分数阶微分方程无穷多点边值问题的正解

Positive Solutions for Infinite-Point Boundary Value Problems of Fractional Differential Equations

摘要: 本文研究了带有积分边界条件的分数阶微分方程无穷多点边值问题的正解。该边值问题正解的存在性是通过Green函数的性质和不动点定理获得的。首先求出非线性系统对于线性系统的Green函数，之后给出Green函数的性质并构造合适的积分算子，然后通过使用不动点定理得到边值问题正解的存在性结果。最后，本文给出具体实例来说明得到定理的实用性。

Abstract: This paper studies the positive solutions of the infinite-point boundary value problem for fractional differential equations with integral boundary conditions. The existence of positive solutions for boundary value problems is obtained through the properties of Green’s function and the fixed point theorem. Firstly, Green’s function for the nonlinear system relative to the linear system is derived. Then, the properties of the Green’s function are presented and a suitable integral operator is constructed. Finally, the existence results of positive solutions for the boundary value problem are obtained by using the fixed point theorem. At the end, specific example is provided to illustrate the practicality of the obtained theorems.

1. 引言

近年来，分数阶微分方程在信号处理与控制、反常扩散和流体力学等领域都有着广泛的应用[1]。目前，对于非线性系统不含有导数项的分数阶微分方程的研究，主要的研究成果有分数阶泛函微分方程[2] [3]、分数阶脉冲微分方程[4] [5]、分数阶模糊微分方程[6]、分数阶随机微分方程[7] [8]和q阶分数阶微分方程[9]-[12]。近年来，许多学者在分数阶微分方程的理论研究中取得了一些成果，在这些研究的成果中，分数阶微分方程边值问题解和正解的存在性是研究的重点[13]-[15]。许多研究成果是将边值问题通过积分变换或拉普拉斯变换转化为等价的积分方程，之后通过应用不动点定理、拓扑度理论和上下解等方法得到边值问题解和正解的存在性[16] [17]。

在2011年，Gao和Han [18]讨论了下列具有非局部边界条件的分数阶微分方程边值问题

$\{\begin{array}{l}{D}_{0+}^{\alpha }u\left(t\right)+a\left(t\right)f\left(t,u\left(t\right)\right)=0,t\in \left(0,1\right),\hfill \\ u\left(0\right)=0,u\left(1\right)={\displaystyle \sum _{i=1}^{\infty }{\alpha }_{i}u\left({\xi }_i\right)}\hfill \end{array}$

正解的存在性，其中$D_{0+}^{\alpha }$ 表示Riemann-Liouville分数阶导数，$1<\alpha \le 2$ ，${\xi }_i\in \left(0,1\right)$ ，${\displaystyle \sum _{i=1}^{\infty }{\alpha }_i{\xi }_i^{\alpha -1}<1}$ ，${\alpha }_i\in \left[0,+\infty \right)$ ，$a\left(t\right)\in C\left(\left[0,1\right],\left[0,+\infty \right)\right)$ 并且在$\left[a,b\right]\subset \left(0,1\right)$ 上，$a\left(t\right)\ne 0$ ，$f\left(t,u\right)\in C(\left[0,1\right]\times \left[0,+\infty \right),\left[0,+\infty \right))$ 。他们利用了Guo-krasnoselskii不动点定理和Leggett-Williams不动点定理。

在2020年，通过使用Guo-Krasnoselskii不动点定理，Shen和Zhou [19]研究了下列具有积分边界条件的分数阶微分方程无穷多点边值问题

$\{\begin{array}{l}{D}_{0+}^{\alpha }u\left(t\right)+h\left(t,u\left(t\right)\right)=0,t\in \left[0,1\right],\hfill \\ u^{\left(i\right)}\left(0\right)=0,i=0,1,\cdots ,n-2,\hfill \\ D_{0+}^{\beta }u\left(1\right)={\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\displaystyle {\int }_0^{{\eta }_i}u\left(s\right)\text{d}s+{\displaystyle \sum _{i=1}^{\infty }{\gamma }_{i}u\left({\eta }_i\right)}}}\hfill \end{array}$

正解的存在性，其中$D_{0+}^{\alpha }$ 表示Riemann-Liouville分数阶导数，$n\in N^{+}$ ，$n\ge 3$ ，$n-1<\alpha \le n$ ，$1\le \beta \le \alpha -1$ ，$0<{\eta }_1<{\eta }_2<\cdots <{\eta }_i<\cdots <1$ ，$\xi =\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}-\frac{1}{\alpha }{\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\eta }_i^{\alpha }-{\displaystyle \sum _{i=1}^{\infty }{\gamma }_i{\eta }_i^{\alpha -1}}}>0$ ，${\beta }_i,{\gamma }_i>0$ ，$\left(i=1,2,\cdots \right)$ ，$h\in C\left(\left[0,1\right]\times \left[0,+\infty \right),\left[0,+\infty \right)\right)$ 。

在上述工作的启发下，本文研究了下列非线性项中含有未知函数导数项的分数阶微分方程边值问题

$\{\begin{array}{l}{D}_{0+}^{\alpha }u\left(t\right)+f\left(t,u\left(t\right),D_{0+}^{p}u\left(t\right)\right)=0,t\in \left[0,1\right],\hfill \\ u^{\left(i\right)}\left(0\right)=0,i=0,1,\cdots ,n-2,\hfill \\ D_{0+}^{\beta }u\left(1\right)={\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\displaystyle {\int }_0^{{\eta }_i}u\left(s\right)\text{d}s+{\displaystyle \sum _{i=1}^{\infty }{\gamma }_{i}u\left({\xi }_i\right)}}}\hfill \end{array}$ (1)

正解的存在性。其中$D_{0+}^{\alpha }$ 表示Riemann-Liouville分数阶导数，$n\in N^{+}$ ，$n\ge 3$ ，$n-1<\alpha \le n$ ，$1\le \beta \le \alpha -1$ ，$0\le p\le \beta -1$ ，$0<{\eta }_1<{\eta }_2<\cdots <{\eta }_i<\cdots <1$ ，$0<{\xi }_1<{\xi }_2<\cdots <{\xi }_i<\cdots <1$ ，${\beta }_i,{\gamma }_i>0\left(i=1,2,\cdots \right)$ ，

$\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}-\frac{1}{\alpha }{\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\eta }_i^{\alpha }-{\displaystyle \sum _{i=1}^{\infty }{\gamma }_i{\xi }_i^{\alpha -1}}}>0$ ，$f\in C$ $(\left[0,1\right]$ $\times \left[0,+\infty \right)\times \left[0,+\infty \right),\left[0,+\infty \right))$ 。

2. 预备知识

首先给出与分数阶微分方程相关的一些基本定义和引理。假设$\mu >0$ 是一个常数并且$\left[\mu \right]$ 表示$\mu$ 的整数部分。

2.1. 定义1 [1]

$\left[0,1\right]$ 上的$\mu$ 阶Riemann-Liouville分数阶积分$I_{0+}^{\mu }u$ 和$I_{1-}^{\mu }u$ 分别为

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

和

$\left(I_{1-}^{\mu }u\right)\left(t\right)=\frac{1}{\Gamma \left(\mu \right)}{\displaystyle {\int }_t^1\frac{u\left(s\right)\text{d}s}{{\left(s-t\right)}^{1-\mu }}}$ .

2.2. 定义2 [1]

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

$\begin{array}{c}\left(D_{0+}^{\mu }u\right)\left(t\right)={\left(\frac{\text{d}}{\text{d}t}\right)}^m\left(I_{0+}^{m-\mu }u\right)\left(t\right)\\ =\frac{1}{\Gamma \left(m-\mu \right)}{\left(\frac{\text{d}}{\text{d}t}\right)}^m{\displaystyle {\int }_0^t\frac{u\left(s\right)\text{d}s}{{\left(t-s\right)}^{\mu -m+1}}}\end{array}$

和

$\begin{array}{c}\left(D_{1-}^{\mu }u\right)\left(t\right)={\left(-\frac{\text{d}}{\text{d}t}\right)}^m\left(I_{1-}^{m-\mu }u\right)\left(t\right)\\ =\frac{1}{\Gamma \left(m-\mu \right)}{\left(-\frac{\text{d}}{\text{d}t}\right)}^m{\displaystyle {\int }_t^1\frac{u\left(s\right)\text{d}s}{{\left(s-t\right)}^{\mu -m+1}}},\end{array}$

其中$m=\left[\mu \right]+1$ 。

2.3. 引理1 [1]

设$u\in C\left[0,1\right]\cap \left[0,1\right]$ 具有$\mu$ 阶导数且$D_{0+}^{\mu }u\in C\left[0,1\right]\cap L\left[0,1\right]$ ，则有

$I_{0+}^{\mu }{D}_{0+}^{\mu }u\left(t\right)=u\left(t\right)+{C^{\prime }}_1{t}^{\mu -1}+{C^{\prime }}_2{t}^{\mu -2}+\cdots +{C^{\prime }}_N{t}^{\mu -N}$ ,

其中，${C^{\prime }}_i\in R,i=1,2,\cdots ,N$ ，

$N=\{\begin{array}{l}\left[\mu \right]+1,\mu \notin N^{+},\hfill \\ \mu ,\mu \in N^{+}.\hfill \end{array}$

2.4. 引理2 [1]

设$\mu >0$ ，$\rho >0$ ，$u\in C\left[0,1\right]\cap L\left[0,1\right]$ ，则有$D_{0+}^{\mu }{I}_{0+}^{\rho }u\left(t\right)=I_{0+}^{\rho -\mu }x\left(t\right)$ 成立。

2.5. 引理3 [20]

设$E$ 是Banach空间，$P$ 是$E$ 中的锥，${\Omega }_1$ 和${\Omega }_2$ 是$E$ 中的有界开子集，并且$0\in {\Omega }_1$ ，${\overline{\Omega }}_1\subset {\Omega }_2$ 。若全连续算子$T:P\cap \left({\overline{\Omega }}_2\backslash {\Omega }_1\right)\to P$ 满足下列条件之一：

(1) $\Vert Tx\Vert \le \Vert x\Vert$ ，$\forall x\in P\cap \partial {\Omega }_1$ 且$\Vert Tx\Vert \ge \Vert x\Vert$ ，$\forall x\in P\cap \partial {\Omega }_2$ ；

(2) $\Vert Tx\Vert \ge \Vert x\Vert$ ，$\forall x\in P\cap \partial {\Omega }_1$ 且$\Vert Tx\Vert \le \Vert x\Vert$ ，$\forall x\in P\cap \partial {\Omega }_2$ ，那么$T$ 在$P\cap \left({\overline{\Omega }}_2\backslash {\Omega }_1\right)$ 中有一个不动点。

2.6. 引理4 [21]

设$P$ 是Banach空间$E$ 上的锥，$\alpha ,\gamma :P\to R^{+}$ 是单调递增且非负的连续函数，$\theta :P\to R^{+}$ 是非负的连续函数，存在$N,a_3$ 使得

(1) 对于所有的$x\in \overline{P}\left(\gamma ,a_3\right)$ ，有$\theta \left(0\right)=0$ ，$\gamma \left(x\right)\le \theta \left(x\right)\le \alpha \left(x\right)$ ，$\Vert x\Vert \le N\gamma \left(x\right)$ ；

(2) 存在$0<a_1<a_2<a_3$ ，对于$\forall \lambda \in \left[0,1\right]$ ，$x\in \partial P\left(\theta ,a_2\right)$ ，有$\theta \left(\lambda x\right)\le \lambda \theta \left(x\right)$ ；

(3) $T:\overline{P}\left(\gamma ,a_3\right)\to P$ 是一个完全连续算子；

(4) 对于$x\in \partial P\left(\gamma ,a_3\right)$ ，有$\gamma \left(Tx\right)>a_3$ ，对于$x\in \partial P\left(\theta ,a_2\right)$ ，有$\theta \left(Tx\right)<a_2$ ，对于$x\in \partial P\left(\alpha ,a_1\right)$ ，有$P\left(\alpha ,a_1\right)\ne \varphi$ 并且$\alpha \left(Tx\right)>a_1$ ，其中$P\left(\alpha ,a_1\right)=\left\{x\in P:\alpha \left(x\right)<a_1\right\}$ 。

那么$T$ 至少有两个不动点$x_1,x_2\in P\left(\gamma ,a_3\right)$ 使得$a_1<\alpha \left(x_1\right)$ ，$\theta \left(x_1\right)<a_2<\theta \left(x_2\right)$ ，$\gamma \left(x_2\right)<a_3$ 。

2.7. 引理5

设$h\in L^1\left[0,1\right]$ ，则下面边值问题

$\{\begin{array}{l}{D}_{0+}^{\alpha }u\left(t\right)+h\left(t\right)=0,t\in \left[0,1\right],\hfill \\ u^{\left(i\right)}\left(0\right)=0,i=0,1,\cdots ,n-2,\hfill \\ D_{0+}^{\beta }u\left(1\right)={\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\displaystyle {\int }_0^{{\eta }_i}u\left(s\right)\text{d}s+{\displaystyle \sum _{i=1}^{\infty }{\gamma }_{i}u\left({\xi }_i\right)}}}\hfill \end{array}$ (2)

有唯一解

$u\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)h\left(s\right)\text{d}s}$ ,

其中

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

$m\left(s\right)=\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}-\frac{1}{\alpha }{{\displaystyle \sum _{s\le {\eta }_i}{\beta }_i\left(\frac{{\eta }_i-s}{1-s}\right)}}^{\alpha }{\left(1-s\right)}^{\beta +1}-{{\displaystyle \sum _{s\le {\xi }_i}{\gamma }_i\left(\frac{{\xi }_i-s}{1-s}\right)}}^{\alpha -1}{\left(1-s\right)}^{\beta },$

所以

$D_{0+}^{p}u\left(t\right)={\displaystyle {\int }_0^{1}H\left(t,s\right)h\left(s\right)\text{d}s}$

其中

$H\left(t,s\right)=\frac{1}{m\left(0\right)\Gamma \left(\alpha -\beta \right)}\{\begin{array}{l}m\left(s\right)t^{\alpha -p-1}{\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(t-s\right)}^{\alpha -p-1},0\le s\le t\le 1,\hfill \\ m\left(s\right)t^{\alpha -p-1}{\left(1-s\right)}^{\alpha -\beta -1},0\le t\le s\le 1.\hfill \end{array}$

证明 令$C_i=-{C^{\prime }}_i$ ，通过2.3引理1和边值问题(2)得到

$\begin{array}{c}u\left(t\right)=-I_{0+}^{\alpha }h\left(t\right)-{C^{\prime }}_1{t}^{\alpha -1}-{C^{\prime }}_2{t}^{\alpha -2}-\cdots -{C^{\prime }}_n{t}^{\alpha -n}\\ =-I_{0+}^{\alpha }h\left(t\right)+C_1{t}^{\alpha -1}+C_2{t}^{\alpha -2}+\cdots +C_n{t}^{\alpha -n}.\end{array}$

由边值条件$u^{\left(i\right)}\left(0\right)=0,i=0,1,\cdots ,n-2$ ，可推出$C_2=C_3=\cdots =C_n=0$ 。那么

$u\left(t\right)=-I_{0+}^{\alpha }h\left(t\right)+C_1{t}^{\alpha -1}$ .

考虑到边值条件

$D_{0+}^{\beta }u\left(1\right)={\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\displaystyle {\int }_0^{{\eta }_i}u\left(s\right)\text{d}s+{\displaystyle \sum _{i=1}^{\infty }{\gamma }_{i}u\left({\xi }_i\right)}}}$ ,

可以得出

$\begin{array}{c}{D}_{0+}^{\beta }u\left(1\right)={\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\displaystyle {\int }_0^{{\eta }_i}u\left(s\right)\text{d}s+{\displaystyle \sum _{i=1}^{\infty }{\gamma }_{i}u\left({\xi }_i\right)}}}\\ =-\frac{1}{\Gamma \left(\alpha -\beta \right)}{{\displaystyle {\int }_0^1\left(1-s\right)}}^{\alpha -\beta -1}h\left(s\right)\text{d}s+C_1\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}\\ ={\displaystyle \sum _{i=1}^{\infty }{\beta }_i\left[-I^{\alpha +1}h\left({\eta }_i\right)+\frac{C_1{\eta }_i^{\alpha }}{\alpha }\right]}+{\displaystyle \sum _{i=1}^{\infty }{\gamma }_i}\left[-I_{0+}^{\alpha }h\left({\xi }_i\right)+C_1{\xi }_i^{\alpha -1}\right],\end{array}$

经过推导可以得出

$\begin{array}{l}{C}_1\left[\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}-{\displaystyle \sum _{i=1}^{\infty }\frac{{\beta }_i{\eta }_i^{\alpha }}{\alpha }-{\displaystyle \sum _{i=1}^{\infty }{\gamma }_i{\xi }_i^{\alpha -1}}}\right]\\ =\frac{1}{\Gamma \left(\alpha -\beta \right)}{\displaystyle {\int }_0^1{\left(1-s\right)}^{\alpha -\beta -1}}h\left(s\right)\text{d}s-{\displaystyle \sum _{i=1}^{\infty }\frac{{\beta }_i}{\Gamma \left(\alpha +1\right)}{\displaystyle {\int }_0^{{\eta }_i}{\left({\eta }_i-s\right)}^{\alpha }h\left(s\right)\text{d}s}}\\ -{\displaystyle \sum _{i=1}^{\infty }\frac{{\gamma }_i}{\Gamma \left(\alpha \right)}}{\displaystyle {\int }_0^{{\xi }_i}{\left({\xi }_i-s\right)}^{\alpha -1}}h\left(s\right)\text{d}s,\end{array}$

所以

$C_1=\frac{1}{m\left(0\right)}\left[{\displaystyle {\int }_0^1\frac{{\left(1-s\right)}^{\alpha -\beta -1}}{\Gamma \left(\alpha -\beta \right)}h\left(s\right)\text{d}s-{\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\displaystyle {\int }_0^{{\eta }_i}\frac{{\left({\eta }_i-s\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}h\left(s\right)\text{d}s-{\displaystyle \sum _{i=1}^{\infty }{\gamma }_i{\displaystyle {\int }_0^{{\xi }_i}\frac{{\left({\xi }_i-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}h\left(s\right)\text{d}s}}}}}\right],$

所以，边值问题(2)的解为

$\begin{array}{c}u\left(t\right)=-I_{0+}^{\alpha }h\left(t\right)+C_1{t}^{\alpha -1}\\ =-{\displaystyle {\int }_0^t\frac{1}{\Gamma \left(\alpha \right)}{\left(t-s\right)}^{\alpha -1}h\left(s\right)\text{d}s}+\frac{1}{\Gamma \left(\alpha \right)m\left(0\right)}[{\displaystyle {\int }_0^1\frac{{\left(1-s\right)}^{\alpha -\beta -1}}{\Gamma \left(\alpha -\beta \right)}}h\left(s\right)\text{d}s\\ -{\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\displaystyle {\int }_0^{{\eta }_i}\frac{{\left({\eta }_i-s\right)}^{\alpha }}{\Gamma \left(\alpha +1\right)}}}h\left(s\right)\text{d}s-{\displaystyle \sum _{i=1}^{\infty }{\gamma }_i{\displaystyle {\int }_0^{{\xi }_i}\frac{{\left({\xi }_i-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)}h\left(s\right)\text{d}s}}]\cdot t^{\alpha -1}\\ =-{\displaystyle {\int }_0^t\frac{1}{\Gamma \left(\alpha \right)}}{\left(t-s\right)}^{\alpha -1}h\left(s\right)\text{d}s+\frac{1}{\Gamma \left(\alpha \right)m\left(0\right)}{\displaystyle {\int }_0^1{\left(1-s\right)}^{\alpha -\beta -1}}[\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}\\ -\frac{1}{\alpha }{\displaystyle \sum _{s\le {\eta }_i}{\beta }_i{\left({\eta }_i-s\right)}^{\alpha }{\left(1-s\right)}^{\beta +1-\alpha }}-{\displaystyle \sum _{s\le {\xi }_i}{\gamma }_i{\left({\xi }_i-s\right)}^{\alpha -1}{\left(1-s\right)}^{\beta +1-\alpha }}]\cdot h\left(s\right)\text{d}s\cdot t^{\alpha -1}\\ =-{\displaystyle {\int }_0^t\frac{1}{\Gamma \left(\alpha \right)}{\left(t-s\right)}^{\alpha -1}h\left(s\right)\text{d}s+{\displaystyle {\int }_0^1\frac{{\left(1-s\right)}^{\alpha -\beta -1}m\left(s\right)h\left(s\right)}{\Gamma \left(\alpha \right)m\left(0\right)}\text{d}s\cdot t^{\alpha -1}}}\\ =-{\displaystyle {\int }_0^t\frac{m\left(0\right){\left(t-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)m\left(0\right)}h\left(s\right)\text{d}s}+{\displaystyle {\int }_0^t\frac{{\left(1-s\right)}^{\alpha -\beta -1}m\left(s\right)t^{\alpha -1}}{\Gamma \left(\alpha \right)m\left(0\right)}h\left(s\right)\text{d}s}+{\displaystyle {\int }_t^1\frac{{\left(1-s\right)}^{\alpha -\beta -1}m\left(s\right)t^{\alpha -1}}{\Gamma \left(\alpha \right)m\left(0\right)}h\left(s\right)\text{d}s}\\ ={\displaystyle {\int }_0^t\frac{m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}{t}^{\alpha -1}-m\left(0\right){\left(t-s\right)}^{\alpha -1}}{\Gamma \left(\alpha \right)m\left(0\right)}h\left(s\right)\text{d}s}+{\displaystyle {\int }_t^1\frac{m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}{t}^{\alpha -1}}{\Gamma \left(\alpha \right)m\left(0\right)}h\left(s\right)\text{d}s}\\ ={\displaystyle {\int }_0^{1}G\left(t,s\right)h\left(s\right)\text{d}s}.\end{array}$

更多地，经过推导有

$\begin{array}{c}{D}_{0+}^{p}u\left(t\right)=-D_{0+}^p{I}_{0+}^{\alpha }h\left(t\right)+C_1{D}_{0+}^p{t}^{\alpha -1}\\ =-\frac{1}{\Gamma \left(\alpha -p\right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\alpha -p-1}h\left(s\right)\text{d}s}\\ +\frac{t^{\alpha -p-1}}{m\left(0\right)\Gamma \left(\alpha -p\right)}{\displaystyle {\int }_0^1{\left(1-s\right)}^{\alpha -\beta -1}}m\left(s\right)h\left(s\right)\text{d}s\\ ={\displaystyle {\int }_0^t\frac{m\left(s\right)t^{\alpha -p-1}{\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(t-s\right)}^{\alpha -p-1}}{m\left(0\right)\Gamma \left(\alpha -p\right)}h\left(s\right)\text{d}s}\\ +{\displaystyle {\int }_t^1\frac{m\left(s\right)t^{\alpha -p-1}{\left(1-s\right)}^{\alpha -\beta -1}}{m\left(0\right)\Gamma \left(\alpha -p\right)}h\left(s\right)\text{d}s}\\ ={\displaystyle {\int }_0^{1}H\left(t,s\right)h\left(s\right)\text{d}s}\end{array}$

证明结束。

2.8. 引理6

如果$m\left(0\right)>0$ ，那么函数$m\left(s\right)$ 在$s\in \left[0,1\right]$ 上是不减的并且$m\left(s\right)>0$ 。

证明 通过公式

$m\left(s\right)=\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}-\frac{1}{\alpha }{{\displaystyle \sum _{s\le {\eta }_i}{\beta }_i\left(\frac{{\eta }_i-s}{1-s}\right)}}^{\alpha }{\left(1-s\right)}^{\beta +1}-{{\displaystyle \sum _{s\le {\xi }_i}{\gamma }_i\left(\frac{{\xi }_i-s}{1-s}\right)}}^{\alpha -1}{\left(1-s\right)}^{\beta },s\in \left[0,1\right]$

可以得出

$m\left(0\right)=\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}-\frac{1}{\alpha }{\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\eta }_i^{\alpha }-{\displaystyle \sum _{i=1}^{\infty }{\gamma }_i{\xi }_i^{\alpha -1}}}>0$

并且

$\begin{array}{l}{m}^{\prime }\left(s\right)={\displaystyle \sum _{s\le {\eta }_i}{\beta }_i{\left({\eta }_i-s\right)}^{\alpha -1}{\left(1-s\right)}^{\beta +1-\alpha }}+\frac{\beta +1-\alpha }{\alpha }{\displaystyle \sum _{s\le {\eta }_i}{\beta }_i{\left({\eta }_i-s\right)}^{\alpha }{\left(1-s\right)}^{\beta -\alpha }}\\ +\left(\alpha -1\right){\displaystyle \sum _{s\le {\xi }_i}{\gamma }_i{\left({\xi }_i-s\right)}^{\alpha -2}{\left(1-s\right)}^{\beta +1-\alpha }}+\left(\beta +1-\alpha \right){\displaystyle \sum _{s\le {\xi }_i}{\gamma }_i{\left({\xi }_i-s\right)}^{\alpha -1}{\left(1-s\right)}^{\beta -\alpha }}\\ ={\displaystyle \sum _{s\le {\eta }_i}{\beta }_i}{\left({\eta }_i-s\right)}^{\alpha -1}{\left(1-s\right)}^{\beta -\alpha }\left[\left(1-{\eta }_i\right)+\frac{\beta +1}{\alpha }\left({\eta }_i-s\right)\right]\\ +{\displaystyle \sum _{s\le \xi {}_i}{\gamma }_i{\left({\xi }_i-s\right)}^{\alpha -2}{\left(1-s\right)}^{\beta -\alpha }}\left[\left(\alpha -1\right)\left(1-{\xi }_i\right)+\beta \left({\xi }_i-s\right)\right]\ge 0,s\in \left[0,1\right],\end{array}$

所以$m\left(s\right)$ 在$\left[0,1\right]$ 上是不减的并且$m\left(s\right)>0$ 。证明结束。

2.9. 引理7

函数$G\left(t,s\right)$ 满足下列条件：

(1) $G\left(t,s\right)\ge 0,\frac{\partial G\left(t,s\right)}{\partial t}\ge 0,0\le t,s\le 1$ ；

(2) $G\left(t,s\right)\ge t^{\alpha -1}G\left(1,s\right),0\le t,s\le 1$ ；

(3) ${\max }_{t\in \left[0,1\right]}G\left(t,s\right)=G\left(1,s\right),0\le s\le 1;$ ；

(4) $H\left(t,s\right)\ge 0,\frac{\partial H\left(t,s\right)}{\partial t}\ge 0,0\le t,s\le 1$ ；

(5) $H\left(t,s\right)\ge t^{\alpha -p-1}H\left(1,s\right),0\le t,s\le 1$ ；

(6) ${\max }_{t\in \left[0,1\right]}H\left(t,s\right)=H\left(1,s\right),0\le s\le 1$ 。

证明 (1) 当$0\le s\le t\le 1$ 时，通过2.7引理5可以得到

$\begin{array}{c}G\left(t,s\right)=\frac{1}{m\left(0\right)\Gamma \left(\alpha \right)}\left[m\left(s\right)t^{\alpha -1}{\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(t-s\right)}^{\alpha -1}\right]\\ =\frac{t^{\alpha -1}}{m\left(0\right)\Gamma \left(\alpha \right)}\left[m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(1-\frac{s}{t}\right)}^{\alpha -1}\right]\\ \ge \frac{m\left(s\right)t^{\alpha -1}}{m\left(0\right)\Gamma \left(\alpha \right)}\left[{\left(1-s\right)}^{\alpha -\beta -1}-{\left(1-\frac{s}{t}\right)}^{\alpha -1}\right]\\ \ge \frac{m\left(s\right)t^{\alpha -1}}{m\left(0\right)\Gamma \left(\alpha \right)}\left[{\left(1-s\right)}^{\alpha -1}-{\left(1-\frac{s}{t}\right)}^{\alpha -1}\right]\\ =\frac{m\left(s\right)}{m\left(0\right)\Gamma \left(\alpha \right)}\left[{\left(t-ts\right)}^{\alpha -1}-{\left(t-s\right)}^{\alpha -1}\right]\ge 0.\end{array}$

同样地，当$0\le t\le s\le 1$ 时可以得到

$G\left(t,s\right)=\frac{1}{m\left(0\right)\Gamma \left(\alpha \right)}\left[m\left(s\right)t^{\alpha -1}{\left(1-s\right)}^{\alpha -\beta -1}\right]\ge 0.$

对Green函数$G\left(t,s\right)$ 关于$t$ 求偏导可以得到

$\frac{\partial G\left(t,s\right)}{\partial t}=\frac{1}{m\left(0\right)\Gamma \left(\alpha \right)}\{\begin{array}{l}\left(\alpha -1\right)m\left(s\right)t^{\alpha -2}{\left(1-s\right)}^{\alpha -\beta -1}-\left(\alpha -1\right)m\left(0\right){\left(t-s\right)}^{\alpha -2},0\le s\le t\le 1,\hfill \\ \left(\alpha -1\right)m\left(s\right)t^{\alpha -2}{\left(1-s\right)}^{\alpha -\beta -1},0\le t\le s\le 1.\hfill \end{array}$

显然$\frac{\partial G\left(t,s\right)}{\partial t}$ 在$\left[0,1\right]\times \left[0,1\right]$ 上是连续的并且当$0\le t\le s\le 1$ 时$\frac{\partial G\left(t,s\right)}{\partial t}\ge 0$ 。接下来只需考虑当$0\le s\le t\le 1$ 时$\frac{\partial G\left(t,s\right)}{\partial t}$ 的性质。当$0\le s\le t\le 1$ 时，

$\begin{array}{c}\frac{\partial G\left(t,s\right)}{\partial t}=\frac{1}{m\left(0\right)\Gamma \left(\alpha \right)}\left[\left(\alpha -1\right)m\left(s\right)t^{\alpha -2}{\left(1-s\right)}^{\alpha -\beta -1}-\left(\alpha -1\right)m\left(0\right){\left(t-s\right)}^{\alpha -2}\right]\\ \ge \frac{\left(\alpha -1\right)m\left(0\right)t^{\alpha -2}}{m\left(0\right)\Gamma \left(\alpha \right)}\left[{\left(1-s\right)}^{\alpha -\beta -1}-{\left(1-\frac{s}{t}\right)}^{\alpha -2}\right]\\ \ge \frac{\left(\alpha -1\right)t^{\alpha -2}}{\Gamma \left(\alpha \right)}\left[{\left(1-s\right)}^{\alpha -2}-{\left(1-\frac{s}{t}\right)}^{\alpha -2}\right]\\ =\frac{\left(\alpha -1\right)}{\Gamma \left(\alpha \right)}\left[{\left(t-ts\right)}^{\alpha -2}-{\left(t-s\right)}^{\alpha -2}\right]\ge 0.\end{array}$

(2) 当$0\le s\le t\le 1$ 时可以得到

$\begin{array}{c}G\left(t,s\right)=\frac{1}{m\left(0\right)\Gamma \left(\alpha \right)}\left[m\left(s\right)t^{\alpha -1}{\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(t-s\right)}^{\alpha -1}\right]\\ =\frac{t^{\alpha -1}}{m\left(0\right)\Gamma \left(\alpha \right)}\left[m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(1-\frac{s}{t}\right)}^{\alpha -1}\right]\\ \ge \frac{t^{\alpha -1}}{m\left(0\right)\Gamma \left(\alpha \right)}\left[m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(1-s\right)}^{\alpha -1}\right]\\ =t^{\alpha -1}G\left(1,s\right).\end{array}$

当$0\le t\le s\le 1$ 时可以得出

$G\left(t,s\right)=\frac{1}{m\left(0\right)\Gamma \left(\alpha \right)}{t}^{\alpha -1}m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}\ge t^{\alpha -1}G\left(1,s\right).$

(3) 由于$\frac{\partial G\left(t,s\right)}{\partial t}\ge 0$ ，所以可以得出$G\left(t,s\right)$ 在$0\le t\le 1$ 上是不减的。所以

${\max }_{t\in \left[0,1\right]}G\left(t,s\right)=G\left(1,s\right)=\frac{1}{m\left(0\right)\Gamma \left(\alpha \right)}[m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(1-s\right)}^{\alpha -1}],0\le s\le 1.$

(4) 当$0\le s\le t\le 1$ 时有

$\begin{array}{c}H\left(t,s\right)=\frac{m\left(s\right)t^{\alpha -p-1}{\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(t-s\right)}^{\alpha -p-1}}{m\left(0\right)\Gamma \left(\alpha -\beta \right)}\\ \ge \frac{m\left(s\right)t^{\alpha -p-1}}{m\left(0\right)\Gamma \left(\alpha -p\right)}\left[{\left(1-s\right)}^{\alpha -\beta -1}-{\left(1-\frac{s}{t}\right)}^{\alpha -p-1}\right]\\ \ge \frac{m\left(s\right)t^{\alpha -p-1}}{m\left(0\right)\Gamma \left(\alpha -p\right)}\left[{\left(1-s\right)}^{\alpha -\beta -1}-{\left(1-s\right)}^{\alpha -p-1}\right]\ge 0.\end{array}$

并且

$\begin{array}{c}\frac{\partial H\left(t,s\right)}{\partial t}=\frac{\left(\alpha -p-1\right)m\left(s\right)t^{\alpha -p-2}{\left(1-s\right)}^{\alpha -\beta -1}-\left(\alpha -p-1\right)m\left(0\right){\left(t-s\right)}^{\alpha -p-2}}{m\left(0\right)\Gamma \left(\alpha -p\right)}\\ \ge \frac{\left(\alpha -p-1\right)m\left(0\right)t^{\alpha -p-2}\left[{\left(1-s\right)}^{\alpha -\beta -1}-{\left(1-s\right)}^{\alpha -p-2}\right]}{m\left(0\right)\Gamma \left(\alpha -p\right)}\ge 0.\end{array}$

当$0\le t\le s\le 1$ 时，$H\left(t,s\right)\ge 0$ 并且$\frac{\partial H\left(t,s\right)}{\partial t}\ge 0$ 是显然的。

(5) 当$0\le t\le s\le 1$ 时有

$H\left(t,s\right)=\frac{m\left(s\right)t^{\alpha -p-1}{\left(1-s\right)}^{\alpha -\beta -1}}{m\left(0\right)\Gamma \left(\alpha -p\right)}\ge t^{\alpha -p-1}H\left(1,s\right).$

当$0\le s\le t\le 1$ 时有

$\begin{array}{c}H\left(t,s\right)=\frac{1}{m\left(0\right)\Gamma \left(\alpha -p\right)}\left[m\left(s\right)t^{\alpha -p-1}{\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(t-s\right)}^{\alpha -p-1}\right]\\ =\frac{t^{\alpha -p-1}}{m\left(0\right)\Gamma \left(\alpha -p\right)}\left[m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(1-\frac{s}{t}\right)}^{\alpha -p-1}\right]\\ \ge \frac{t^{\alpha -p-1}}{m\left(0\right)\Gamma \left(\alpha -p\right)}\left[m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(1-s\right)}^{\alpha -p-1}\right]\\ =t^{\alpha -p-1}H\left(1,s\right).\end{array}$

(6) 通过(4)的性质可以得到

${\max }_{t\in \left[0,1\right]}H\left(t,s\right)=H\left(1,s\right)=\frac{1}{m\left(0\right)\Gamma \left(\alpha -p\right)}\left[m\left(s\right){\left(1-s\right)}^{\alpha -\beta -1}-m\left(0\right){\left(1-s\right)}^{\alpha -p-1}\right],0\le s\le 1.$

3. 主要结果

在本节中，得到了关于问题(1)正解的存在性的一些主要结果。

在后续部分，定义空间$E=\left\{u|u\in C\left[0,1\right],D_{0+}^{p}u\in C\left[0,1\right]\right\}$ ，则$E$ 是一个巴拿赫空间[22]。其上的范数为

$\Vert u\Vert ={\Vert u\Vert }_1+{\Vert u\Vert }_2={\max }_{t\in \left[0,1\right]}\left|u\left(t\right)\right|+{\max }_{t\in \left[0,1\right]}\left|D_{0+}^{p}u\left(t\right)\right|.$

在该空间上定义锥为

$P=\left\{u\left(t\right)\in E|u\left(t\right)\ge 0,u\left(t\right)+D_{0+}^{p}u\left(t\right)\ge {\left(\frac{1}{4}\right)}^{\alpha -1}\Vert u\Vert ,t\in \left[\frac{1}{4},1\right]\right\}.$

之后，定义算子$T:P\to P$ 为

$\left(Tu\right)\left(t\right)={\displaystyle \underset{0}{\overset{1}{\int }}G\left(t,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s},t\in \left[\frac{1}{4},1\right].$

根据2.7引理5可以得到

$D_{0+}^p\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{1}H\left(t,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s},t\in \left[\frac{1}{4},1\right].$

在接下来的证明过程中，定义

$\Vert Tu\Vert ={\Vert Tu\Vert }_1+{\Vert Tu\Vert }_2={\max }_{t\in \left[\frac{1}{4},1\right]}\left|\left(Tu\right)\left(t\right)\right|+{\max }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^p\left(Tu\right)\left(t\right)\right|.$

那么，讨论边值问题(1)正解的存在性就转变为讨论算子T的不动点问题。由于使用不动点定理可以得到算子T的不动点的存在性，算子T的不动点就是边值问题(1)的正解，故边值问题(1)正解的存在性就可以得出，也即本文讨论的问题。本节主要应用了Guo-Krasnoselskii不动点定理和Twin不动点定理得到了边值问题(1)正解的存在性。

3.1. 定理1

$T:P\to P$ 是完全连续算子。

证明 对于$\forall u\in P$ 都有

$\begin{array}{c}\left(Tu\right)\left(t\right)+D_{0+}^p\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(t,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}}\\ \ge t^{\alpha -1}{\displaystyle {\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s+t^{\alpha -p-1}{\displaystyle {\int }_0^{1}H\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}}\\ \ge t^{\alpha -1}{[\max }_{t\in \left[\frac{1}{4},1\right]}\left|\left(Tu\right)\left(t\right)\right|+{\max }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^p\left(Tu\right)\left(t\right)\right|]=t^{\alpha -1}\Vert Tu\Vert ,\end{array}$

当$t\in \left[\frac{1}{4},1\right]$ 时可以得到

$\left(Tu\right)\left(t\right)+D_{0+}^p\left(Tu\right)\left(t\right)\ge {\left(\frac{1}{4}\right)}^{\alpha -1}\Vert Tu\Vert ,$

所以$Tu\in P$ 并且$T$ 是一个$P\to P$ 的算子。

接下来，证明$T$ 在$P$ 是有界的。

令

$k_1={\displaystyle {\int }_0^{1}G\left(1,s\right)\text{d}s},k_2={\displaystyle {\int }_0^{1}H\left(1,s\right)\text{d}s}.$

若$Z$ 是$P$ 上的任意有界集，那么$\exists L\ge 0,$ 对于$\forall u\in Z$ ，使得$\Vert u\Vert \le L$ ，由于$f$ 是连续函数，所以对于$\forall u\in Z$ ，$\left(t,u\left(t\right),D_{0+}^{p}u\left(t\right)\right)\in \left[0,1\right]\times \left[0,L\right]\times \left[0,L\right]$ ，$\exists M>0$ ，使得$f\left(t,u\left(t\right),D_{0+}^{p}u\left(t\right)\right)\le M$ 。所以可以推出

${\Vert Tu\Vert }_1={\max }_{t\in \left[\frac{1}{4},1\right]}\left|\left(Tu\right)\left(t\right)\right|={\displaystyle \underset{0}{\overset{1}{\int }}G\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}\le k_{1}M,$

${\Vert Tu\Vert }_2={\max }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^p\left(Tu\right)\left(t\right)\right|={\displaystyle \underset{0}{\overset{1}{\int }}H\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}\le k_{2}M.$

那么

$\begin{array}{c}\Vert Tu\Vert ={\Vert Tu\Vert }_1+{\Vert Tu\Vert }_2={\max }_{t\in \left[\frac{1}{4},1\right]}\left|\left(Tu\right)\left(t\right)\right|+{\max }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^p\left(Tu\right)\left(t\right)\right|\\ ={\displaystyle \underset{0}{\overset{1}{\int }}G\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{1}H}\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s\\ \le M\left(k_1+k_2\right),\end{array}$

所以$T\left(Z\right)$ 是一致有界的。

接下来，证明$T$ 是等度连续的。由于$G\left(t,s\right)$ 和$H\left(t,s\right)$ 在$\left[0,1\right]\times \left[0,1\right]$ 上是连续的，令$t_1,t_2,s\in \left[\frac{1}{4},1\right]$ ，对于$\forall \epsilon >0$ ，$\exists \delta >0$ ，当$\left|t_2-t_1\right|<\delta$ 时有

$\left|G\left(t_2,s\right)-G\left(t_1,s\right)\right|<\frac{\epsilon }{2M+1},\left|H\left(t_2,s\right)-H\left(t_1,s\right)\right|<\frac{\epsilon }{2M+1}.$

那么

$\begin{array}{c}\Vert \left(Tu\right)\left(t_2\right)-\left(Tu\right)\left(t_1\right)\Vert ={\Vert \left(Tu\right)\left(t_2\right)-\left(Tu\right)\left(t_1\right)\Vert }_1+{\Vert \left(Tu\right)\left(t_2\right)-\left(Tu\right)\left(t_1\right)\Vert }_2\\ \le {\displaystyle {\int }_0^1\left|G\left(t_2,s\right)-G\left(t_1,s\right)\right|}\left|f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\right|\text{d}s\\ +{\displaystyle {\int }_0^1\left|H\left(t_2,s\right)-H\left(t_1,s\right)\right|}\left|f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\right|\text{d}s\\ <\frac{\epsilon }{2M+1}M+\frac{\epsilon }{2M+1}M<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .\end{array}$

由此可以得到算子T满足Arzela-Ascoli定理的所有条件，根据Arzela-Ascoli定理可以得到算子T是相对紧的，这就说明算子T是完全连续的，证明结束。

在本章剩余部分，记

$f_{\infty }={\lim }_{\left(u,v\right)\to +\infty }{\inf }_{t\in \left[0,1\right]}\frac{f\left(t,u,v\right)}{u+v},f^0={\lim }_{\left(u,v\right)\to 0}{\sup }_{t\in \left[0,1\right]}\frac{f\left(t,u,v\right)}{u+v}$

$L_1=\frac{1}{{\displaystyle {\int }_0^{1}G\left(1,s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(1,s\right)\text{d}s}}},L_2=\frac{1}{{\displaystyle {\int }_0^{1}G\left(\frac{1}{4},s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(\frac{1}{4},s\right)\text{d}s}}}$

3.2. 定理2

对于$f\in C\left(\left[0,1\right]\times \left[0,+\infty \right)\times \left[0,+\infty \right),\left[0,+\infty \right)\right)$ ，若$f^0=0$ ，$f_{\infty }=+\infty$ ，那么边值问题(1)有一个正解。

证明 由于$f_{\infty }=+\infty$ ，所以可以得到$\exists r_2>0$ ，使得

$f\left(t,u,v\right)\ge {\delta }_1\left(u+v\right)\ge {\delta }_1{\left(\frac{1}{4}\right)}^{\alpha -1}\Vert u\Vert$

对于$\left(t,u,v\right)\in \left[0,1\right]\times \left[r_2,+\infty \right)\times \left[r_2,+\infty \right)$ ，其中${\delta }_1\ge 4^{\alpha -1}{\left[{\displaystyle {\int }_0^{1}G\left(1,s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(1,s\right)\text{d}s}}\right]}^{-1}$ 。

同样地，由于$f^0=0$ ，那么$\exists r_2>r_1>0$ 使得

$f\left(t,u,v\right)\le {\delta }_2\left(u+v\right)\le {\delta }_2\Vert u\Vert$

对于$\left(t,u,v\right)\in \left[0,1\right]\times \left[0,r_1\right]\times \left[0,r_1\right]$ ，其中${\delta }_2\le {\left[{\displaystyle {\int }_0^{1}G\left(1,s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(1,s\right)\text{d}s}}\right]}^{-1}$ 。

令${\Omega }_1=\left\{u\in P:\Vert u\Vert <r_1\right\}$ ，${\Omega }_2=\left\{u\in P:\Vert u\Vert <r_2\right\}$ 。

当$u\in P\cap \partial {\Omega }_2$ 时，可以得到

$\begin{array}{c}\Vert Tu\Vert ={\max }_{t\in \left[\frac{1}{4},1\right]}\left|\left(Tu\right)\left(t\right)\right|+{\max }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^p\left(Tu\right)\left(t\right)\right|\\ ={\displaystyle \underset{0}{\overset{1}{\int }}G\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{1}H}\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s\\ \ge {\delta }_1{\left(\frac{1}{4}\right)}^{\alpha -1}\Vert u\Vert \left[{\displaystyle {\int }_0^{1}G\left(1,s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(1,s\right)\text{d}s}}\right]\ge \Vert u\Vert .\end{array}$

当$u\in P\cap \partial {\Omega }_1$ 时，可以推出

$\begin{array}{c}\Vert Tu\Vert ={\max }_{t\in \left[\frac{1}{4},1\right]}\left|\left(Tu\right)\left(t\right)\right|+{\max }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^p\left(Tu\right)\left(t\right)\right|\\ ={\displaystyle \underset{0}{\overset{1}{\int }}G\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{1}H}\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s\\ \le {\delta }_2\Vert u\Vert \left[{\displaystyle {\int }_0^{1}G\left(1,s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(1,s\right)\text{d}s}}\right]\le \Vert u\Vert .\end{array}$

根据Guo-Krasnoselskii不动点定理，边值问题(1)在$P\cap \left({\overline{\Omega }}_2\backslash {\Omega }_1\right)$ 上有一个正解，证明完成。

3.3. 定理3

对于$f\in C\left(\left[0,1\right]\times \left[0,+\infty \right)\times \left[0,+\infty \right),\left[0,+\infty \right)\right)$ ，如果存在三个常数$a_1,a_2,a_3$ 满足$0<a_1<a_2<a_3$ ，并且下列条件也成立：

(1) $f\left(t,u,v\right)>a_1{L}_1$ ，对于$\left(t,u,v\right)\in \left[0,1\right]\times \left[0,a_1\right]\times \left[0,a_1\right]$ ；

(2) $f\left(t,u,v\right)>a_3{L}_2$ ，对于$\left(t,u,v\right)\in \left[\frac{1}{4},1\right]\times \left[0,N{a}_3\right]\times \left[0,N{a}_3\right]$ ；

(3) $f\left(t,u,v\right)<a_2{L}_1$ ，对于$\left(t,u,v\right)\in \left[0,1\right]\times \left[0,N{a}_2\right]\times \left[0,N{a}_2\right]$ ；

那么，边值问题(1)至少有两个正解$u_1,u_2$ 并且满足$a_1<\alpha \left(u_1\right)$ ，$\theta \left(u_1\right)<a_2<\theta \left(u_2\right)$ ，$\gamma \left(u_2\right)<a_3$ 。

证明 在证明的过程中，

$\overline{P}\left(\gamma ,a_1\right)=\left\{u\in P:\gamma \left(u\right)\le a_1\right\},P\left(\gamma ,a_1\right)=\left\{u\in P:\gamma \left(u\right)<a_1\right\}.$

令

$\gamma \left(u\right)={\min }_{t\in \left[\frac{1}{4},1\right]}\left|u\left(t\right)\right|+{\min }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^{p}u\left(t\right)\right|,$

$\theta \left(u\right)=\alpha \left(u\right)={\max }_{t\in \left[0,1\right]}\left|u\left(t\right)\right|+{\max }_{t\in \left[0,1\right]}\left|D_{0+}^{p}u\left(t\right)\right|.$

显然$\gamma \left(u\right)$ 和$\alpha \left(u\right)$ 是非负递增的连续函数并且$\theta \left(u\right)$ 是非负的连续函数，且$\gamma \left(u\right)\le \theta \left(u\right)=\alpha \left(u\right)$ ，$\theta \left(0\right)=0$ 。

另外，可以推出

$\begin{array}{c}\gamma \left(u\right)={\min }_{t\in \left[\frac{1}{4},1\right]}\left|u\left(t\right)\right|+{\min }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^{p}u\left(t\right)\right|\\ ={\displaystyle \underset{0}{\overset{1}{\int }}G\left(\frac{1}{4},s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{1}H}\left(\frac{1}{4},s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s\\ \ge {\left(\frac{1}{4}\right)}^{\alpha -1}\left[{\displaystyle {\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{1}H\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}\right]\\ ={\left(\frac{1}{4}\right)}^{\alpha -1}\Vert u\Vert ,\end{array}$

所以对于所有$u\in \overline{P}\left(\gamma ,a_3\right)$ ，$\Vert u\Vert \le 4^{\alpha -1}\gamma \left(u\right)=N\gamma \left(u\right)$ 。上述证明表明该定理满足2.6引理4的条件(1)。

对于$\forall \lambda \in \left[0,1\right]$ ，$u\in \partial P\left(\theta ,a_2\right)$ 可以得到

$\begin{array}{c}\theta \left(\lambda u\right)={\max }_{t\in \left[0,1\right]}\left|\lambda u\left(t\right)\right|+{\max }_{t\in \left[0,1\right]}\left|D_{0+}^p\lambda u\left(t\right)\right|\\ =\lambda {\max }_{t\in \left[0,1\right]}\left|u\left(t\right)\right|+\lambda {\max }_{t\in \left[0,1\right]}\left|D_{0+}^{p}u\left(t\right)\right|=\lambda \theta \left(u\right),\end{array}$

这表明该定理满足2.6引理4的条件(2)。

运用和3.1定理1同样的证明方法可以得到$T:\overline{P}\left(\gamma ,a_3\right)\to P$ 是一个完全连续算子，这表明该定理满足2.6引理4的条件(3)。

对于$\forall u\in \partial P\left(\alpha ,a_1\right)$ ，可以得到$\alpha \left(u\right)=a_1$ ，$0\le u\left(t\right)\le a_1$ ，$0\le v=D_{0+}^{p}u\left(t\right)\le a_1$ ，并且通过该定理的条件(1)可以得到

$\begin{array}{c}\alpha \left(Tu\right)={\max }_{t\in \left[0,1\right]}\left|Tu\right|+{\max }_{t\in \left[0,1\right]}\left|D_{0+}^{p}Tu\right|\\ ={\displaystyle {\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{1}H\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}\\ >a_1{L}_1\left[{\displaystyle {\int }_0^{1}G\left(1,s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(1,s\right)\text{d}s}}\right]=a_1,\end{array}$

这表明$\alpha \left(Tu\right)>a_1$ 对于$\forall u\in \partial P\left(\alpha ,a_1\right)$ 。令$u=\frac{a_1}{3}$ ，那么$\alpha \left(u\right)=\alpha \left(\frac{a_1}{3}\right)=\frac{a_1}{3}<a_1$ 。这表面$P\left(\alpha ,a_1\right)\ne \varphi$ 。对于$\forall u\in \partial P\left(\gamma ,a_3\right)$ ，可以得到$\gamma \left(u\right)=a_3$ ，$0\le u\left(t\right)\le \Vert u\Vert \le N\gamma \left(u\right)\le N{a}_3$ ，$0\le v=D_{0+}^{p}u\left(t\right)\le N{a}_3$ ，通过该定理的条件(2)可以得到

$\begin{array}{c}\gamma \left(Tu\right)={\min }_{t\in \left[\frac{1}{4},1\right]}\left|Tu\right|+{\min }_{t\in \left[\frac{1}{4},1\right]}\left|D_{0+}^{p}Tu\right|\\ ={\displaystyle {\int }_0^{1}G\left(\frac{1}{4},s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{1}H\left(\frac{1}{4},s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}\\ >a_3{L}_2\left[{\displaystyle {\int }_0^{1}G\left(\frac{1}{4},s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(\frac{1}{4},s\right)\text{d}s}}\right]=a_3,\end{array}$

所以得出$\gamma \left(Tu\right)>a_3$ 对于$\forall u\in \partial P\left(\gamma ,a_3\right)$ 。

同样地，对于$\forall u\in \partial P\left(\theta ,a_2\right)$ ，可以得到$\theta \left(u\right)=a_2$ ，$0\le u=u\left(t\right)\le \Vert u\Vert \le N{a}_2$ ，$0\le v=D_{0+}^{p}u\left(t\right)\le N{a}_2$ ，并且通过该定理的条件(3)可以得出

$\begin{array}{c}\theta \left(Tu\right)={\max }_{t\in \left[0,1\right]}\left|Tu\right|+{\max }_{t\in \left[0,1\right]}\left|D_{0+}^{p}Tu\right|\\ ={\displaystyle {\int }_0^{1}G\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}+{\displaystyle {\int }_0^{1}H\left(1,s\right)f\left(s,u\left(s\right),D_{0+}^{p}u\left(s\right)\right)\text{d}s}\\ <a_2{L}_1\left[{\displaystyle {\int }_0^{1}G\left(1,s\right)\text{d}s+{\displaystyle {\int }_0^{1}H\left(1,s\right)\text{d}s}}\right]=a_2,\end{array}$

这表明$\theta \left(Tu\right)<a_2$ 对于$\forall u\in \partial P\left(\theta ,a_2\right)$ ，这表明该定理满足2.6引理4的条件(4)。

综上，2.6引理4的所有条件都满足。所以边值问题(1)至少有两个正解$u_1,u_2$ 并且满足$a_1<\alpha \left(u_1\right)$ ，$\theta \left(u_1\right)<a_2<\theta \left(u_2\right)$ ，$\gamma \left(u_2\right)<a_3$ 。

证明完成。

4. 实例分析

在这一部分，给出该实例来说明得到定理的实用性。

考虑下面的边值问题

$\{\begin{array}{l}{D}_{0+}^{\frac{9}{2}}u\left(t\right)+f\left(t,u\left(t\right),D_{0+}^{\frac{3}{2}}u\left(t\right)\right)=0,t\in \left[0,1\right],\hfill \\ u\left(0\right)=u^{\left(1\right)}\left(0\right)=\cdots =u^{\left(n-2\right)}\left(0\right)=0,\hfill \\ D_{0+}^{\frac{5}{2}}u\left(1\right)={\displaystyle \sum _{i=1}^{\infty }\frac{1}{2^i}{\displaystyle {\int }_0^{1-\frac{1}{i+1}}u\left(s\right)\text{d}s+{\displaystyle \sum _{i=1}^{\infty }\frac{1}{3^i}u\left(1-\frac{1}{i+2}\right),}}}\hfill \end{array}$ (3)

其中

$\begin{array}{l}\alpha =\frac{9}{2},1<\beta =\frac{5}{2}\le \alpha -1=\frac{7}{2},0<p=\frac{3}{2}=\beta -1=\frac{3}{2},\\ {\eta }_i=1-\frac{1}{i+1},0<{\eta }_1<{\eta }_2<\cdots <{\eta }_i<\cdots <1,\\ {\xi }_i=1-\frac{1}{i+2},0<{\xi }_1<{\xi }_2<\cdots <{\xi }_i<\cdots <1,\\ {\beta }_i=\frac{1}{2^i}>0,{\gamma }_i=\frac{1}{3^i}>0,\end{array}$

并且

$\begin{array}{l}m\left(0\right)=\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\beta \right)}-\frac{1}{\alpha }{\displaystyle \sum _{i=1}^{\infty }{\beta }_i{\eta }_i^{\alpha }-{\displaystyle \sum _{i=1}^{\infty }{\gamma }_i{\xi }_i^{\alpha -1}}}\\ =\frac{\Gamma \left(\frac{9}{2}\right)}{\Gamma \left(2\right)}-\frac{2}{9}{\displaystyle \sum _{i=1}^{\infty }\frac{1}{2^i}}{\left(1-\frac{1}{i+1}\right)}^{\frac{9}{2}}-{{\displaystyle \sum _{i=1}^{\infty }\frac{1}{3^i}\left(1-\frac{1}{i+2}\right)}}^{\frac{7}{2}}\\ \ge \frac{105\sqrt{\pi }}{16}-\frac{2}{9}\left(\frac{1}{2}\times 1^{\frac{9}{2}}+\frac{1}{2^2}\times 1^{\frac{9}{2}}+\cdots +\frac{1}{2^n}\times 1^{\frac{9}{2}}+\cdots \right)\\ -\left(\frac{1}{3}\times 1^{\frac{7}{2}}+\frac{1}{3^2}\times 1^{\frac{7}{2}}+\cdots +\frac{1}{3^n}\times 1^{\frac{7}{2}}+\cdots \right)\\ \approx 10.91>0,\end{array}$

令

$f\left(t,u\left(t\right),D_{0+}^{\frac{3}{2}}u\left(t\right)\right)=\frac{{\left(u^{\frac{3}{2}}+u\right)}^2}{1+\sin t},$

对于

$\left(t,u\left(t\right),D_{0+}^{\frac{3}{2}}u\left(t\right)\right)\in \left[0,1\right]\times \left[0,+\infty \right)\times \left[0,+\infty \right).$

容易得出$f^0={\lim }_{\left(u,u^{\frac{3}{2}}\right)\to 0}{\sup }_{t\in \left[0,1\right]}\frac{u^{\frac{3}{2}}+u}{1+\sin t}=0$ ，$f_{\infty }={\lim }_{\left(u,u^{\frac{3}{2}}\right)\to +\infty }{\inf }_{t\in \left[0,1\right]}\frac{u^{\frac{3}{2}}+u}{1+\sin t}=+\infty$ 。所以根据3.2定理2得出边值问题(3)有一个正解。