一类无穷区间上的多点分数阶边值问题解的存在性和唯一性

Existence and Uniqueness of Solutions to a Class of Multi-Point Fractional Boundary Value Problem on the Infinite Interval

摘要:

本文讨论一类无穷区间上的多点分数阶边值问题，的可解性。通过应用Leray-Schauder非线性抉择定理和Banach压缩映像原理，得到解的存在性和唯一性。最后给出例子说明定理的适用性。

In this paper, we consider the following multi-point boundary value problem of fractional differ-ential equation on the infinite interval ,

By using Leray-Schauder Nonlinear Alternative theorem and Banach fixed point theorem, some results on the existence and uniqueness of solutions can be established.

Abstract:

1. 引言

由于分数计算理论和应用的快速发展，分数阶微分方程引起数学爱好者的极大兴趣。其主要应用于流体力学、分数控制系统、神经分数模型、力学、物理学、黏弹力学、化学工程和经济等方面。分数计算理论是解决微分、积分方程及其它特殊方程的有效工具。

但是分数阶微分方程边值问题的研究还处于初级阶段，尤其是无穷区间上的分数阶微分方程边值问题尚不多见。在以往的参考文献中，多是以下积分边值条件的模型。 [1] 中，赵和葛研究了无穷区间上的分数阶边值问题

$D_{0+}^{\alpha }u\left(t\right)+f\left(t,u\left(t\right)\right)=0,t\in \left(0,\infty \right),\alpha \in \left(1,2\right),$

$u\left(0\right)=0,\underset{t\to \infty }{\lim }{D}_{0+}^{\alpha -1}u\left(t\right)=\beta u\left(\xi \right),$

其中， $0<\xi <\infty$ ， $D_{0+}^{\alpha }$ 是Riemann-Liouvill分数阶导数。

[2] 中，Nieto研究了以下边值问题

$D_{0+}^{\alpha }u\left(t\right)+f\left(t,u\left(t\right)\right)=0,t\in \left[0,1\right],n-1<\alpha <n,n\in N,$

$u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(0\right)=\cdots =u^{n-2}\left(0\right)=0,u\left(1\right)={\displaystyle \underset{i=1}{\overset{m-2}{\sum }}{a}_{i}u\left({\eta }_i\right)},$

其中， $n\ge 2,a_i>0,0<{\eta }_1<{\eta }_2<\cdots <{\eta }_{m-2}<1,f\in C\left(\left[0,1\right]\times R,R\right)$ 。 ${}_{R}D{}_{0+}^{\alpha },{}_{C}D{}_{0+}^{\alpha }$ 分别是Riemann-Liouvill分数阶导数和Caputo分数阶导数。

然而，据作者所知，到目前还没有文献研究以下无穷区间上的分数阶边值问题

$D_{0+}^{\alpha }u\left(t\right)=f\left(t,u\left(t\right),D_{0+}^{\alpha -1}u\left(t\right)\right),t\in J:=\left(0,\infty \right),n-1<\alpha <n,$ (1.1)

$u\left(0\right)=D_{0+}^{\alpha -2}u\left(0\right)=D_{0+}^{\alpha -3}u\left(0\right)=\cdots =D_{0+}^{\alpha -\left(n-1\right)}u\left(0\right)=0,D_{0+}^{\alpha -1}u\left(\infty \right)={\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i{D}_{0+}^{\alpha -1}u\left({\xi }_i\right)}$ (1.2)

其中 $n-1<\alpha \le n,n\in N,n>2,0<{\xi }_1<{\xi }_2<\cdots <{\xi }_m<+\infty ,{\alpha }_i>0,i=1,2,\cdots ,m$ 。 $f\in C\left(J\times R\times R,R\right)$ 。 $D_{0+}^{\alpha }$ 和 $I_{0+}^{\alpha }$ 分别是Riemann-Liouvill分数阶导数和分数阶积分。

当 $\alpha =n$ 时，问题(1.1) (1.2)是 $n$ 阶多点边值问题，很多文献已做过研究，如 [3] [4] [5] [6] 。

本文应用非线性抉择定理和Banach压缩映像原理研究边值问题(1.1) (1.2)的解的存在性和唯一性。

本文结构如下：第二部分给出背景材料和预备知识；第三部分给出所研究问题(1.1) (1.2)的解的存在性和唯一性；最后给出例子说明我们的主要结论。

2. 预备知识

定义2.1 ( [7] )函数 $y:\left(0,+\infty \right)\to R$ 的 $\alpha$ 阶Riemann-Liouville分数阶积分为

$I_{0+}^{\alpha }y\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}y\left(s\right)\text{d}s},t>0,$

其中 $\alpha >0$ ， $\Gamma (\cdot )$ 为gamma函数。

定义2.2 ( [7] )函数 $y:\left(0,+\infty \right)\to R$ 的 $\alpha$ 阶Riemann-Liouville分数阶导数为

$D_{0+}^{\alpha }y\left(t\right)=\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{\text{d}}{\text{d}t}\right)}^n{\displaystyle {\int }_0^t\frac{y\left(s\right)}{{\left(t-s\right)}^{\alpha -n+1}}\text{d}s},$

其中 $\alpha >0$ ， $\Gamma (\cdot )$ 为gamma函数， $n=\left[\alpha \right]+1$ 。

引理2.1 ( [7] )设 $f\in C\left[0,1\right],q\ge p\ge 0$ ，则

$D_{0+}^p{I}_{0+}^{q}f\left(t\right)=I_{0+}^{q-p}f\left(t\right).$

引理2.2 ( [7] )若 $\alpha >0$ ，则分数阶微分方程 $D_{0+}^{\alpha }u\left(t\right)=0$ 当且仅当

$u\left(t\right)=c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+\cdots +c_n{t}^{\alpha -n},$

其中 $c_i\in R,i=1,2,\cdots ,n$ ，其中 $n$ 是大于等于 $\alpha$ 的最小整数。

引理2.3 ( [8] )若 $\alpha >0,u\in C\left(0,1\right)\cap L\left(0,1\right),{}^{c}D{}_{0+}^{\alpha }u\left(t\right)\in C\left(0,1\right)\cap L\left(0,1\right)$ ，存在 $c_i,i=1,2,\cdots ,N$ 使得

$I_{0+}^{\alpha }{D}_{0+}^{\alpha }u\left(t\right)=u\left(t\right)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+\cdots +c_N{t}^{\alpha -N},$

其中 $c_i\in R,i=0,1,2,\cdots ,N-1$ ， $N=\left[\alpha \right]+1$ 。

定义空间

$X=\left\{u\left(t\right)\in C\left(J,R\right):\underset{t\in J}{\sup }\frac{\left|u\left(t\right)\right|}{1+t^{\alpha -1}}<+\infty ,D_{0+}^{\alpha -1}u\left(t\right)\in C\left(J,R\right),\underset{t\in J}{\sup }\left|D_{0+}^{\alpha -1}u\left(t\right)\right|<+\infty \right\}$

模为 $\Vert u\Vert =\max \left\{\underset{t\in J}{\sup }\frac{\left|u\left(t\right)\right|}{1+t^{\alpha -1}},\underset{t\in J}{\sup }\left|D_{0+}^{\alpha -1}u\left(t\right)\right|\right\}$ 。

定理2.1 ( [9] )设 $X$ 是一个实Banach空间， $\Omega$ 是 $X$ 中的有界开子集， $0\in \Omega ,F:\bar{\Omega }\to X$ 是一个全连续算子。则 $\exists x\in \partial \Omega ,\lambda >1,s.t.F\left(x\right)=\lambda x$ ，或存在不动点 $x^{\ast }\in \bar{\Omega }$ 。

引理2.4 ( [9] ) $\left(X,\Vert ·\Vert \right)$ 是Banach空间。

证明：设 ${\left\{u_n\right\}}_{n=1}^{\infty }$ 是空间 $\left(X,\Vert ·\Vert \right)$ 中的Cauchy序列，则对 $\forall \epsilon >0,\exists N>0$ ， $s.t.$ 当 $\forall t\in J$ ， $n,m>N$ ，有 $\left|\frac{u_n\left(t\right)}{1+t^{\alpha -1}}-\frac{u_m\left(t\right)}{1+t^{\alpha -1}}\right|<\epsilon$ ，则 ${\left\{\frac{u_n\left(t\right)}{1+t^{\alpha -1}}\right\}}_{n=1}^{\infty }$ 一致收敛于 $\frac{u\left(t\right)}{1+t^{\alpha -1}}$ 及 $u\left(t\right)\in X$ 。并且 ${\left\{D_{0+}^{\alpha -1}{u}_n\right\}}_{n=1}^{\infty }$ 一致收敛于 $v\in X$ ，并且 $\underset{t\in J}{\sup }\left|v\left(t\right)\right|<+\infty$ 。

接下来证明 $v=D_{0+}^{\alpha -1}u$ 。

令 $\underset{t\in J}{\sup }\frac{\left|u\left(t\right)\right|}{1+t^{\alpha -1}}=\frac{M_0}{2}$ ，则对于常数 $\frac{M_0}{2}>0,t\in J,\exists N>0$ ， $s.t.$ 当 $n>N$ 时，有

$\left|\frac{u_n\left(t\right)}{1+t^{\alpha -1}}-\frac{u\left(t\right)}{1+t^{\alpha -1}}\right|<\frac{M_0}{2}.$

令 $M_i=\underset{t\in J}{\sup }\frac{\left|u_i\left(t\right)\right|}{1+t^{\alpha -1}},i=1,2,\cdots ,N$ ， $M=\max \left\{M_i,i=1,2,\cdots ,N\right\}$ ，易得 $\frac{\left|u_n\left(t\right)\right|}{1+t^{\alpha -1}}\le M,n=1,2,\cdots$ 。于是对 $\forall t\in J,n-1<\alpha <n$ ，有

$\begin{array}{l}\left|{\displaystyle {\int }_0^1{\left(t-s\right)}^{n-1-\alpha }\left(1+s^{\alpha -1}\right)\frac{\left|u_n\left(s\right)\right|}{1+s^{\alpha -1}}\text{d}s}\right|\le M{\displaystyle {\int }_0^t{\left(t-s\right)}^{n-1-\alpha }\left(1+s^{\alpha -1}\right)\text{d}s}\\ =M\left[t^{n-\alpha }{\displaystyle {\int }_0^1{\left(1-\tau \right)}^{n-1-\alpha }\text{d}\tau }+t^{n-1}{\displaystyle {\int }_0^1{\left(1-\tau \right)}^{n-1-\alpha }{\tau }^{\alpha -1}\text{d}\tau }\right]\\ =\frac{M}{n-\alpha }{t}^{n-\alpha }+B\left(\alpha ,n-\alpha \right)M{t}^{n-1},\end{array}$

其中， $B\left(\alpha ,n-\alpha \right)$ 是Beta函数。根据 ${\left\{D_{0+}^{\alpha -1}{u}_n\left(t\right)\right\}}_{n=1}^{\infty }$ 的一致收敛性及Lebesgue控制收敛定理，可得

$\begin{array}{c}v\left(t\right)=\underset{n\to +\infty }{\lim }{D}_{0+}^{\alpha -1}{u}_n\left(t\right)=\underset{n\to +\infty }{\lim }\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{\text{d}}{\text{d}t}\right)}^{n-1}{\displaystyle {\int }_0^t{\left(t-s\right)}^{n-1-\alpha }\left(1+s^{\alpha -1}\right)\frac{u_n\left(s\right)}{1+s^{\alpha -1}}\text{d}s}\\ =\frac{1}{\Gamma \left(n-\alpha \right)}{\left(\frac{\text{d}}{\text{d}t}\right)}^{n-1}{\displaystyle {\int }_0^t{\left(t-s\right)}^{n-1-\alpha }\left(1+s^{\alpha -1}\right)\frac{u\left(s\right)}{1+s^{\alpha -1}}\text{d}s}\\ =D_{0+}^{\alpha -1}u\left(t\right).\end{array}$

当 $\alpha =n$ 时，有

$v\left(t\right)=\underset{n\to +\infty }{\lim }{u}_n^{\left(n-1\right)}\left(t\right)={\left(\frac{\text{d}}{\text{d}t}\right)}^{n-1}\underset{n\to +\infty }{\lim }{u}_n\left(t\right)=u^{\left(n-1\right)}\left(t\right).$

所以， $\left(X,\Vert ·\Vert \right)$ 是Banach空间。

注意到Arzela-Ascoli定理不能在空间 $X$ 中使用，为此，引入以下改进的紧凑型标准。

引理2.5 ( [10] )设 $Z\subseteq Y$ 是有界集，那么，当下述条件成立时， $Z$ 在 $Y$ 中是相对紧的：

(i) 对 $\forall u\left(t\right)\in Z$ ， $\frac{u\left(t\right)}{1+t^{\alpha -1}}$ 和 $D_{0+}^{\alpha -1}u\left(t\right)$ 在 $J$ 的任何紧区间上是等度连续的。

(ii) 对于给定的 $\epsilon >0$ ， $\exists$ 常数 $T=T\left(\epsilon \right)>0$ ， $s.t.$ 对 $\forall t_1,t_2\ge T,u\left(t\right)\in Z$ ，有

$\left|\frac{u\left(t_1\right)}{1+t_1{}^{\alpha -1}}-\frac{u\left(t_2\right)}{1+t_2{}^{\alpha -1}}\right|<\epsilon ,\left|D_{0+}^{\alpha -1}u\left(t_1\right)-D_{0+}^{\alpha -1}u\left(t_2\right)\right|<\epsilon .$

3. 主要结论

令 $A=\left(1+\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\right){\displaystyle {\int }_0^{\infty }\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}$ $+\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s,}$

$B=\left(1+\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\right){\displaystyle {\int }_0^{\infty }c\left(s\right)\text{d}s}+\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}c\left(s\right)\text{d}s.}$

假设以下条件成立：

(H₁)存在非负函数 $a\left(t\right),b\left(t\right),c\left(t\right)\in L^1\left(J\right),s.t.$

$\left|f\left(t,x,y\right)\right|\le a\left(t\right)\left|x\right|+b\left(t\right)\left|y\right|+c\left(t\right),{\displaystyle {\int }_0^{+\infty }\left(\left(1+t^{\alpha -1}\right)a\left(t\right)+b\left(t\right)\right)\text{d}t<\Gamma \left(\alpha \right)},$ ${\displaystyle {\int }_0^{+\infty }c\left(t\right)\text{d}t}<+\infty .$

(H₂)假设 ${\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}<1,A<1$ 。

引理3.1 假设(H₁) (H₂)成立。问题(1.1) (1.2)等价于积分方程

$\begin{array}{c}u\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}\\ +\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}[{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}{\displaystyle {\int }_0^{{\xi }_i}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}\\ -{\displaystyle {\int }_0^{\infty }f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}]t^{\alpha -1}.\end{array}$ (3.1)

证明：由条件(H₁)，

${\displaystyle {\int }_0^{\infty }\left|f\left(t,u\left(t\right),D_{0+}^{\alpha -1}u\left(t\right)\right)\right|\text{d}t}\le \Vert u\Vert {\displaystyle {\int }_0^{\infty }\left(\left(1+t^{\alpha -1}\right)a\left(t\right)+b\left(t\right)\right)\text{d}t}+{\displaystyle {\int }_0^{\infty }c\left(t\right)\text{d}t}<+\infty .$

所以(3.1)定义有意义。

根据引理2.3，

$u\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+\cdots +c_n{t}^{\alpha -n}.$ (3.2)

由条件(1.2)，可得

$c_2=c_3=\cdots =c_n=0.$ (3.3)

$c_1=\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\left[{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}{\displaystyle {\int }_0^{{\xi }_i}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}-{\displaystyle {\int }_0^{\infty }f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}\right].$ (3.4)

将(3.3) (3.4)带入(3.2)，得

$\begin{array}{c}u\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}\\ +\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}[{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}{\displaystyle {\int }_0^{{\xi }_i}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}\\ -{\displaystyle {\int }_0^{\infty }f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}]t^{\alpha -1}.\end{array}$ (3.5)

定义积分算子 $T:X\to X$

$\begin{array}{c}Tu\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}\\ +\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}[{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}{\displaystyle {\int }_0^{{\xi }_i}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}\\ -{\displaystyle {\int }_0^{\infty }f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}]t^{\alpha -1}.\end{array}$ (3.6)

引理3.1证明算子 $T$ 的不动点就是问题(1.1) (1.2)的解。

引理3.2 假设(H₁) (H₂)成立。则 $T:\bar{\Omega }\to X$ 一致连续。其中 $\Omega =\left\{u\in X,\Vert u\Vert <R\right\}$ ，

$R\le \frac{B}{1-A},A<1.$

证明：第一步：证明 $T:\bar{\Omega }\to X$ 是相对紧的。

方便起见，在这一步中，记

$G=\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\left|{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}{\displaystyle {\int }_0^{{\xi }_i}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s-{\displaystyle {\int }_0^{\infty }f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}}\right|$

设 $V$ 是 $\bar{\Omega }$ 中的子集， $I\subset J$ 是一个紧区间， $t_1,t_2\in I,t_1<t_2$ ，则对 $\forall u\left(t\right)\in V$ ，有

$\begin{array}{c}\left|\frac{Tu\left(t_2\right)}{1+t_2^{\alpha -1}}-\frac{Tu\left(t_1\right)}{1+t_1^{\alpha -1}}\right|\le |\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{t_2}\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}-\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{t_1}\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}}\\ \cdot f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{t_1}\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}\\ -\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{t_1}\frac{{\left(t_1-s\right)}^{\alpha -1}}{1+t_1^{\alpha -1}}f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\text{d}s}|+G\cdot \left|\frac{t_2{}^{\alpha -1}}{1+t_2^{\alpha -1}}-\frac{t_1{}^{\alpha -1}}{1+t_1^{\alpha -1}}\right|\\ \le \frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_{t_1}^{t_2}\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{t_1}|\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}}\\ -\frac{{\left(t_1-s\right)}^{\alpha -1}}{1+t_1^{\alpha -1}}|\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s+G\cdot \left|\frac{t_2{}^{\alpha -1}}{1+t_2^{\alpha -1}}-\frac{t_1{}^{\alpha -1}}{1+t_1^{\alpha -1}}\right|,\end{array}$

且有

$\left|D_{0+}^{\alpha -1}Tu\left(t_2\right)-D_{0+}^{\alpha -1}Tu\left(t_1\right)\right|\le {\displaystyle {\int }_{t_1}^{t_2}\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s}.$

注意到对 $\forall u\left(t\right)\in V$ ，有 $f\left(t,u\left(t\right),D_{0+}^{\alpha -1}u\left(t\right)\right)$ 在 $I$ 中有界。

故得 $\frac{Tu\left(t\right)}{1+t^{\alpha -1}},D_{0+}^{\alpha -1}Tu\left(t\right)$ 在 $I$ 中一致连续。

接下来证明对 $\forall u\left(t\right)\in V$ ，有 $f\left(t,u\left(t\right),D_{0+}^{\alpha -1}u\left(t\right)\right)$ 满足引理2.5的条件(ii)。

根据条件(H₁)，

${\displaystyle {\int }_0^{\infty }\left|f\left(t,u\left(t\right),D_{0+}^{\alpha -1}u\left(t\right)\right)\right|\text{d}t\le \Vert u\Vert {\displaystyle {\int }_0^{\infty }\left(\left(1+t^{\alpha -1}\right)a\left(t\right)+b\left(t\right)\right)\text{d}t}}$ $+{\displaystyle {\int }_0^{\infty }c\left(t\right)\text{d}t}<A\Vert u\Vert +B\le R,$

于是，对 $\forall \epsilon >0$ ，存在常数 $L>0$ ，使得

${\displaystyle {\int }_L^{\infty }\left|f\left(t,u\left(t\right),D_{0+}^{\alpha -1}u\left(t\right)\right)\right|\text{d}t<\epsilon .}$ (3.7)

另一方面，因为 $\underset{t\to +\infty }{\lim }\frac{t^{\alpha -1}}{1+t^{\alpha -1}}=1$ ，故 $\exists T_1>0,s.t.\forall t_1,t_2\ge T_1$ ，有

$\left|\frac{t_2^{\alpha -1}}{1+t_2^{\alpha -1}}-\frac{t_1^{\alpha -1}}{1+t_1^{\alpha -1}}\right|\le \left|1-\frac{t_2^{\alpha -1}}{1+t_2^{\alpha -1}}\right|+\left|1-\frac{t_1^{\alpha -1}}{1+t_1^{\alpha -1}}\right|<\epsilon .$ (3.8)

类似地，因为 $\underset{t\to +\infty }{\lim }\frac{{\left(t-L\right)}^{\alpha -1}}{1+t^{\alpha -1}}=1$ ，故 $\exists T_2>L>0,s.t.\forall t_1,t_2\ge T_2,0\le s\le L$ ，有

$\begin{array}{c}\left|\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}-\frac{{\left(t_1-s\right)}^{\alpha -1}}{1+t_1^{\alpha -1}}\right|\le \left|1-\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}\right|+\left|1-\frac{{\left(t_1-s\right)}^{\alpha -1}}{1+t_1^{\alpha -1}}\right|\\ \le \left|1-\frac{{\left(t_2-L\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}\right|+\left|1-\frac{{\left(t_1-L\right)}^{\alpha -1}}{1+t_1^{\alpha -1}}\right|<\epsilon .\end{array}$ (3.9)

令 $T>\max \left\{T_1,T_2\right\}$ ，则对 $\forall t_1,t_2\ge T$ ，根据(3.7)~(3.9)，可得

$\begin{array}{c}\left|\frac{Tu\left(t_2\right)}{1+t_2^{\alpha -1}}-\frac{Tu\left(t_1\right)}{1+t_1^{\alpha -1}}\right|\le \frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^L\left|\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}-\frac{{\left(t_1-s\right)}^{\alpha -1}}{1+t_1^{\alpha -1}}\right|}\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s\\ +\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_L^{t_1}\frac{{\left(t_1-s\right)}^{\alpha -1}}{1+t_1^{\alpha -1}}}\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s\\ +\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_L^{t_2}\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}}\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s+G\left|\frac{t_2^{\alpha -1}}{1+t_2^{\alpha -1}}-\frac{t_1^{\alpha -1}}{1+t_1^{\alpha -1}}\right|\\ \le \frac{\max \left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^L\left|\frac{{\left(t_2-s\right)}^{\alpha -1}}{1+t_2^{\alpha -1}}-\frac{{\left(t_1-s\right)}^{\alpha -1}}{1+t_1^{\alpha -1}}\right|\text{d}s}\end{array}$

$\begin{array}{l}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_L^{\infty }\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s}\\ +\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_L^{\infty }\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s}+G\left|\frac{t_2{}^{\alpha -1}}{1+t_2^{\alpha -1}}-\frac{t_1{}^{\alpha -1}}{1+t_1^{\alpha -1}}\right|\\ \le \frac{\max \left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|}{\Gamma \left(\alpha \right)}L\epsilon +\frac{2\epsilon }{\Gamma \left(\alpha \right)}+G\epsilon \\ =\left(\frac{\max \left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|}{\Gamma \left(\alpha \right)}L+\frac{2}{\Gamma \left(\alpha \right)}+G\right)\epsilon ,\end{array}$

且有

$\left|D_{0+}^{\alpha -1}Tu\left(t_2\right)-D_{0+}^{\alpha -1}Tu\left(t_1\right)\right|\le {\displaystyle {\int }_{t_1}^{t_2}\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s}$ $\le {\displaystyle {\int }_L^{\infty }\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s<\epsilon .}$

根据引理2.5知， $TV$ 是相对紧的。

第二步：证明 $T:\bar{\Omega }\to X$ 连续。

设 $u_n,u\in \bar{\Omega },n=1,2,\cdots$ ，且当 $n\to +\infty$ 时，有 $\Vert u_n-u\Vert \to 0$ ，则

$\begin{array}{c}\left|\frac{T{u}_n\left(t\right)}{1+t^{\alpha -1}}-\frac{Tu\left(t\right)}{1+t^{\alpha -1}}\right|\le \frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t\left|f\left(s,u_n\left(s\right),D_{0+}^{\alpha -1}{u}_n\left(s\right)\right)\right|\text{d}s}+\frac{1}{\Gamma (\alpha )}{\displaystyle {\int }_0^t\left|f\left(\mathrm{s,u}(s),D_{0+}^{\mathrm{\alpha -1}}u\left(s\right)\right)\right|\text{d}s}\\ +\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}\left|f\left(s,u_n\left(s\right),D_{0+}^{\alpha -1}{u}_n\left(s\right)\right)\right|\text{d}s}+\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s}\\ +\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{\infty }\left|f\left(s,u_n\left(s\right),D_{0+}^{\alpha -1}{u}_n\left(s\right)\right)\right|\text{d}s}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{\infty }\left|f\left(s,u\left(s\right),D_{0+}^{\alpha -1}u\left(s\right)\right)\right|\text{d}s}\end{array}$

$\begin{array}{l}\le \frac{2R}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+\frac{2}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{t}c\left(s\right)\text{d}s}\\ +\frac{2R{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+\frac{2{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}c\left(s\right)\text{d}s}\\ +\frac{2}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{\infty }c\left(s\right)\text{d}s}+\frac{2R}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{\infty }\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}\\ \le \frac{2R}{\Gamma \left(\alpha \right)}A+\frac{2}{\Gamma \left(\alpha \right)}B<+\infty .\end{array}$

且有

$\begin{array}{c}\left|\frac{T{u}_n\left(t\right)}{1+t^{\alpha -1}}-\frac{Tu\left(t\right)}{1+t^{\alpha -1}}\right|\le 2R{\displaystyle {\int }_0^t\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+2{\displaystyle {\int }_0^{t}c\left(s\right)\text{d}s}\\ +\frac{2R{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+2\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}c\left(s\right)\text{d}s}\\ +\frac{2}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{\infty }\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+\frac{2}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{\infty }c\left(s\right)\text{d}s}\\ \le 2RA+2B<+\infty .\end{array}$

根据Lebesgue控制收敛定理， $T$ 连续。

综上， $T$ 全连续。

定理3.1 设(H₁)，(H₂)成立，则边值问题(1.1) (1.2)至少有一个解 $u\in X$ 。

证明： $\Omega$ 如引理3.2中定义。设 $u\in \partial \Omega ,\lambda >1,s.t.Tu=\lambda u$ ，则

$\begin{array}{c}\frac{\left|Tu\left(t\right)\right|}{1+t^{\alpha -1}}\le \frac{1}{\Gamma \left(\alpha \right)}\Vert u\Vert {\displaystyle {\int }_0^t\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{t}c\left(s\right)\text{d}s}\\ +\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\Vert u\Vert {\displaystyle {\int }_0^{\infty }\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{\infty }c\left(s\right)\text{d}s}\\ +\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\Vert u\Vert {\displaystyle {\int }_0^{{\xi }_i}\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}c\left(s\right)\text{d}s}\end{array}$

$\begin{array}{l}\le \left(\frac{1}{\Gamma \left(\alpha \right)}+\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\right)\Vert u\Vert {\displaystyle {\int }_0^{\infty }\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}\\ +\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\Vert u\Vert {\displaystyle {\int }_0^{{\xi }_i}\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}\\ +\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}c\left(s\right)\text{d}s}+\left(\frac{1}{\Gamma \left(\alpha \right)}+\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\right){\displaystyle {\int }_0^{\infty }c\left(s\right)\text{d}s}\\ \le AR+B\le A\frac{B}{1-A}+B=R,\end{array}$

且有

$\begin{array}{c}\left|D_{0+}^{\alpha -1}Tu\left(t\right)\right|\le \Vert u\Vert {\displaystyle {\int }_0^t\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{t}c\left(s\right)\text{d}s}\\ +\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\Vert u\Vert {\displaystyle {\int }_0^{\infty }\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{\infty }c\left(s\right)\text{d}s}\\ +\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\Vert u\Vert {\displaystyle {\int }_0^{{\xi }_i}\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}+\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}c\left(s\right)\text{d}s}\end{array}$

$\begin{array}{l}\le \left(1+\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\right)\Vert u\Vert {\displaystyle {\int }_0^{\infty }\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}\\ +\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\Vert u\Vert {\displaystyle {\int }_0^{{\xi }_i}\left(\left(1+s^{\alpha -1}\right)a\left(s\right)+b\left(s\right)\right)\text{d}s}\\ +\frac{{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}c\left(s\right)\text{d}s}+\left(1+\frac{1}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}\right){\displaystyle {\int }_0^{\infty }c\left(s\right)\text{d}s}\\ \le AR+B=R.\end{array}$

故

$\lambda R=\lambda \Vert u\Vert =\Vert Tu\Vert =\max \left\{\sup \frac{\left|Tu\left(t\right)\right|}{1+t^{\alpha -1}},\sup \left|D_{0+}^{\alpha -1}Tu\left(t\right)\right|\right\}\le R.$

得到 $\lambda \le 1,$ 这与 $\lambda >1$ 矛盾。根据引理1.1， $\bar{\Omega }$ 中存在一个不动点。故问题(1.1) (1.2)在 $X$ 中至少有一个解。

定理3.2 设(H₁)，(H₂)以及下面的(H₃)成立，则边值问题(1.1) (1.2)有唯一解 $u\in X$ 。

(H₃)存在非负函数 $l_1\left(t\right),l_2\left(t\right)\in L^1\left(J\right)$ 使得

$\left|f\left(t,x_1,y_1\right)-f\left(t,x_2,y_2\right)\right|\le l_1\left(t\right)\left|x_1-x_2\right|+l_2\left(t\right)\left|y_1-y_2\right|$

及

${\displaystyle {\int }_0^{\infty }\max \left\{\left(1+t^{\alpha -1}\right)l_1\left(t\right),l_2\left(t\right)\right\}\text{d}t}<\Gamma \left(\alpha \right),{\displaystyle {\int }_0^{{\xi }_i}\max \left\{\left(1+t^{\alpha -1}\right)l_1\left(t\right),l_2\left(t\right)\right\}\text{d}t}<+\infty ,$

$k=2P{\displaystyle {\int }_0^{\infty }\max \left\{\left(1+t^{\alpha -1}\right)l_1\left(t\right),l_2\left(t\right)\right\}\text{d}t}+2Q{\displaystyle {\int }_0^{{\xi }_i}\max \left\{\left(1+t^{\alpha -1}\right)l_1\left(t\right),l_2\left(t\right)\right\}\text{d}t}<1,$

成立。其中 $P=\frac{1}{\Gamma \left(\alpha \right)}+\frac{2}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}+1,Q=\frac{2\left|{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right|}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}$ 。

证明：对 $\forall u_1,u_2\in X$ ，有

$\begin{array}{c}\Vert T{u}_1-T{u}_2\Vert \le \underset{t\in J}{\sup }\frac{\left|T{u}_1-T{u}_2\right|}{1+t^{\alpha -1}}+\underset{t\in J}{\sup }\left|D_{0+}^{\alpha -1}T{u}_1-D_{0+}^{\alpha -1}T{u}_2\right|\\ \le \frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t\left|f\left(s,u_1\left(s\right),D_{0+}^{\alpha -1}{u}_1\left(s\right)\right)-f\left(s,u_2\left(s\right),D_{0+}^{\alpha -1}{u}_2\left(s\right)\right)\right|\text{d}s}\\ +\frac{2{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{{\xi }_i}\left|f\left(s,u_1\left(s\right),D_{0+}^{\alpha -1}{u}_1\left(s\right)\right)-f\left(s,u_2\left(s\right),D_{0+}^{\alpha -1}{u}_2\left(s\right)\right)\right|\text{d}s}\\ +\frac{2}{\Gamma \left(\alpha \right)\left(1-{\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i}\right)}{\displaystyle {\int }_0^{\infty }\left|f\left(s,u_1\left(s\right),D_{0+}^{\alpha -1}{u}_1\left(s\right)\right)-f\left(s,u_2\left(s\right),D_{0+}^{\alpha -1}{u}_2\left(s\right)\right)\right|\text{d}s}\\ +{\displaystyle {\int }_0^t\left|f\left(s,u_1\left(s\right),D_{0+}^{\alpha -1}{u}_1\left(s\right)\right)-f\left(s,u_2\left(s\right),D_{0+}^{\alpha -1}{u}_2\left(s\right)\right)\right|\text{d}s}\end{array}$

$\begin{array}{l}\le P{\displaystyle {\int }_0^{\infty }\left|f\left(s,u_1\left(s\right),D_{0+}^{\alpha -1}{u}_1\left(s\right)\right)-f\left(s,u_2\left(s\right),D_{0+}^{\alpha -1}{u}_2\left(s\right)\right)\right|\text{d}s}\\ +Q{\displaystyle {\int }_0^{{\xi }_i}\left|f\left(s,u_1\left(s\right),D_{0+}^{\alpha -1}{u}_1\left(s\right)\right)-f\left(s,u_2\left(s\right),D_{0+}^{\alpha -1}{u}_2\left(s\right)\right)\right|\text{d}s}\\ \le 2P\Vert u_1-u_2\Vert {\displaystyle {\int }_0^{\infty }\max \left\{\left(1+t^{\alpha -1}\right)l_1\left(t\right),l_2\left(t\right)\right\}\text{d}t}\end{array}$

$\begin{array}{l}\le 2Q\Vert u_1-u_2\Vert {\displaystyle {\int }_0^{{\xi }_i}\max \left\{\left(1+t^{\alpha -1}\right)l_1\left(t\right),l_2\left(t\right)\right\}\text{d}t}\\ \le \{2P{\displaystyle {\int }_0^{\infty }\max \left\{\left(1+t^{\alpha -1}\right)l_1\left(t\right),l_2\left(t\right)\right\}\text{d}t}+2Q{\displaystyle {\int }_0^{{\xi }_i}\max \left\{\left(1+t^{\alpha -1}\right)l_1\left(t\right),l_2\left(t\right)\right\}\text{d}t}\}\Vert u_1-u_2\Vert \\ =k\Vert u_1-u_2\Vert .\end{array}$

因为 $k<1$ ，所以 $T$ 收敛。根据Banach不动点定理，可得 $T$ 有唯一不动点。故边值问题(1.1) (1.2)有唯一解。

4. 应用

例4.1 考虑边值问题

$D_{0+}^{\frac{5}{2}}u\left(t\right)=\frac{\sqrt{\left|u\left(t\right)D_{0+}^{\frac{3}{2}}u\left(t\right)\right|}}{4{\text{e}}^{t^{\frac{3}{2}}}},t\in J:=\left(0,\infty \right),$

$u\left(0\right)=D_{0+}^{\frac{1}{2}}u\left(0\right)=0,D_{0+}^{\frac{3}{2}}u\left(\infty \right)={\displaystyle \underset{i=1}{\overset{m}{\sum }}{\alpha }_i{D}_{0+}^{\frac{3}{2}}u\left({\xi }_i\right),}$

其中， $\alpha =\frac{5}{2},f\left(t,x,y\right)=\frac{\sqrt{\left|u\left(t\right)D_{0+}^{\frac{3}{2}}u\left(t\right)\right|}}{4{\text{e}}^{t^{\frac{3}{2}}}}.$

因为

$\left|f\left(t,x,y\right)\right|\le \frac{1}{8{\text{e}}^{t^{\frac{3}{2}}}}\left|x\right|+\frac{1}{8{\text{e}}^{t^{\frac{3}{2}}}}\left|y\right|,$

通过简单计算可知，

${\displaystyle {\int }_0^{\infty }\left(\left(1+s^{\frac{3}{2}}\right)\frac{1}{8{\text{e}}^{t^{\frac{3}{2}}}}+\frac{1}{8{\text{e}}^{t^{\frac{3}{2}}}}\right)\text{d}s}=\frac{1}{3}\Gamma \left(\frac{5}{3}\right)<\Gamma \left(\frac{2n-1}{2}\right)=\frac{3}{4}\Gamma \left(\frac{1}{2}\right)=\frac{3}{4}\sqrt{\text{π}}.$

故定理3.1的条件成立。所以边值问题(1.1) (1.2)至少有一个解。

基金项目

安徽省自然科学基金项目支持(KJ2016A071)；校质量工程项目支持(2016xjjyxm04)。