一类具积分边值条件的有序分数阶微分方程解

Solution of Sequential Fractional Differential Equations with Integral Boundary Value Conditions

摘要: 本文研究了一类带有非局部积分边值条件的 阶有序分数阶微分方程，其中 。借助于Banach压缩映像原理，非线性压缩和Leray-Schauder度理论，得到该问题解的存在性若干结果，并给出一个例子来说明所得结果的应用性。

Abstract: In this paper, we investigate a class of nonlocal integral boundary value problems of sequential fractional differential equations of order with . By means of Banach’s contraction mapping principle, nonlinear contraction and Leray-Schauder degree theory, some existence result of solutions are obtained. An illustrative example is given to show the applicability of our results.

文章引用：周宗福, 蒋伟. 一类具积分边值条件的有序分数阶微分方程解[J]. 应用数学进展, 2018, 7(8): 1095-1104. https://doi.org/10.12677/AAM.2018.78127

1. 引言

近年来，数学工作者对分数阶微分方程初值问题和边值问题进行了广泛的研究。一些学者研究了带有多点的，积分的，反周期的，非局部的等类型的边值条件的分数阶微分方程，得到了一些很好的结果 [1] - [13] 。分数阶微分方程边值问题方面还有其他很多研究成果，参见文献 [14] - [20] 。

本文研究带有积分边界条件的有序分数阶微分方程：

${}^{c}D{}_{0^{+}}^q\left(D+\lambda \right)u\left(t\right)=f\left(t,u\left(t\right)\right),t\in \left[0,1\right],1<q\le 2,$ (1.1)

$u\left(0\right)=0,u^{\prime }\left(0\right)=\vartheta {\displaystyle {\int }_0^{\xi }{u}^{\prime }}\left(s\right)\text{d}s,u\left(1\right)=0,$ (1.2)

其中 ${}^{c}D{}_{0^{+}}^q$ 是Caputo分数阶导数，，D是正常导数， $f:\left[0,1\right]\times R\to R$ 连续， $\lambda ,\vartheta$ 为正的实常数， $0<\xi <1$ 。

本文主要目标是讨论边值问题(1.1)~(1.2)解的存在性。

在文献 [21] 中，研究了如下的有序分数阶微分方程的边值问题：

$\{\begin{array}{l}{}^{c}D{}_{0^{+}}^q\left(D+\lambda \right)u\left(t\right)=f\left(t,u\left(t\right)\right),t\in \left[0,1\right],1<q\le 2,\\ u\left(0\right)=0,u^{\prime }\left(0\right)=0,u\left(1\right)=\beta u\left(\eta \right),0<\eta <1,\end{array}$

其中 $f:\left[0,1\right]\times R\to R$ ， $\lambda$ 为正实数，为实数。

在文献 [22] 中，研究了如下具边值条件的分数阶微分方程：

$\{\begin{array}{l}{}^{c}D{}_{0^{+}}^{q}u\left(t\right)=f\left(t,u\left(t\right)\right),t\in [0,1],1<q\le 2，\\ u^{\prime }\left(0\right)=\alpha {\displaystyle {\int }_0^{\xi }{u}^{\prime }\left(s\right)\text{d}s},u\left(1\right)=\beta g\left(u^{\prime }\left(\eta \right)\right),0\le \eta ,\xi \le 1,\end{array}$

其中 $f:\left[0,1\right]\times R\to R$ ， $g:R\to R$ 连续， $\alpha ,\beta$ 为常实数。

受文献 [21] 和 [22] 的启发，本文研究边值问题(1.1)~(1.2)，运用Banach压缩映像原理，非线性压缩，以及Leray-Schauder度理论，得到边值问题(1.1)~(1.2)解存在的若干结果。

本文的内容安排如下：第二部分给出分数阶微积分的一些定义及有关引理。第三部分给出三个主要结论：第一个结论是根据Banach压缩映像原理得出的，第二个结论是根据非线性压缩得出的，第三个结论是根据Leray-Schauder度理论得出的。第四部分给出实例，说明我们主要结果的应用性。

2. 预备知识

在这部分，我们介绍一些分数阶微积分定义和引理，为后面我们建立主要结果做准备。

定义2.1 函数 $f:\left(0,+\infty \right)\to R$ 的q阶分数阶积分定义为

$I_{0^{+}}^{q}f\left(t\right)=\frac{1}{\Gamma \left(q\right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}f\left(s\right)\text{d}s},t>0,n-1<q\le n.$

其中右边是逐点定义在 $\left(0,+\infty \right)$ 上。

定义2.2 函数 $f:\left(0,+\infty \right)\to R$ 的q阶Riemann-Liouville分数阶导数定义为

$D_{0^{+}}^{q}f\left(t\right)=\frac{1}{\Gamma \left(n-q\right)}\frac{{\text{d}}^n}{\text{d}{t}^n}{\displaystyle {\int }_0^t{\left(t-s\right)}^{n-q-1}}f\left(s\right)\text{d}s,t>0,n-1<q\le n.$

定义2.3 函数 $f:\left(0,+\infty \right)\to R$ 的q阶Caputo分数阶导数定义为

${}^{c}D{}_{0^{+}}^{q}f\left(t\right)=\frac{1}{\Gamma \left(n-q\right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{n-q-1}{f}^{\left(n\right)}\left(s\right)\text{d}s},t>0,n-1<q<n.$

注2.4 (i) 若，则

${}^{c}D{}_{0^{+}}^{q}f\left(t\right)={}^{c}D{}_{0^{+}}^q\left[f\left(t\right)-{\displaystyle \underset{k=0}{\overset{n-1}{\sum }}\frac{t_k}{k!}{f}^{\left(k\right)}\left(0\right)}\right],t>0,n-1<q<n.$

(ii) 常数的Caputo分数阶导数为零。

引理2.5 [1] 对于 $q>0$ ，分数阶微分方程 ${}^{c}D{}_{0^{+}}^{q}u\left(t\right)=0$ 的通解为

$u\left(t\right)=C_0+C_{1}t+C_2{t}^2+\cdots +C_{n-1}{t}^{n-1},$

其中 $C_i\in ℝ$ ， $i=0,1,2,\cdots ,n-1$ ， $n=\left[q\right]+1$ ， $\left[q\right]$ 表示实数q的整数部分。

引理2.6 [1] 对任意 $q>0$ ，有下式成立：

$I_{0^{+}}^q\left({}^{c}D{}_{0^{+}}^{q}u\right)\left(t\right)=u\left(t\right)+C_0+C_{1}t+C_2{t}^2+\cdots +C_{n-1}{t}^{n-1},$

其中 $C_i\in ℝ$ ， $i=0,1,2,\cdots ,n-1$ ， $n=\left[q\right]+1$ 。

在本文的以下部分，我们假设：

$\gamma \beta -\alpha \eta >0,$

其中 $\alpha =\lambda \xi +{\text{e}}^{-\lambda \xi }-1$ ， $\gamma =\lambda -1+{\text{e}}^{-\lambda }$ ， $$ ， $\eta =1-{\text{e}}^{-\lambda }$ 。

对 $h\in C\left[0,1\right]$ ，引入线性方程：

${}^{c}D{}_{0^{+}}^q\left(D+\lambda \right)u\left(t\right)=h\left(t\right),t\in \left[0,1\right],1<q\le 2.$ (2.1)

考虑带有边界条件(1.2)的有序分数阶微分方程(2.1)，即考虑边值问题(2.1)~(1.2)。

引理2.7 边值问题(2.1)~(1.2)的唯一解为

$\begin{array}{c}u\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda t}\right)-\eta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\right)\text{d}\tau }\\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -u\right)}^{q-1}h\left(u\right)\text{d}u}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda t}\right)-\beta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\right)\right].\end{array}$ (2.2)

证 对方程(2.1)两端作用 $I^q$ ，我们得到

$\left(D+\lambda \right)u\left(t\right)=I_{0^{+}}^{q}h\left(t\right)+C_0+C_{1}t,$

也可以写成

$D\left({\text{e}}^{\lambda t}u\left(t\right)\right)=\left[I_{0^{+}}^{q}h\left(t\right)+C_0+C_{1}t\right]{\text{e}}^{\lambda t},$

上式两端从0到t积分，结合(1.2)中的 $u\left(0\right)=0$ ，可得

$u\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-u\right)}^{q-1}}h\left(u\right)\text{d}u\right)\text{d}s}+\frac{C_0}{\lambda }\left(1-{\text{e}}^{-\lambda t}\right)+\frac{C_1}{{\lambda }^2}\left(\lambda t-1+{\text{e}}^{-\lambda t}\right).$ (2.3)

再由边值条件(1.2)可得

$\begin{array}{c}{C}_0=\frac{1}{\lambda \left(\beta \gamma -\alpha \eta \right)}\{\lambda \gamma [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\right)\text{d}\tau \\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -u\right)}^{q-1}h\left(u\right)\text{d}u}\right)\text{d}\tau }]+\alpha {\text{e}}^{-\lambda }{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\},\end{array}$

$\begin{array}{c}{C}_1=\frac{1}{\lambda \left(\beta \gamma -\alpha \eta \right)}\{-\lambda \eta [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\right)\text{d}\tau \\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -u\right)}^{q-1}h\left(u\right)\text{d}u}\right)\text{d}\tau }]-\lambda \beta {\text{e}}^{-\lambda }{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\},\end{array}$

将 $C_0$ ， $C_1$ 代入到(2.3)，即得到(2.2)。证毕。

3. 主要结果

令 $E=C\left[0,1\right]$ ，定义其中的范数为 $\Vert u\Vert =\sup \left\{\left|u\left(t\right)\right|:0\le t\le 1\right\}$ ，则E为一个Banach空间。

由引理2.7，我们可将边值问题(1.1)~(1.2)转换成 $u=Tu,T:\mathbb{E}\to \mathbb{E}$ 定义为：

$\begin{array}{c}Tu\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda t}\right)-\eta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau }\\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda t}\right)-\beta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s\right)\right].\end{array}$ (3.1)

显然，u是边值问题(1.1)~(1.2)的解当且仅当u为算子T的不动点。

首先，通过Banach压缩映像原理来研究边值问题(1.1)~(1.2)解的存在唯一性。

定理3.1 设f连续且满足：

$\left|f\left(t,u\right)-f\left(t,v\right)\right|\le L\left|u-v\right|,\forall t\in \left[0,1\right],u,v\in R$ ,

其中Lipschitz常数 $L>0$ 。若 $L<\frac{1}{B}$ ，则边值问题(1.1)~(1.2)存在唯一解，其中

证 第一步. T如(3.1)定义，令

$M=\underset{t\in \left[0,1\right]}{\sup }\left|f\left(t,0\right)\right|,r\ge \frac{MB}{1-LB}.$

又令 $B_r=\left\{u\in E:\Vert u\Vert \le r\right\}$ 。对 $\forall u\in B_r$ ， $t\in \left[0,1\right]$ ，有

$\begin{array}{c}\left|Tu\left(t\right)\right|=|{\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda t}\right)-\eta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda t}\right)-\beta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s\right)\right]|\end{array}$

$\begin{array}{l}\le {\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,0\right)\right|+\left|f\left(w,0\right)\right|\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,0\right)\right|+\left|f\left(w,0\right)\right|\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,0\right)\right|+\left|f\left(w,0\right)\right|\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)-\left|\beta \right|\lambda }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,0\right)\right|+\left|f\left(w,0\right)\right|\right)\text{d}w}\right)\text{d}s\right)\right]\end{array}$

$\begin{array}{l}\le \left(L\Vert u\Vert +M\right)(\frac{1-{\text{e}}^{-\lambda }}{\lambda \Gamma \left(q+1\right)}\left[1+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\right]\\ +\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[\frac{\xi }{\Gamma \left(q+1\right)}+\frac{{\xi }^{q+1}}{\Gamma \left(q+2\right)}\right])\\ \le \left(Lr+M\right)B\le r.\end{array}$

于是，得到 $Tu\in B_r$ 。因此 $T{B}_r\subset B_r$ 。

第二步，对于 $u,v\in B_r$ ，，有

$\begin{array}{l}\left|Tu\left(t\right)-Tv\left(t\right)\right|\le {\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s\\ +\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}[\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau }\\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}\tau }]\\ +\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)-\left|\beta \right|\lambda }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}s\right)\right]\\ \le L\Vert u-v\Vert \cdot B=LB\Vert u-v\Vert .\end{array}$

即有 $\Vert Tu-Tv\Vert \le BL\Vert u-v\Vert$ 。由于 $BL<1$ ，因此T在上是压缩的。由压缩映像原理可知，定理的结论成立。证毕。

接下来我们利用非线性压缩的一个著名定理来分析(1.1)~(1.2)解的存在问题。

定义3.2 令F为一个Banach空间， $A:F\to F$ 为一个映射。如果存在一个单调不减的连续函数 $\phi :R^{+}\to R^{+}$ ，使得 $\phi \left(0\right)=0$ ，对任意 $k>0$ ，有 $\phi \left(k\right)<k$ ，且满足：

$\Vert Au-Av\Vert \le \phi \left(\Vert u-v\Vert \right),\forall u,v\in F,$

则称 $\mathfrak{A}$ 是非线性压缩算子。

引理3.3 [23] 设F为一个Banach空间， $A:F\to F$ 是非线性压缩算子，则 $\mathfrak{A}$ 在F上存在唯一的不动点。

定理3.4 设 $$ 连续，且设

$\left|f\left(t,u\right)-f\left(t,v\right)\right|\le y\left(t\right)\frac{\left|u-v\right|}{H^{\ast }+\left|u-v\right|},\forall t\in \left[0,1\right],u,v\in R,$

其中是连续的，且正常数 $H^{\ast }$ 由下式给出：

$\begin{array}{c}{H}^{\ast }={\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w\right)\text{d}s\\ +\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}[\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}\tau }]\\ +\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[{\text{e}}^{-\lambda }{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}s}\right].\end{array}$

则边值问题(1.1)~(1.2)存在唯一解。

证 令 $T:E\to E$ 为(3.1)定义的算子，定义一个单调不减的连续函数 $\phi :R^{+}\to R^{+}$ ，

$\phi \left(k\right)=\frac{H^{\ast }k}{H^{\ast }+k},\forall k\ge 0.$

注意到函数 $\phi$ 满足 $\phi \left(0\right)=0$ ，且对任意 $k>0$ ，有 $\phi \left(k\right)<k$ 。

对任意 $u,v\in \mathbb{E}$ ，且任意 $t\in \left[0,1\right]$ ，我们有

$\begin{array}{l}\left|Tu\left(t\right)-Tv\left(t\right)\right|\\ \le {\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)-\left|\beta \right|\lambda }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}s\right)\right]\end{array}$

$\begin{array}{l}\le \frac{\phi \left(\Vert u-v\Vert \right)}{H^{\ast }}\{{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}s}\right]\le \phi \left(\Vert u-v\Vert \right),\end{array}$

故有 $\Vert Tu-Tv\Vert \le \phi \left(\Vert u-\nu \Vert \right)$ ，因此T是非线性压缩算子，从而由引理3.3知，算子T存在唯一不动点，

即边值问题(1.1)~(1.2)存在唯一解。证毕。

现在利用Leray-Schauder度理论，讨论边值问题(1.1)~(1.2)的解的存在性。

定理3.5 设 $f:\left[0,1\right]\times R\to R$ 连续，且存在常数 $0\le {\eta }_0<\frac{1}{B}$ (这里的B即为定理3.1中的B)， $K>0$ ，使得

$\left|f\left(t,u\right)\right|\le {\eta }_0\left|u\right|+K,\forall t\in \left[0,1\right],u\in R,$

则边值问题(1.1)~(1.2)至少存在一个解。

证 考虑算子方程 $$ ，下面证明此方程至少存在一个解。

取 $B_R=\left\{u\in \mathbb{E}:\Vert u\Vert <R\right\},R>0$ ，R将在后面给出。我们将证明 $T:\overline{B_R}\to C\left[0,1\right]$ 满足条件： $u\ne \theta Tu,\forall u\in \partial B_R,\forall \theta \in \left[0,1\right]$ 。由Arzela-Ascoli定理易知，T在 $\overline{B_R}$ 上是全连续的。定义一个连续映射 $h_{\theta }$ ：

$h_{\theta }\left(u\right)=u-\theta Tu,\theta \in \left[0,1\right].$

通过拓扑度的同伦不变性，可得出：

$\mathrm{deg}\left(h_{\theta },B_R,0\right)=\mathrm{deg}\left(I-\theta T,B_R,0\right)=\mathrm{deg}\left(h_1,B_R,0\right)=\mathrm{deg}\left(h_0,B_R,0\right)=1\ne 0$ ，其中I是恒等算子。

再由Leray-Schauder度的非零性质可知，至少存在一个 $u\in B_R$ 使 $h_1\left(u\right)=u-Tu=0$ ，可见u即为T的不动点。

假设 $\exists \theta \in \left[0,1\right]$ ， $u\in E$ ，使得 $u=\theta Tu$ ，那么对 $\forall t\in \left[0,1\right],u\left(t\right)=\theta Tu\left(t\right)$ ，则

$\begin{array}{l}\left|u\left(t\right)\right|=\left|\theta Tu\left(t\right)\right|\le \left|Tu\left(t\right)\right|\\ \le {\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s}\right)\text{d}\tau }\\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)-\left|\beta \right|\lambda }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s\right)\right]\end{array}$

$\begin{array}{l}\le \left({\eta }_0\Vert u\Vert +K\right)(\frac{1-{\text{e}}^{-\lambda }}{\lambda \Gamma \left(q+1\right)}\left[1+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2(\beta \gamma -\alpha \eta )}\right]\\ +\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[\frac{\xi }{\Gamma \left(q+1\right)}+\frac{{\xi }^{q+1}}{\Gamma \left(q+2\right)}\right])\\ =\left({\eta }_0\Vert u\Vert +K\right)B\end{array}$

从而 $\Vert u\Vert \le \left({\eta }_0\Vert u\Vert +K\right)B$ ，即有 $\Vert u\Vert \le \frac{BK}{1-{\eta }_{0}B}$ 。

取 $R=\frac{BK}{1-{\eta }_{0}B}+1$ ，则 $\forall u\in \partial B_R,\forall \theta \in \left[0,1\right]$ ，有 $u\ne \theta Tu$ 。因此边值问题(1.1)~(1.2)至少存在一个解。证毕。

4. 举例

例4.1 考虑如下边值问题

$$ (4.1)

其中 $f\left(t,u\right)=\frac{1}{2\text{π}}\sin \frac{\text{π}}{2}u+1+\frac{u}{u^2+1},q=\frac{3}{2},\xi =\frac{1}{3},\vartheta =1,\lambda =3$ 。

直接计算易知

$B=\frac{1-{\text{e}}^{-\lambda }}{\lambda \Gamma \left(q+1\right)}\left[1+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\right]+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[\frac{\xi }{\Gamma \left(q+1\right)}+\frac{{\xi }^{q+1}}{\Gamma \left(q+2\right)}\right]\le \mathrm{0.7715.}$

取

$\eta =\frac{5}{4}<\frac{1}{B}\approx 1.29613,K=1,$

则

$\left|f\left(t,u\right)\right|=\left|\frac{1}{2\text{π}}\sin \frac{\text{π}}{2}u+1+\frac{u}{u^2+1}\right|\le \frac{1}{\text{2π}}\left|\frac{\text{π}}{2}u\right|+1+\left|\frac{u}{u^2+1}\right|\le \frac{5}{4}\left|u\right|+1\le \eta \left|u\right|+K.$

因此定理3.5的条件都满足，所以由定理3.5知边值问题(4.1)至少存在一个解。

致谢

本文得到国家自然科学基金项目(编号：11371027)和安徽省自然科学基金项目(编号：1608085MA12)资助。

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1. 引言

近年来，数学工作者对分数阶微分方程初值问题和边值问题进行了广泛的研究。一些学者研究了带有多点的，积分的，反周期的，非局部的等类型的边值条件的分数阶微分方程，得到了一些很好的结果 [1] - [13] 。分数阶微分方程边值问题方面还有其他很多研究成果，参见文献 [14] - [20] 。

本文研究带有积分边界条件的有序分数阶微分方程：

${}^{c}D{}_{0^{+}}^q\left(D+\lambda \right)u\left(t\right)=f\left(t,u\left(t\right)\right),t\in \left[0,1\right],1<q\le 2,$ (1.1)

$u\left(0\right)=0,u^{\prime }\left(0\right)=\vartheta {\displaystyle {\int }_0^{\xi }{u}^{\prime }}\left(s\right)\text{d}s,u\left(1\right)=0,$ (1.2)

其中 ${}^{c}D{}_{0^{+}}^q$ 是Caputo分数阶导数，，D是正常导数， $f:\left[0,1\right]\times R\to R$ 连续， $\lambda ,\vartheta$ 为正的实常数， $0<\xi <1$ 。

本文主要目标是讨论边值问题(1.1)~(1.2)解的存在性。

在文献 [21] 中，研究了如下的有序分数阶微分方程的边值问题：

$\{\begin{array}{l}{}^{c}D{}_{0^{+}}^q\left(D+\lambda \right)u\left(t\right)=f\left(t,u\left(t\right)\right),t\in \left[0,1\right],1<q\le 2,\\ u\left(0\right)=0,u^{\prime }\left(0\right)=0,u\left(1\right)=\beta u\left(\eta \right),0<\eta <1,\end{array}$

其中 $f:\left[0,1\right]\times R\to R$ ， $\lambda$ 为正实数，为实数。

在文献 [22] 中，研究了如下具边值条件的分数阶微分方程：

$\{\begin{array}{l}{}^{c}D{}_{0^{+}}^{q}u\left(t\right)=f\left(t,u\left(t\right)\right),t\in [0,1],1<q\le 2，\\ u^{\prime }\left(0\right)=\alpha {\displaystyle {\int }_0^{\xi }{u}^{\prime }\left(s\right)\text{d}s},u\left(1\right)=\beta g\left(u^{\prime }\left(\eta \right)\right),0\le \eta ,\xi \le 1,\end{array}$

其中 $f:\left[0,1\right]\times R\to R$ ， $g:R\to R$ 连续， $\alpha ,\beta$ 为常实数。

受文献 [21] 和 [22] 的启发，本文研究边值问题(1.1)~(1.2)，运用Banach压缩映像原理，非线性压缩，以及Leray-Schauder度理论，得到边值问题(1.1)~(1.2)解存在的若干结果。

本文的内容安排如下：第二部分给出分数阶微积分的一些定义及有关引理。第三部分给出三个主要结论：第一个结论是根据Banach压缩映像原理得出的，第二个结论是根据非线性压缩得出的，第三个结论是根据Leray-Schauder度理论得出的。第四部分给出实例，说明我们主要结果的应用性。

2. 预备知识

在这部分，我们介绍一些分数阶微积分定义和引理，为后面我们建立主要结果做准备。

定义2.1 函数 $f:\left(0,+\infty \right)\to R$ 的q阶分数阶积分定义为

$I_{0^{+}}^{q}f\left(t\right)=\frac{1}{\Gamma \left(q\right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{q-1}f\left(s\right)\text{d}s},t>0,n-1<q\le n.$

其中右边是逐点定义在 $\left(0,+\infty \right)$ 上。

定义2.2 函数 $f:\left(0,+\infty \right)\to R$ 的q阶Riemann-Liouville分数阶导数定义为

$D_{0^{+}}^{q}f\left(t\right)=\frac{1}{\Gamma \left(n-q\right)}\frac{{\text{d}}^n}{\text{d}{t}^n}{\displaystyle {\int }_0^t{\left(t-s\right)}^{n-q-1}}f\left(s\right)\text{d}s,t>0,n-1<q\le n.$

定义2.3 函数 $f:\left(0,+\infty \right)\to R$ 的q阶Caputo分数阶导数定义为

${}^{c}D{}_{0^{+}}^{q}f\left(t\right)=\frac{1}{\Gamma \left(n-q\right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{n-q-1}{f}^{\left(n\right)}\left(s\right)\text{d}s},t>0,n-1<q<n.$

注2.4 (i) 若，则

${}^{c}D{}_{0^{+}}^{q}f\left(t\right)={}^{c}D{}_{0^{+}}^q\left[f\left(t\right)-{\displaystyle \underset{k=0}{\overset{n-1}{\sum }}\frac{t_k}{k!}{f}^{\left(k\right)}\left(0\right)}\right],t>0,n-1<q<n.$

(ii) 常数的Caputo分数阶导数为零。

引理2.5 [1] 对于 $q>0$ ，分数阶微分方程 ${}^{c}D{}_{0^{+}}^{q}u\left(t\right)=0$ 的通解为

$u\left(t\right)=C_0+C_{1}t+C_2{t}^2+\cdots +C_{n-1}{t}^{n-1},$

其中 $C_i\in ℝ$ ， $i=0,1,2,\cdots ,n-1$ ， $n=\left[q\right]+1$ ， $\left[q\right]$ 表示实数q的整数部分。

引理2.6 [1] 对任意 $q>0$ ，有下式成立：

$I_{0^{+}}^q\left({}^{c}D{}_{0^{+}}^{q}u\right)\left(t\right)=u\left(t\right)+C_0+C_{1}t+C_2{t}^2+\cdots +C_{n-1}{t}^{n-1},$

其中 $C_i\in ℝ$ ， $i=0,1,2,\cdots ,n-1$ ， $n=\left[q\right]+1$ 。

在本文的以下部分，我们假设：

$\gamma \beta -\alpha \eta >0,$

其中 $\alpha =\lambda \xi +{\text{e}}^{-\lambda \xi }-1$ ， $\gamma =\lambda -1+{\text{e}}^{-\lambda }$ ， $$ ， $\eta =1-{\text{e}}^{-\lambda }$ 。

对 $h\in C\left[0,1\right]$ ，引入线性方程：

${}^{c}D{}_{0^{+}}^q\left(D+\lambda \right)u\left(t\right)=h\left(t\right),t\in \left[0,1\right],1<q\le 2.$ (2.1)

考虑带有边界条件(1.2)的有序分数阶微分方程(2.1)，即考虑边值问题(2.1)~(1.2)。

引理2.7 边值问题(2.1)~(1.2)的唯一解为

$\begin{array}{c}u\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda t}\right)-\eta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\right)\text{d}\tau }\\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -u\right)}^{q-1}h\left(u\right)\text{d}u}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda t}\right)-\beta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\right)\right].\end{array}$ (2.2)

证 对方程(2.1)两端作用 $I^q$ ，我们得到

$\left(D+\lambda \right)u\left(t\right)=I_{0^{+}}^{q}h\left(t\right)+C_0+C_{1}t,$

也可以写成

$D\left({\text{e}}^{\lambda t}u\left(t\right)\right)=\left[I_{0^{+}}^{q}h\left(t\right)+C_0+C_{1}t\right]{\text{e}}^{\lambda t},$

上式两端从0到t积分，结合(1.2)中的 $u\left(0\right)=0$ ，可得

$u\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-u\right)}^{q-1}}h\left(u\right)\text{d}u\right)\text{d}s}+\frac{C_0}{\lambda }\left(1-{\text{e}}^{-\lambda t}\right)+\frac{C_1}{{\lambda }^2}\left(\lambda t-1+{\text{e}}^{-\lambda t}\right).$ (2.3)

再由边值条件(1.2)可得

$\begin{array}{c}{C}_0=\frac{1}{\lambda \left(\beta \gamma -\alpha \eta \right)}\{\lambda \gamma [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\right)\text{d}\tau \\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -u\right)}^{q-1}h\left(u\right)\text{d}u}\right)\text{d}\tau }]+\alpha {\text{e}}^{-\lambda }{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\},\end{array}$

$\begin{array}{c}{C}_1=\frac{1}{\lambda \left(\beta \gamma -\alpha \eta \right)}\{-\lambda \eta [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\right)\text{d}\tau \\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -u\right)}^{q-1}h\left(u\right)\text{d}u}\right)\text{d}\tau }]-\lambda \beta {\text{e}}^{-\lambda }{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-u\right)}^{q-1}h\left(u\right)\text{d}u\right)\text{d}s\},\end{array}$

将 $C_0$ ， $C_1$ 代入到(2.3)，即得到(2.2)。证毕。

3. 主要结果

令 $E=C\left[0,1\right]$ ，定义其中的范数为 $\Vert u\Vert =\sup \left\{\left|u\left(t\right)\right|:0\le t\le 1\right\}$ ，则E为一个Banach空间。

由引理2.7，我们可将边值问题(1.1)~(1.2)转换成 $u=Tu,T:\mathbb{E}\to \mathbb{E}$ 定义为：

$\begin{array}{c}Tu\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda t}\right)-\eta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau }\\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda t}\right)-\beta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s\right)\right].\end{array}$ (3.1)

显然，u是边值问题(1.1)~(1.2)的解当且仅当u为算子T的不动点。

首先，通过Banach压缩映像原理来研究边值问题(1.1)~(1.2)解的存在唯一性。

定理3.1 设f连续且满足：

$\left|f\left(t,u\right)-f\left(t,v\right)\right|\le L\left|u-v\right|,\forall t\in \left[0,1\right],u,v\in R$ ,

其中Lipschitz常数 $L>0$ 。若 $L<\frac{1}{B}$ ，则边值问题(1.1)~(1.2)存在唯一解，其中

证 第一步. T如(3.1)定义，令

$M=\underset{t\in \left[0,1\right]}{\sup }\left|f\left(t,0\right)\right|,r\ge \frac{MB}{1-LB}.$

又令 $B_r=\left\{u\in E:\Vert u\Vert \le r\right\}$ 。对 $\forall u\in B_r$ ， $t\in \left[0,1\right]$ ，有

$\begin{array}{c}\left|Tu\left(t\right)\right|=|{\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda t}\right)-\eta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ -{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda t}\right)-\beta \left(\lambda t-1+{\text{e}}^{-\lambda t}\right)}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}f\left(w,u\left(w\right)\right)\text{d}w}\right)\text{d}s\right)\right]|\end{array}$

$\begin{array}{l}\le {\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,0\right)\right|+\left|f\left(w,0\right)\right|\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,0\right)\right|+\left|f\left(w,0\right)\right|\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,0\right)\right|+\left|f\left(w,0\right)\right|\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)-\left|\beta \right|\lambda }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,0\right)\right|+\left|f\left(w,0\right)\right|\right)\text{d}w}\right)\text{d}s\right)\right]\end{array}$

$\begin{array}{l}\le \left(L\Vert u\Vert +M\right)(\frac{1-{\text{e}}^{-\lambda }}{\lambda \Gamma \left(q+1\right)}\left[1+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\right]\\ +\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[\frac{\xi }{\Gamma \left(q+1\right)}+\frac{{\xi }^{q+1}}{\Gamma \left(q+2\right)}\right])\\ \le \left(Lr+M\right)B\le r.\end{array}$

于是，得到 $Tu\in B_r$ 。因此 $T{B}_r\subset B_r$ 。

第二步，对于 $u,v\in B_r$ ，，有

$\begin{array}{l}\left|Tu\left(t\right)-Tv\left(t\right)\right|\le {\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s\\ +\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}[\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau }\\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}\tau }]\\ +\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)-\left|\beta \right|\lambda }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}s\right)\right]\\ \le L\Vert u-v\Vert \cdot B=LB\Vert u-v\Vert .\end{array}$

即有 $\Vert Tu-Tv\Vert \le BL\Vert u-v\Vert$ 。由于 $BL<1$ ，因此T在上是压缩的。由压缩映像原理可知，定理的结论成立。证毕。

接下来我们利用非线性压缩的一个著名定理来分析(1.1)~(1.2)解的存在问题。

定义3.2 令F为一个Banach空间， $A:F\to F$ 为一个映射。如果存在一个单调不减的连续函数 $\phi :R^{+}\to R^{+}$ ，使得 $\phi \left(0\right)=0$ ，对任意 $k>0$ ，有 $\phi \left(k\right)<k$ ，且满足：

$\Vert Au-Av\Vert \le \phi \left(\Vert u-v\Vert \right),\forall u,v\in F,$

则称 $\mathfrak{A}$ 是非线性压缩算子。

引理3.3 [23] 设F为一个Banach空间， $A:F\to F$ 是非线性压缩算子，则 $\mathfrak{A}$ 在F上存在唯一的不动点。

定理3.4 设 $$ 连续，且设

$\left|f\left(t,u\right)-f\left(t,v\right)\right|\le y\left(t\right)\frac{\left|u-v\right|}{H^{\ast }+\left|u-v\right|},\forall t\in \left[0,1\right],u,v\in R,$

其中是连续的，且正常数 $H^{\ast }$ 由下式给出：

$\begin{array}{c}{H}^{\ast }={\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w\right)\text{d}s\\ +\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}[\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}\tau }]\\ +\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[{\text{e}}^{-\lambda }{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}s}\right].\end{array}$

则边值问题(1.1)~(1.2)存在唯一解。

证 令 $T:E\to E$ 为(3.1)定义的算子，定义一个单调不减的连续函数 $\phi :R^{+}\to R^{+}$ ，

$\phi \left(k\right)=\frac{H^{\ast }k}{H^{\ast }+k},\forall k\ge 0.$

注意到函数 $\phi$ 满足 $\phi \left(0\right)=0$ ，且对任意 $k>0$ ，有 $\phi \left(k\right)<k$ 。

对任意 $u,v\in \mathbb{E}$ ，且任意 $t\in \left[0,1\right]$ ，我们有

$\begin{array}{l}\left|Tu\left(t\right)-Tv\left(t\right)\right|\\ \le {\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)-\left|\beta \right|\lambda }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)-f\left(w,v\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}s\right)\right]\end{array}$

$\begin{array}{l}\le \frac{\phi \left(\Vert u-v\Vert \right)}{H^{\ast }}\{{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }}\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}s\right)\text{d}\tau \\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }{\displaystyle {\int }_0^1{\text{e}}^{\lambda s}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}{\left(s-w\right)}^{q-1}y\left(w\right)\text{d}w}\right)\text{d}s}\right]\le \phi \left(\Vert u-v\Vert \right),\end{array}$

故有 $\Vert Tu-Tv\Vert \le \phi \left(\Vert u-\nu \Vert \right)$ ，因此T是非线性压缩算子，从而由引理3.3知，算子T存在唯一不动点，

即边值问题(1.1)~(1.2)存在唯一解。证毕。

现在利用Leray-Schauder度理论，讨论边值问题(1.1)~(1.2)的解的存在性。

定理3.5 设 $f:\left[0,1\right]\times R\to R$ 连续，且存在常数 $0\le {\eta }_0<\frac{1}{B}$ (这里的B即为定理3.1中的B)， $K>0$ ，使得

$\left|f\left(t,u\right)\right|\le {\eta }_0\left|u\right|+K,\forall t\in \left[0,1\right],u\in R,$

则边值问题(1.1)~(1.2)至少存在一个解。

证 考虑算子方程 $$ ，下面证明此方程至少存在一个解。

取 $B_R=\left\{u\in \mathbb{E}:\Vert u\Vert <R\right\},R>0$ ，R将在后面给出。我们将证明 $T:\overline{B_R}\to C\left[0,1\right]$ 满足条件： $u\ne \theta Tu,\forall u\in \partial B_R,\forall \theta \in \left[0,1\right]$ 。由Arzela-Ascoli定理易知，T在 $\overline{B_R}$ 上是全连续的。定义一个连续映射 $h_{\theta }$ ：

$h_{\theta }\left(u\right)=u-\theta Tu,\theta \in \left[0,1\right].$

通过拓扑度的同伦不变性，可得出：

$\mathrm{deg}\left(h_{\theta },B_R,0\right)=\mathrm{deg}\left(I-\theta T,B_R,0\right)=\mathrm{deg}\left(h_1,B_R,0\right)=\mathrm{deg}\left(h_0,B_R,0\right)=1\ne 0$ ，其中I是恒等算子。

再由Leray-Schauder度的非零性质可知，至少存在一个 $u\in B_R$ 使 $h_1\left(u\right)=u-Tu=0$ ，可见u即为T的不动点。

假设 $\exists \theta \in \left[0,1\right]$ ， $u\in E$ ，使得 $u=\theta Tu$ ，那么对 $\forall t\in \left[0,1\right],u\left(t\right)=\theta Tu\left(t\right)$ ，则

$\begin{array}{l}\left|u\left(t\right)\right|=\left|\theta Tu\left(t\right)\right|\le \left|Tu\left(t\right)\right|\\ \le {\displaystyle {\int }_0^t{\text{e}}^{-\lambda \left(t-s\right)}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot [\lambda {\displaystyle {\int }_0^{\xi }{\text{e}}^{-\lambda \tau }\left({\displaystyle {\int }_0^{\tau }{\text{e}}^{\lambda s}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s}\right)\text{d}\tau }\\ +{\displaystyle {\int }_0^{\xi }\left({\displaystyle {\int }_0^{\tau }\frac{1}{\Gamma \left(q\right)}{\left(\tau -w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)\right|\right)\text{d}w}\right)\text{d}\tau }]+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)-\left|\beta \right|\lambda }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\\ \cdot \left[{\text{e}}^{-\lambda }\left({\displaystyle {\int }_0^1{\text{e}}^{\lambda s}}\left({\displaystyle {\int }_0^s\frac{1}{\Gamma \left(q\right)}}{\left(s-w\right)}^{q-1}\left(\left|f\left(w,u\left(w\right)\right)\right|\right)\text{d}w\right)\text{d}s\right)\right]\end{array}$

$\begin{array}{l}\le \left({\eta }_0\Vert u\Vert +K\right)(\frac{1-{\text{e}}^{-\lambda }}{\lambda \Gamma \left(q+1\right)}\left[1+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2(\beta \gamma -\alpha \eta )}\right]\\ +\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[\frac{\xi }{\Gamma \left(q+1\right)}+\frac{{\xi }^{q+1}}{\Gamma \left(q+2\right)}\right])\\ =\left({\eta }_0\Vert u\Vert +K\right)B\end{array}$

从而 $\Vert u\Vert \le \left({\eta }_0\Vert u\Vert +K\right)B$ ，即有 $\Vert u\Vert \le \frac{BK}{1-{\eta }_{0}B}$ 。

取 $R=\frac{BK}{1-{\eta }_{0}B}+1$ ，则 $\forall u\in \partial B_R,\forall \theta \in \left[0,1\right]$ ，有 $u\ne \theta Tu$ 。因此边值问题(1.1)~(1.2)至少存在一个解。证毕。

4. 举例

例4.1 考虑如下边值问题

$$ (4.1)

其中 $f\left(t,u\right)=\frac{1}{2\text{π}}\sin \frac{\text{π}}{2}u+1+\frac{u}{u^2+1},q=\frac{3}{2},\xi =\frac{1}{3},\vartheta =1,\lambda =3$ 。

直接计算易知

$B=\frac{1-{\text{e}}^{-\lambda }}{\lambda \Gamma \left(q+1\right)}\left[1+\frac{\alpha \left(1-{\text{e}}^{-\lambda }\right)+\lambda \left|\beta \right|}{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\right]+\frac{\gamma \left(1-{\text{e}}^{-\lambda }\right)+\lambda \eta }{{\lambda }^2\left(\beta \gamma -\alpha \eta \right)}\left[\frac{\xi }{\Gamma \left(q+1\right)}+\frac{{\xi }^{q+1}}{\Gamma \left(q+2\right)}\right]\le \mathrm{0.7715.}$

取

$\eta =\frac{5}{4}<\frac{1}{B}\approx 1.29613,K=1,$

则

$\left|f\left(t,u\right)\right|=\left|\frac{1}{2\text{π}}\sin \frac{\text{π}}{2}u+1+\frac{u}{u^2+1}\right|\le \frac{1}{\text{2π}}\left|\frac{\text{π}}{2}u\right|+1+\left|\frac{u}{u^2+1}\right|\le \frac{5}{4}\left|u\right|+1\le \eta \left|u\right|+K.$

因此定理3.5的条件都满足，所以由定理3.5知边值问题(4.1)至少存在一个解。

致谢

本文得到国家自然科学基金项目(编号：11371027)和安徽省自然科学基金项目(编号：1608085MA12)资助。