太原师范学院，数学与统计学院，山西 晋中

收稿日期：2023年2月27日；录用日期：2023年3月24日；发布日期：2023年3月31日

摘要

本文研究了一类带有p-Laplacian算子的分数阶积分边值问题正解的存在性，通过研究格林函数的性质，运用锥拉伸锥压缩不动点定理以及Leggett-Williams不动点定理，获得了该边值问题至少存在一个正解及三个正解的充分条件，并给出实例验证所得结论。

关键词

p-Laplacian算子，分数阶微分方程，边值问题

Multiple Positive Solutions for Fractional Integral Boundary Value Problems with p-Laplacian Operators

Yu Wu^*, Chang Liu

School of Mathematics and Statistics, Taiyuan Normal University, Jinzhong Shanxi

Received: Feb. 27^th, 2023; accepted: Mar. 24^th, 2023; published: Mar. 31^st, 2023

ABSTRACT

Studying the properties of Green’s function and using the cone stretching cone compression fixed point theorem and the Leggett-Williams fixed point theorem, this paper studies the existence of positive solutions for a class of fractional integral boundary value problems with p-Laplacian operators, obtains sufficient conditions for the existence of at least one positive solution and three positive solutions to the boundary value problems, and gives some examples illustrating the results obtained.

Keywords:p-Laplacian Operators, Fractional Differential Equations, Boundary Value Problems

Copyright © 2023 by author(s) and Hans Publishers Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

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

1. 引言

分数阶微积分是由整数阶微积分推广而来的，已广泛应用于控制系统、空气动力学、流体力学等众多工程分支中。因此，分数阶微分方程理论研究近年来得到了快速发展，研究成果非常丰富 [1] [2] [3] [4] [5] 。其中：文 [4] 运用锥理论得到了一类分数阶无穷点边值问题存在一个局部正解以及多个局部正解的充分条件；文 [5] 利用Schauder不动点定理及凹算子不动点定理建立了高阶分数阶多点边值问题正解的存在性及唯一正解的存在性结果。

带有p-Laplacian算子的方程模型能更好地描述多孔介质中的湍流、弹性理论等诸多领域中的实际问题，目前，带有p-Laplacian算子的分数阶微分方程受到众多学者的关注 [6] [7] [8] [9] [10] 。其中：文 [9] 研究了如下带有p-Laplacian算子的分数阶微分方程

$\{\begin{array}{l}{D}_{0+}^{\beta }\left({\varphi }_p\left(D_{0+}^{\alpha }u\left(t\right)\right)\right)+f\left(t,u\left(t\right)\right)=0,0<t<1,\\ u\left(0\right)=u^{\prime }\left(0\right)=u\left(1\right)=D_{0+}^{\alpha }u\left(0\right)=0,\\ D_{0+}^{\alpha }u\left(1\right)=\lambda D_{0+}^{\alpha }u\left(\xi \right),\end{array}$

其中： $D_{0+}^{\alpha }$ 和 $D_{0+}^{\beta }$ 是标准的Riemann-Liouville分数阶导数， $2<\alpha \le 3$ ， $1<\beta \le 2$ ， $0<\xi <1$ ， $\lambda \ge 0$ ， ${\varphi }_p\left(s\right)={\left|s\right|}^{p-2}s$ ， $p>1$ ， ${\varphi }_p^{-1}={\varphi }_q$ ， $1/p+1/q=1$ 。作者应用锥上的不动点定理获得了上述边值问题至少存在一个和两个正解的充分条件。文 [10] 研究了如下带有p-Laplacian算子的分数阶积分边值问题

$\{\begin{array}{l}-D_{0+}^{\alpha }\left({\varphi }_p\left(-D_{0+}^{\beta }x\left(t\right)\right)\right)=f\left(t,x\left(t\right),D_{0+}^{\beta }x\left(t\right)\right),0<t<1,\\ D_{0+}^{\beta }x(t)=0,\begin{array}{cc}& \end{array}{\left({\varphi }_p\left(-D_{0+}^{\beta }x\left(0\right)\right)\right)}^{\prime }=0,\\ D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }x\left(1\right)\right)\right)=I_{0+}^v\left({\varphi }_p\left(-D_{0+}^{\beta }x\left(\eta \right)\right)\right),\\ x\left(0\right)=0,\begin{array}{cc}& \end{array}{D}_{0+}^{\beta -1}x\left(1\right)=I_{0+}^{w}g\left(\xi ,x\left(\xi \right)\right)+k,\end{array}$

其中： $D_{0+}^{\alpha }$ ， $D_{0+}^{\beta }$ 和 $D_{0+}^{\gamma }$ 是标准的Riemann-Liouville分数阶导数， $I_{0+}^v$ ， $I_{0+}^w$ 是标准的Riemann-Liouville分数阶积分， $0<\gamma <1<\beta <2<\alpha <3$ ， $v,w>0$ ， $0<\eta ,\xi <1$ ， $k\in R$ ， $f\in C\left(\left[0,1\right]\times R\times R,R\right)$ ， $g\in C\left(\left[0,1\right]\times R,R\right)$ ， ${\varphi }_p\left(s\right)={\left|s\right|}^{p-2}s$ ， $p>1$ ， ${\varphi }_p^{-1}={\varphi }_q$ ， $1/p+1/q=1$ 。作者应用上下解方法和单调迭代方法得到了上述边值问题极值解的存在性的充分条件。

受上述文献启发，本文讨论如下带有p-Laplacian算子的分数阶积分边值问题

$\{\begin{array}{l}-D_{0+}^{\alpha }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=\lambda f\left(t,u\left(t\right),\left(Tu\right)\left(t\right)\right),0<t<1,\\ u\left(0\right)=D_{0+}^{\beta }u\left(0\right)=D_{0+}^{\beta -1}u\left(1\right)={\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(0\right)\right)\right)}^{\prime }=0,\\ D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(1\right)\right)\right)=I_{0+}^w\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(\eta \right)\right)\right),\end{array}$ (1)

其中： $D_{0+}^{\alpha }$ ， $D_{0+}^{\beta }$ ， $D_{0+}^{\gamma }$ 是Riemann-Liouville分数阶导数， $I_{0+}^w$ 是Riemann-Liouville分数阶积分， $0<\gamma <1<\beta <2<\alpha <3$ ， $0<\eta <1$ ， $\lambda ,w>0$ ， $f\in C\left(\left(0,1\right)\times \left[0,\infty \right)\times \left[0,\infty \right),\left[0,\infty \right)\right)$ ， $\left(Tu\right)\left(t\right)={\displaystyle {\int }_0^{t}K\left(t,s\right)u\left(s\right)\text{d}s}$ ，且 $K\in C\left(D,\left[0,+\infty \right)\right)$ ， $D=\left\{\left(t,s\right)\in \left[0,1\right]\times \left[0,1\right]:t\ge s\right\}$ ， ${\varphi }_p\left(s\right)={\left|s\right|}^{p-2}s$ ， $p>1$ ， ${\varphi }_p^{-1}={\varphi }_q$ ， $1/p+1/q=1$ 。本文运用锥拉伸锥压缩不动点定理以及Leggett-Williams不动点定理研究该边值问题正解的存在性。

2. 预备知识和引理

定义1 [1] 函数 $f:\left(0,+\infty \right)\to R$ 的 $\alpha >0$ 阶Riemann-Liouville分数阶积分定义为

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

其中：等式的右端在 $\left(0,+\infty \right)$ 有定义。

定义2 [1] 连续函数 $f:\left(0,+\infty \right)\to R$ 的 $\alpha >0$ 阶Riemann-Liouville分数阶导数定义为

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

其中：等式的右端在 $\left(0,+\infty \right)$ 有定义， $n=\min \left\{m\in Z:m\ge \alpha \right\}$ ， $\Gamma (\cdot )$ 为Gamma函数。

引理1 [1] 设 $\alpha >0$ ， $u\in C\left(0,1\right)\cap L\left(0,1\right)$ ， $D_{0+}^{\alpha }u\in C\left(0,1\right)\cap L\left(0,1\right)$ ，则

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

其中： $n=\min \left\{m\in Z:m\ge \alpha \right\}$ ， $c_i\in R$ ， $i=1,2,\cdots ,n$ 。

引理2 若 $y\left(t\right)\in C\left[0,1\right]$ ，则分数阶微分方程边值问题

$\{\begin{array}{l}-D_{0+}^{\alpha }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=y\left(t\right),0<t<1,\\ u\left(0\right)=D_{0+}^{\beta }u\left(0\right)=D_{0+}^{\beta -1}u\left(1\right)={\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(0\right)\right)\right)}^{\prime }=0,\\ D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(1\right)\right)\right)=I_{0+}^w\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(\eta \right)\right)\right),\end{array}$

有唯一解

$u\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)y\left(\tau \right)\text{d}\tau }\right)\text{d}s}$ .

其中：

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

$H\left(s,\tau \right)=\frac{1}{\Lambda }\{\begin{array}{ll}\begin{array}{l}{\left(s-\tau \right)}^{\alpha -1}\left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}\right]\\ \begin{array}{cc}& \end{array}+\left[\Gamma \left(\alpha +w\right){\left(1-\tau \right)}^{\alpha -\gamma -1}-\Gamma \left(\alpha -\gamma \right){\left(\eta -\tau \right)}^{\alpha +w-1}\right]s^{\alpha -1},\end{array}\hfill & 0\le \tau \le \eta ,s\le 1,\hfill \\ \left[\Gamma \left(\alpha +w\right){\left(1-\tau \right)}^{\alpha -\gamma -1}-\Gamma \left(\alpha -\gamma \right){\left(\eta -\tau \right)}^{\alpha +w-1}\right]s^{\alpha -1},\hfill & 0\le s\le \tau \le \eta \le 1,\hfill \\ {\left(s-\tau \right)}^{\alpha -1}\left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}\right]+\Gamma \left(\alpha +w\right){\left(1-\tau \right)}^{\alpha -\gamma -1}{s}^{\alpha -1},\hfill & 0\le \eta \le \tau \le s\le 1,\hfill \\ \Gamma \left(\alpha +w\right){\left(1-\tau \right)}^{\alpha -\gamma -1}{s}^{\alpha -1},\hfill & 0\le \eta ,s\le \tau \le 1,\hfill \end{array}$

其中： $\Lambda =\Gamma \left(\alpha \right)\left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}\right]$ 。

证明 由 ${\varphi }_p^{-1}={\varphi }_q$ 以及引理1可得， $-D_{0+}^{\alpha }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=y\left(t\right)$ 等价于

${\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)=-I_{0+}^{\alpha }y\left(t\right)+c_1{t}^{\alpha -1}+c_2{t}^{\alpha -2}+c_3{t}^{\alpha -3}$ ， $c_1,c_2,c_3\in R$ ，

由边界条件 $D_{0+}^{\beta }u\left(0\right)=0$ ， ${\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(0\right)\right)\right)}^{\prime }=0$ 可得 $c_2=c_3=0$ 。故

${\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)=-I_{0+}^{\alpha }y\left(t\right)+c_1{t}^{\alpha -1}$ (2)

对(2)式两边求 $\gamma$ 阶导数得

$D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=-\frac{1}{\Gamma \left(\alpha -\gamma \right)}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{\alpha -\gamma -1}}y\left(\tau \right)\text{d}\tau +c_1\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha -\gamma \right)}{t}^{\alpha -\gamma -1}$

对(2)式两边求w阶积分得

$I_{0+}^w\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)\right)=-\frac{1}{\Gamma \left(\alpha +w\right)}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{\alpha +w-1}}y\left(\tau \right)\text{d}\tau +c_1\frac{\Gamma \left(\alpha \right)}{\Gamma \left(\alpha +w\right)}{t}^{\alpha +w-1}$

由边界条件 $D_{0+}^{\gamma }\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(1\right)\right)\right)=I_{0+}^w\left({\varphi }_p\left(-D_{0+}^{\beta }u\left(\eta \right)\right)\right)$ 可得

$c_1=\frac{\Gamma \left(\alpha +w\right){\displaystyle {\int }_0^1{\left(1-\tau \right)}^{\alpha -\gamma -1}y\left(\tau \right)\text{d}\tau }-\Gamma \left(\alpha -\gamma \right){\displaystyle {\int }_0^{\eta }{\left(\eta -\tau \right)}^{\alpha +w-1}y\left(\tau \right)\text{d}\tau }}{\Gamma \left(\alpha \right)\left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}\right]}$ (3)

将(3)式代入(2)式可得

${\varphi }_p\left(-D_{0+}^{\beta }u\left(t\right)\right)=-\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_0^t{\left(t-\tau \right)}^{\alpha -1}y\left(\tau \right)\text{d}\tau }+c_1{t}^{\alpha -1}={\displaystyle {\int }_0^{1}H\left(t,\tau \right)y\left(\tau \right)\text{d}\tau }$ (4)

令 ${\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(t,\tau \right)y\left(\tau \right)\text{d}\tau }\right)=h\left(t\right)$ ，可得 $D_{0+}^{\beta }u\left(t\right)=-h\left(t\right)$ ，再由引理1可得

$\begin{array}{c}u\left(t\right)=-I_{0+}^{\beta }h\left(t\right)+d_1{t}^{\beta -1}+d_2{t}^{\beta -2}\\ =-\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}h\left(s\right)\text{d}s+d_1{t}^{\beta -1}+d_2{t}^{\beta -2},d_1,d_2\in R.\end{array}$ (5)

由边界条件 $u\left(0\right)=0$ 得 $d_2=0$ 。再由定义2对(5)式两边求 $\beta -1$ 阶导数可得

$D_{0+}^{\beta -1}u\left(t\right)=-D_{0+}^{\beta -1}{I}_{0+}^{\beta }h\left(t\right)+d_1{D}_{0+}^{\beta -1}{t}^{\beta -1}=-{\displaystyle {\int }_0^{t}h\left(s\right)\text{d}s}+d_1\Gamma \left(\beta \right)$ .

由边界条件 $D_{0+}^{\beta -1}u\left(1\right)=0$ 得

$d_1=\frac{{\displaystyle {\int }_0^{1}h\left(s\right)\text{d}s}}{\Gamma (\; \beta \; )}$

于是边值问题有唯一解

$\begin{array}{c}u\left(t\right)=-\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}h\left(s\right)\text{d}s+d_1{t}^{\beta -1}\\ =-\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^t{\left(t-s\right)}^{\beta -1}}h\left(s\right)\text{d}s+\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{t}^{\beta -1}}h\left(s\right)\text{d}s\\ ={\displaystyle {\int }_0^{1}G\left(t,s\right)h\left(s\right)\text{d}s}.\end{array}$ (6)

证毕。

引理3 函数 $G\left(t,s\right)$ ， $H\left(t,s\right)$ 满足如下性质：

1) 对任意的 $t,s\in \left[0,1\right]$ ， $G\left(t,s\right)\ge 0$ ， $H\left(t,s\right)\ge 0$ ；

2) $\underset{t\in \left[0,1\right]}{\max }G\left(t,s\right)=G\left(s,s\right)=\frac{1}{\Gamma \left(\beta \right)}{s}^{\beta -1}\le \frac{1}{\Gamma \left(\beta \right)}$ ， $s\in \left[0,1\right]$ ；

3) $\underset{t\in \left[1/4,3/4\right]}{\min }G\left(t,s\right)=:m\left(s\right)=\frac{1}{\Gamma \left(\beta \right)}\{\begin{array}{ll}{\left(\frac{3}{4}\right)}^{\beta -1}-{\left(\frac{3}{4}-s\right)}^{\beta -1},\hfill & s\in \left[0,\frac{1}{4}\right];\hfill \\ \min \left\{{\left(\frac{3}{4}\right)}^{\beta -1}-{\left(\frac{3}{4}-s\right)}^{\beta -1},\frac{1}{4^{\beta -1}}\right\},\hfill & s\in \left(\frac{1}{4},\frac{3}{4}\right);\hfill \\ \frac{1}{4^{\beta -1}},\hfill & s\in \left[\frac{3}{4},1\right].\hfill \end{array}$

证明 1) 由函数 $G\left(t,s\right)$ 的表达式易知，对任意的 $t,s\in \left[0,1\right]$ ， $G\left(t,s\right)\ge 0$ 。

对于函数 $H\left(t,s\right)$ ，注意到 $\alpha -\gamma -1<\alpha +w-1$ 以及 $1<\alpha -\gamma <2<\alpha +w$ ，则有

$\Gamma \left(\alpha -\gamma \right)<1<\Gamma \left(\alpha +w\right)$

${\left(1-s\right)}^{\alpha -\gamma -1}>{\left(\eta -s\right)}^{\alpha -\gamma -1}>{\left(\eta -s\right)}^{\alpha +w-1}$

进而有 $\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right){\eta }^{\alpha +w-1}>\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right)>0$ ，以及

$\Gamma \left(\alpha +w\right){\left(1-s\right)}^{\alpha -\gamma -1}-\Gamma \left(\alpha -\gamma \right){\left(\eta -s\right)}^{\alpha +w-1}\ge \left[\Gamma \left(\alpha +w\right)-\Gamma \left(\alpha -\gamma \right)\right]{\left(\eta -s\right)}^{\alpha +w-1}>0$

进而容易得到对任意的 $t,s\in \left[0,1\right]$ ， $H\left(t,s\right)\ge 0$ 。

2) 对于函数 $G\left(t,s\right)$ ，通过简单计算，可得当 $0<t\le s$ 时，

$\frac{\partial }{\partial t}G\left(t,s\right)=\frac{1}{\Gamma \left(\beta \right)}\left(\beta -1\right)t^{\beta -2}>0$

当 $0\le s<t$ 时，

$\frac{\partial }{\partial t}G\left(t,s\right)=\frac{1}{\Gamma \left(\beta \right)}\left(\beta -1\right)\left[t^{\beta -2}-{\left(t-s\right)}^{\beta -2}\right]<0$

即 $G\left(t,s\right)$ 在 $t\le s$ 上是关于t的增函数，在 $s<t$ 上是关于t的减函数。进而有

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

3) 令

$g_1\left(t,s\right)=\frac{t^{\beta -1}-{\left(t-s\right)}^{\beta -1}}{\Gamma \left(\beta \right)}$ , $g_2\left(t,s\right)=\frac{t^{\beta -1}}{\Gamma (\; \beta \; )}$

可得

$\begin{array}{c}\underset{t\in \left[1/4,3/4\right]}{\min }G\left(t,s\right)=\{\begin{array}{ll}{g}_1\left(\frac{3}{4},s\right),\hfill & s\in \left[0,\frac{1}{4}\right];\hfill \\ \min \left\{g_1\left(\frac{3}{4},s\right),g_2\left(\frac{1}{4},s\right)\right\},\hfill & s\in \left(\frac{1}{4},\frac{3}{4}\right);\hfill \\ g_2\left(\frac{1}{4},s\right),\hfill & s\in \left[\frac{3}{4},1\right];\hfill \end{array}\\ =\frac{1}{\Gamma \left(\beta \right)}\{\begin{array}{ll}{\left(\frac{3}{4}\right)}^{\beta -1}-{\left(\frac{3}{4}-s\right)}^{\beta -1},\hfill & s\in \left[0,\frac{1}{4}\right];\hfill \\ \min \left\{{\left(\frac{3}{4}\right)}^{\beta -1}-{\left(\frac{3}{4}-s\right)}^{\beta -1},\frac{1}{4^{\beta -1}}\right\},\hfill & s\in \left(\frac{1}{4},\frac{3}{4}\right);\hfill \\ \frac{1}{4^{\beta -1}},\hfill & s\in \left[\frac{3}{4},1\right];\hfill \end{array}\\ =:m\left(s\right),\end{array}$

证毕。

设E是实Banach空间， $\theta$ 为E中的零元素。P为E中的非空凸闭子集，若

$x\in P,\lambda \ge 0\Rightarrow \lambda x\in P;x\in P,-x\in P\Rightarrow x=\theta ,$

则称P为E中的锥。

引理4 [3] 设E是一个Banach空间，P为E中的一个锥， ${\Omega }_1,{\Omega }_2$ 是E中的两个有界开集，并且 ${\Omega }_1\subset {\Omega }_2$ 。假设 $T:P\cap \left({\Omega }_2\backslash {\Omega }_1\right)\to P$ 是全连续算子，若下列条件二者成立其一：

(K1) $\Vert Tx\Vert \le \Vert x\Vert$ ， $x\in P\cap \partial {\Omega }_1$ ； $\Vert Tx\Vert \ge \Vert x\Vert$ ， $x\in P\cap \partial {\Omega }_2$

(K2) $\Vert Tx\Vert \ge \Vert x\Vert$ ， $x\in P\cap \partial {\Omega }_1$ ； $\Vert Tx\Vert \le \Vert x\Vert$ ， $x\in P\cap \partial {\Omega }_2$

那么T在 $P\cap \left({\Omega }_2\backslash {\Omega }_1\right)$ 中至少有一个不动点。

定义3 [3] 设P为Banach空间E中的锥。若映射 $\psi :P\to \left[0,\infty \right)$ 连续且 $\forall x,y\in P$ ， $\forall \eta \in \left[0,1\right]$ ，有 $\psi \left(\eta x+\left(1-\eta \right)y\right)\ge \eta \psi \left(x\right)+\left(1-\eta \right)\psi \left(y\right)$ ，则称 $\psi$ 是P上非负连续凹泛函。

引理5 [3] 设常数 $0<a<b<d\le c$ ， $T:\overline{P_c}\to \overline{P_c}$ 是全连续算子， $\psi$ 是P上的一个非负连续凹泛函，且对任意的 $x\in P_c$ ，有 $\psi \left(x\right)\le \Vert x\Vert$ 。又设

(L1) $\left\{x\in P\left(\psi ,b,d\right)|\psi \left(x\right)>b\right\}\ne \varnothing$ 且对 $x\in P\left(\psi ,b,d\right)$ ，有 $\psi \left(Tx\right)>b$ ；

(L2) 当 $x\in \overline{P_a}$ 时， $\Vert Tx\Vert <a$ ；

(L3) 当 $x\in P\left(\psi ,b,c\right)$ 且 $\Vert Tx\Vert >d$ 时， $\psi \left(Tx\right)>b$ ．

则算子T至少有三个不动点 $x_1,x_2,x_3$ ，满足 $\Vert x_1\Vert <a$ ， $b<\psi \left(x_2\right)$ ， $\Vert x_3\Vert >a$ ， $\psi \left(x_3\right)<b$ 。

3. 主要结果

设 $E=C\left[0,1\right]$ 。对任意的 $u\in E$ ，定义其范数为

$\Vert u\Vert =\underset{t\in \left[0,1\right]}{\max }\left|u\left(t\right)\right|$

容易证明E为Banach空间。再令

$P=\left\{u\in E|u\left(t\right)\ge 0,t\in \left[0,1\right]\right\}$

则P是E中的锥。定义泛函

$\psi \left(u\right)=\underset{t\in \left[1/4,3/4\right]}{\min }u\left(t\right)$ ，

则它是P中的非负连续凹泛函。定义算子A如下：对任意的 $u\in P$ ，

$\left(Au\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s$ , $t\in \left[0,1\right]$ (7)

引理6 算子 $A:P\to P$ 是全连续算子。

证明 根据引理3的1)以及函数f的非负连续性可得，对任意的 $u\in P$ 有

$\left(Au\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\ge 0$ , $t\in \left[0,1\right]$

所以 $A\left(P\right)\subset P$ 。以下证明A为紧算子。设 $\Omega$ 是锥P中的有界集，即存在 $R_1>0$ ，使得

$\Vert x\Vert \le R_1$ , $\forall x\in \Omega$

设 $W_{R_1}=\sup \left\{f\left(t,u_1,u_2\right):\left(t,u_1,u_2\right)\in \left(\left[0,1\right]\times \left[0,R_1\right]\times \left[0,k^{*}{R}_1\right]\right)\right\}$ ，其中

$k^{*}=\underset{t\in \left[0,1\right]}{\sup }{\displaystyle {\int }_0^{t}K\left(t,s\right)\text{d}s}$

进而对任意的 $u\in \Omega$ ，由引理3的2)得

$\begin{array}{c}\left|\left(Au\right)\left(t\right)\right|=\left|{\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\right|\\ \le \frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\\ \le \frac{{\varphi }_q\left(\lambda W_{R_1}\right)}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s,\end{array}$

由此可知 $A\left(\Omega \right)$ 为E中一致有界的子集合。

再证 $A\left(\Omega \right)$ 为E中等度连续的子集合。对任意的 $0\le t_1<t_2\le 1$ ， $u\in \Omega$ 有

$\begin{array}{c}\left|\left(Au\right)\left(t_2\right)-\left(Au\right)\left(t_1\right)\right|\le {\displaystyle {\int }_0^1\left|G\left(t_2,s\right)-G\left(t_1,s\right)\right|{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}\\ \le {\varphi }_q\left(\lambda W_{R_1}\right){\displaystyle {\int }_0^1\left|G\left(t_2,s\right)-G\left(t_1,s\right)\right|{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s,\end{array}$

又 $G\left(t,s\right)$ 在 $\left[0,1\right]\times \left[0,1\right]$ 上连续，从而一致连续，故 $A\left(\Omega \right)$ 为E中等度连续的子集合。进而集合 $A\left(\Omega \right)$ 是一致有界且等度连续的，根据Arzela-Ascoli定理可知 $A\left(\Omega \right)$ 是E中的相对紧集。

最后证明 $A:P\to P$ 是连续算子。设 $\left\{u_n\right\}\subset \Omega$ ， $u\in \Omega$ ，当 $n\to +\infty$ 时，有 $\Vert u_n-u\Vert \to 0$ ，故存在常数 $R_0>1$ ，使得 $\Vert u\Vert <R_0$ ， $\Vert u_n\Vert <R_0$ ， $n=1,2,\cdots$ 。于是对 $s\in \left[0,1\right]$ ，

$\left|{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)-{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\right|\to 0,n\to +\infty .$

$\begin{array}{l}|{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}-{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}|\\ \le 2{\varphi }_q\left(\lambda W_{R_0}\right){\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s}<+\infty ,\end{array}$

其中 $W_{R_0}=\sup \left\{f\left(t,u_1,u_2\right):\left(t,u_1,u_2\right)\in \left(\left[0,1\right]\times \left[0,R_0\right]\times \left[0,k^{*}{R}_0\right]\right)\right\}$ 。由Lebesgue控制收敛定理可知，

$|{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}-{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}|\to 0,n\to +\infty .$

进而根据引理3的2)得

$\begin{array}{c}\left|\left(A{u}_n\right)\left(t\right)-\left(Au\right)\left(t\right)\right|=|{\displaystyle {\int }_0^{1}G\left(t,s\right)}[{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)\\ -{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)]\text{d}s|\\ \le \frac{1}{\Gamma \left(\beta \right)}|{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u_n\left(\tau \right),\left(T{u}_n\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}\\ -{\displaystyle {\int }_0^1{\varphi }_q\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s}|\\ \to 0,n\to +\infty .\end{array}$

则 $A:P\to P$ 是连续算子。证毕。

为了方便叙述，记：

$M={\left[\frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left(\lambda {\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\right]}^{-1}$

$N={\left[{\displaystyle {\int }_0^{1}m\left(s\right){\varphi }_q}\left(\lambda {\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\right]}^{-1}$

定理1 若存在两个常数 $r_2>r_1>0$ ，使得

(H1) $\forall \left(t,u,v\right)\in \left[1/4,3/4\right]\times \left[0,r_1\right]\times \left[0,k^{*}{r}_1\right]$ ， $f\left(t,u,v\right)\ge {\left(N{r}_1\right)}^{p-1}$ ，

(H2) $\forall \left(t,u,v\right)\in \left[0,1\right]\times \left[0,r_2\right]\times \left[0,k^{*}{r}_2\right]$ ， $f\left(t,u,v\right)\le {\left(M{r}_2\right)}^{p-1}$ ，

则边值问题(1)至少有一个正解u，使得 $r_1\le \Vert u\Vert \le r_2$ 。

证明 令 ${\Omega }_1=\left\{u\in P|\Vert u\Vert <r_1\right\}$ ，当 $u\in \partial {\Omega }_1$ 时，有 $0\le u\left(t\right)\le r_1$ ， $0\le \left(Tu\right)\left(t\right)\le k^{*}{r}_1$ ， $t\in \left[0,1\right]$ 。由(H1)及引理3的3)可得

$\begin{array}{c}\left(Au\right)\left(t\right)={\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\\ \ge {\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\\ \ge {\displaystyle {\int }_0^1\underset{t\in \left[1/4,3/4\right]}{\min }G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\lambda {\varphi }_p\left(N{r}_1\right)\text{d}\tau }\right)\text{d}s\\ =N{r}_1{\displaystyle {\int }_0^{1}m\left(s\right){\varphi }_q}\left(\lambda {\displaystyle {\int }_{1/4}^{3/4}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\\ =r_1=\Vert u\Vert ,\forall t\in \left[1/4,3/4\right],\end{array}$

因而当 $u\in \partial {\Omega }_1$ 时，有 $\Vert Au\Vert \ge \Vert u\Vert$ 。

另一方面，令 ${\Omega }_2=\left\{u\in P|\Vert u\Vert <r_2\right\}$ ，当 $u\in \partial {\Omega }_2$ 时，有 $0\le u\left(t\right)\le r_2$ ， $0\le \left(Tu\right)\left(t\right)\le k^{*}{r}_2$ ， $t\in \left[0,1\right]$ 。由(H2)及引理3的2)可得

$\begin{array}{c}\Vert Au\Vert =\underset{t\in \left[0,1\right]}{\max }\left|\left(Au\right)\left(t\right)\right|\\ =\underset{t\in \left[0,1\right]}{\max }\left|{\displaystyle {\int }_0^{1}G\left(t,s\right){\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda f\left(\tau ,u\left(\tau \right),\left(Tu\right)\left(\tau \right)\right)\text{d}\tau }\right)\text{d}s\right|\\ \le \frac{1}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left({\displaystyle {\int }_0^{1}H\left(s,\tau \right)\lambda {\varphi }_p\left(M{r}_2\right)\text{d}\tau }\right)\text{d}s\\ =\frac{M{r}_2}{\Gamma \left(\beta \right)}{\displaystyle {\int }_0^1{\varphi }_q}\left(\lambda {\displaystyle {\int }_0^{1}H\left(s,\tau \right)\text{d}\tau }\right)\text{d}s\\ =r_2=\Vert u\Vert ,\forall t\in \left[0,1\right],\end{array}$