]=0. \end{equation} $

由(2.9)和(2.10)式得

$ \begin{eqnarray*} &&\sum\limits_{j=1}^kl_{j}^{-2}\left[\frac{1}{\Gamma(\alpha-2)} \int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s\right]-2l_{i}^{-2}C_{i}^{(3)}\\ &=&\sum\limits_{j=1\atop j\neq i}^{k} sl_{j}^{-2}\left[\frac{1}{\Gamma(\alpha-2)} \int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s+2C_{i}^{(3)}\right], \\ &&\sum\limits_{j=1}^kl_{j}^{-2}\left[\frac{1}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s\right]\\ &&-\sum\limits_{j=1\atop j\neq i}^{k}l_{j}^{-2}\left[\frac{1}{\Gamma(\alpha-2)}\int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s\right] =2\sum\limits_{j=1}^kl_{i}^{-2}C_{i}^{(3)}. \end{eqnarray*} $

故

$ \begin{eqnarray*} C_{i}^{(3)}&=&\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}} {2\sum\limits_{j=1}^{k}l_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s\\ &&-\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}} {2\sum\limits_{j=1}^{k}l_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s, i=1, 2, \cdots , k, \\ C_{i}^{(1)}&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h_{i}(s){\rm d}s-\frac{1}{\Gamma(\alpha-2)} \sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{{j=1}}^{k}l_{j}^{-2}}\Big) \int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s\\ &&+\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2 \sum\limits_{{j=1}}^{k}l_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s, i=1, 2, \cdots , k. \end{eqnarray*} $

因此可得

$ \begin{eqnarray*} u_{i}(t)&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h_{i}(s){\rm d}s-\frac{1}{\Gamma(\alpha)} \int_{0}^{1}(1-s)^{\alpha-1}h_{i}(s){\rm d}s\\ &&+\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}} \Big)\int_{0}^{1}(1-s)^{\alpha-3}(1-t^{2})h_{j}(s){\rm d}s\\ &&-\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}} {2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(1-t^{2})(h_{j}(s)-h_{i}(s)){\rm d}s. \end{eqnarray*} $

因此微分方程(2.7)的解一定是积分方程(2.8)的解.

下面证明积分方程(2.8)的解一定是微分方程(2.7)的解.

记

$ \begin{eqnarray*} u_{i}(t)&= &\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h_{i}(s){\rm d}s-\frac{1}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h_{i}(s){\rm d}s\\ &&+\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(1-t^{2})h_{j}(s){\rm d}s\\ &&-\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(1-t^{2})(h_{j}(s)-h_{i}(s)){\rm d}s. \end{eqnarray*} $

则有

$ \begin{eqnarray*} u'_{i}(t)&= &\frac{1}{\Gamma(\alpha-1)}\int_{0}^{t}(t-s)^{\alpha-2}h_{i}(s){\rm d}s- \frac{2t}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s\\ &&+\frac{2t}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s, \\ u''_{i}(t)&= &\frac{1}{\Gamma(\alpha-2)}\int_{0}^{t}(t-s)^{\alpha-3}h_{i}(s){\rm d}s- \frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s\\ &&+\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s. \end{eqnarray*} $

当$ 2<\alpha<3 $时, 有

$ \begin{eqnarray*} u'''_{i}(t)&=&\frac{\rm d}{{\rm d}t}\Big[\frac{1}{\Gamma(\alpha-2)}\int_{0}^{t}(t-s)^{\alpha-3}h_{i}(s){\rm d}s\\ &&- \frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}h_{j}(s){\rm d}s\\ &&+\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(h_{j}(s)-h_{i}(s)){\rm d}s\Big]\\ &=&\frac{\rm d}{{\rm d}t}\Big[\frac{1}{\Gamma(\alpha-2)}\int_{0}^{t}(t-s)^{\alpha-3}h_{i}(s){\rm d}s\Big]-0+0\\ &=&\frac{\rm d}{{\rm d}t}\Big[\frac{-1}{\Gamma(\alpha-2)}\int_{0}^{t}\frac{1}{(\alpha-2)}h_{i}(s){\rm d}{(t-s)^{\alpha-2}}\Big]\\ &=&\frac{\rm d}{{\rm d}t}\Big[\frac{t^{\alpha-2}h_{i}(0)}{\Gamma(\alpha-2)(\alpha-2)}+\frac{1}{\Gamma(\alpha-1)}\int_{0}^{t}(t-s)^{\alpha-2}h'_{i}(s){\rm d}s\Big]\\ &=&\frac{t^{\alpha-3}h_{i}(0)}{\Gamma(\alpha-2)}+\frac{\rm d}{{\rm d}t}\Big[\frac{1}{\Gamma(\alpha-1)}\int_{0}^{t}(t-s)^{\alpha-2}h'_{i}(s){\rm d}s\Big]\\ &=&\frac{t^{\alpha-3}h_{i}(0)}{\Gamma(\alpha-2)}+\frac{\rm d}{{\rm d}t}D_{0, t}^{1-\alpha}h'_{i}(t) =\frac{t^{\alpha-3}h_{i}(0)}{\Gamma(\alpha-2)}+\frac{\rm d}{{\rm d}t}D_{0, t}^{-1}D_{0, t}^{2-\alpha}h'_{i}(t) \\ & =&\frac{t^{\alpha-3}h_{i}(0)}{\Gamma(\alpha-2)}+D_{0, t}^{2-\alpha}h'_{i}(t). \end{eqnarray*} $

因此

$ \begin{eqnarray*} {}_{C}D_{0, t}^{\alpha}u_{i}(t)&=&\frac{1}{\Gamma(3-\alpha)}\Big(\int_{0}^{t}(t-s)^{3-\alpha-1}u'''_{i}(s){\rm d}s\Big)\\ &=&D_{0, t}^{\alpha-3}u'''_{i}(t) =D_{0, t}^{\alpha-3}\Big(\frac{t^{\alpha-3}h_{i}(0)}{\Gamma(\alpha-2)}+D_{0, t}^{2-\alpha}h'_{i}(t)\Big)\\ &=&D_{0, t}^{\alpha-3}\Big(\frac{t^{\alpha-3}h_{i}(0)}{\Gamma(\alpha-2)}\Big)+D_{0, t}^{\alpha-3}D_{0, t}^{2-\alpha}h'_{i}(t)\\ &=&\frac{1}{\Gamma(3-\alpha)}\int_{0}^{t}(t-s)^{2-\alpha}\frac{s^{\alpha-3}}{\Gamma(\alpha-2)}h_{i}(0){\rm d}s+D_{0, t}^{-1}h'_{i}(t)\\ &=&\frac{1}{\Gamma(3-\alpha)\Gamma(\alpha-2)}\int_{0}^{1}t^{2-\alpha}(1-\tau)^{2-\alpha}t^{\alpha-3}\tau^{\alpha-3}th_{i}(0){\rm d}\tau+h_{i}(t)-h_{i}(0)\\ &=&\frac{1}{\Gamma(3-\alpha)\Gamma(\alpha-2)}\frac{\Gamma(3-\alpha)\Gamma(\alpha-2)}{\Gamma(1)}h_{i}(0)+h_{i}(t)-h_{i}(0)\\ &=&h_{i}(0)+h_{i}(t)-h_{i}(0) =h_{i}(t). \end{eqnarray*} $

当$ \alpha=3 $时, 有

$ \begin{eqnarray*} {}_{C}D_{0, t}^{3}u_{i}(t)&=&u'''_{i}(t)=\frac{\rm d}{{\rm d}t}\Big[\frac{1}{\Gamma(1)}\int_{0}^{t}h_{i}(s){\rm d}s- \frac{2}{\Gamma(1)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}h_{j}(s){\rm d}s\\ &&+\frac{2}{\Gamma(1)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(h_{j}(s)-h_{i}(s)){\rm d}s\Big]\\ &=&\frac{\rm d}{{\rm d}t}\Big(\int_{0}^{t}h_{i}(s){\rm d}s\Big) =h_{i}(t). \end{eqnarray*} $

因此积分方程(2.8)的解一定是微分方程(2.7)的解.

注2.9  从上面的证明知道, 在证明微分方程(2.7)的解是积分方程(2.8)的解时要求$ h_{i}\in C[0, 1] $即可. 在证明积分方程(2.8)的解一定是微分方程(2.7)的解时, 当$ \alpha=3 $时$ h_{i}\in C[0, 1] $即可, 但是当$ 2<\alpha<3 $时要求$ h_{i}\in C^{1}[0, 1] $, 所以在引理2.8中要求$ h_{i}\in C^{1}[0, 1] $. 当把线性项$ h_{i}(t) $换成非线性项$ l_{i}^{\alpha}f_{i}(t, u_{i}(t), l_{i}^{-\beta}\cdot {}_{C}D_{0, t}^{\beta}u_{i}(t)) $是, 对函数$ f_{i} $要求关于三个变元连续可微.

由引理2.8, 定义算子$ T: X^{k}\rightarrow X^{k} $为

$ T(u_{1}, u_{2}, \cdots , u_{k})(t)=(T_{1}(u_{1}, u_{2}, \cdots , u_{k})(t), \cdots , T_{k}(u_{1}, u_{2}, \cdots , u_{k})(t)), $

其中

(2.11) $ \begin{eqnarray} &&T_{i}(u_{1}, u_{2}, \cdots , u_{k})(t){}\\ &=&l_{i}^{\alpha}\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)){\rm d}s\\ &&-l_{i}^{\alpha}\frac{1}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)){\rm d}s {}\\ &&+\frac{1-t^{2}}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s)){\rm d}s{} \\ &&-\frac{1-t^{2}}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s)){\rm d}s{} \\ &&+\frac{1-t^{2}}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)){\rm d}s. \end{eqnarray} $

则有

$ \begin{eqnarray*} &&T'_{i}(u_{1}, u_{2}, \cdots , u_{k})(t)\\ &=&l_{i}^{\alpha}\frac{1}{\Gamma(\alpha-1)}\int_{0}^{t}(t-s)^{\alpha-2}f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)){\rm d}s\nonumber\\ &&-\frac{2t}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s)){\rm d}s{\nonumber} \\ &&+\frac{2t}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s)){\rm d}s{\nonumber} \\ &&-\frac{2t}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)){\rm d}s. \end{eqnarray*} $

进一步有

$ \begin{eqnarray*} &&T''_{i}(u_{1}, u_{2}, \cdots , u_{k})(t)\\ &=&l_{i}^{\alpha}\frac{1}{\Gamma(\alpha-2)}\int_{0}^{t}(t-s)^{\alpha-3}f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)){\rm d}s\nonumber\\ &&-\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s)){\rm d}s{\nonumber} \\ &&+\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s)){\rm d}s{\nonumber} \\ &&-\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)){\rm d}s. \end{eqnarray*} $

定理3.1   假设条件(H2)成立, 如果满足$ (\sum\limits_{i=1}^kM_{i})(\sum\limits_{i=1}^k\omega_{i})<1, $则边值问题(2.3)–(2.6)存在唯一解.

证  令$ u=(u_{1}, u_{2}, \cdots , u_{k}), v=(v_{1}, v_{2}, \cdots , v_{k})\in X^{k}, t\in[0, 1], $则由(2.11)式可得

$ \begin{eqnarray*} &&|T_{i}u(t)-T_{i}v(t)|\\ &\leq&\frac{l_{i}^{\alpha}}{\Gamma(\alpha)}\int_{0}^{t}\left[|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))-f_{i}(s, v_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}v_{i}(s))|\right](t-s)^{\alpha-1}{\rm d}s\\ &&+\frac{l_{i}^{\alpha}}{\Gamma(\alpha)}\int_{0}^{1}\left[|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))-f_{i}(s, v_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}v_{i}(s))|\right](1-s)^{\alpha-1}{\rm d}s\\ &&+2\frac{1-t^{2}}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big[|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))\\ &&-f_{j}(s, v_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}v_{j}(s))|\Big]{\rm d}s+\frac{1-t^{2}}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}\\ &&\times\Big[|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))-f_{i}(s, v_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}v_{i}(s))|\Big]{\rm d}s. \end{eqnarray*} $

因为$ 1-t^{2}<1, (t<1), (\frac{l_{j}^{-2}}{\sum\limits_{j=1}^{k}l_{j}^{-2}})<1, j=1, 2, \cdots , k, $故由(H2)可知

$ \begin{eqnarray*} |T_{i}u(t)-T_{i}v(t)| &\leq&\frac{2l_{i}^{\alpha}}{\Gamma(\alpha)}\omega_{i}||u_{i}-v_{i}||+\frac{2l_{i}^{\alpha-\beta}}{\Gamma(\alpha)}\omega_{i}||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\\ &&+2\sum\limits_{j=1}^k\Big(\frac{l_{j}^{\alpha}}{\Gamma(\alpha-2)}\omega_{j}||u_{j}-v_{j}||+\frac{l_{j}^{\alpha-\beta}}{\Gamma(\alpha-2)}\omega_{j}||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{i}^{\alpha}}{\Gamma(\alpha-2)}\omega_{i}||u_{i}-v_{i}||+\frac{l_{i}^{\alpha-\beta}}{\Gamma(\alpha-2)}\omega_{i}||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big)\\ &\leq&\frac{2}{\Gamma(\alpha)}\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big)\\ &&+2\sum\limits_{j=1}^k\frac{1}{\Gamma(\alpha-2)}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\omega_{j}\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+\sum\limits_{j=1\atop j\neq i}^{k}\frac{1}{\Gamma(\alpha-2)}\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big), \\ |T'_{i}u(t)-T'_{i}v(t)|& \leq&\frac{2l_{i}^{\alpha}}{\Gamma(\alpha-1)}\omega_{i}||u_{i}-v_{i}||+\frac{2l_{i}^{\alpha-\beta}}{\Gamma(\alpha-1)}\omega_{i}||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\\ &&+4\sum\limits_{j=1}^k\Big(\frac{l_{j}^{\alpha}}{\Gamma(\alpha-2)}\omega_{j}||u_{j}-v_{j}||+\frac{l_{j}^{\alpha-\beta}}{\Gamma(\alpha-2)}\omega_{j}||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+2\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{i}^{\alpha}}{\Gamma(\alpha-2)}\omega_{i}||u_{i}-v_{i}||+\frac{l_{i}^{\alpha-\beta}}{\Gamma(\alpha-2)}\omega_{i}||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big)\\ &\leq&\frac{2}{\Gamma(\alpha-1)}\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big)\\ &&+4\sum\limits_{j=1}^k\frac{1}{\Gamma(\alpha-2)}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\omega_{j}\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+2\sum\limits_{j=1\atop j\neq i}^{k}\frac{1}{\Gamma(\alpha-2)}\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big), \\ |T''_{i}u(t)-T''_{i}v(t)| &\leq&\frac{2l_{i}^{\alpha}}{\Gamma(\alpha-2)}\omega_{i}||u_{i}-v_{i}||+\frac{2l_{i}^{\alpha-\beta}}{\Gamma(\alpha-2)}\omega_{i}||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\\ &&+4\sum\limits_{j=1}^k\Big(\frac{l_{j}^{\alpha}}{\Gamma(\alpha-2)}\omega_{j}||u_{j}-v_{j}||+\frac{l_{j}^{\alpha-\beta}}{\Gamma(\alpha-2)}\omega_{j}||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+2\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{i}^{\alpha}}{\Gamma(\alpha-2)}\omega_{i}||u_{i}-v_{i}||+\frac{l_{i}^{\alpha-\beta}}{\Gamma(\alpha-2)}\omega_{i}||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big)\\ &\leq&\frac{2}{\Gamma(\alpha-2)}\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big)\\ &&+4\sum\limits_{j=1}^k\frac{1}{\Gamma(\alpha-2)}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\omega_{j}\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+2\sum\limits_{j=1\atop j\neq i}^{k}\frac{1}{\Gamma(\alpha-2)}\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big). \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&\|T_{i}u-T_{i}v\|\\ &\leq&\Big(\frac{2}{\Gamma(\alpha)}+\frac{2}{\Gamma(\alpha-1)}+\frac{2}{\Gamma(\alpha-2)}\Big)\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big)\nonumber\\ &&+\frac{10}{\Gamma(\alpha-2)}\sum^k_{j=1}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\omega_{j}\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+\frac{6}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big). \end{eqnarray*} $

另一方面, 有

$ {}_{C}D_{0, t}^{\beta}(t^{r})=\frac{\Gamma(\gamma+1)}{\Gamma(\gamma+1-\beta)}t^{\gamma-\beta}. $

从而

$ \begin{eqnarray*} &&|_{C}D_{0, t}^{\beta}T_{i}u(t)-_{C}D_{0, t}^{\beta}T_{i}v(t)|\\ &\leq&\frac{l_{i}^{\alpha}}{\Gamma(\alpha-\beta)}\int_{0}^{t}(t\!-\!s)^{\alpha-\beta-1} \Big(|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)) \!-\!f_{i}(s, v_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}v_{i}(s))|\Big){\rm d}s\\ &&+\frac{2t^{2-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big) l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}\\ &&\times\Big(|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s)) -f_{j}(s, v_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}v_{j}(s))|\Big){\rm d}s\\ &&+\frac{2t^{2-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big) l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}\\ &&\times\Big(|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s)) -f_{j}(s, v_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}v_{j}(s))|\Big){\rm d}s\\ &&+\frac{2t^{2-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big) l_{i}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}\\ &&\times\Big(|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s)) -f_{i}(s, v_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}v_{i}(s))|\Big){\rm d}s, \\ &&|_{C}D_{0, t}^{\beta}T_{i}u(t)-_{C}D_{0, t}^{\beta}T_{i}v(t)| \\ &\leq&\frac{l_{i}^{\alpha}}{\Gamma(\alpha-\beta)}\omega_{i}||u_{i}-v_{i}||+\frac{l_{i}^{\alpha-\beta}}{\Gamma(\alpha-\beta)}\omega_{i}||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\\ &&+\sum\limits_{j=1}^k\Big(\frac{2l_{j}^{\alpha}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\omega_{j}||u_{j}-v_{j}||+\frac{2l_{j}^{\alpha-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)} \omega_{j}||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ && +\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{2l_{j}^{\alpha}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\omega_{j}||u_{j}-v_{j}|| +\frac{2l_{j}^{\alpha-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)} \omega_{j}||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big) \\ &&+\frac{2l_{i}^{\alpha}}{\Gamma(\alpha-2)\Gamma(3-\beta)} \omega_{i}||u_{i}-v_{i}||+\frac{2l_{i}^{\alpha-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\omega_{i}||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}|| \\ &\leq&\frac{(l_{i}^{\alpha}+l_{i}^{\alpha-\beta})}{\Gamma(\alpha-\beta)}\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big)\\ &&+\frac{2}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1}^k\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\omega_{j}\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+\frac{2}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\omega_{j}\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\\ &&+\frac{2(l_{i}^{\alpha}+l_{i}^{\alpha-\beta})}{\Gamma(\alpha-2)\Gamma(3-\beta)}\omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big). \end{eqnarray*} $

因此

$ \begin{eqnarray*} &&||_{C}D_{0, t}^{\beta}T_{i}u-_{C}D_{0, t}^{\beta}T_{i}v||\\ &\leq&\Big(\frac{1}{\Gamma(\alpha-\beta)}+\frac{4}{\Gamma(\alpha-2)\Gamma(3-\beta)}\Big)\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big) \omega_{i}\Big(||u_{i}-v_{i}||+||_{C}D_{0, s}^{\beta}u_{i}-_{C}D_{0, s}^{\beta}v_{i}||\Big) \\ &&+\frac{4}{\Gamma(\alpha-2)\Gamma(3-\beta)} \sum\limits_{j=1\atop j\neq i}^{k}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big) \omega_{j}\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big). \end{eqnarray*} $

因此

(3.1) $ \begin{eqnarray} ||T_{i}u-T_{i}v||_{X^{k}} &\leq&\Big[\Big(l_{i}^{\alpha}+l_{i}^{\alpha-\beta}\Big)(\frac{11}{\Gamma(\alpha-1)}+\frac{2}{\Gamma(\alpha+1)}+\frac{1}{\Gamma(\alpha)}+\frac{1}{\Gamma(\alpha-\beta)} {}\\ &&+\frac{4}{\Gamma(\alpha-2)\Gamma(3-\beta)})\Big]\Big(\sum\limits_{i=1}^k\omega_{i}\Big)\Big(\sum\limits_{j=1}^k\Big(||u_{j}-v_{j}||+ ||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\Big){} \\ &&+\frac{10}{\Gamma(\alpha-2)}\sum^k_{j=1}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\Big(\sum\limits_{j=1}^k\omega_{j}\Big)\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big){}\\ &&+\sum\limits_{j=1\atop j\neq i}^{k}\Big(l_{j}^{\alpha}+l_{j}^{\alpha-\beta}\Big)\Big(\frac{6}{\Gamma(\alpha-2)}+\frac{4}{\Gamma(\alpha-2)\Gamma(3-\beta)}\Big){} \\ &&\times\Big(\sum\limits_{i=1}^k\omega_{i}\Big)\Big(\sum\limits_{j=1}^k\Big(||u_{j}-v_{j}||+||_{C}D_{0, s}^{\beta}u_{j}-_{C}D_{0, s}^{\beta}v_{j}||\Big)\Big){} \\ &=&M_{i}\Big(\sum\limits_{i=1}^k\omega_{i}\Big)||u-v||_{X^{k}}. \end{eqnarray} $

由(3.1)式得

$ \begin{eqnarray*} ||T_{i}u-T_{i}v||_{X^{k}} =\sum\limits_{i=1}^k||T_{i}u-T_{i}v||_{X} \leq\Big(\sum\limits_{i=1}^kM_{i}\Big)\Big(\sum\limits_{i=1}^k\omega_{i}\Big)||u-v||_{X^{k}}. \end{eqnarray*} $

由$ \Big(\sum\limits_{i=1}^kM_{i}\Big)\Big(\sum\limits_{i=1}^k\omega_{i}\Big)<1 $可知$ T $是压缩的, 故由引理2.4可得问题(2.3)–(2.6)存在唯一解. 因此原问题(1.1)–(1.4)存在唯一解.

定理3.2  假设(H1)–(H3)成立, 则边值问题(2.3)–(2.6)至少有一个解.

证  由函数$ f_{i} $的连续性知$ T:X^{k}\rightarrow X^{k} $是连续算子. 设$ \Omega\subset X^{k} $是有界集, 故对任意的$ u=(u_{1}, u_{2}, \cdots , u_{k})\in\Omega, t\in[0, 1], $有

$ \begin{eqnarray*} |T_{i}u(t)|&\leq&\frac{l_{i}^{\alpha}}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|{\rm d}s\\ &&+\frac{l_{i}^{\alpha}}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|{\rm d}s\\ &&+\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1}^kl_{j}^{\alpha}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(1-t^{2}) |f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))|{\rm d}s\\ &&+\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}l_{j}^{\alpha}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(1+t^{2}) |f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))|{\rm d}s\\ &&+\frac{1}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}l_{i}^{\alpha}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)\int_{0}^{1}(1-s)^{\alpha-3}(1+t^{2})|f_{j}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|{\rm d}s, \\ |T_{i}'u(t)|&\leq&\frac{l_{i}^{\alpha}}{\Gamma(\alpha-1)}\int_{0}^{t}(t-s)^{\alpha-2}|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|{\rm d}s\\ &&+\frac{2t}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))|{\rm d}s\\ &&+\frac{2t}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))|{\rm d}s\\ &&+\frac{2t}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|{\rm d}s, \\ |T_{i}''u(t)|&\leq&\frac{l_{i}^{\alpha}}{\Gamma(\alpha-2)}\int_{0}^{t}(t-s)^{\alpha-3}|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|{\rm d}s\\ &&+\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))|{\rm d}s\\ &&+\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))|{\rm d}s\\ &&+\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|{\rm d}s. \end{eqnarray*} $

由条件(H3), 有

(3.2) $ \begin{equation} |T_{i}u(t)|\leq\frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha)}+\sum\limits_{j=1}^k\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)} +\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)}\Big), \end{equation} $

(3.3) $ \begin{equation} |T_{i}'u(t)|\leq\frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-1)} +\sum\limits_{j=1}^k\frac{4L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)} +\sum\limits_{j=1\atop j\neq i}^{k}\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)}, \end{equation} $

(3.4) $ \begin{equation} |T_{i}''u(t)|\leq\frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-2)} +4\sum\limits_{j=1}^k\frac{L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)} +\sum\limits_{j=1\atop j\neq i}^{k}\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)}. \end{equation} $

因为

$ \begin{eqnarray*} &&|_{C}D_{0, t}^{\beta}T_{i}u(t)|\\ &\leq&\frac{l_{i}^{\alpha}}{\Gamma(\alpha-\beta)}\int_{0}^{t}(t-s)^{\alpha-\beta-1}\Big(|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|\Big){\rm d}s\\ &&+\frac{2t^{2-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big) l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}\Big(|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))|\Big){\rm d}s\\ &&+\frac{2t^{2-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big) l_{j}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}\Big(|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))|\Big){\rm d}s\\ &&+\frac{2t^{2-\beta}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big) l_{i}^{\alpha}\int_{0}^{1}(1-s)^{\alpha-3}\Big(|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))|\Big){\rm d}s, \end{eqnarray*} $

因为$ t^{2-\beta}<1\ (0<2-\beta<1) $且$ \Big(\frac{l_{j}^{-2}}{\sum\limits_{j=1}^kl_{j}^{-2}}\Big)<1, j=1, 2, \cdots , k, $故由(H3)可知

$ |_{C}D_{0, t}^{\beta}T_{i}u(t)|\leq\frac{L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-\beta)} +\sum\limits_{j=1}^k\frac{6L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)\Gamma(3-\beta)}. $

由(3.2)–(3.4)式可得

$ \begin{eqnarray*} &&||T_{i}u||+||T_{i}'u||+||T_{i}''u||+||_{C}D_{0, t}^{\beta}T_{i}u||\\ &\leq& L_{i}l_{i}^{\alpha}\Big(\frac{2}{\Gamma(\alpha)}+\frac{2}{\Gamma(\alpha-1)}+\frac{2}{\Gamma(\alpha-2)}+ \frac{1}{\Gamma(\alpha-\beta)}\Big)+\frac{2}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}L_{j}l_{j}^{\alpha}\\ &&+\sum^k_{j=1}L_{j}l_{j}^{\alpha}\Big(\frac{10}{\Gamma(\alpha-2)}+\frac{6}{\Gamma(\alpha-2)\Gamma(3-\beta)}\Big)\\ &=&P_{i}\Big(\sum\limits_{j=1}^kL_{j}\Big). \end{eqnarray*} $

因此

(3.5) $ \begin{equation} ||Tu||_{X^{k}}=\sum\limits_{i=1}^k||T_{i}u||_{X}\leq\Big(\sum\limits_{i=1}^kP_{i}\Big)\Big(\sum\limits_{i=1}^kL_{j}\Big)<\infty. \end{equation} $

所以由(3.5)式知$ T $是一致有界的.

下证$ T $是等度连续的. 对任意的$ u=(u_{1}, u_{2}, \cdots , u_{k})\in\Omega, t_{1}, t_{2}\in[0, 1], t_{1}<t_{2}, $有

$ \begin{eqnarray*} &&|T_{i}u(t_{2})-T_{i}u(t_{1})|\\ &\leq& l_{i}^{\alpha}\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}} \Big((t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1}\Big)\Big|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))\Big|{\rm d}s\\ &&+l_{i}^{\alpha}\frac{1}{\Gamma(\alpha)}\int_{ t_{1}}^{t_{2}}\Big((t_{2}-s)^{\alpha-1}\Big)\Big|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))\Big|{\rm d}s\\ &&+\frac{(t_{2}^{2}-t_{1}^{2})}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))\Big|{\rm d}s\\ &&+\frac{(t_{2}^{2}-t_{1}^{2})}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))\Big|{\rm d}s\\ &&+\frac{(t_{2}^{2}-t_{1}^{2})}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))\Big|{\rm d}s, \\ &&|T_{i}'u(t_{2})-T_{i}'u(t_{1})|\\ &\leq& l_{i}^{\alpha}\frac{1}{\Gamma(\alpha-1)}\int_{0}^{t_{1}}\Big((t_{2}-s)^{\alpha-2} -(t_{1}-s)^{\alpha-2}\Big)\Big|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))\Big|{\rm d}s\\ &&+l_{i}^{\alpha}\frac{1}{\Gamma(\alpha-1)}\int_{ t_{1}}^{t_{2}}\Big((t_{2}-s)^{\alpha-2}\Big)\Big|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))\Big|{\rm d}s\\ &&+\frac{2(t_{2}-t_{1})}{\Gamma(\alpha-2)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))\Big|{\rm d}s\\ &&+\frac{2(t_{2}-t_{1})}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))\Big|{\rm d}s\\ &&+\frac{2(t_{2}-t_{1})}{\Gamma(\alpha-2)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))\Big|{\rm d}s. \end{eqnarray*} $

由(H3)得

$ \begin{eqnarray*} &&|T_{i}u(t_{2})-T_{i}u(t_{1})|\\ &\leq& \frac{L_{i}l_{i}^{\alpha}}{\Gamma(\alpha+1)}(t_{2}^{\alpha}-t_{1}^{\alpha}) +\sum\limits_{j=1}^k\frac{L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-1)}(t_{2}^{2}-t_{1}^{2})\nonumber\\ &&+\sum\limits_{j=1\atop j\neq i}^{k}\frac{L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-1)}(t_{2}^{2}-t_{1}^{2}) +\frac{L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-1)}(t_{2}^{2}-t_{1}^{2}){\nonumber} \\ &=&\frac{L_{i}l_{i}^{\alpha}}{\Gamma(\alpha+1)}(t_{2}^{\alpha}-t_{1}^{\alpha})+ L_{i}l_{i}^{\alpha}(t_{2}^{2}-t_{1}^{2})\frac{2}{\Gamma(\alpha-1)}{\nonumber} +\sum\limits_{j=1\atop j\neq i}^{k} L_{j}l_{j}^{\alpha}(t_{2}^{2}-t_{1}^{2})\frac{2}{\Gamma(\alpha-1)}, {\nonumber} \\ &&|T_{i}'u(t_{2})-T_{i}'u(t_{1})|\\ &\leq& \frac{L_{i}l_{i}^{\alpha}}{\Gamma(\alpha)}(t_{2}^{\alpha-1}-t_{1}^{\alpha-1}) +\frac{L_{i}l_{i}^{\alpha}(t_{2}-t_{1})^{\alpha-1}}{\Gamma(\alpha)}+\sum\limits_{j=1}^k\frac{2L_{j} l_{j}^{\alpha}}{\Gamma(\alpha-1)}(t_{2}-t_{1})\nonumber\\ &&+\sum\limits_{j=1\atop j\neq i}^{k}\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-1)}(t_{2}-t_{1}) +\frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-1)}(t_{2}-t_{1}). \end{eqnarray*} $

$ |T_{i}''u(t_{2})-T_{i}''u(t_{1})| \leq \frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-1)}(t_{2}^{\alpha-2}-t_{1}^{\alpha-2}). $

同理

$ \begin{eqnarray*} &&|_{C}D_{0, t}^{\beta}T_{i}u(t_{2})-_{C}D_{0, t}^{\beta}T_{i}u(t_{1})|\\ &\leq& \frac{l_{i}^{\alpha}}{\Gamma(\alpha-\beta)}\int_{0}^{t_{1}}\Big((t_{2}-s)^{\alpha-\beta-1}-(t_{1}-s)^{\alpha-\beta-1}\Big)\Big|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))\Big|{\rm d}s\\ &&+\frac{2(t_{2}^{2-\beta}-t_{1}^{2-\beta})}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1}^k\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))\Big|{\rm d}s\\ &&+\frac{2(t_{2}^{2-\beta}-t_{1}^{2-\beta})}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{j}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{j}(s, u_{j}(s), l_{j}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{j}(s))\Big|{\rm d}s\\ &&+\frac{2(t_{2}^{2-\beta}-t_{1}^{2-\beta})}{\Gamma(\alpha-2)\Gamma(3-\beta)}\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{l_{j}^{-2}}{2\sum\limits_{j=1}^kl_{j}^{-2}}\Big)l_{i}^{\alpha} \int_{0}^{1}(1-s)^{\alpha-3}\Big|f_{i}(s, u_{i}(s), l_{i}^{-\beta}\cdot {}_{C}D_{0, s}^{\beta}u_{i}(s))\Big|{\rm d}s. \end{eqnarray*} $

由(H3)得

$ \begin{eqnarray*} &&|_{C}D_{0, t}^{\beta}T_{i}u(t_{2})-_{C}D_{0, t}^{\beta}T_{i}u(t_{1})|\nonumber\\ &\leq&\frac{L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-\beta)}(t_{2}^{\alpha-\beta}-t_{1}^{\alpha-\beta})+\sum\limits_{j=1}^k\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-1)\Gamma(3-\beta)}(t_{2}^{\alpha-\beta}-t_{1}^{\alpha-\beta}){\nonumber} \\ &&+\sum\limits_{j=1\atop j\neq i}^{k}\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-1)\Gamma(3-\beta)}(t_{2}^{2-\beta}-t_{1}^{2-\beta}) +\frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-1)\Gamma(3-\beta)}(t_{2}^{2-\beta}-t_{1}^{2-\beta}){\nonumber} \\ &=&\frac{L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-\beta)}(t_{2}^{\alpha-\beta}-t_{1}^{\alpha-\beta})+\frac{4L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-1)\Gamma(3-\beta)}(t_{2}^{2-\beta}-t_{1}^{2-\beta})\\ &&+\sum\limits_{j=1\atop j\neq i}^{k}\frac{4L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-1)\Gamma(3-\beta)}(t_{2}^{\alpha-\beta} -t_{1}^{\alpha-\beta}). \end{eqnarray*} $

因此$ ||T_{i}u(t_{2})-T_{i}u(t_{1})||_{X}\rightarrow 0, (t_{2}\rightarrow t_{1}) $且$ ||Tu(t_{2})-Tu(t_{1})||_{X^{k}}\rightarrow 0, (t_{2}\rightarrow t_{1}). $

综上可得$ T $在$ X $上等度连续, 且由Arzela-Ascoli定理知$ T $是全连续算子.

定义集合

$ Q=\{(u_{1}, u_{2}, \cdots , u_{k})\in X^{k}:(u_{1}, u_{2}, \cdots , u_{k})=\mu T(u_{1}, u_{2}, \cdots , u_{k}), 0<\mu<1\}. $

当$ (u_{1}, u_{2}, \cdots , u_{k})\in Q $时, 有$ (u_{1}, u_{2}, \cdots , u_{k})=\mu T(u_{1}, u_{2}, \cdots , u_{k}), t\in[0, 1]. $即$ u_{i}(t)=\mu T_{i}(u_{1}, u_{2}, \cdots , u_{k}), i=1, 2, \cdots , k. $因此由(3.2)–(3.4)式及(H3)得

$ ||u_{i}||\leq \mu\Big[\frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha)}+\sum\limits_{j=1}^k\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)} +\sum\limits_{j=1\atop j\neq i}^{k}\Big(\frac{L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)}\Big)\Big], $

$ ||u_{i}'|| \leq\mu\Big[\frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-1)} +\sum\limits_{j=1}^k\frac{4L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)} +\sum\limits_{j=1\atop j\neq i}^{k}\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)}\Big], $

$ ||u''|| \leq\mu\Big[\frac{2L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-2)} +4\sum\limits_{j=1}^k\frac{L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)} +\sum\limits_{j=1\atop j\neq i}^{k}\frac{2L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)}\Big]. $

又

$ |_{C}D_{0, t}^{\beta}u_{i}|\leq \mu\Big[\frac{L_{i}l_{i}^{\alpha}}{\Gamma(\alpha-\beta)} +\sum\limits_{j=1}^k\frac{6L_{j}l_{j}^{\alpha}}{\Gamma(\alpha-2)\Gamma(3-\beta)}\Big]. $

因此

$ ||u||_{X^{k}}\leq\Big(\sum\limits_{i=1}^kP_{i}\Big)\Big(\sum\limits_{j=1}^kL_{j}\Big)<\infty. $

故$ Q $有界, 由引理2.3知, 算子$ T $至少有一个不动点, 即边值问题(2.3)–(2.6)在$ [0, 1] $上至少有一个解, 因此边值问题(1.1)–(1.4)也至少有一个解.