一类分数阶三点边值问题解的存在性

Existence of Solutions for a Class of Fractional Three-Point Boundary Value Problems

摘要: 本文运用Leray-Schauder二择一定理，研究了一类分数阶微分方程的三点边值问题 得到该问题解的存在性。进一步证明了限制非线性项的函数k(t)，在被形如Bt^μ的函数控制后，该问题至少存在一个解。最后，通过实例验证了结论的有效性。

Abstract: In this paper, three-point boundary value problem for a class of fractional differential equations are studied by using Leray-Schauder nonlinear alternative. The existence of the solution is obtained. Furthermore, it is proved that the function k(t), which is limited to the nonlinear term, exists at least one solution of the problem when it is controlled by a function of the form Bt^μ. Finally, an example is given to verify the validity of the conclusion.

1. 引言

近年来，分数阶微分方程已经对数学模型的研究过程和其他学科领域的研究产生了深远的影响，例如电化学、热传导、地下水流动系统、多孔介质等。目前，分数阶边值问题的研究已取得了很多结果 [1] [2] [3] [4] [5]。

三点边值问题在微分方程理论中运用广泛 [6] [7]。庞杨等在文献 [8] 利用半序集上的不动点定理得到了Caputo分数阶三点边值问题

$\{\begin{array}{l}{}^c{D}_{0^{+}}^{\alpha }u\left(t\right)+f\left(t,u\left(t\right)\right)=0,0<t<1,\\ u\left(0\right)=u^{″}\left(0\right)=u^{‴}\left(0\right)=0,u^{\prime }\left(1\right)=\beta u(\; \eta \; )\end{array}$

解的存在性，其中 $\text{3}<\alpha \le 4$， $\text{0}<\beta \le 1$， $\text{0}<\eta <1$， ${}^c{D}_{0^{+}}^{\alpha }$ 是标准Caputo分数阶导数。

Yan Sun在文献 [9] 等利用锥拉伸与压缩不动点定理得到了四阶三点边值问题

$\{\begin{array}{l}{u}^{(4)}\left(t\right)+f\left(t,u\left(t\right)\right)=0,0\le t\le 1,\\ u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(0\right)=0,u^{″}\left(1\right)-\alpha u^{″}\left(\eta \right)=\lambda \end{array}$

解的存在性，其中 $\eta \in \left(0,1\right)$， $\alpha \in \left[0,\frac{1}{\eta }\right)$， $\lambda \in \left[0,\infty \right)$。

2015年，Batarfi和Jorge Losada在文献 [10] 运用Banach不动点定理和Krasnoselskii定理研究了分数阶三点边值问题

$\{\begin{array}{l}{D}^{\alpha }\left(D+\lambda \right)x\left(t\right)=f\left(t,x\left(t\right)\right),t\in [0,1],\\ x\left(0\right)=0,x^{\prime }\left(0\right)=0,x\left(1\right)=\beta x\left(\eta \right)\end{array}$ (1.1)

解的存在性，其中 $D^{\alpha }$ 为 $\alpha \in \left(1,2\right]$ 的一致分数阶导数， $D$ 是一阶导数， $f:\left[0,1\right]\times R\to R$ 连续函数， $\lambda ,\beta \in R$， $\lambda >0$ 且 $\eta \in \left(0,1\right)$。

受上述文献的启发，本文运用Leray-Schauder非线性抉择研究问题(1.1)解的存在性，其中参数 $\lambda >\text{1}$ 且 $\beta ,\eta \in \left(0,1\right)$。

2. 预备知识

定义1 [11] 设 $\alpha \in \left(0,1\right)$ 并且 $f:\left[0,\infty \right)\to R$， $f$ 的 $\alpha$ 阶分数导数可定义为

$D^{\alpha }x\left(t\right)=\underset{\epsilon \to 0}{\lim }\frac{x\left(t+\epsilon t^{1-\alpha }\right)-x\left(t\right)}{\epsilon },\forall t>0$。

如果 $f$ 在 $t\in \left(0,a\right)$ 是 $\alpha -$ 可微的， $a>0$ 且 $\underset{t\to 0^{+}}{\lim }{D}^{\alpha }x\left(t\right)$ 存在，定义 $D^{\alpha }x\left(\text{0}\right)=\underset{t\to 0^{+}}{\lim }{D}^{\alpha }x\left(t\right)$。

定义 2 [11] 设 $\alpha \in \left(n,n+1\right]$ 并且 $x$ 在 $t>0$ 是 $n-$ 可微的，则 $\alpha$ 阶导数为

$D^{\alpha }x\left(t\right)=\underset{\epsilon \to 0}{\lim }\frac{x^{\left(\left[\alpha \right]-1\right)}\left(t+\epsilon t^{\left(\left[\alpha \right]-\alpha \right)}\right)-x^{\left(\left[\alpha \right]-1\right)}\left(t\right)}{\epsilon },$

其中 $\left[\alpha \right]$ 是大于或等于 $\alpha$ 的最小整数。如果 $f$ 在 $t\in \left(0,a\right)$ 是 $\alpha -$ 可微的， $a>0$ 且 $\underset{t\to 0^{+}}{\lim }{D}^{\alpha }x\left(t\right)$ 存在，定义 $D^{\alpha }x\left(\text{0}\right)=\underset{t\to 0^{+}}{\lim }{D}^{\alpha }x\left(t\right)$。

定义3 [11] 设 $\alpha \in \left(0,1\right)$，对于 $t\ge 0$， $\alpha$ 阶积分记为 $I^{\alpha }x\left(t\right)$，可定义为

$D^{-\alpha }x\left(t\right)\equiv I^{\alpha }x\left(t\right)=I^1\left(t^{\alpha -1}x\right)\left(t\right)={\displaystyle {\int }_0^t\frac{x\left(s\right)}{s^{1-\alpha }}ds}$。

定义 4 [11] 设 $\alpha \in \left(n-1,n\right)$， $\alpha$ 阶积分定义为

$D^{-\alpha }x\left(t\right)\equiv I^{\alpha }x\left(t\right)=I^n\left(t^{\alpha -\left[\alpha \right]}x\right)\left(t\right)$，

其中 $I^n$ 表示普通积分 $I^{\text{1}}$ 进行 $n$ 次积分。

引理1 [10] $\alpha \in \left(1,2\right]$， $x$ 是定义在 $I^{\alpha }$ 上的连续函数，当 $t\ge 0$ 时， $D^{\alpha }{I}^{\alpha }x\left(t\right)=x\left(t\right)$。

引理2 [12] 设 $E$ 为Banach空间， $\Omega$ 是 $E$ 的有界开子集，且 $\text{0}\in \Omega$。 $T:\bar{\Omega }\to E$ 是全连续算子，则必有下列结论之一成立。

1) 存在 $u\in \partial \Omega$ 及 $\theta >1$，使得 $Tu=\theta u$ ；

2) 存在 $T$ 上的不动点 $u*\in \bar{\Omega }$。

引理3 [2] 对于三点边值问题

$\{\begin{array}{l}{D}^{\alpha }\left(D+\lambda \right)x\left(t\right)=\sigma \left(t\right),t\in \left[0,1\right],\\ x\left(0\right)=0,x^{\prime }\left(0\right)=0,x\left(1\right)=\beta x\left(\eta \right),\end{array}$

其中 $\alpha \in \left(1,2\right]$， $\sigma \in C\left[0,1\right]$。当 $\beta \ne \frac{\lambda +{\text{e}}^{-\lambda }-1}{\lambda \eta +{\text{e}}^{-\lambda \eta }-1}$ 时，则该问题有唯一解

$\begin{array}{c}x\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{-\lambda (t-s)}\left({\displaystyle {\int }_0^s\sigma \left(u\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}+A\left(t\right)[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left({\displaystyle {\int }_0^s\sigma \left(u\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}\\ -{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^s\sigma \left(u\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}],\end{array}$

其中 $A\left(t\right)=\frac{1}{\Delta }\left(\lambda t+{\mathrm{e}}^{-\lambda t}-1\right)$， $\Delta =\lambda +{\text{e}}^{-\lambda }-1-\beta \left(\lambda \eta +{\text{e}}^{-\lambda \eta }-1\right)\ne \text{0}$。

设 $E=C\left[0,1\right]$，对 $x\in E$， $E$ 在范数 $\Vert x\Vert =\underset{\begin{array}{l}t\in [0,1]\\ x\in \bar{\Omega }\end{array}}{\max }\left|x\left(t\right)\right|$ 下成为Banach空间。

定义算子 $T:E\to E$，

$\begin{array}{c}\left(Tx\right)\left(t\right)={\displaystyle {\int }_0^t{\text{e}}^{-\lambda (t-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}+A\left(t\right)[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}\\ -{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}]\end{array}$。

引理4 $T$ 是 $E\to E$ 的全连续算子。

证明：设 $\Omega \subset E$ 为有界开集，由 $f$ 的连续性可知 $f\left(t,x\right)$ 在 $\left[0,1\right]\times \bar{\Omega }$ 上有界，记 $R=\underset{t\in [0,1]}{\max }\left|f\left(t,x\left(t\right)\right)\right|$。

对 $\forall x\in \bar{\Omega }$，由算子 $T$ 的定义可得

$\begin{array}{c}\Vert Tx\Vert =\underset{t\in [0,1]}{\max }|{\displaystyle {\int }_0^t{\text{e}}^{-\lambda (t-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}+A\left(t\right)[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}\\ -{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}]|\\ \le {\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (\text{1}-s)}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)ds}+A\left(\text{1}\right)[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)\text{d}u}\right)}\text{d}s\\ +{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)ds}]\end{array}$

$\begin{array}{l}\le R\left[\left(1+A\left(1\right)\right){\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (\text{1}-s)}\left({\displaystyle {\int }_0^s{u}^{\alpha -2}\left(s-u\right)du}\right)ds}+A\left(\text{1}\right)\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left({\displaystyle {\int }_0^s{u}^{\alpha -2}\left(s-u\right)\text{d}u}\right)}\text{d}s\right]\\ \le \frac{R}{\alpha \left(\alpha -1\right)}\left(\left(1+A\left(1\right)\right){\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (\text{1}-s)}{s}^{\alpha }ds}+A\left(\text{1}\right)\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}{s}^{\alpha }}\text{d}s\right)\\ \le \frac{R\left(\left(1+A\left(1\right)\right)\left(1-{\text{e}}^{-\lambda }\right)+A\left(\text{1}\right)\beta {\eta }^{\alpha }\left(1-{\text{e}}^{-\lambda \eta }\right)\right)}{\alpha \left(\alpha -1\right)}.\end{array}$

从而可知 $T\left(\Omega \right)$ 为一致有界的。

进一步，对任意的 $x\in \Omega$ 和任意的 $t_1$， $t_2\in \left[0,1\right]$ 且 $t_1<t_2$ 有

$\begin{array}{l}\left|\left(Tx\right)\left(t_{\text{2}}\right)-\left(Tx\right)\left(t_1\right)\right|=|{\displaystyle {\int }_0^{t_2}{\text{e}}^{-\lambda (t_2-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}\\ +A\left(t_2\right)[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)}}{u}^{\alpha -2}\left(s-u\right)du)ds-{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}]\\ -{\displaystyle {\int }_0^{t_1}{\text{e}}^{-\lambda (t_1-s)}}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds\\ -A\left(t_1\right)[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}}}\left(s-u\right)du)ds-{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}]|\end{array}$

$\begin{array}{l}\le |{\displaystyle {\int }_0^{t_2}{\text{e}}^{-\lambda (t_2-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}-{\displaystyle {\int }_0^{t_1}{\text{e}}^{-\lambda (t_2-s)}({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du)ds}}|\\ +|{\displaystyle {\int }_0^{t_{\text{1}}}{\text{e}}^{-\lambda (t_2-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}-{\displaystyle {\int }_0^{t_1}{\text{e}}^{-\lambda (t_1-s)}}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds|\\ +\left|A\left(t_2\right)-A\left(t_1\right)\right||[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du)ds}}\\ -{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}]|\end{array}$

$\begin{array}{l}\le {\displaystyle {\int }_{t_1}^{t_2}{\text{e}}^{-\lambda (t_2-s)}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)ds}+{\displaystyle {\int }_0^{t_1}\left|{\text{e}}^{-\lambda (t_2-s)}-{\text{e}}^{-\lambda (t_2-s)}\right|}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)ds\\ +\frac{2\lambda \left(t_2-t_1\right)}{\Delta }[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|}}{u}^{\alpha -2}\left(s-u\right)du)ds+{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)ds}]\\ \le \frac{R}{\alpha \left(\alpha -1\right)}[{\displaystyle {\int }_{t_1}^{t_2}{\text{e}}^{-\lambda (t_2-s)}{s}^{\alpha }ds}+{\displaystyle {\int }_0^{t_1}\left|{\text{e}}^{-\lambda (t_2-s)}-{\text{e}}^{-\lambda (t_2-s)}\right|s^{\alpha }ds}+\frac{\text{2}\lambda \left(t_2-t_1\right)}{\Delta }\left(\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}{s}^{\alpha }ds}+{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}{s}^{\alpha }ds}\right)]\\ =\frac{R}{\alpha \left(\alpha -1\right)}[\frac{t_2^{\alpha }-t_1^{\alpha }}{\lambda }\left(1-{\text{e}}^{-\lambda t_2}\right)+\frac{t_1^{\alpha }}{\lambda }\left({\text{e}}^{-\lambda (t_2-s)}-{\text{e}}^{-\lambda (t_2-s)}\right)+\frac{\text{2}\lambda \left(t_2-t_1\right)}{\Delta }\frac{\beta {\eta }^{\alpha }\left(1-{\text{e}}^{-\lambda \eta }\right)+1-{\text{e}}^{-\lambda }}{\lambda }]\\ \le \frac{R}{\alpha \left(\alpha -1\right)}\left[\frac{t_2^{\alpha }-t_1^{\alpha }}{\lambda }+\left(t_2-t_1\right)+\frac{\text{2}\left(\beta {\eta }^{\alpha }\left(1-{\text{e}}^{-\lambda \eta }\right)+1-{\text{e}}^{-\lambda }\right)}{\Delta }\left(t_2-t_1\right)\right].\end{array}$

再根据 $t$， $t^{\alpha }$ 在 $\left[0,1\right]$ 上的一致连续性，得 $T\left(\Omega \right)$ 是等度连续的。利用Arzela-Ascoli定理，可知上述引理成立。

3. 主要结果及证明

本文的主要结果是：

定理1设 $f\left(t,0\right)\ne 0$， $\lambda >1$， $0<\beta <1$，且存在非负函数 $k$， $h\in L^1[0,1]$，使下列条件成立：

(C1) $\left|f\left(t,x\right)\right|\le k\left(t\right)\left|x\right|+h\left(t\right)$， $a.e.\left(t,x\right)\in \left[0,1\right]\times R$ ；

(C2) $\frac{\text{2}}{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^1\left(1-s\right)s^{\alpha -2}k\left(s\right)ds}+\frac{\beta }{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^{\eta }\left(\eta -s\right)s^{\alpha -2}k\left(s\right)ds}<1$，

则问题(1.1)至少有一个解 $u*\in C\left[0,1\right]$。

证明：记

$M=\frac{\text{2}}{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^1\left(1-s\right)s^{\alpha -2}k\left(s\right)ds}+\frac{\beta }{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^{\eta }\left(\eta -s\right)s^{\alpha -2}k\left(s\right)ds}$，

$N=\frac{\text{2}}{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^1\left(1-s\right)s^{\alpha -2}h\left(s\right)ds}+\frac{\beta }{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^{\eta }\left(\eta -s\right)s^{\alpha -2}h\left(s\right)ds}$，

由条件(C2)可知， $M<1$。因为 $f\left(t,0\right)\ne 0$，则存在一个区间 $\left[a,b\right]\subset \left[0,1\right]$，使得 $\underset{a\le t\le b}{\min }\left|f\left(t,0\right)\right|>0$。当

$h\left(t\right)\ge \left|f\left(t,0\right)\right|$，对 $a.e.t\in \left[0,1\right]$，得到 $N>0$。

令 $r=N{\left(1-M\right)}^{-1}$， $\Omega =\left\{x\in E:\Vert x\Vert <r\right\}$。假设 $x\in \partial \Omega$， $\theta >1$，使得 $Tx=\theta x$。于是有

$\begin{array}{c}\theta r=\theta \Vert x\Vert =\Vert Tx\Vert =\underset{t\in [0,1]}{\max }|{\displaystyle {\int }_0^t{\text{e}}^{-\lambda (t-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}\\ +A\left(t\right)[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds-{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^{s}f\left(u,x\left(u\right)\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}]|\\ \le {\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (\text{1}-s)}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)ds}+A\left(\text{1}\right)[\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)}\text{d}s\\ +{\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (1-s)}\left({\displaystyle {\int }_0^s\left|f\left(u,x\left(u\right)\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)ds}]\end{array}$

$\begin{array}{c}\le \left(1+A\left(1\right)\right){\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (\text{1}-s)}\left({\displaystyle {\int }_0^{s}k\left(u\right)\left|x\left(u\right)\right|u^{\alpha -2}\left(s-u\right)du}\right)ds}+A\left(\text{1}\right)\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left({\displaystyle {\int }_0^{s}k\left(u\right)\left|x\left(u\right)\right|u^{\alpha -2}\left(s-u\right)\text{d}u}\right)\text{d}s}\\ +\left(1+A\left(1\right)\right){\displaystyle {\int }_0^{\text{1}}{\text{e}}^{-\lambda (\text{1}-s)}\left({\displaystyle {\int }_0^{s}h\left(u\right)u^{\alpha -2}\left(s-u\right)du}\right)ds}+A\left(\text{1}\right)\beta {\displaystyle {\int }_0^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left({\displaystyle {\int }_0^{s}h\left(u\right)u^{\alpha -2}\left(s-u\right)\text{d}u}\right)\text{d}s}\\ \le \frac{2}{1-\beta }{\displaystyle {\int }_0^{\text{1}}\left({\displaystyle {\int }_u^1{\text{e}}^{-\lambda (\text{1}-s)}\left(s-u\right)ds}\right)k\left(u\right)\left|x\left(u\right)\right|u^{\alpha -2}du}+\frac{\beta }{1-\beta }{\displaystyle {\int }_0^{\eta }\left({\displaystyle {\int }_u^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left(s-u\right)ds}\right)k\left(u\right)\left|x\left(u\right)\right|u^{\alpha -2}du}\\ +\frac{2}{1-\beta }{\displaystyle {\int }_0^{\text{1}}\left({\displaystyle {\int }_u^1{\text{e}}^{-\lambda (\text{1}-s)}\left(s-u\right)ds}\right)h\left(u\right)u^{\alpha -2}du}+\frac{\beta }{1-\beta }{\displaystyle {\int }_0^{\eta }\left({\displaystyle {\int }_u^{\eta }{\text{e}}^{-\lambda (\eta -s)}\left(s-u\right)ds}\right)h\left(u\right)u^{\alpha -2}du}\end{array}$

$\begin{array}{c}\le \frac{2}{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^{\text{1}}\left(1-s\right)s^{\alpha -2}k\left(s\right)\left|x\left(s\right)\right|ds}+\frac{\beta }{1-\beta }{\displaystyle {\int }_0^{\eta }\left(1-s\right)s^{\alpha -2}k\left(s\right)\left|x\left(s\right)\right|ds}\\ +\frac{\text{2}}{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^1\left(1-s\right)s^{\alpha -2}h\left(s\right)ds}+\frac{\beta }{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^{\eta }\left(\eta -s\right)s^{\alpha -2}h\left(s\right)ds}\\ =M\Vert x\Vert +N.\end{array}$

进一步，有

$\theta \le M+\frac{N}{r}=M+\frac{N}{N{\left(1-M\right)}^{-1}}=M+\left(1-M\right)=1$。

这与 $\theta >1$ 产生矛盾。由引理2可知， $T$ 有不动点 $u*\in \bar{\Omega }$。另一方面， $f\left(t,0\right)\ne 0$，则问题(1.1)有非平凡解 $u*\in C\left[0,1\right]$。

定理2设 $f\left(t,0\right)\ne 0$， $0<\beta <1$，且存在非负函数 $k$， $h\in L^1\left[0,1\right]$，使条件(C1)，(C3)满足，

(C3)存在一个常数 $\mu >0$，使得

$k\left(s\right)\le B{s}^{\mu },a.e.s\in \left[0,1\right],$

其中

$B=\frac{\lambda \left(1-\beta \right)\left(\alpha -1+\mu \right)\left(\alpha +\mu \right)}{2+\beta {\eta }^{\alpha +\mu }}$。

则问题(1.1)至少有一个解 $u*\in E$。

证明：由定理1的条件可知，要证明定理2，只需要证明 $M<1$ 即可。

$\begin{array}{c}M=\frac{\text{2}}{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^1\left(1-s\right)s^{\alpha -2}k(s)ds}+\frac{\beta }{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^{\eta }\left(\eta -s\right)s^{\alpha -2}k\left(s\right)ds}\\ \le B\left(\frac{\text{2}}{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^1\left(1-s\right)s^{\alpha -2+\mu }ds}+\frac{\beta }{\lambda \left(1-\beta \right)}{\displaystyle {\int }_0^{\eta }\left(\eta -s\right)s^{\alpha -2+\mu }ds}\right)\\ \le B\left(\frac{\text{2}}{\lambda \left(1-\beta \right)}\frac{\text{1}}{\left(\alpha -1+\mu \right)\left(\alpha +\mu \right)}+\frac{\beta }{\lambda \left(1-\beta \right)}\frac{{\eta }^{\alpha +\mu }}{\left(\alpha -1+\mu \right)\left(\alpha +\mu \right)}\right)\\ \le B\frac{\text{2}+\beta {\eta }^{\alpha +\mu }}{\lambda \left(1-\beta \right)\left(\alpha -1+\mu \right)\left(\alpha +\mu \right)}\\ =1.\end{array}$

当 $M<1$ 时，有 $\theta \le 1$。则结论得证。

4. 示例

考虑下面的分数阶微分方程边值问题:

$\{\begin{array}{l}{D}^{\frac{\text{4}}{\text{3}}}\left(D+\text{5}\right)x\left(t\right)=\frac{t}{5}\left|x\right|\cos \sqrt[3]{x}+2t+1,t\in \left[0,1\right],\\ x\left(0\right)=0,x^{\prime }\left(0\right)=0,x\left(1\right)=\frac{1}{2}x(\; 1\; 3\; )\end{array}$

解的存在性，这里 $\alpha =\frac{4}{3}$， $\lambda =5$， $\beta =\frac{1}{2}$， $\eta =\frac{\text{1}}{\text{3}}$。显然，

$f\left(t,x\right)=\frac{t}{5}\left|x\right|\cos \sqrt[3]{x}+2t+1,$

$k\left(t\right)=\frac{t}{2},h\left(t\right)=1$。

容易证得 $k$， $h\in L^1\left[0,1\right]$ 为非负函数，且有

$\left|f\left(t,x\right)\right|\le k\left(t\right)\left|x\right|+h\left(t\right),a.e.\left(t,x\right)\in \left[0,1\right]\times R,$

进一步，有

$\begin{array}{c}M=\frac{\text{2}}{\text{5}\times \left(1-\frac{\text{1}}{\text{2}}\right)}{\displaystyle {\int }_0^1\left(1-s\right)s^{-\frac{2}{3}}\cdot \frac{s}{2}ds}+\frac{\frac{1}{2}}{5\times \left(1-\frac{1}{2}\right)}{\displaystyle {\int }_0^{\frac{1}{3}}\left(\frac{1}{3}-s\right)s^{-\frac{2}{3}}\cdot \frac{s}{2}ds}\\ \approx \text{0}\text{.13105}<1.\end{array}$

故由定理1可知，该分数阶微分方程边值问题有解。

基金项目

国家自然科学基金(11561063)。