分数阶微分方程边值问题正解的存在性

Existence of Positive Solutions for Boundary Value Problem of Fractional Differential Equations

摘要: 分数阶导数是整数阶导数的推广，分数阶导数有Riemann-Liouville分数阶导数、Marchaud分数阶导数、Caputo分数阶导数等。分数阶微分方程模型具有深刻的物理背景和丰富的理论内涵，在诸多领域应用广泛，如血液流动问题、化学工程、热弹性、地下水流动、人口动力学等。分数阶微分方程边值问题正解的性质是近几年研究的热点之一。在本文中，首先，构造相应线性边值问题的格林函数，其次，分析格林函数的性质，构造合适的锥，再次，利用Guo-Krasnoselskii不动点定理得到了带积分边界条件的分数阶微分方程边值问题正解的存在性结果，最后，通过一个实例说明了结果的合理性。

Abstract: Fractional derivatives are generalizations for derivative of integral order. There are several kinds of fractional derivatives, such as Riemann-Liouville fractional derivative, Marchaud fractional deriva-tive Caputo fractional derivative, etc. Fractional differential equation model has profound physical background and rich theoretical connotation. It is widely used in many fields, such as blood flow problem, chemical engineering, thermoelasticity, groundwater flow, population dynamics and so on. The properties of positive solutions for boundary value problems of fractional differential equations are one of the hot topics in recent years. In this paper, firstly, the Green’s function of the corre-sponding linear boundary value problem is constructed. Secondly, the properties of the Green’s function are analyzed, a suitable cone is constructed. Thirdly, by using Guo-Krasnoselskii fixed point theorem, the existence of positive solutions for boundary value problems of fractional differential equations with integral boundary conditions is obtained. Finally, an example is given to illustrate the rationality of the results.

1. 引言

近几十年来，由于分数阶微分方程及其边值问题在工程、地质科学、控制、物理学、材料科学等领域的广泛应用，已经成为诸多学者研究的热点。分数阶微积分算子具有非局部特性，用其去刻画分形介质中的湍流，弥散，半导体物理，粘弹性材料等具体事物的“记忆”和“遗传”特性将会更加精细准确，这也是分数阶微积分相较于整数阶微积分的最大优势 [1] [2] [3] 。最近许多学者利用Banach压缩映射原理 [4] 、非线性抉择 [5] 、Guo-Krasnoselskii不动点定理 [6] [7] [8] [9] [10] 、单调迭代法 [11] 等研究了分数阶微分方程边值问题解的相关性质。

在文献 [12] 中，作者利用Guo-Krasnoselskii不动点定理，讨论了如下边值问题正解的存在性：

$\{\begin{array}{l}\left(D_{0+}^{\alpha }u\right)\left(t\right)+f\left(t,u\left(t\right)\right)=0,t\in \left(0,1\right),\hfill \\ u\left(0\right)=u^{\prime }\left(0\right)=0,u\left(1\right)=\lambda {\displaystyle {\int }_0^{1}u\left(s\right)\text{d}s}\hfill \end{array}$

其中 $2<\alpha \le 3$ ， $\lambda >0$ ， $\lambda \ne \alpha$ 和 $f:\left[0,1\right]\times \left[0,+\infty \right)\to \left[0,+\infty \right)$ 是连续的。

在文献 [13] 中，作者利用Leray-Schauder非线性抉择和Guo-Krasnoselskii不动点定理研究了下列边值问题正解的存在性结果：

$\{\begin{array}{l}\left({}^{C}D{}_{0+}^{\alpha }u\right)\left(t\right)+f\left(t,u\left(t\right)\right)=0,t\in \left(0,1\right),\hfill \\ \begin{array}{l}{u}^{″}\left(0\right)=u^{‴}\left(0\right)=0,\\ u^{\prime }\left(0\right)=u\left(1\right)=\eta {\displaystyle {\int }_0^{1}u\left(s\right)\text{d}s}\end{array}\hfill \end{array}$

其中， $3<\alpha <4$ ， $0<\eta <2$ 和 $f:\left[0,1\right]\times \left[0,+\infty \right)\to \left[0,+\infty \right)$ 是连续的。

本文受到上述参考文献的启发，研究了下列带有积分边值条件的分数阶微分方程边值问题正解的存在性：

$\{\begin{array}{l}\left({}^{C}D{}_{0+}^{\alpha }u\right)\left(t\right)+f\left(t,u\left(t\right)\right)=0,t\in \left[0,1\right],\hfill \\ \begin{array}{l}{u}^{″}\left(0\right)=u^{‴}\left(0\right)=0,\\ u^{\prime }\left(0\right)={\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s},u^{\prime }\left(0\right)={\displaystyle {\int }_0^1{\eta }_2\left(s\right)u\left(s\right)\text{d}s,}\end{array}\hfill \end{array}$ (1)

其中， $3<\alpha <4$ ， $f:\left[0,1\right]\times \left[0,+\infty \right)\to \left[0,+\infty \right)$ 和 ${\eta }_i\left(i=1,2\right):\left[0,1\right]\to \left[0,+\infty \right)$ 是连续的。

2. 预备知识

下面将介绍与本文相关的定义和引理，见文献 [14] [15] [16] 。设 $\mathbb{N}=\left\{1,2,3,\cdots \right\},p,q>0$ ，且 $\left[q\right]$ 表示q的整数部分。

2.1. 定义1 [14]

$\left[0,1\right]$ 上的q阶Riemann-Liouville分数阶积分 $I_{0+}^{q}u$ 定义为

$\left(I_{0+}^{q}u\right)\left(t\right):=\frac{1}{\Gamma \left(q\right)}{\displaystyle {\int }_0^t\frac{u\left(s\right)\text{d}s}{{\left(t-s\right)}^{1-q}}}.$

2.2. 定义2 [14]

$\left[0,1\right]$ 上的q阶Riemann-Liouville分数阶导数 $D_{0+}^{q}u$ 定义为

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

其中 $n=\left[q\right]+1$ 。

2.3. 定义2 [14]

令 $D_{0+}^q\left[u\left(s\right)\right]\left(t\right)\equiv \left(D_{0+}^{q}u\right)\left(t\right)$ 为q阶的Riemann-Liouville分数阶导数，那么 $\left[0,1\right]$ 上的q阶Caputo分数阶导数 ${}^{C}D{}_{0+}^{q}u$ 通过以上Riemann-Liouville分数阶导数定义为

$\left({}^{C}D{}_{0+}^{q}u\right)\left(t\right):=\left(D_{0+}^q\left[u\left(s\right)-{\displaystyle \underset{k=0}{\overset{n-1}{\sum }}\frac{u^{\left(k\right)}\left(0\right)}{k!}{s}^k}\right]\right)\left(t\right),$

其中

$n=\{\begin{array}{l}\left[q\right]+1,q\notin \left\{1,2,3,\cdots \right\},\hfill \\ q,q\in \left\{1,2,3,\cdots \right\}.\hfill \end{array}$ (2)

2.4. 引理1 [14]

设n是由(2)给出。如果 $u\in C^n\left[0,1\right]$ ，那么

$\left(I_{0+}^q{}^{C}D{}_{0+}^{q}u\right)\left(t\right)=u\left(t\right)+c_0+c_{1}t+c_2{t}^2+\cdots +c_{n-1}{t}^{n-1},$

其中， $c_i\in ℝ,i=0,1,\cdots ,n-1$ 。

2.5. 引理2 [15]

如果 $p>q$ ， $u\in C\left[0,1\right]$ ，那么对于任意的 $t\in \left[0,1\right]$ 方程 $\left({}^{C}D{}_{0+}^q{I}_{0+}^{p}u\right)\left(t\right)=\left(I_{0+}^{p-q}u\right)\left(t\right)$ 成立。

2.6. 引理3 [15]

设n由(2)给出，那么有以下关系式成立：

(1) 当 $k\in \left\{0,1,2,\cdots ,n-1\right\}$ 时， ${}^{C}D{}_{0+}^q{t}^k=0$ ；

(2) 如果 $p>n$ ，那么 ${}^{C}D{}_{0+}^q{t}^{p-1}=\frac{\Gamma \left(p\right)}{\Gamma \left(p-q\right)}{t}^{p-q-1}$ 。

2.7. 引理4 [16]

(Guo-Krasnoselskii不动点定理)设E为实Banach空间，K是E中的锥， ${\Omega }_1$ 和 ${\Omega }_2$ 是E中的有界开子集，并且 $\theta \in {\Omega }_1$ ， ${\bar{\Omega }}_1\subset {\Omega }_2$ 。若全连续算子 $T:K\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)\to K$ 满足下述条件之一：

(1) $\Vert Tu\Vert \le \Vert u\Vert ,u\in K\cap \partial {\Omega }_1$ 且 $\Vert Tu\Vert \ge \Vert u\Vert ,u\in K\cap \partial {\Omega }_2$ ；

(2) $\Vert Tu\Vert \ge \Vert u\Vert ,u\in K\cap \partial {\Omega }_1$ 且 $\Vert Tu\Vert \le \Vert u\Vert ,u\in K\cap \partial {\Omega }_2$ ，

则T在 $K\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)$ 中有一个不动点。

3. 主要结果

设 $C\left[0,1\right]$ 是定义在 $\left[0,1\right]$ 上的所有连续函数组成的Banach空间，定义其范数为

$\Vert u\Vert =\underset{0\le t\le 1}{\max }\left|u\left(t\right)\right|$ .

为方便起见，在下文中定义

$P_i={\displaystyle {\int }_0^1\left(s-1\right){\eta }_i\left(s\right)\text{d}s},i=1,2;Q_i={\displaystyle {\int }_0^1{\eta }_i\left(s\right)\text{d}s},i=1,2.$

3.1. 引理5

设 $\left(1-P_1\right)\left(1-Q_2\right)-P_2{Q}_1\ne 0$ ，那么对给定的 $y\in C\left[0,1\right]$ ，则边值问题

$\{\begin{array}{l}\left({}^{C}D{}_{0+}^{\alpha }u\right)\left(t\right)+y\left(t\right)=0,t\in \left[0,1\right],\hfill \\ \begin{array}{l}{u}^{″}\left(0\right)=u^{‴}\left(0\right)=0,\\ u^{\prime }\left(0\right)={\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s}\text{,}{u}^{\prime }\left(0\right)={\displaystyle {\int }_0^1{\eta }_2\left(s\right)u\left(s\right)\text{d}s}\end{array}\hfill \end{array}$ (3)

有唯一解

$u\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)y\left(s\right)\text{d}s}\text{,}t\in \left[0,1\right]$

其中

$G\left(t,s\right)=H\left(t,s\right)+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i\left(t\right){\displaystyle {\int }_0^{1}H\left(\tau ,s\right){\eta }_i\left(\tau \right)\text{d}\tau }},\left(t,s\right)\in \left[0,1\right]\times \left[0,1\right],$

且

$H\left(t,s\right)=\frac{1}{\Gamma \left(\alpha \right)}\{\begin{array}{l}\begin{array}{cc}\left[{\left(1-s\right)}^{\alpha -1}-{\left(t-s\right)}^{\alpha -1}\right],& 0\le s\le t\le 1,\end{array}\hfill \\ \begin{array}{cc}{\left(1-s\right)}^{\alpha -1},& 0\le t\le s\le 1,\end{array}\hfill \end{array}$

${\varphi }_1\left(t\right)=\frac{P_2+\left(t-1\right)\left(1-Q_2\right)}{\left(1-P_1\right)\left(1-Q_2\right)-P_2{Q}_1},t\in \left[0,1\right]$

和

${\varphi }_1\left(t\right)=\frac{1-P_1+\left(t-1\right)Q_1}{\left(1-P_1\right)\left(1-Q_2\right)-P_2{Q}_1},t\in \left[0,1\right].$

证明 由(3)中的微分方程和引理1可得

$u\left(t\right)=-\left(I_{0+}^{\alpha }y\right)\left(t\right)-c_0-c_{1}t-c_2{t}^2-c_3{t}^3,t\in \left[0,1\right],$

因此

$u^{\prime }\left(t\right)=-\left(I_{0+}^{\alpha -1}y\right)\left(t\right)-c_1-2c_{2}t-3c_3{t}^2,t\in \left[0,1\right],$

$u^{″}\left(t\right)=-\left(I_{0+}^{\alpha -2}y\right)\left(t\right)-2c_2-6c_{3}t,t\in \left[0,1\right]$

和

$u^{‴}\left(t\right)=-\left(I_{0+}^{\alpha -3}y\right)\left(t\right)-6c_3,t\in \left[0,1\right].$

由上式和(3)中的边值条件可得

$c_2=c_3=0$ ，

$c_0=-\left(I_{0+}^{\alpha }y\right)\left(1\right)+{\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s}-{\displaystyle {\int }_0^1{\eta }_2\left(s\right)u\left(s\right)\text{d}s}$

和

$c_1={\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s}.$

从而

$\begin{array}{c}u\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}\left\{{\displaystyle {\int }_0^t\left[{\left(1-s\right)}^{\alpha -1}-{\left(t-s\right)}^{\alpha -1}\right]y\left(s\right)\text{d}s}+{\displaystyle {\int }_t^1{\left(1-s\right)}^{\alpha -1}y\left(s\right)\text{d}s}\right\}+\left(t-1\right){\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s}+{\displaystyle {\int }_0^1{\eta }_2\left(s\right)u\left(s\right)\text{d}s}\\ ={\displaystyle {\int }_0^{1}G\left(t,s\right)y\left(s\right)\text{d}s}+\left(t-1\right){\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s}+{\displaystyle {\int }_0^1{\eta }_2\left(s\right)u\left(s\right)\text{d}s}.\end{array}$ (4)

由(4)可得

$\left(1-P_1\right){\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s}-Q_1{\displaystyle {\int }_0^1{\eta }_2\left(s\right)u\left(s\right)\text{d}s}={\displaystyle {\int }_0^1{\eta }_1\left(s\right){\displaystyle {\int }_0^{1}H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau \text{d}s}}$

和

$-P_2{\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s}+\left(1-Q_2\right){\displaystyle {\int }_0^1{\eta }_2\left(s\right)u\left(s\right)\text{d}s}={\displaystyle {\int }_0^1{\eta }_2\left(s\right){\displaystyle {\int }_0^{1}H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau \text{d}s}}.$

由此可以得到

${\displaystyle {\int }_0^1{\eta }_1\left(s\right)u\left(s\right)\text{d}s}=\frac{\left(1-Q_2\right){\displaystyle {\int }_0^1{\eta }_1\left(s\right){\displaystyle {\int }_0^{1}H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau \text{d}s}}+Q_1{\displaystyle {\int }_0^1{\eta }_2\left(s\right){\displaystyle {\int }_0^{1}H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau \text{d}s}}}{\left(1-P_1\right)\left(1-Q_2\right)-P_2{Q}_1}$

和

${\displaystyle {\int }_0^1{\eta }_2\left(s\right)u\left(s\right)\text{d}s}=\frac{P_2{\displaystyle {\int }_0^1{\eta }_1\left(s\right){\displaystyle {\int }_0^{1}H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau \text{d}s}}+\left(1-P_1\right){\displaystyle {\int }_0^1{\eta }_2\left(s\right){\displaystyle {\int }_0^{1}H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau \text{d}s}}}{\left(1-P_1\right)\left(1-Q_2\right)-P_2{Q}_1},$

再结合(4)可得

$\begin{array}{l}u\left(t\right)={\displaystyle {\int }_0^{1}H\left(t,s\right)y\left(s\right)\text{d}s}+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i}\left(t\right){\displaystyle {\int }_0^1{\eta }_i\left(s\right)H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau \text{d}s}\\ ={\displaystyle {\int }_0^{1}H\left(t,s\right)y\left(s\right)\text{d}s}+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i}\left(t\right){\displaystyle {\int }_0^1{\eta }_i\left(\tau \right)H\left(\tau ,s\right)y\left(s\right)\text{d}s\text{d}\tau }\\ ={\displaystyle {\int }_0^{1}H\left(t,s\right)y\left(s\right)\text{d}s}+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i}\left(t\right){\displaystyle {\int }_0^{1}y\left(s\right){\displaystyle {\int }_0^{1}H\left(\tau ,s\right){\eta }_i\left(\tau \right)\text{d}\tau \text{d}\tau }}\\ ={\displaystyle {\int }_0^1\left[H\left(t,s\right)+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i}\left(t\right){\displaystyle {\int }_0^{1}H\left(\tau ,s\right){\eta }_i\left(\tau \right)\text{d}\tau }\right]y\left(s\right)\text{d}s}\\ ={\displaystyle {\int }_0^{1}G\left(t,s\right)y\left(s\right)\text{d}s.}\end{array}$

3.2. 引理6

$H\left(t,s\right)$ 满足如下性质：

(1) $H\left(t,s\right)\le H\left(s,s\right),\left(t,s\right)\in \left[0,1\right]\times \left[0,1\right]$ ；

(2) 存在 $a,b\in ℝ$ ，满足 $0<a<b<1$ ，使得 $H\left(t,s\right)\ge \lambda H\left(s,s\right),\left(t,s\right)\in \left[0,1\right]\times \left[0,1\right]$ ，其中

$\lambda =\underset{0\le a\le t\le b<1}{\min }\left\{1-t^{\alpha -1}\right\}$

证明 (1)显然成立，下面证明(2)成立。

当 $s\le t$ 时，

$\begin{array}{c}H\left(t,s\right)=\frac{1}{\Gamma \left(\alpha \right)}\left[{\left(1-s\right)}^{\alpha -1}-{\left(t-s\right)}^{\alpha -1}\right]\\ \ge \frac{1}{\Gamma \left(\alpha \right)}\left[{\left(1-s\right)}^{\alpha -1}-{\left(t-ts\right)}^{\alpha -1}\right]\\ =\frac{1}{\Gamma \left(\alpha \right)}{\left(1-s\right)}^{\alpha -1}\left(1-t^{\alpha -1}\right)\\ \ge \lambda H\left(s,s\right),\end{array}$

当 $s\ge t$ 时，(2)显然成立。

在后文中，总是假设下列条件成立：

$P_1\le 1$ ， $Q_2\ge 1$ 和 $\left(1-P_1\right)\left(1-Q_2\right)>P_2{Q}_1$ .

3.3. 引理7

$G\left(t,s\right)$ 满足如下性质：

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

(2) $G\left(t,s\right)\ge \lambda mH\left(s,s\right),\left(t,s\right)\in \left[a,b\right]\times \left[0,1\right]$ ，其中 $\lambda$ 由引理6给出且

$m=1+{\displaystyle \underset{i=1}{\overset{2}{\sum }}\underset{0\le t\le 1}{\min }{\varphi }_t\left(t\right){\displaystyle {\int }_0^1{\eta }_i\left(\tau \right)\text{d}\tau }}$

和

$M=1+{\displaystyle \underset{i=1}{\overset{2}{\sum }}\underset{0\le t\le 1}{\max }{\varphi }_t\left(t\right){\displaystyle {\int }_0^1{\eta }_i\left(\tau \right)\text{d}\tau }}.$

证明 由引理6中的(1)可知

$\begin{array}{l}G\left(t,s\right)=H\left(t,s\right)+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i\left(t\right){\displaystyle {\int }_0^{1}H\left(\tau ,s\right){\eta }_i\left(\tau \right)\text{d}\tau }}\\ \le H\left(s,s\right)+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i\left(t\right){\displaystyle {\int }_0^{1}H\left(s,s\right){\eta }_i\left(\tau \right)\text{d}\tau }}\\ =\left(1+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i\left(t\right){\displaystyle {\int }_0^1{\eta }_i\left(\tau \right)\text{d}\tau }}\right)H\left(s,s\right)\\ \le MH\left(s,s\right),\left(t,s\right)\in \left[0,1\right]\times \left[0,1\right],\end{array}$

由引理6中的(2)可知

$\begin{array}{l}G\left(t,s\right)=H\left(t,s\right)+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i\left(t\right){\displaystyle {\int }_0^{1}H\left(\tau ,s\right){\eta }_i\left(\tau \right)\text{d}\tau }}\\ \ge \lambda H\left(s,s\right)+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i\left(t\right){\displaystyle {\int }_0^1\lambda H\left(s,s\right){\eta }_i\left(\tau \right)\text{d}\tau }}\\ =\left(1+{\displaystyle \underset{i=1}{\overset{2}{\sum }}{\varphi }_i\left(t\right){\displaystyle {\int }_0^1{\eta }_i\left(\tau \right)\text{d}\tau }}\right)\lambda H\left(s,s\right)\\ \ge \lambda mH\left(s,s\right),\left(t,s\right)\in \left[a,b\right]\times \left[0,1\right].\end{array}$

令 $E=C\left[0,1\right]$ ，定义其范数 $\Vert u\Vert =\underset{0\le t\le 1}{\max }\left|u\left(t\right)\right|$ ，并且 $K=\left\{u\in E:u\left(t\right)\ge \omega \Vert u\Vert ,t\in \left[a,b\right]\right\}$ ，其中 $\omega =\frac{\lambda m}{M}$ ，不难得到K是E中的锥。

在K上定义算子T如下：

$\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)\text{d}s},u\in K,t\in \left[0,1\right],$

显然，如果u是算子T的不动点，则u是边值问题(1)的非负解。

3.4. 定理8

假设 $f:\left[0,1\right]\times \left[0,+\infty \right)\to \left[0,+\infty \right)$ 连续且满足下面条件：

(H1) 存在一个常数 $r>0$ 使得

$f\left(t,x\right)\le G_{1}r,\left(t,x\right)\in \left[0,1\right]\times \left[0,r\right]$ ，其中 $G_1=\frac{1}{M{\displaystyle {\int }_0^{1}H\left(s,s\right)\text{d}s}}$ ；

(H2) 存在一个常数 $R>0$ 且 $R\ne r$ 使得

$f\left(t,x\right)\ge G_{2}R,\left(t,x\right)\in \left[a,b\right]\times \left[\frac{\lambda m}{M}R,R\right]$ ，其中 $G_2=\frac{1}{\lambda m\omega {\displaystyle {\int }_a^{b}H\left(s,s\right)\text{d}s}}$ ，

则边值问题(1)存在一个正解满足 $\min \left\{r,R\right\}\le \Vert u\Vert \le \max \left\{r,R\right\}$ 。

证明对任意的 $u\in K$ ，由引理7可知

$\begin{array}{c}\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)\text{d}s}\\ \le M{\displaystyle {\int }_0^{1}H\left(s,s\right)f\left(s,u\left(s\right)\right)\text{d}s},\end{array}$

因此

$\Vert Tu\Vert \le M{\displaystyle {\int }_0^{1}H\left(s,s\right)f\left(s,u\left(s\right)\right)\text{d}s},$

结合引理7可知

$\begin{array}{c}\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)\text{d}s}\\ \ge \lambda M{\displaystyle {\int }_0^{1}H\left(s,s\right)f\left(s,u\left(s\right)\right)\text{d}s}\\ \ge \frac{\lambda m}{M}\Vert Tu\Vert ,\end{array}$

这意味着 $Tu\in K$ ，此外，通过Arzela-Ascoli定理容易得到 $T:K\to K$ 是全连续算子。

设 ${\Omega }_1=\left\{u\in E:\Vert u\Vert <r\right\}$ ， ${\Omega }_2=\left\{u\in E:\Vert u\Vert <R\right\}$ 。

一方面，对任意的 $u\in K\cap \partial {\Omega }_1$ ，由引理6和(H1)可得

$\begin{array}{c}\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)\text{d}s}\\ \le M{G}_1{\displaystyle {\int }_0^{1}H\left(s,s\right)u\left(s\right)\text{d}s}\\ \le M{G}_1\Vert u\Vert {\displaystyle {\int }_0^{1}H\left(s,s\right)\text{d}s}\\ \le \Vert u\Vert ,t\in \left[0,1\right],\end{array}$

这意味着

$\Vert Tu\Vert \le \Vert u\Vert ,u\in K\cap \partial {\Omega }_1,$ (5)

另一方面，对任意的 $u\in K\cap \partial {\Omega }_2$ ，由引理6和(H2)可得

$\begin{array}{c}\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right)\right)\text{d}s}\\ \ge \lambda m{G}_2{\displaystyle {\int }_0^{1}H\left(s,s\right)u\left(s\right)\text{d}s}\\ \ge \lambda m{G}_2\omega \Vert u\Vert {\displaystyle {\int }_0^{1}H\left(s,s\right)\text{d}s}\\ \ge \Vert u\Vert ,t\in \left[a,b\right],\end{array}$

这意味着

$\Vert Tu\Vert \ge \Vert u\Vert ,u\in K\cap \partial {\Omega }_2.$ (6)

由(5)，(6)和引理4可知算子T有一个不动点 $u\in K\cap \left({\bar{\Omega }}_2\backslash {\Omega }_1\right)$ ，显然这个不动点是边值问题(1)的正解。

4. 实例分析

考虑边值问题

$\{\begin{array}{l}\left({}^{C}D{}_{0+}^{\frac{7}{2}}u\right)\left(t\right)+\frac{5\sqrt{\pi t}}{4}{u}^2\left(t\right)=0,t\in \left[0,1\right],\hfill \\ \begin{array}{l}{u}^{″}\left(0\right)=u^{‴}\left(0\right)=0,\\ u^{\prime }\left(0\right)={\displaystyle {\int }_0^{1}su\left(s\right)\text{d}s},u\left(1\right)={\displaystyle {\int }_0^1\left(1-s\right)u\left(s\right)\text{d}s.}\end{array}\hfill \end{array}$ (7)

由于 ${\eta }_1\left(s\right)=s$ ， ${\eta }_2\left(s\right)=1-s$ ， $s\in \left[0,1\right]$ 。经过简单计算可得 $P_1=-\frac{1}{6}$ ， $P_2=-\frac{1}{3}$ ， $Q_1=\frac{1}{2}$ ， $Q_2=\frac{1}{2}$ ，显然， $\left(1-P_1\right)\left(1-Q_2\right)-P_2{Q}_1\ne 0$ 。

又因为

$\underset{0\le t\le 1}{\max }{\varphi }_1\left(t\right)=-\frac{4}{9}$ ， $\underset{0\le t\le 1}{\min }{\varphi }_1\left(t\right)=-\frac{10}{9}$

和

$\underset{0\le t\le 1}{\max }{\varphi }_2\left(t\right)=\frac{14}{9}$ ， $\underset{0\le t\le 1}{\min }{\varphi }_2\left(t\right)=\frac{8}{9}$ .

因此可得 $M=\frac{14}{9}$ ， $m=\frac{8}{9}$ 。

由于 $\alpha =\frac{7}{2}$ ，若选择 $a=\frac{1}{4}$ ， $b=\frac{3}{4}$ ，则计算可得 $\lambda =\frac{32-9\sqrt{3}}{32}$ ， $G_1=\frac{135\sqrt{\pi }}{32}$ 和 $G_2=\frac{47040\sqrt{\pi }}{1331\sqrt{\pi }-1868}$ 。

设 $f\left(t,x\right)=\frac{5\sqrt{\pi t}{x}^2}{4},\left(t,x\right)\in \left[0,1\right]\times \left[0,+\infty \right)$ ，如果取 $r=\frac{8}{27}$ ， $R=\frac{2940}{1331\sqrt{\pi }-1868}$ ，则可以得到

$f\left(t,x\right)\le f\left(1,\frac{8}{27}\right)=G_{1}r,\left(t,x\right)\in \left[0,1\right]\times \left[0,\frac{8}{27}\right]$

和

$f\left(t,x\right)\ge f\left(\frac{1}{4},\frac{2940}{1331\sqrt{\pi }-1868}\right)=G_{2}R,\left(t,x\right)\in \left[\frac{1}{4},\frac{3}{4}\right]\times \left[\frac{3360-945\sqrt{3}}{2662\sqrt{\pi }-3736},\frac{2940}{1331\sqrt{\pi }-1868}\right].$

这意味着定理8的所有条件均满足，所以，由定理8可知，边值问题(7)存在正解。