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

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。

Abstract: Fractional calculus is a kind of differential and integral which can deal with any order. It is a gen-eralization of integral calculus. In this paper, we mainly study the positive solutions for the frac-tional 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.

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。