非瞬时脉冲分数阶迭代微分方程正解的存在性和唯一性

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

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。

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.

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

$\left|Tu\left({t^{\prime }}_1\right)-Tu\left({t^{\prime }}_2\right)\right|=\left|g_i\left({t^{\prime }}_2\right)-g_i\left({t^{\prime }}_1\right)\right|<ϵ.$

所以， $T\left(\Omega \right)$ 在J上等度连续，由Arzela-Ascoli定理知T是紧的。综上所述，T是全连续算子。

3. 边值问题解的存在唯一性

记 $m_f=\underset{\left(t,u,v\right)\in J\times J\times J}{\min }f\left(t,u,v\right)$， $m_g=\underset{1\le i\le m}{\min }\left\{\underset{t\in \left(t_i,s_i\right]}{\inf }{g}_i\left(t\right)\right\}$， $m_Q=\underset{1\le i\le m}{\min }\left\{\underset{\left(t,u\right)\in J\times J}{\inf }\left|Q_{i+1}\left(t,u\right)\right|\right\}$。

假设下列条件成立：

(H1)存在常数 $0<N\le 1$ 使得下式成立

$M_g+M_Q+\frac{{\gamma }_1^{\alpha }}{\text{Γ}\left(\alpha +1\right)}\left(M-m\right)\le N.$

(H2) $m_g-\frac{M{\gamma }_1^{\alpha }}{\text{Γ}\left(\alpha +1\right)}+m_Q\ge 0$。

定理3.1 假设条件(H1)(H2)成立，则非瞬时脉冲分数阶迭代微分方程非局部边值问题(1.1)至少存在一个解。

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

由条件(H1)(H2)可知，当 $t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m$ 时，

$\begin{array}{c}Tu\left(t\right)\le \frac{1}{t_{i+1}-s_i}(\frac{1}{\Gamma \left(\alpha \right)}\left(M\left(t_{i+1}-s_i\right){\displaystyle {\int }_{s_i}^t{\left(t-s\right)}^{\alpha -1}\text{d}s}-m\left(t-s_i\right){\displaystyle {\int }_{s_i}^{t_{i+1}}{\left(t_{i+1}-s\right)}^{\alpha -1}\text{d}s}\right)\\ \stackrel{{}^{}}{}+M_g\left(t_{i+1}-s_i\right)+M_Q\left(t-s_i\right))\\ \le \frac{1}{t_{i+1}-s_i}\left(\frac{1}{\Gamma \left(\alpha +1\right)}\left(M\left(t_{i+1}-s_i\right){\left(t-s_i\right)}^{\alpha -1}-m\left(t-s_i\right){\left(t_{i+1}-s_i\right)}^{\alpha }\right)+M_Q\right)\left(t-s_i\right)+M_g\left(t_{i+1}-s_i\right)\\ \le \frac{{\left(t_{i+1}-s_i\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}\left(M-m\right)+M_g+M_Q\\ \le \frac{{\gamma }_1^{\alpha }}{\Gamma \left(\alpha +1\right)}\left(M-m\right)+M_g+M_Q\le 1,\end{array}$

$\begin{array}{c}Tu\left(t\right)\ge \frac{1}{t_{i+1}-s_i}(\frac{1}{\Gamma \left(\alpha \right)}\left(m\left(t_{i+1}-s_i\right){\displaystyle {\int }_{s_i}^t{\left(t-s\right)}^{\alpha -1}\text{d}s}-M\left(t-s_i\right){\displaystyle {\int }_{s_i}^{t_{i+1}}{\left(t_{i+1}-s\right)}^{\alpha -1}\text{d}s}\right)\\ \stackrel{{}^{}}{}+m_g\left(t_{i+1}-s_i\right)+m_Q\left(t-s_i\right))\\ \ge \frac{1}{t_{i+1}-s_i}\left(\frac{1}{\Gamma \left(\alpha +1\right)}\left(m\left(t_{i+1}-s_i\right){\left(t-s_i\right)}^{\alpha -1}-M{\left(t_{i+1}-s_i\right)}^{\alpha }\right)+m_Q+m_g\right)\left(t-s_i\right)\\ \ge \frac{1}{t_{i+1}-s_i}\left(\frac{-M{\left(t_{i+1}-s_i\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}+m_Q+m_g\right)\left(t-s_i\right)\\ \ge \frac{1}{t_{i+1}-s_i}\left(\frac{-M{\gamma }_1^{\alpha }}{\Gamma \left(\alpha +1\right)}+m_Q+m_g\right)\left(t-s_i\right)\ge 0.\end{array}$

又因为当 $t\in \left(t_i,s_i\right],i=1,2,\cdots ,m$ 时，有 $0\le Tu\left(t\right)=g_i\left(t\right)\le 1$，所以对任意的 $t\in J$ 有

$0\le Tu\left(t\right)\le 1.$

令 ${\Omega }_N=\left\{u:J\to J|u\in P{C}_{\eta }\left(J,J\right),\Vert u\Vert \le N\right\}$，由条件(H1)及引理2.6的证明过程可知对任意的 ${t^{\prime }}_1<{t^{\prime }}_2\in J_0$，任意的 $u\in {\Omega }_N$ 有 $\left|Tu\left({t^{\prime }}_2\right)-Tu\left({t^{\prime }}_1\right)\right|\le \eta \left|{t^{\prime }}_2-{t^{\prime }}_1\right|$。所以 $T:{\Omega }_N\to P{C}_{\eta }\left(J,J\right)$。再由引理2.6可得 $T:{\Omega }_N\to P{C}_{\eta }\left(J,J\right)$ 是全连续算子。

综上所述，由Schauder不动点定理可知算子T 在 $P{C}_{\eta }\left(J,J\right)$ 上有不动点，故边值问题(1.1)至少有一个解。

假设下列条件成立：

(H3)存在常数 $L_1,L_2\ge 0$，使得对任意 $t\in J$ 及任取 $u_1,u_2,v_1,v_2,u,v\in \left[0,1\right]$，有

$\left|f\left(t,u_1,v_1\right)-f\left(t,u_2,v_2\right)\right|\le L_1\left|u_1-u_2\right|,$

$\left|Q_i\left(t,u\right)-Q_i\left(t,v\right)\right|\le L_2\left|u-v\right|,i=1,2,\cdots ,m.$

定理3.2 假设条件(H1)~(H3)成立，且有

$\frac{2L_1{\gamma }_1^{\alpha }}{\text{Γ}\left(\alpha +1\right)}+L_2<1,$

则非瞬时脉冲分数阶迭代微分方程非局部边值问题(1.1)存在唯一解 $u^{*}\in P{C}_{\eta }\left(J,J\right)$。

证明：由定理3.1的条件可知算子 $T:P{C}_{\eta }\left(J,J\right)\to P{C}_{\eta }\left(J,J\right)$ 取 $\beta =\frac{2L_1{\gamma }_1^{\alpha }}{\text{Γ}\left(\alpha +1\right)}+L_2$，对任意的 $u,v\in P{C}_{\eta }\left(J,J\right)$，由条件(H1) (H2)可得，当 $t\in \left(s_i,t_{i+1}\right],i=0,1,\cdots ,m$ 时有

$\begin{array}{l}\left|\left(Tu\right)\left(t\right)-\left(Tv\right)\left(t\right)\right|\\ \le \frac{1}{t_{i+1}-s_i}(\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right)}{\left(t-s\right)}^{\alpha -1}\left|f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)-f\left(s,v\left(s\right),v\left(v\left(s\right)\right)\right)\right|\text{d}s\\ +\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}\left|f\left(s,u\left(s\right),u\left(u\left(s\right)\right)\right)-f\left(s,v\left(s\right),v\left(v\left(s\right)\right)\right)\right|\text{d}s\\ +\left|Q_{i+1}\left(t_{i+1},u\left(t_{i+1}\right)\right)-Q_{i+1}\left(t_{i+1},v\left(t_{i+1}\right)\right)\right|\left(t-s_i\right)\end{array}$

$\begin{array}{l}\le \frac{1}{t_{i+1}-s_i}(\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right)}{\left(t-s\right)}^{\alpha -1}{L}_1\left|u\left(s\right)-v\left(s\right)\right|\text{d}s+L_2\left|u\left(t_{i+1}\right)-v\left(t_{i+1}\right)\right|\left(t-s_i\right)\\ +\frac{1}{\text{Γ}\left(\alpha \right)}{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}{L}_1\left|u\left(s\right)-v\left(s\right)\right|\text{d}s\\ \le \frac{1}{t_{i+1}-s_i}\left(\frac{L_1}{\text{Γ}\left(\alpha \right)}\left({\displaystyle {\int }_{s_i}^t\left(t_{i+1}-s_i\right)}{\left(t-s\right)}^{\alpha -1}\text{d}s+{\displaystyle {\int }_{s_i}^{t_{i+1}}\left(t-s_i\right)}{\left(t_{i+1}-s\right)}^{\alpha -1}\text{d}s\right)+L_2\left(t-s_i\right)\right)\Vert u-v\Vert \\ \le \frac{1}{t_{i+1}-s_i}\left(\frac{L_1}{\text{Γ}\left(\alpha +1\right)}\left(\left(t_{i+1}-s_i\right){\left(t-s_i\right)}^{\alpha }+\left(t-s_i\right){\left(t_{i+1}-s_i\right)}^{\alpha }\right)+L_2\left(t-s_i\right)\right)\Vert u-v\Vert \\ \le \left(\frac{2L_1{\gamma }_1^{\alpha }}{\text{Γ}\left(\alpha +1\right)}+L_2\right)\Vert u-v\Vert .\end{array}$

当 $t\in \left(t_i,s_i\right],i=1,2,\cdots ,m$ 时有 $\left|Tu\left(t\right)-Tv\left(t\right)\right|=0$。

综上所述，对任意的 $t\in J$ 都有 $\Vert Tu-Tv\Vert \le \beta \Vert u-v\Vert$，因为 $\frac{2L_1{\gamma }_1^{\alpha }}{\text{Γ}\left(\alpha +1\right)}+L_2<1$，所以 $\beta \in \left[0,1\right)$，从而可得

T为压缩映射。由Banach压缩映射原理可得，存在唯一一个元素 $u^{*}\in \Omega$，使得 $T{u}^{*}=u^{*}$，即问题(1.1)存在唯一解。证毕。

4. 结论

本文首先得到了所研究问题的等价积分方程，然后定义了相应算子并利用Arzela-Ascoli定理证明了算子的全连续性，最后利用Schauder不动点定理和压缩映射原理分别得到了分数阶迭代微分方程边值问题解的存在性和唯一性结果。

参考文献