ABSTRACT

Fractional calculus is a kind of differential and integral which can deal with any order. It is a generalization of integral calculus. In this paper, we mainly study the positive solutions for the fractional differential equation with integral boundary value problem. By using the Banach fixed point theorem, we obtain the uniqueness of the positive solutions of the equation.

Keywords:Fractional Differential Equation, Positive Solutions, Integral Boundary Value Problem, The Fixed Point Theorem

分数阶微分方程积分边值问题正解的唯一性

卢云雪，王颖^*，吴英昭，李大伟

临沂大学数学与统计学院，山东 临沂

收稿日期：2019年12月6日；录用日期：2019年12月24日；发布日期：2019年12月31日

摘 要

分数阶微分是可以处理任意阶数的微分、积分，是整数阶微积分的推广。本文主要研究分数阶微分方程积分边值问题的正解，应用Banach不动点定理，得到了方程正解的唯一性。

关键词 :分数阶微分方程，正解，积分边值问题，不动点定理

Copyright © 2020 by author(s) and Hans Publishers Inc.

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

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

1. 引言

分数阶微分方程在科学和工程领域的应用，特别是在流变学、电子网络、流体力学、粘弹性以及化学物理学等方面的应用，使得对分数阶微分方程的研究已经变得越来越重要，已成为人们研究的热点 [1] - [9]，本文研究分数阶微分方程积分边值问题(BVP)：

$\{\begin{array}{l}{D}_{0^{+}}^{\alpha }x\left(t\right)+\lambda f\left(t,x\left(t\right)\right)=0,0<t<1,\\ x\left(0\right)=x^{\prime }\left(0\right)=\cdots =x^{\left(n-2\right)}=0,D_{0^{+}}^{\alpha -1}x\left(t\right)={\displaystyle {\int }_0^{1}h\left(t\right)x\left(t\right)\text{d}A\left(t\right)}\end{array}$ (1.1)

其中 $n-1<\alpha \le n,n\ge 2$ 中， $D_{0^{+}}^{\alpha }$ 是Riemann-Liouville微分。 $\lambda >0$ 是参数， $h:\left(0,1\right)\to \left[0,+\infty \right)$ 是连续的并且 $h\in L^1\left(0,1\right)$，${\displaystyle {\int }_0^{1}h\left(s\right)x\left(s\right)\text{d}A\left(s\right)}$ 表示具有广义测度的Riemann-Stieltjes积分， $A:\left(0,1\right)\to \left(-\infty ,+\infty \right)$ 是有界变差函数， ${\displaystyle {\int }_0^{1}h\left(t\right)t^{\alpha -1}\text{d}A\left(t\right)}<\text{Γ}\left(\alpha \right)$。 $f:\left[0,1\right]\times \left[0,+\infty \right)\to \left[0,+\infty \right)$ 是连续函数。

2. 预备知识

定义2.1： [10] [11] (Riemann-Liouville) $\alpha$ 阶积分定义为

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

其中 $n-1<\alpha \le n$，n为整数。

定义2.2： [10] [11] (Riemann-Liouville) $\alpha$ 阶导数定义为

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

其中 $n-1<\alpha \le n$，n为整数。

引理2.1： [10] [11] 若 $\alpha >0$，$x\in L\left(0,1\right)$，$D_{0^{+}}^{\alpha }x\in L\left(0,1\right)$，则

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

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

引理2.2：假设 $y\in C\left(0,1\right)\cap L^1\left(0,1\right)$，则分数阶微分方程

$\{\begin{array}{l}{D}_{0^{+}}^{\alpha }x\left(t\right)+y\left(t\right)=0,t\in \left(0,1\right),n-1<\alpha \le n,n\ge 2,\\ x\left(0\right)=x^{\prime }\left(0\right)=\cdots =x^{\left(n-2\right)}=0,D_{0^{+}}^{\alpha -1}x\left(1\right)={\displaystyle {\int }_0^{1}h\left(t\right)x\left(t\right)\text{d}A\left(t\right)}\end{array}$ (2.1)

有解

$x\left(t\right)={\displaystyle {\int }_0^{\infty }G\left(t,s\right)y\left(s\right)\text{d}s},$

其中

$G\left(t,s\right)=G_0\left(t,s\right)+G_1\left(t,s\right),$ (2.2)

$G_0\left(t,s\right)=\frac{1}{\Gamma \left(\alpha \right)}\{\begin{array}{l}{t}^{\alpha -1}-{\left(t-s\right)}^{\alpha -1},0\le s\le t\le 1,\\ t^{\alpha -1},0\le t\le s\le 1,\end{array}$

$G_1\left(t,s\right)=\frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{1}h\left(t\right)t^{\alpha -1}\text{d}A\left(t\right)}}{\displaystyle {\int }_0^{1}h\left(t\right)G_0\left(t,s\right)\text{d}A\left(t\right)}.$

证明：由引理2.1，(2.1)中的方程可转化为等价于的积分方程

$x\left(t\right)=-I_{0^{+}}^{\alpha }y\left(t\right)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+\cdots +c_n{t}^{\alpha -n},c_i\in \left(-\infty ,+\infty \right),i=1,2,\cdots ,n,$

即

$x\left(t\right)=-\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}y\left(s\right)\text{d}s}+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+\cdots +c_n{t}^{\alpha -n},c_i\in \left(-\infty ,+\infty \right),i=1,2,\cdots ,n,$

由于 $x\left(0\right)=x^{\prime }\left(0\right)=\cdots =x^{\left(n-2\right)}=0$，得 $c_2=c_3=\cdots =c_n=0$，因此有

$x\left(t\right)=-\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}y\left(s\right)\text{d}s}+c_1{t}^{\alpha -1},$

$D_{0^{+}}^{\alpha -1}x\left(t\right)=c_1\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{t}y\left(s\right)\text{d}s}.$

又有 $D_{0^{+}}^{\alpha -1}x\left(1\right)={\displaystyle {\int }_0^{1}h\left(t\right)x\left(t\right)\text{d}A\left(t\right)}$ 及

$D_{0^{+}}^{\alpha -1}x\left(1\right)=c_1\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{1}y\left(s\right)\text{d}s}.$

可得

$c_1=\frac{1}{\Gamma \left(\alpha \right)}\left({\displaystyle {\int }_0^{1}h\left(t\right)x\left(t\right)\text{d}A\left(t\right)}+{\displaystyle {\int }_0^{1}y\left(s\right)\text{d}s}\right),$

所以

$\begin{array}{c}x\left(t\right)=-\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -1}y\left(s\right)\text{d}s}+t^{\alpha -1}\frac{1}{\Gamma \left(\alpha \right)}\left({\displaystyle {\int }_0^{1}h\left(t\right)x\left(t\right)\text{d}A\left(t\right)}+{\displaystyle {\int }_0^{1}y\left(s\right)\text{d}s}\right)\\ ={\displaystyle {\int }_0^1{G}_0\left(t,s\right)y\left(s\right)\text{d}s}+\frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{1}h\left(t\right)x\left(t\right)\text{d}A\left(t\right)}.\end{array}$ (2.3)

对(2.3)式两端乘以 $h\left(t\right)$ 并且求关于 $A\left(t\right)$ 的积分，有

${\displaystyle {\int }_0^{1}h\left(t\right)x\left(t\right)\text{d}A\left(t\right)}=\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{1}h\left(t\right)t^{\alpha -1}\text{d}A\left(t\right)}}{\displaystyle {\int }_0^{1}h\left(t\right)}{\displaystyle {\int }_0^1{G}_0\left(t,s\right)y\left(s\right)\text{d}s\text{d}A\left(t\right)},$

所以

$\begin{array}{c}x\left(t\right)={\displaystyle {\int }_0^1{G}_0\left(t,s\right)y\left(s\right)\text{d}s}+\frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^{1}h\left(t\right)x\left(t\right)\text{d}A\left(t\right)}\\ ={\displaystyle {\int }_0^1{G}_0\left(t,s\right)y\left(s\right)\text{d}s}+\frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{1}h\left(t\right)t^{\alpha -1}\text{d}A\left(t\right)}}{\displaystyle {\int }_0^{1}h\left(t\right)}{\displaystyle {\int }_0^1{G}_0\left(t,s\right)y\left(s\right)\text{d}s\text{d}A\left(t\right)}\\ ={\displaystyle {\int }_0^1{G}_0\left(t,s\right)y\left(s\right)\text{d}s}+\frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{1}h\left(t\right)t^{\alpha -1}\text{d}A\left(t\right)}}{\displaystyle {\int }_0^{1}h\left(t\right)}{\displaystyle {\int }_0^1{G}_0\left(t,s\right)y\left(s\right)\text{d}s\text{d}A\left(t\right)}\\ ={\displaystyle {\int }_0^1{G}_0\left(t,s\right)y\left(s\right)\text{d}s}+{\displaystyle {\int }_0^1{G}_1\left(t,s\right)y\left(s\right)\text{d}s}={\displaystyle {\int }_0^{1}G\left(t,s\right)y\left(s\right)\text{d}s}.\end{array}$

引理2.3：由(2.2)定义的 $G\left(t,s\right)$ 有下列性质：

1) $G\left(t,s\right)\ge 0$，$\left(t,s\right)\in \left[0,1\right]\times \left[0,1\right]$。

2) $G\left(t,s\right)$ 在 $\left[0,1\right]\times \left[0,1\right]$ 上连续。

3) $G\left(t,s\right)\le \omega$，$\omega =\max \left\{\frac{1}{\Gamma \left(\alpha \right)},\frac{{\displaystyle {\int }_0^{1}h\left(t\right)\text{d}A\left(t\right)}}{\Gamma \left(\alpha \right)\left(\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{1}h\left(t\right)t^{\alpha -1}\text{d}A\left(t\right)}\right)}\right\}$。

证明：根据 $G\left(t,s\right)$ 的定义，只需证明(3)成立。由于

$G_0\left(t,s\right)\le \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)}\le \frac{1}{\Gamma \left(\alpha \right)},$

$\begin{array}{c}{G}_1\left(t,s\right)\le \frac{t^{\alpha -1}}{\Gamma \left(\alpha \right)\left(\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{1}h\left(t\right)t^{\alpha -1}\text{d}A\left(t\right)}\right)}{\displaystyle {\int }_0^{1}h\left(t\right)\text{d}A\left(t\right)}\\ \le \frac{1}{\Gamma \left(\alpha \right)\left(\Gamma \left(\alpha \right)-{\displaystyle {\int }_0^{1}h\left(t\right)t^{\alpha -1}\text{d}A\left(t\right)}\right)}{\displaystyle {\int }_0^{1}h\left(t\right)\text{d}A\left(t\right)},\end{array}$

所以 $G\left(t,s\right)=G_0\left(t,s\right)+G_1\left(t,s\right)\le \omega$。

3. 主要结果

设 $X=C\left[0,1\right]$，定义范数 $\Vert x\Vert ={\max }_{0\le t\le 1}\left|x\left(t\right)\right|$。则X是Banach空间，记

$K=\left\{x\in X:x\left(t\right)\ge 0,t\in \left[0,1\right]\right\},$

因此K是X的一个锥。本文，我们假设下面的条件(H₁)成立。

(H₁) $f:\left[0,1\right]\times \left[0,+\infty \right)\to \left[0,+\infty \right)$ 是连续函数。

由(H₁)，定义积分算子 $T:K\to X$ ：

$\left(Tx\right)\left(t\right)=\lambda {\displaystyle {\int }_0^{1}G\left(t,s\right)f\left(s,x\left(s\right)\right)\text{d}s},t\in \left[0,1\right]$ (3.1)

显然BVP(1.1)有解x当且仅当 $x\in K$ 是由(3.1)定义的算子T的不动点。

引理3.1：假设条件(H₁)成立，则 $T:K\to K$ 是全连续算子。

定理3.1：假设条件(H₁)成立，并且存在函数 $m\left(t\right)\ge 0$ 满足下列条件。

(H₂) $\left|f\left(t,x_2\right)-f\left(t,x_1\right)\right|\le m\left(t\right)\left|x_2-x_1\right|,t\in \left[0,1\right],x_1,x_2\in \left[0,+\infty \right)$。

若

${\displaystyle {\int }_0^{1}m\left(s\right)\text{d}s}<\frac{1}{\lambda \omega },$

则BVP(1.1)有唯一正解。

证明：对任意的 $x\in K$，由于 $G\left(t,s\right)\ge 0$，$f\left(t,x\right)\ge 0$，可得 $Tx\left(t\right)\ge 0$，因而 $T\left(K\right)\subseteq K$。

$\begin{array}{c}\Vert T{x}_2-T{x}_1\Vert =\underset{t\in \left[0,1\right]}{\max }\left|T{x}_2\left(t\right)-T{x}_1\left(t\right)\right|\\ =\underset{t\in \left[0,1\right]}{\max }\lambda {\displaystyle {\int }_0^{1}G\left(t,s\right)\left|f\left(s,x_2\left(s\right)\right)-f\left(s,x_1\left(s\right)\right)\right|\text{d}s}\\ \le \lambda {\displaystyle {\int }_0^1\omega m\left(s\right)\left|x_2\left(s\right)-x_1\left(s\right)\right|\text{d}s}\\ \le \lambda \omega {\displaystyle {\int }_0^{1}m\left(s\right)\text{d}s}\Vert x_2-x_1\Vert \\ <\Vert x_2-x_1\Vert \end{array}$

由引理3.1， $T:K\to K$ 是全连续算子，根据Banach不动点理论，算子T在K中有唯一不动点，即为BVP(1.1)的唯一正解。

基金项目

本文受到临沂大学大学生创新创业训练计划项目(X201910452076)部分资助。

文章引用

卢云雪,王 颖,吴英昭,李大伟. 分数阶微分方程积分边值问题正解的唯一性
Uniqueness of Positive Solutions for the Fractional Differential Equation with Integral Boundary Value Problem[J]. 理论数学, 2020, 10(01): 6-10. https://doi.org/10.12677/PM.2020.101002

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