上海理工大学，理学院，上海

收稿日期：2022年2月28日；录用日期：2022年3月23日；发布日期：2022年3月30日

摘要

本文中，我们研究了一类带有非瞬时脉冲的分数阶迭代微分方程边值问题，运用Schauder不动点定理证明了解的存在性结果，利用压缩映射原理证明了解的唯一性。

关键词

迭代微分方程，非瞬时脉冲，压缩映射，不动点

Existence and Uniqueness of Positive Solutions for Non-Instantaneous Impulsive Fractional Iterative Differential Equations

Lili Cao, Xiping Liu

College of Science, University of Shanghai for Science and Technology, Shanghai

Received: Feb. 28^th, 2022; accepted: Mar. 23^rd, 2022; published: Mar. 30^th, 2022

ABSTRACT

In this paper, we study a class of boundary value problems for fractional iterative differential equations with non-instantaneous impulses. The existence of the solutions is obtained by using Schauder fixed point theorem. The uniqueness of the solutions is obtained by using the principle of contraction mapping.

Keywords:Iterative Differential Equation, Non-Instantaneous Impulses, Contraction Mapping, Fixed Point

Copyright © 2022 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

反馈是自动控制系统中常见的现象，在电动力学、流行病学、生物学中的许多研究都与迭代微分方程密切相关。近年来迭代整数阶微分方程已经被许多研究者研究，Buica等在文 [1] 中研究了一类泛函微分方程解的存在性和连续依赖性问题

$\{\begin{array}{l}{u}^{\prime }\left(x\right)=f\left(x,u\left(u\left(x\right)\right)\right),a\le x\le b,\\ u\left(x_0\right)=u_0,\end{array}$

其中 $x_0,u_0\in \left[a,b\right]$,$f\in C^1\left(\left[a,b\right],\left[a,b\right]\right)$，用Schauder不动点定理和压缩映射原理，得到了其解的存在性和唯一性，Kaufmann等在文 [2] 中研究了一类二阶迭代边值问题

$\{\begin{array}{l}{x}^{″}\left(t\right)=f\left(t,x\left(x\left(t\right)\right)\right),a\le t\le b,\\ x\left(a\right)=a,x\left(b\right)=b,\end{array}$

应用Schauder不动点定理得到了解的存在性。

分数阶微分方程是微分方程理论中的一个重要研究分支，一直受到人们的关注。它们作为一种有价值的工具出现在科学和工程各个领域的许多现象的模型中。我们可以在物理、化学、生物学等领域找到许多应用。作为整数阶的推广，分数阶微分方程可以更准确的描述某些事物发展的过程，一些杰出的专著为分数阶微分方程的定性分析提供了主要的理论工具，同时也展示了整数阶微分模型与分数阶微分模型之间的相互联系和对比 [3] [4]。Wang等在文 [5] 中研究了一类具有参数线性修正迭代的分数阶微分方程

$\{\begin{array}{l}{D}_t^{q}x\left(t\right)=f\left(t,x\left(t\right),x\left(\lambda t\right),x\left(\lambda x\left(\lambda t\right)\right)\right),0<\lambda <1,t\in \left[0,1\right],\\ x\left(0\right)=0,x^{\prime }\left(0\right)=0,\end{array}$

其中 $D_t^q$ 是Caputo导数 $1<q<2$，f是Carathéodory函数。应用Picard算子理论研究了解的存在唯一性。Deng等在文 [6] 中研究了分数阶迭代微分方程

$D_t^{q}x\left(t\right)=f\left(t,x\left(t\right),x\left(x^v\left(t\right)\right)\right),v\in R\backslash \left\{0\right\},t\in J:=\left[a,b\right],1<q<2,$

$x\left(a\right)=x_0,x^{\prime }\left(a\right)=0,$

其中 $x_0\in J$。应用非扩张映射和不动点方法研究了分数阶迭代微分方程解的存在性和逼近性。利用Chebyshev范数、Bielecki范数，在两个不同的工作空间中得到了近似解的不动点迭代法的存在性定理和收敛性定理。

受以上问题及文 [7] 的启发，本文主要研究一类非瞬时脉冲分数阶微分方程局部边值问题

$\{\begin{array}{l}{D}_{s_i^{+}}^{\alpha }u\left(t\right)=f\left(t,u\left(t\right),u\left(u\left(t\right)\right)\right),t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m,\\ u\left(t\right)=g_i\left(t\right),t\in \left(t_i,s_i\right],i=1,2,\cdots ,m,\\ \Delta u\left(s_i\right)=0,\Delta u\left(t_i\right)=Q_i\left(t_i,u\left(t_i\right)\right),i=1,2,\cdots ,m,\\ u\left(0\right)=0,u\left(1\right)=u_0,\end{array}$ (1.1)

的解的存在性和唯一性，其中 $D_{s_i^{+}}^{\alpha }$ 为Caputo型分数阶导数， $1<\alpha <2$，令 $J=\left[0,1\right]$，$J_0=J\backslash \left\{t_1,t_2,\cdots ,t_m\right\}$，

$0=s_0<t_1<s_1<\cdots <s_i<t_{i+1}<\cdots <t_{m+1}=1$，$f\in C\left(J^3,ℝ\right)$，$Q_i\in C\left(J^2,{ℝ}^{-}\right)$，$i=1,2,\cdots ,m$。 $u\left(s_i^{+}\right),u\left(s_i^{-}\right)$ 分别是 $u\left(t\right)$ 在 $s_i$ 的右极限和左极限， $u\left(t_i^{+}\right),u\left(t_i^{-}\right)$ 分别是 $u\left(t\right)$ 在 $t_i$ 的右极限和左极限， $\Delta u\left(s_i\right)=u\left(s_i^{+}\right)-u\left(s_i^{-}\right)$，$\Delta u\left(t_i\right)=u\left(t_i^{+}\right)-u^{\prime }\left(t_i^{-}\right)$。

现有的研究迭代微分方程的文章大多数集中于研究整数阶的迭代微分方程，研究分数阶迭代微分方程的文章很少，研究带有非瞬时脉冲的分数阶微分方程更不多见。分数阶迭代微分方程能更准确的描述某些事物发展过程中的变化趋势，比如在流行病治疗过程中药物对病情的影响，所以研究这个是非常有意义的。

本文的主要结构如下：在第二部分我们给出了一些关于分数阶微分方程的引理以及要用到的不动点定理；在第三部分证明了解的存在唯一性结果。

2. 准备工作

在这一部分，我们回忆一些关于分数阶微分积分定义。

定义2.1 (见 [3])连续函数 $u:\left(a,+\infty \right)\to ℝ$ 的 $\alpha >0$ 阶Riemann-Liouvile型分数阶积分定义为

$I_{a^{+}}^{\alpha }u\left(t\right)=\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_a^t{\left(t-s\right)}^{\alpha -1}u\left(s\right)\text{d}s},t>a,$

其中 $\text{Γ}\left(\alpha \right)$ 为Gamma函数， $n-1<\alpha <n$。

定义2.2 (见 [3])连续函数 $u:\left(a,+\infty \right)\to R$ 的 $\alpha >0$ 阶Caputo型分数阶导数定义为

$D_{a^{+}}^{\alpha }u\left(t\right)=\frac{1}{\text{Γ}\left(n-\alpha \right)}{\displaystyle {\int }_a^t{\left(t-s\right)}^{n-\alpha -1}{u}^{\left(n\right)}\left(s\right)\text{d}s},t>a,$

其中 $\text{Γ}\left(n-\alpha \right)$ 为Gamma函数， $n-1<\alpha <n$。

引理2.1 (见 [3])设 $\alpha >0,u\in C\left(0,1\right)\cap L\left(0,1\right)$ 则

$I_{a^{+}}^{\alpha }\left(D_{a^{+}}^{\alpha }\right)u\left(t\right)=u\left(t\right)+c_0+c_{1}t+c_2{t}^2+c_{n-1}{t}^{n-1},$

其中 $c_i\in R\left(i=0,1,2,\cdots ,n-1\right)$，$n-1<\alpha <n$。

引理2.2 (Schauder不动点定理) (见 [8])设E是一个实赋范线性空间D是Banach空间E中的一个非空有界闭凸子集，又设映射 $A:D\to D$ 是全连续算子，则A在D中必有不动点。

引理2.3 (Banach压缩映像原理) (见 [9])设E是一个Banach 空间， $\Omega \subseteq E$ 是一个非空闭集， $A:\Omega \to \Omega$，若存在 $\beta \in \left[0,1\right)$ 使得对任意的 $u,v\in \Omega$ 有 $\Vert Au-Av\Vert \le \beta \Vert u-v\Vert$，则有唯一一个元素 $u^{*}\in \Omega$ 使得 $A{u}^{*}=u^{*}$ 即 $u^{*}$ 是算子A的唯一不动点。

为了方便，记 $g_0\left(0\right)=g_0\left(s_0\right)=0$，$g_{m+1}\left(1\right)=u_0$，$q_{m+1}=0$。

引理2.4 设 $h\in C\left(J,ℝ\right)$，$g_i\in C\left(J,J\right)$，$q_i\in ℝ$，$i=1,2,\cdots ,m$，那么线性分数阶非瞬时脉冲微分方程非局部边值问题

$\{\begin{array}{l}{D}_{s_i^{+}}^{\alpha }u\left(t\right)=h\left(t\right),t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m,\\ u\left(t\right)=g_i\left(t\right),t\in \left(t_i,s_i\right],i=1,2,\cdots ,m,\\ \Delta u\left(s_i\right)=0,\Delta u\left(t_i\right)=Q_i\left(t_i,u\left(t_i\right)\right),i=1,2,\cdots ,m,\\ u\left(0\right)=0,u\left(1\right)=u_0,\end{array}$ (2.1)

有唯一解

$u\left(t\right)=\{\begin{array}{l}\frac{1}{t_{i+1}-s_i}(g_{i+1}\left(t_{i+1}\right)\left(t-s_i\right)+g_i\left(s_i\right)\left(t_{i+1}-t\right)+\frac{1}{\Gamma \left(\alpha \right)}({\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right){\left(t-s\right)}^{\alpha -1}h\left(s\right)\text{d}s}\\ \stackrel{{}^{}}{}-{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right){\left(t_{i+1}-s\right)}^{\alpha -1}h\left(s\right)\text{d}s})-q_{i+1}\left(t-s_i\right)),t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m,\\ g_i\left(t\right),t\in \left(t_i,s_i\right],i=1,2,\cdots ,m.\end{array}$

证明：假设 $u=u\left(t\right)$ 是问题(2.1)的解，则存在常数 $c_{i1},c_{i2},i=0,1,\cdots ,m$，使得对任意的 $t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m$，有

$u\left(t\right)=c_{i1}+c_{i2}t+\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_i}^t{\left(t-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s.$ (2.2)

对任意的 $t\in \left(t_i,s_i\right],i=1,2,\cdots ,m$，有

$u\left(t\right)=g_i\left(t\right),$

$u\left(t_i^{+}\right)=g_i\left(t_i\right).$ (2.3)

对任意的 $t\in \left[0,t_1\right]$ 有

$u\left(t\right)=c_{01}+c_{02}t+\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s,$ (2.4)

将初值条件 $u\left(0\right)=0$ 代入(2.4)得 $c_{01}=0$，

$u\left(t\right)=c_{02}t+\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s,$ (2.5)

$u\left(t_1^{-}\right)=c_{02}{t}_1+\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s,$

由 $u\left(t_1^{+}\right)-u\left(t_1^{-}\right)=q_1$ 结合(2.3)得

$c_{02}=\frac{1}{t_1}\left(g_1\left(t_1\right)-\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_0^{t_1}{\left(t_1-s\right)}^{\alpha -1}h\left(s\right)\text{d}s}-q_1\right),$

将上式代入(2.5)得

$u\left(t\right)=\frac{1}{t_1}\left(g_1\left(t_1\right)t-\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_0^{t_1}t}{\left(t_1-s\right)}^{\alpha -1}h\left(s\right)\text{d}s-q_{1}t\right)+\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s.$

对任意的 $t\in \left(s_1,t_2\right]$ 有

$u\left(t\right)=c_{11}+c_{12}t+\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_1}^t{\left(t-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s,$

由 $u\left(s_1\right)=g_1\left(s_1\right)$ 得 $c_{11}=g_1\left(s_1\right)-c_{12}{s}_1$，所以

$u\left(t\right)=g_1\left(s_1\right)+c_{12}\left(t-s_1\right)+\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_1}^t{\left(t-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s,$ (2.6)

$u\left(t_2^{-}\right)=g_1\left(s_1\right)+c_{12}\left(t_2-s_1\right)+\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_1}^{t_2}{\left(t_2-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s,$

由 $u\left(t_2^{+}\right)-u\left(t_2^{-}\right)=q_2$，结合(2.3)得

$c_{12}=\frac{1}{t_2-s_1}\left(g_2\left(t_2\right)-\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_1}^{t_2}{\left(t_2-s\right)}^{\alpha -1}h}\left(s\right)\text{d}s-g_1\left(s_1\right)-q_2\right),$

将上式代入(2.5)整理得

$\begin{array}{c}u\left(t\right)=\frac{1}{t_2-s_1}(g_2\left(t_2\right)\left(t-s_1\right)+g_1\left(s_1\right)\left(t_2-t\right)+\frac{1}{\text{Γ}\left(\alpha \right)}({\displaystyle {\int }_{s_1}^t\left(t_2-s_1\right)}{\left(t-s\right)}^{\alpha -1}h\left(s\right)\text{d}s\\ \stackrel{{}^{}}{}-{\displaystyle {\int }_{s_1}^{t_2}\left(t-s_1\right)}{\left(t_2-s\right)}^{\alpha -1}h\left(s\right)\text{d}s)-q_2\left(t-s_1\right)),\end{array}$

重复上述过程可得对任意的 $t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m$ 有

$\begin{array}{c}u\left(t\right)=\frac{1}{t_{i+1}-s_i}(g_{i+1}\left(t_{i+1}\right)\left(t-s_i\right)+g_i\left(s_i\right)\left(t_{i+1}-t\right)+\frac{1}{\Gamma \left(\alpha \right)}({\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right){\left(t-s\right)}^{\alpha -1}h\left(s\right)\text{d}s}\\ \stackrel{{}^{}}{}-{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right){\left(t_{i+1}-s\right)}^{\alpha -1}h\left(s\right)\text{d}s})-q_{i+1}\left(t-s_i\right)).\end{array}$

综上所述结论得证。

记 $Q_{m+1}\left(t_{m+1},u\left(t_{m+1}\right)\right)=0$，由上述引理的证明过程可知下面引理成立。

引理2.5 若 $f\in C\left(J\times J\times J,ℝ\right)$，$g_i\in C\left(J,J\right),i=0,1,2,\cdots ,m$，那么问题(1.1)等价于以下积分方程

$u\left(t\right)=\{\begin{array}{l}\frac{1}{t_{i+1}-s_i}(g_i\left(s_i\right)\left(t_{i+1}-t\right)+\frac{1}{\text{Γ}\left(\alpha \right)}({\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right){\left(t-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s}\\ -{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right){\left(t_{i+1}-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s})\\ \stackrel{{}^{}}{}+\left(g_{i+1}\left(t_{i+1}\right)-Q_{i+1}\left(t_{i+1},u\left(t_{i+1}\right)\right)\right)\left(t-s_i\right)),t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m,\\ g_i\left(t\right),t\in \left(t_i,s_i\right],i=1,2,\cdots ,m.\end{array}$

定义空间 $PC\left(J,ℝ\right)=\left\{u:J\to ℝ|u\in C\left(J_0,ℝ\right),u\left(t_i^{+}\right)和u\left(t_i^{-}\right)存在且u\left(t_i\right)=u\left(t_i^{-}\right),i=1,2,\cdots ,m\right\}$ 为分段连续函数空间，并赋予范数 $\Vert u\Vert ={\sup }_{t\in \left[0,1\right]}\left|u\left(t\right)\right|$。

取 ${\eta }_1=\frac{2M_f}{\Gamma \left(\alpha \right)}{t}_1^{\alpha -1}+\frac{M_g+M_Q}{t_1}$,${\eta }_3=\frac{2M_f}{\Gamma \left(\alpha \right)}{\gamma }_1^{\alpha -1}+\frac{M_g+M_Q}{{\gamma }_2}$，设有 ${\eta }_2\ge 0$ 使得对任意的 ${t^{\prime }}_1,{t^{\prime }}_2\in \left(t_i,s_i\right]$，

$i=1,2,\cdots ,m$，有 $\left|g_i\left({t^{\prime }}_1\right)-g_i\left({t^{\prime }}_2\right)\right|\le {\eta }_2\left|{t^{\prime }}_1-{t^{\prime }}_2\right|$，取 $\eta =\max \left\{{\eta }_1,{\eta }_2,{\eta }_3\right\}$。

记集合： $P{C}_{\eta }\left(J,J\right)=\{u\in PC\left(J,ℝ\right)|0\le u\left(t\right)\le 1$，$\left|u\left({\tau }_2\right)-u\left({\tau }_1\right)\right|\le \eta \left|{\tau }_2-{\tau }_1\right|$，任意 ${\tau }_1,{\tau }_2\in \left[0,t_1\right],\left(s_i,t_{i+1}\right]$ 或 $\left(t_i,s_i\right],i=1,2,\cdots ,m\}$。

定义算子 $T:P{C}_{\eta }\left(J,J\right)\to PC\left(J,ℝ\right)$ ：

$Tu\left(t\right)=\{\begin{array}{l}\frac{1}{t_{i+1}-s_i}(g_i\left(s_i\right)\left(t_{i+1}-t\right)+\frac{1}{\text{Γ}\left(\alpha \right)}({\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right){\left(t-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s}\\ -{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right){\left(t_{i+1}-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s})\\ \stackrel{{}^{}}{}+\left(g_{i+1}\left(t_{i+1}\right)-Q_{i+1}\left(t_{i+1},u\left(t_{i+1}\right)\right)\right)\left(t-s_i\right)),t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m,\\ g_i\left(t\right),t\in \left(t_i,s_i\right],i=1,2,\cdots ,m.\end{array}$

为了方便记

$M_f=\underset{\left(t,u,v\right)\in J\times J\times J}{\max }f\left(t,u,v\right),M_g=\underset{1\le i\le m}{\max }\left\{\underset{t\in \left(t_i,s_i\right]}{\sup }{g}_i\left(t\right)\right\},M_Q=\underset{1\le i\le m}{\max }\left\{\underset{(t,u)\in J\times J}{\sup }\left|Q_{i+1}\left(t,u\right)\right|\right\},$

$M=\max \left\{0,M_f\right\},m=\min \left\{0,m_f\right\},{\gamma }_1=\underset{0\le i\le m}{\max }\left\{t_{i+1}-s_i\right\},{\gamma }_2=\underset{0\le i\le m}{\max }\left\{t_{i+1}-s_i\right\}\text{.}$

引理2.6 算子 $T:P{C}_{\eta }\left(J,J\right)\to PC\left(J,ℝ\right)$ 全连续

证明：首先，证明T为连续算子。设 $u_n,u\in P{C}_{\eta }\left(J,J\right)\left(n=1,2,\cdots \right)$，且 $\Vert u_n-u\Vert \to 0\left(n\to \infty \right)$，即

对任意的 $t\in J$，当 $n\to \infty$ 时有 $u_n\left(t\right)\to u\left(t\right)$。由于 $f,Q_i$ 是连续函数，从而满足 $n\to \infty$ 时， $\left|f\left(s,u_n\left(s\right),u_n\left(u_n\left(s\right)\right)\right)-f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\right|\to 0$，$\left|Q_i\left(t_i{u}_n\left(t_i\right)\right)-Q_i\left(t_i,u\left(t_i\right)\right)\right|\to 0\left(i=0,1,2,\cdots ,m\right)$。由

Lebesgue控制收敛定理可以知道，当 $n\to \infty$ 时 ${\displaystyle {\int }_{s_i}^t\left|f\left(s,u_n\left(s\right),u_n\left(u_n\left(s\right)\right)\right)-f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\right|}\to 0$。因此，

当 $t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m$ 时可得

$\begin{array}{c}\left|T{u}_n\left(t\right)-Tu\left(t\right)\right|\le \frac{1}{\Gamma \left(\alpha \right)}({\displaystyle {\int }_{s_i}^t{\left(t-s\right)}^{\alpha -1}}\left|f\left(s,u_n\left(s\right),u_n\left(u_n\left(s\right)\right)\right)-f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\right|\text{d}s\\ +{\displaystyle {\int }_{s_i}^{t_{i+1}}{\left(t_{i+1}-s\right)}^{\alpha -1}}\left|f\left(s,u_n\left(s\right),u_n\left(u_n\left(s\right)\right)\right)-f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\right|\text{d}s\\ +\left|Q_{i+1}\left(t_{i+1},u_n\left(t_{i+1}\right)\right)-Q_{i+1}\left(t_{i+1},u\left(t_{i+1}\right)\right)\right|\\ +\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_{s_i}^t{\left(t-s\right)}^{\alpha -1}}\left|f\left(s,u_n\left(s\right),u_n\left(u_n\left(s\right)\right)\right)-f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\right|\text{d}s.\end{array}$

当 $t\in \left(t_i,s_i\right],i=1,2,\cdots ,m$ 时，可得 $\left|T{u}_n\left(t\right)-Tu\left(t\right)\right|=0$。所以， $\Vert T{u}_n\left(t\right)-Tu\left(t\right)\Vert \to 0$,$\left(n\to \infty \right)$，因此T连续。

其次，证明T是紧的，令

$\Omega =\left\{u:J\to J|u\in P{C}_{\eta }\left(J,J\right),\Vert u\Vert \le r\right\}.$

当 $t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m$ 时，对任意的 $u\in \Omega$ 有

$\begin{array}{c}\left|Tu\left(t\right)\right|=\frac{1}{t_{i+1}-s_i}|g_{i+1}\left(t_{i+1}\right)\left(t-s_i\right)+g_i\left(s_i\right)\left(t_{i+1}-t\right)+\frac{1}{\text{Γ}\left(\alpha \right)}({\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right)}{\left(t-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s\\ -{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s)-Q_{i+1}\left(t_{i+1},u\left(t_{i+1}\right)\right)\left(t-s_i\right)|\\ \le \frac{1}{t_{i+1}-s_i}(M_g\left(t_{i+1}-s_i\right)+|\frac{1}{\text{Γ}\left(\alpha \right)}({\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right)}{\left(t-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s\\ \stackrel{{}^{}}{}-{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s|-M_Q\left(t-s_i\right))\end{array}$

$\begin{array}{l}\le M_g+M_Q+\frac{1}{t_{i+1}-s_i}\frac{1}{\text{Γ}\left(\alpha \right)}\left|M{\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right){\left(t-s\right)}^{\alpha -1}\text{d}s}-m{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}\text{d}s\right|\\ \le M_g+M_Q+\frac{1}{\text{Γ}\left(\alpha \right)}\left|M{\displaystyle {\int }_{s_i}^t{\left(t-s\right)}^{\alpha -1}\text{d}s}-m{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}\text{d}s\right|\\ \le M_g+M_Q+\frac{{\left(t_{i+1}-s_i\right)}^{\alpha }}{\text{Γ}\left(\alpha \right)}\left(M-m\right).\end{array}$

当 $t\in \left(t_i,s_i\right],i=1,2,\cdots ,m$ 时，对任意的 $u\in \Omega$ 有 $\left|Tu\left(t\right)\right|=\left|g_i\left(t\right)\right|\le M_g$。因此， $T\left(\Omega \right)$ 一致有界。

对任意的 $ϵ>0$，取 ${\delta }_1=\frac{\Gamma (\alpha ){\gamma }_2ϵ}{2M_f{\gamma }_1{}^{\alpha -1}{\gamma }_2+\Gamma \left(\alpha \right)\left(M_g+M_Q\right)}$。由于 $g_i\left(t\right)$ 在 $\left[t_i,s_i\right]$ 上连续，所以 $g_i\left(t\right)$ 在 $\left[t_i,s_i\right]$

上一致连续，从而存在常数 ${\delta }_2>0$ 使得当 ${t^{\prime }}_1<{t^{\prime }}_2\in \left(t_i,s_i\right]$ 且 $\left|{t^{\prime }}_1-{t^{\prime }}_2\right|<{\delta }_2$ 时，有 $\left|g_i\left({t^{\prime }}_2\right)-g_i\left({t^{\prime }}_1\right)\right|<ϵ$，

取 $\delta =\min \left\{{\delta }_1,{\delta }_2\right\}$。

当 ${t^{\prime }}_1<{t^{\prime }}_2\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m$，且 $\left|{t^{\prime }}_1-{t^{\prime }}_2\right|<\delta$ 时，对任意的 $u\in \Omega$ 有

$\begin{array}{l}\left|Tu\left({t^{\prime }}_1\right)-Tu\left({t^{\prime }}_2\right)\right|\\ \le \frac{1}{t_{i+1}-s_i}(\left(g_{i+1}\left(t_{i+1}\right)+g_i\left(s_i\right)\right)\left|{t^{\prime }}_1-{t^{\prime }}_2\right|+\frac{1}{\Gamma \left(\alpha \right)}(|{\displaystyle {\int }_{s_i}^{{t^{\prime }}_2}\left(t_{i+1}-s_i\right)}{\left({t^{\prime }}_2-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s\\ -{\displaystyle {\int }_{s_i}^{{t^{\prime }}_1}\left(t_{i+1}-s_i\right)}{\left({t^{\prime }}_1-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s|+|{\displaystyle {\int }_{s_i}^{t_{i+1}}\left({t^{\prime }}_2-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s\\ \stackrel{{}^{}}{}-{\displaystyle {\int }_{s_i}^{t_{i+1}}\left({t^{\prime }}_1-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)\text{d}s|)+\left|Q_{i+1}\left(t_{i+1},u\left(t_{i+1}\right)\right)\right|\left|{t^{\prime }}_1-{t^{\prime }}_2\right|)\\ \le \frac{1}{t_{i+1}-s_i}(\left(2M_g+M_Q\right)\left|{t^{\prime }}_1-{t^{\prime }}_2\right|+\frac{M}{\Gamma \left(\alpha \right)}(|{\displaystyle {\int }_{s_i}^{{t^{\prime }}_1}\left(t_{i+1}-s_i\right)}\left({\left({t^{\prime }}_2-s\right)}^{\alpha -1}-{\left({t^{\prime }}_1-s\right)}^{\alpha -1}\right)\text{d}s\end{array}$

$\begin{array}{l}\stackrel{{}^{}}{}-{\displaystyle {\int }_{{t^{\prime }}_1}^{{t^{\prime }}_2}\left(t_{i+1}-s_i\right)}{\left({t^{\prime }}_2-s\right)}^{\alpha -1}\text{d}s|+{\displaystyle {\int }_{s_i}^{t_{i+1}}{\left(t_{i+1}-s\right)}^{\alpha -1}\text{d}s}\left|{t^{\prime }}_1-{t^{\prime }}_2\right|))\\ \le \frac{1}{t_{i+1}-s_i}\left(2M_g+M_Q\right)\left|{t^{\prime }}_1-{t^{\prime }}_2\right|+\frac{M}{\Gamma \left(\alpha \right)}(\left|{\displaystyle {\int }_{s_i}^{{t^{\prime }}_1}\left({\left({t^{\prime }}_2-s\right)}^{\alpha -1}-{\left({t^{\prime }}_1-s\right)}^{\alpha -1}\right)\text{d}s}+{\displaystyle {\int }_{{t^{\prime }}_1}^{{t^{\prime }}_2}{\left({t^{\prime }}_2-s\right)}^{\alpha -1}\text{d}s}\right|\stackrel{{}^{}}{}\\ +\frac{1}{t_{i+1}-s_i}\underset{s_i}{\overset{t_{i+1}}{{\displaystyle \int }}}{\left(t_{i+1}-s\right)}^{\alpha -1}\text{d}s\left|{t^{\prime }}_1-{t^{\prime }}_2\right|)\\ \le \frac{1}{t_{i+1}-s_i}\left(2M_g+M_Q\right)\left|{t^{\prime }}_1-{t^{\prime }}_2\right|+\frac{M}{\Gamma \left(\alpha +1\right)}\left|{\left({t^{\prime }}_2-s_i\right)}^{\alpha }-{\left({t^{\prime }}_1-s_i\right)}^{\alpha }\right|+\frac{M}{\Gamma \left(\alpha +1\right)}{\left(t_{i+1}-s_i\right)}^{\alpha -1}\left|{t^{\prime }}_1-{t^{\prime }}_2\right|\\ \le \left(\frac{2M}{\Gamma \left(\alpha \right)}{\gamma }_1^{\alpha -1}+\frac{2M_g+M_Q}{{\gamma }_2}\right)\left|{t^{\prime }}_1-{t^{\prime }}_2\right|<ϵ.\end{array}$

当 ${t^{\prime }}_1<{t^{\prime }}_2\in \left(t_i,s_i\right],i=1,2,\cdots ,m$ 且 $\left|{t^{\prime }}_1-{t^{\prime }}_2\right|<\delta$ 时，对任意的 $u\in \Omega$ 有