数学物理学报, 2023, 43(1): 53-68

高阶Riemann-Liouville型分数阶脉冲微分方程积分边值问题的正解

徐家发,*, 杨志春,

重庆师范大学 数学科学学院 重庆 401331

Positive Solutions for a High Order Riemann-Liouville Type Fractional Impulsive Differential Equation Integral Boundary Value Problem

Xu Jiafa,*, Yang Zhichun,

School of Mathematical Sciences, Chongqing Normal University, Chongqing 401331

通讯作者: *徐家发, E-mail: xujiafa292@sina.com

收稿日期: 2021-12-9   修回日期: 2022-10-17  

基金资助: 国家自然科学基金项目(11971081)
重庆市教育委员会科技研究重大项目(KJZD-M202000502)

Received: 2021-12-9   Revised: 2022-10-17  

Fund supported: The NSFC(11971081)
Science and Technology Research Program of Chongqing Municipal Education Commission(KJZD-M202000502)

作者简介 About authors

杨志春,E-mail:yangzhch@126.com

摘要

该文研究了具有半正非线性项和脉冲项的高阶Riemann-Liouville型分数阶脉冲微分方程积分边值问题. 利用不动点指数理论, 在超线性增长和次线性增长等条件下获得了该问题正解的存在性结论, 推广了近期这方面一些已有的成果.

关键词: 分数阶微分方程; 边值问题; 脉冲; 不动点指数; 正解

Abstract

In this paper, we study a high order Riemann-Liouville type fractional impulsive differential equation integral boundary value problem involving semipositone the nonlinear and impulsive terms. By virtue of the fixed point index, we obtain the positive solutions theorems under some appropriate superlinear and sublinear growth conditions. The results here extend the existing study.

Keywords: Fractional differential equations; Boundary value problems; Impulse; Fixed point index; Positive solutions

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本文引用格式

徐家发, 杨志春. 高阶Riemann-Liouville型分数阶脉冲微分方程积分边值问题的正解[J]. 数学物理学报, 2023, 43(1): 53-68

Xu Jiafa, Yang Zhichun. Positive Solutions for a High Order Riemann-Liouville Type Fractional Impulsive Differential Equation Integral Boundary Value Problem[J]. Acta Mathematica Scientia, 2023, 43(1): 53-68

1 引言

本文研究如下具有半正非线性项和脉冲项的高阶Riemann-Liouville型分数阶脉冲微分方程积分边值问题

$\begin{equation}\label{pro} \left\{\begin{array}{l} -D_{0+}^{\alpha} u(t)=f(t, u(t)), \ t \in(0,1) \backslash\left\{t_{k}\right\}_{k=1}^{m}, \\ \Delta u\left(t_{k}\right)=I_{k}\left(u\left(t_{k}\right)\right), \ t=t_{k}, k=1,2,\cdots,m,\\ u(0)=u^{\prime}(0)=\cdots=u^{(n-2)}(0), \ u^{\prime}(1)=\int_{0}^{1} u(t) {\rm d}\eta(t), \end{array}\right. \end{equation}$

其中$\alpha\in (n-1,n]$, $n$为正整数且$n\ge 3$, $D_{0+}^{\alpha}$$\alpha$ 阶Riemann-Liouville型分数阶导数, 脉冲点列$\{t_k\}_{k=1}^m$ 满足: $0<t_{1}<\cdots<t_{m}<1$, $\Delta u\left(t_{k}\right)=u\left(t_{k}^{+}\right)-u\left(t_{k}^{-}\right)$, $u\left(t_{k}^{-}\right)=u\left(t_{k}\right)$, $u\left(t_{k}^{+}\right)=\lim\limits_{h \rightarrow 0} u\left(t_{k}+h\right)$, $u\left(t_{k}^{-}\right)=\lim\limits_{h \rightarrow 0} u\left(t_{k}-h\right)$ 分别是$u(t)$在点$t=t_k$的右、左极限. $f,I_k,\eta$满足以下条件

(H$f$) $f\in C([0,1]\times \Bbb R ^+, \Bbb R )$, 且存在正常数$M>0$使得$f(t,u)\ge -M$, $\forall t\in [0,1], u\in \Bbb R ^+$,

(H$I_k$) $I_{k} \in C\left(\Bbb R ^{+}, \Bbb R \right)$, 且存在$M_k>0$使得$I_k(u)\ge -M_k,$$\forall u\in \Bbb R ^+, k=1,2,\cdots,m$,

(H$\eta$) $\eta$$[0,1]$上的有界变差函数, 且$\int_0^1 t^{\alpha-1}{\rm d}\eta(t)\in (0,\alpha-1)$.

注1.1$\int_{0}^{1} u(t) {\rm d}\eta(t)$被称为Riemann-Stieltjes积分, 其在研究各类边值问题中起着重要作用. 若$\eta$可导, 则

$\int_{0}^{1} u(t) {\rm d}\eta(t)=\int_{0}^{1} u(t)\eta'(t){\rm d}t, $

此为Riemann积分; 若存在$\{\xi_i\}_{i=1}^{m-2}$满足: $0<\xi_1<\xi_2<\cdots<\xi_{m-2}<1$, $m$是正整数且$m\ge 3$使得 $u'(1)=\sum\limits_{i=1}^{m-2}a_iu(\xi_i)$, 此为多点边值问题. 若令

$\eta(t)=\left\{\begin{array}{ll}0,t\in [t\in [\xi_{m-2},1],\end{array}\right.$

$\sum\limits_{i=1}^{m-2}a_iu(\xi_i)=\int_0^1 u(t){\rm d}\eta(t)$, 即多点边值问题亦可用Riemann-Stieltjes积分表示.

分数阶微积分已有300多年历史, 但在发展的早期并没有引起足够的兴趣. 然而最近几十年, 研究者们发现它们在工程和物理学、力学、化学、经济学和生物学等学科中发挥着不可替代的作用, 对该类问题的研究愈发受到广泛关注, 涌现出一大批优秀的研究成果[1-29].

众所周知, 非线性泛函分析中的拓扑方法是研究微分方程最有力的工具, 运用该方法处理的各类问题中人们普遍使用一个基本的限制条件: 方程的非线性项是非负的. 这显然局限了研究的视角和创新. 然而在刻画自然现象时, 许多不确定的物理变量、参数以及扰动因素等常常会影响到整个系统的稳定性, 非负性显然不足以描述此类问题. 例如上世纪70年代荷兰化学家Aris在研究化学反应时发现一些惰性材料、 催化剂等常常对整个反应起到加速或抑制的作用, 这样在刻画反应的微分方程中必含有某个扰动项, 即非线性项具有形式 $f(t, x) \geq-M, M>0$, 这就是数学模型下的半正问题. 近年来亦有很多文献研究分数阶半正问题[1-11], 例如在文献[1]中作者研究了如下抽象的 HIV-1 人口受感染的分数阶动力学模型

$ \left\{\begin{array}{l} D^{\alpha} u(t)+\lambda f\left(t, u(t), D^{\beta} u(t), v(t)\right)=0, \\ D^{\gamma} v(t)+\lambda g(t, u(t))=0, \\ D^{\beta} u(0)=D^{\beta+1} u(0)=0, D^{\beta} u(1)=\int_{0}^{1} D^{\beta} u(s) {\rm d} A(s), \\ v(0)=v^{\prime}(0)=0, v(1)=\int_{0}^{1} v(s) {\rm d} B(s), \end{array}\right. $

其中 $2<\alpha, \beta \leq 3,1<\beta<1$, $u$ 是末受感染的 ${\rm CD} 4+{\rm T}$ 细胞, $v$ 是已感染的 ${\rm CD} 4+{\rm T}$ 细胞, $f$$g$ 允许变号的, 且满足半正型条件, 作者借助锥上的不动点定理, 在一定的参数范围内获得该问题正解的存在性.

在文献[2]中作者运用Leggett-Williams 和 Guo-Krasnosel'skii不动点定理研究了如下带有$p$-Laplacian的高阶分数阶微分方程积分边值问题多正解的存在性

$\left\{ \begin{array}{ll} -D_{0+}^{\mu}\left(\varphi_{p}\left(-D_{0+}^{\alpha} u(t)\right)\right)=f(t, u(t)), \ 0<t<1, \\ u(0)=u^{\prime}(0)=\cdots=u^{(n-2)}(0)=0, \ D_{0+}^{\alpha} u(0)=D_{0+}^{\alpha} u(1)=0, \\ D_{0+}^{\beta} u(1)=\lambda \int_{0}^{\eta} D_{0+}^{\gamma} u(t) {\rm d} A(t), \end{array}\right. $

其中非线性项$f$满足半正型和有界性条件.

另一方面, 我们注意到由于脉冲微分系统充分考虑瞬时突变现象对状态的影响, 它比不带脉冲的系统更能精确地反映事物的规律, 其研究日益受到人们的重视. 近年来, 针对于分数阶脉冲微分方程边值问题的研究已有涉猎[12-25]. 然而正如文献[12]中所说的那样, 仅有少量的文献研究分数阶脉冲微分方程边值问题(尤其是积分边值问题), 其困难在于该类方程所对应的等价方程形式过于复杂, 很难用Green函数表达成Hammerstein型积分方程的形式. 例如在文献[13]中作者研究了如下Caputo型分数阶脉冲边值问题

$ \left\{\begin{array}{l} { }^{C} D^{\alpha} u(t)=f(t, u(t)), \quad 1<\alpha \leq 2, t \in J^{\prime}, \\ \Delta u\left(t_{k}\right)=I_{k}\left(u\left(t_{k}\right)\right), \quad \Delta u^{\prime}\left(t_{k}\right)=I_{k}^{*}\left(u\left(t_{k}\right)\right), \quad k=1,2, \cdots, p, \\ T u^{\prime}(0)=-a u(0)-b u(T), \quad T u^{\prime}(T)=c u(0)+{\rm d} u(T), \quad a, b, c, d \in \Bbb R, \end{array}\right. $

其等价方程为

$u(t)=\left\{\begin{array}{ll} \int_{0}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)){\rm d}s+\frac{(b+d) T+(a d-b c) t}{\Lambda T} \int_{t_{p}}^{T} \frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)){\rm d}s\\ -\frac{(b+1) T+(a+b) t}{\Lambda} \int_{t_{p}}^{T} \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)){\rm d}s+{\cal A}, \quad t \in J_{0};\\ \int_{t_{k}}^{t} \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)){\rm d}s+\frac{(b+d) T+(a d-b c) t}{\Lambda T} \int_{t_{p}}^{T} \frac{(T-s)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)){\rm d}s\\ -\frac{(b+1) T+(a+b) t}{\Lambda} \int_{t_{p}}^{T} \frac{(T-s)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)) {\rm d} s\\ +\sum_{i=1}^{k}\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)) {\rm d} s+I_{i}\left(u\left(t_{i}\right)\right)\right]\\ +\sum_{i=1}^{k-1}\left(t_{k}-t_{i}\right)\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)){\rm d} s+I_{i}^{*}\left(u\left(t_{i}\right)\right)\right]\\ +\sum_{i=1}^{k}\left(t-t_{k}\right)\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)) {\rm d} s+I_{i}^{*}\left(u\left(t_{i}\right)\right)\right]+{\cal A}, \\ \qquad t \in J_{k}, k=1,2, \cdots, p,\end{array}\right.$

其中

$\begin{eqnarray*} {\cal A}&=& \frac{(b+d) T+(a d-b c) t}{\Lambda T} \sum_{i=1}^{p}\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-1}}{\Gamma(\alpha)} f(s,u(s)) {\rm d} s+I_{i}\left(u\left(t_{i}\right)\right)\right] \\ &&+\frac{(b+d) T+(a d-b c) t}{\Lambda T} \sum_{i=1}^{p-1}\left(t_{p}-t_{i}\right)\left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-2}}{\Gamma(\alpha-1)} f(s,u(s)) {\rm d} s+I_{i}^{*}\left(u\left(t_{i}\right)\right)\right] \\ &&-\sum_{i=1}^{p}\left[\frac{(1-d)(T+a t)+b(c+1) t}{\Lambda}+\frac{[(b+d) T+(a d-b c) t] t_{p}}{\Lambda T}\right]\\ &&\times \left[\int_{t_{i-1}}^{t_{i}} \frac{\left(t_{i}-s\right)^{\alpha-2}}{\Gamma(\alpha-1)}f(s,u(s)){\rm d} s+I_{i}^{*}\left(u\left(t_{i}\right)\right)\right]. \end{eqnarray*} $

如此的庞然大物处理起来非常困难, 只能针对非线性项提一些简单的条件, 如有界性条件, Lipschitz条件. 但是我们注意到, 并不是所有的分数阶脉冲微分方程边值问题都只能表达成如此繁复的式子, 例如文献[14]中作者研究的分数阶脉冲问题

$ \left\{\begin{array}{l} { }^{C} D^{q} u(t)=f(t, u(t)), \quad \mbox{ a.e. }\ t \in[0,1], \\ \Delta u\left(t_{k}\right)=I_{k}\left(u\left(t_{k}\right)\right), \quad \Delta u^{\prime}\left(t_{k}\right)=J_{k}\left(u\left(t_{k}\right)\right), \quad k=1,2, \cdots, m, \\ a u(0)-b u(1)=0, \quad a u^{\prime}(0)-b u^{\prime}(1)=0 \end{array}\right. $

就可以用Green函数表示其等价的方程

$ u(t)=\int_{0}^{1} G_{1}(t, s) f(s,u(s)){\rm d}s+\sum_{i=1}^{m} G_{2}\left(t, t_{i}\right) J_{i}\left(u\left(t_{i}\right)\right)+\sum_{i=1}^{m} G_{3}\left(t, t_{i}\right) I_{i}\left(u\left(t_{i}\right)\right). $

在本文中作者运用不动点指数理论, 在非线性项和脉冲项非负等条件下获得了正解的存在性.

本文受上述文献的启发, 在非线性项和脉冲项满足半正型条件下研究更为复杂的高阶Riemann-Liouville型分数阶脉冲微分方程积分边值问题. 首先运用半正型条件将其转化为等价的非负算子方程, 构造合适的工作空间和锥, 然后运用锥上的不动点指数获得正解的存在性. 最后给出一些具体的实例验证本文结论的有效性.

2 基础知识

限于篇幅, 我们仅给出Riemann-Liouville型分数阶导数的定义, 细节参见专著[27-29].

定义2.1 连续函数 $f:(0,+\infty) \rightarrow(-\infty,+\infty)$的 Riemann-Liouville 型分数阶($\alpha$阶)导数定义为

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

右端在 $(0,+\infty)$ 上逐点有定义, 其中 $n=[\alpha]+1,[\alpha]$$\alpha$ 的整数部分.

引理2.2 ([15,引理2.4]) 令$h\in C[0,1]$, $\delta_{\alpha,\eta}=\alpha-1-\int_0^1 t^{\alpha-1}{\rm d}\eta(t)$, 则边值问题

$ \left\{\begin{array}{l} -D_{0+}^{\alpha} u(t)=h(t), \ t \in(0,1) \backslash\left\{t_{k}\right\}_{k=1}^{m}, \\ \Delta u\left(t_{k}\right)=I_{k}\left(u\left(t_{k}\right)\right), \ t=t_{k}, k=1,2,\cdots,m,\\ u(0)=u^{\prime}(0)=\cdots=u^{(n-2)}(0), \ u^{\prime}(1)=\int_{0}^{1} u(t) {\rm d}\eta(t) \end{array}\right. $

的解可以表示为

$ u(t)=\int_{0}^{1} G(t, s) h(s){\rm d}s+t^{\alpha-1} \sum_{t \leq t_{k}<1} t_{k}^{1-\alpha} I_{k}\left(u\left(t_{k}\right)\right), \ t \in[0,1], $

其中

$ G(t, s)=g(t, s)+\frac{t^{\alpha-1}}{\delta_{\alpha,\eta}} \int_{0}^{1} g(\tau, s) {\rm d}\eta(\tau), $
$ g(t, s)= \left\{\begin{array}{ll} \frac{t^{\alpha-1}(1-s)^{\alpha-2}-(t-s)^{\alpha-1}}{\Gamma(\alpha)}, & 0 \leq s \leq t \leq 1, \\ \frac{t^{\alpha-1}(1-s)^{\alpha-2}}{\Gamma(\alpha)}, & 0 \leq t \leq s \leq 1.\end{array}\right. $

引理2.3$\varphi(t)=t(1-t)^{\alpha-2},t\in [0,1]$, 则

$t^{\alpha-1} \varphi(s) \left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right]\le \Gamma(\alpha) G(t, s)\le \varphi(s)\left[1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right],\forall t,s\in [0,1].$

根据文献[26,引理2.8]和文献[引理4], $g$满足如下不等式

$ t^{\alpha-1} s(1-s)^{\alpha-2} \le \Gamma(\alpha) g(t, s) \le s(1-s)^{\alpha-2}, \forall t, s \in[0,1].$

从而由$G$$g$的关系, 我们有

$\begin{eqnarray*} t^{\alpha-1} s(1-s)^{\alpha-2} \left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right] &\le& \Gamma(\alpha) G(t, s)\\ &\le &s(1-s)^{\alpha-2} \left[1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right],\forall t,s \in [0,1]. \end{eqnarray*}$

证毕.

引理2.4

$\kappa_1=\frac{\alpha \Gamma(\alpha-1)}{\Gamma(2 \alpha)}\left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right], \kappa_2=\frac{1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}{\alpha(\alpha-1) \Gamma(\alpha)}, $

$\kappa_1 \varphi(s)\le \int_0^1 G(t,s)\varphi(t){\rm d}t \le \kappa_2 \varphi(s),\forall s\in [0,1].$

利用引理2.3很容易获得该不等式, 故而略去其证明.

$PC[0,1]=\{u\in [0,1]\to \Bbb R ^+: u$$ t\not =t_k$上连续, 且 $ u(t_k^-), u(t_k^+)$存在, $ u(t_k^-)=u(t_k),1\le k\le m \}$, 则$PC[0,1]$ 带上范数$\|u\|=\sup\limits_{t\in [0,1]}|u(t)|$构成一Banach 空间. 令$P=\{u\in PC[0,1]: u(t)\ge 0,t\in [0,1]\}$, 则$P$$PC[0,1]$上的锥. 根据引理2.2知问题(1.1)等价于如下的积分方程

$\begin{matrix}\label{equ1} u(t)&=&\int_{0}^{1} G(t, s) f(s,u(s)){\rm d}s+t^{\alpha-1} \sum_{t \leq t_{k}<1} t_{k}^{1-\alpha} I_{k}\left(u\left(t_{k}\right)\right) \\ & = &\int_{0}^{1} G(t, s) f(s,u(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) I_{k}\left(u\left(t_{k}\right)\right), \end{matrix}$

其中

$H(t,t_k)=\left\{\begin{array}{ll}t^{\alpha-1}t_{k}^{1-\alpha}, & 0\le t \leq t_{k}<1,\\ 0, & 0<t_k< t \le 1.\end{array}\right.$

根据条件(H$f$)和(H$I_k$), 我们不能直接由方程(2.1)构造算子方程(有可能是变号的), 故而需要考虑如下的辅助问题

$\begin{equation}\label{w} \left\{\begin{array}{l} -D_{0+}^{\alpha} w(t)=M, \ t \in(0,1) \backslash\left\{t_{k}\right\}_{k=1}^{m}, \\ \Delta w\left(t_{k}\right)=M_k, \ t=t_{k}, k=1,2,\cdots,m,\\ w(0)=w^{\prime}(0)=\cdots=w^{(n-2)}(0), \ w^{\prime}(1)=\int_{0}^{1} w(t) {\rm d}\eta(t), \end{array}\right. \end{equation}$

其中$M,M_k$见条件(H$f$)和(H$I_k$). 根据引理2.2知(2.2)式的解表示为

$\begin{equation}\label{w1}w(t)=\int_{0}^{1} G(t, s) M{\rm d}s+ \sum_{k=1}^m H(t,t_k) M_k,t\in [0,1].\end{equation}$

引理2.5 定义

$\widetilde{f}(t,u)=\left\{\begin{array}{ll}f(t,u)+M,& u\ge 0,\\ f(t,0)+M, & u<0,\end{array}\right. \widetilde{I}_k(u)=\left\{\begin{array}{ll}I_k(u)+M_k,& u\ge 0,\\ I_k(0)+M_k, & u<0,\end{array}\right. $

以及积分方程

$\begin{equation}\label{equ2}u(t)=\int_{0}^{1} G(t, s) \widetilde{f}(s,u(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) \widetilde{I}_{k}\left(u\left(t_{k}\right)-w(t_k)\right),\forall t\in [0,1].\end{equation}$

下述两条成立: (i) 若$u^*$是方程(2.1)的正解, 则$u^*+w$是(2.4)式的解;

(ii) 若$u^*$是方程(2.4)的解且$u^*(t)\ge w(t),t\in [0,1]$, 则$u^*-w$是方程(2.1)的解.

已知$u^*$是方程(2.1)的正解, 则将$u^*+w$代入(2.4)式, 我们有

$\begin{eqnarray*} u^*(t)+w(t)& =&\int_{0}^{1} G(t, s) \widetilde{f}(s,u^*(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) \widetilde{I}_{k}\left(u^*\left(t_{k}\right)\right)\\ & =&\int_{0}^{1} G(t, s) [{f}(s,u^*(s))+M]{\rm d}s+ \sum_{k=1}^m H(t,t_k) [{I}_{k}\left(u^*\left(t_{k}\right)\right)+M_k]\\ & = &\int_{0}^{1} G(t, s) {f}(s,u^*(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) {I}_{k}\left(u^*\left(t_{k}\right)\right)\\ && + \int_{0}^{1} G(t, s) M {\rm d}s+ \sum_{k=1}^m H(t,t_k) M_k. \end{eqnarray*}$

由(2.3)式可知该式成立.

另一方面, 已知$u^*$是(2.4)式的解且$u^*(t)\ge w(t),t\in [0,1]$, 则将$u^*-w$代入方程(2.1)可得

$ u^*(t)-w(t) =\int_{0}^{1} G(t, s) f(s,u^*(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) I_{k}\left(u^*\left(t_{k}\right)-w(t_k)\right), $

从而

$\begin{eqnarray*} u^*(t)& =&\int_{0}^{1} G(t, s) f(s,u^*(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) I_{k}\left(u^*\left(t_{k}\right)-w(t_k)\right) +w(t) \\ & =&\int_{0}^{1} G(t, s) [f(s,u^*(s)-w(s)) +M]{\rm d}s+ \sum_{k=1}^m H(t,t_k) [I_{k}\left(u^*\left(t_{k}\right)-w(t_k)\right)+M_k]\\ & =&\int_{0}^{1} G(t, s) \widetilde{f}(s,u^*(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) \widetilde{I}_{k}\left(u^*\left(t_{k}\right)-w(t_k)\right). \end{eqnarray*}$

此式即为(2.4)式. 证毕.

根据引理2.5可知, 仅需找寻(2.4)式的且超过$w$的解即可获得原问题(1.1)的正解, 即$u^*$ 是(2.4)式的解且$u^*(t)\ge w(t),t\in [0,1]$, 则$u^*-w$是方程(1.1)的正解.

注意到$\widetilde{f},\widetilde{I}_k$的非负性, 从而我们可以定义算子$A:P\to P$ 如下

$\begin{equation}\label{opA}(Au)(t)=\int_{0}^{1} G(t, s) \widetilde{f}(s,u(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) \widetilde{I}_{k}\left(u\left(t_{k}\right)-w(t_k)\right),\forall t\in [0,1].\end{equation}$

运用文献[15,引理2.9], 结合Ascoli-Arzela定理可证明算子$A$是一全连续算子, 且若存在$u^*$$A$的不动点且不小于$w$, 则$u^*-w$是方程(1.1)的正解.

引理2.6 定义

$P_0=\bigg\{u\in P: u(t)\ge t^{\alpha-1}\frac{\kappa_1\Gamma(2\alpha)}{\alpha^2(\alpha-1)^2\kappa_2\Gamma^2(\alpha-1)}\|u\|,t\in [0,1]\bigg\}, $

$A(P)\subset P_0$.

对任意的$u\in P$, 根据引理2.3右侧不等式可得

$ (Au)(t)\le \int_{0}^{1} \frac{\varphi(s)}{\Gamma(\alpha)}\left[1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right] \widetilde{f}(s,u(s)-w(s)){\rm d}s+ \sum_{k=1}^m t_k^{1-\alpha} \widetilde{I}_{k}\left(u\left(t_{k}\right)-w(t_k)\right). $

另一方面, 再由引理2.3中左侧不等式可知

$\begin{eqnarray*} (Au)(t)& \ge& \int_{0}^{1} t^{\alpha-1} \frac{\varphi(s)}{\Gamma(\alpha)} \left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right] \widetilde{f}(s,u(s)-w(s)){\rm d}s\\ &&+ \sum_{k=1}^m t^{\alpha-1}t_k^{1-\alpha} \widetilde{I}_{k}\left(u\left(t_{k}\right)-w(t_k)\right)\\ & \ge & t^{\alpha-1}\frac{1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}} {1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}\left[\int_{0}^{1} \frac{\varphi(s)}{\Gamma(\alpha)}\left[1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right] \widetilde{f}(s,u(s)-w(s)){\rm d}s\right.\\ &&\left.+\sum_{k=1}^m t_k^{1-\alpha} \widetilde{I}_{k}\left(u\left(t_{k}\right)-w(t_k)\right)\right]\\ & \ge &t^{\alpha-1}\frac{1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}} {1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}} \|Au\|.\end{eqnarray*}$

利用$\kappa_1,\kappa_2$的定义知$\frac{1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}{1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}=\frac{\kappa_1\Gamma(2\alpha)}{\alpha^2(\alpha-1)^2\kappa_2\Gamma^2(\alpha-1)}$. 证毕.

为了方便书写, 我们记

$\frac{\kappa_1\Gamma(2\alpha)}{\alpha^2(\alpha-1)^2\kappa_2\Gamma^2(\alpha-1)}=\omega_{\kappa_1,\kappa_2}.$

注意到我们的任务是找寻$A$超过$w$的不动点, 结合引理2.6, 若$u^*$符合这一要求, 则

$\begin{eqnarray*} u^*(t)-w(t)& \ge& t^{\alpha-1} \omega_{\kappa_1,\kappa_2} \|u^*\|- \int_{0}^{1} \frac{t^{\alpha-1}(1-s)^{\alpha-2}}{\Gamma(\alpha)} M {\rm d} s- \sum_{k=1}^m t^{\alpha-1}t_k^{1-\alpha} M_k\\ & \ge &t^{\alpha-1} \left[\omega_{\kappa_1,\kappa_2} \|u^*\|-\frac{M}{(\alpha-1)\Gamma(\alpha)}-\sum_{k=1}^m t_k^{1-\alpha} M_k\right]\\ & \ge& 0. \end{eqnarray*}$

若上式成立, 仅需

$\|u^*\|\ge \omega_{\kappa_1,\kappa_2}^{-1}\left[\frac{M}{(\alpha-1)\Gamma(\alpha)}+\sum_{k=1}^m t_k^{1-\alpha} M_k\right]:=\Theta_{\kappa_1,\kappa_2,M,M_k}.$

此式亦表明若能找到$A$的不动点$u^*$, 且其范数不小于$\Theta_{\kappa_1,\kappa_2,M,M_k}$, 则$u^*-w$ 是问题(1.1)的正解.

引理2.7[30]$E$是一Banach空间, $B_R$$E$中的有界开子集, $P$$E$上的锥. 若$A:\overline{B_R}\cap P\to P$是一全连续算子, 且$u\not = Au+\lambda \phi, \lambda\ge 0, \phi\in E, u\in \partial B_R\cap P$, 则$i(A, B_R\cap P, P)=0,$ 其中$i$是锥$P$上的不动点指数.

引理2.8[30]$E$是一Banach空间, $B_r$$E$中的有界开子集, $P$$E$上的锥, $0\in B_r$.$A:\overline{B_r}\cap P\to P$是一全连续算子, 且$u\not = \lambda Au, \lambda\in [0,1], u\in \partial B_r\cap P$, 则$i(A, B_r\cap P, P)=1$.

3 主要结论

我们先给出本节需要使用的条件

(H1) 存在$b\ge 0$, $l_k\ge 0(\not \equiv 0)$, $k=1,2,\cdots,m$使得当$ b<\kappa_1^{-1}$时,

$\frac{b\kappa_1-1}{\alpha(\alpha-1)} + \omega_{\kappa_1,\kappa_2} \sum\limits_{k=1}^m l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t >0, $$\liminf_{u\to +\infty}\frac{I_k(u)}{u}\ge l_k, k=1,2,\cdots,m,\ \liminf_{u\to +\infty}\frac{f(t,u)}{u}\ge b,$

对任意的$t\in [0,1]$一致成立;

(H2) 存在$Q(t):[0,1]\to \Bbb R ^+(\not \equiv 0)$$N_k>0\ (k=1,2,\cdots,m)$使得

$ \widetilde{f}(t,u)\le \frac{\Gamma(\alpha)}{1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}\frac{\Theta_{\kappa_1,\kappa_2,M,M_k} Q(t)}{3\int_0^1 Q(t)\varphi(t){\rm d}t},$
$ \widetilde{I}_k(u)\le \frac{\Theta_{\kappa_1,\kappa_2,M,M_k} N_k}{2\sum\limits_{k=1}^m t_k^{1-\alpha}N_k}, \forall t\in [0,1], u\in [0,\Theta_{\kappa_1,\kappa_2,M,M_k}], k=1,2,\cdots,m; $

(H3) 存在$d\ge 0,$$e_k\ge 0(\not \equiv 0)$, $k=1,2,\cdots,m$使得

$d<\kappa_2^{-1}\Longrightarrow \frac{\omega_{\kappa_1,\kappa_2}(1-d\kappa_2)\Gamma(\alpha+1)\Gamma(\alpha-1)}{\Gamma(2\alpha)}- \sum\limits_{k=1}^m t_k^{1-\alpha} e_k \int_0^{t_k}t^\alpha (1-t)^{\alpha-2}{\rm d}t>0,$
$\limsup_{u\to +\infty}\frac{I_k(u)}{u}\le e_k, k=1,2,\cdots,m, \ \limsup_{u\to +\infty}\frac{f(t,u)}{u}\le d,$

对任意的$t\in [0,1]$一致成立;

(H4) 存在$t_0\in (0,1]$, $\widetilde{Q}(t):[0,1]\to \Bbb R ^+(\not \equiv 0)$$\widetilde{N}_k>0\ (k=1,2,\cdots,m)$使得

$ \widetilde{f}(t,u)\ge \frac{\widetilde{Q}(s)\Theta_{\kappa_1,\kappa_2,M,M_k}\Gamma(\alpha)}{t_0^{\alpha-1}\left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right]\int_0^1 \varphi(s)\widetilde{Q}(s){\rm d}s },$
$ \widetilde{I}_k(u)\ge \frac{\Theta_{\kappa_1,\kappa_2,M,M_k} \widetilde{N}_k}{t_0^{\alpha-1}\sum\limits_{t \leq t_{k}<1} t_k^{1-\alpha} \widetilde{N}_k }, \forall t\in [0,1], u\in [0,\Theta_{\kappa_1,\kappa_2,M,M_k}], k=1,2,\cdots,m.$

定义$B_\rho=\{u\in P: \|u\|<\rho\}$, $\rho>0$, 则$B_\rho$$P$中的开球, 且

$\overline{B_\rho}=\{u\in P: \|u\|\le\rho\}, \partial B_\rho=\{u\in P: \|u\|=\rho\}.$

定理3.1 若(H$f$), (H$I_k$), (H$\eta$), (H1)-(H2)成立, 则问题(1.1)至少有一个正解.

第一步: 存在正常数$R_1>\Theta_{\kappa_1,\kappa_2,M,M_k}$使得

$\begin{equation}\label{que}u\not = Au+\lambda \phi, \lambda\ge 0, u\in \partial B_{R_1}\cap P,\end{equation}$

其中$\phi$$P_0$中的一固定元素. 反证法, 若(3.1)式不成立, 则存在$u\in \partial B_{R_1}\cap P$, $\lambda\ge 0$使得

$\begin{equation}\label{antique}u = Au+\lambda \phi.\end{equation}$

结合引理2.6知 $u\in P_0$. 注意到$u\in \partial B_{R_1}\cap P,R_1>\Theta_{\kappa_1,\kappa_2,M,M_k}$, 从而根据(H1)可知

$\liminf_{u\to +\infty}\frac{I_k(u)+M_k}{u}\ge l_k, k=1,2,\cdots,m,\ \liminf_{u\to +\infty}\frac{f(t,u)+M}{u}\ge b,$

对任意的$t\in [0,1]$一致成立, 则存在$c>0,c_k>0\ (k=1,2,\cdots,m)$使得

$\widetilde{f}(t,u)\ge bu-c,\ \widetilde{I}_k(u)\ge l_k u-c_k, \forall u\in \Bbb R ^+, t\in [0,1], k=1,2,\cdots,m.$

运用(3.2)式可得

$\begin{eqnarray*} u(t)& \ge& (Au)(t)\\ & \ge& \int_{0}^{1} G(t, s) [b(u(s)-w(s))-c]{\rm d}s+ \sum_{k=1}^m H(t,t_k) [l_k\left(u\left(t_{k}\right)-w(t_k)\right)-c_k] \\ & \ge & b \int_{0}^{1} G(t, s) (u(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t,t_k) l_k\left(u\left(t_{k}\right)-w(t_k)\right)-\sum_{k=1}^m t_k^{1-\alpha}c_k-c\kappa_2. \end{eqnarray*}$

在上式两端乘上$\varphi(t)$并在[0,1]上积分, 利用引理2.4并经整理可得

$\begin{matrix} \label{gujishi} && \int_0^1 u(t)\varphi(t){\rm d}t+\frac{\sum\limits_{k=1}^m t_k^{1-\alpha}c_k+c\kappa_2}{\alpha(\alpha-1)}+b \kappa_2\int_{0}^{1} \varphi( s) w(s){\rm d}s \\ && +\sum_{k=1}^m t_k^{1-\alpha}l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t w(t_k) \\ & \ge& b\kappa_1 \int_{0}^{1} u(t)\varphi(t){\rm d}t+ \sum_{k=1}^m t_k^{1-\alpha}l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t u\left(t_{k}\right). \end{matrix} $

注意到$u\in P_0$, 则

$t^{\alpha-1}\omega_{\kappa_1,\kappa_2}\|u\|\le u(t)\le \|u\|,\forall t\in [0,1], $

此时若$t=t_k$, 则

$u(t_k)\ge t_k^{\alpha-1}\omega_{\kappa_1,\kappa_2}\|u\|.$

以下分两种情形讨论

1) $b\kappa_1\ge 1$. 根据(3.3)式得

$\begin{eqnarray*} &&(b\kappa_1 -1)\int_{0}^{1} t^{\alpha-1}\omega_{\kappa_1,\kappa_2}\|u\| \varphi(t){\rm d}t+ \sum_{k=1}^m t_k^{1-\alpha}l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t \cdot t_k^{\alpha-1}\omega_{\kappa_1,\kappa_2}\|u\|\\ & \le &\frac{\sum\limits_{k=1}^m t_k^{1-\alpha}c_k+c\kappa_2}{\alpha(\alpha-1)}+b \kappa_2\int_{0}^{1} \varphi( s) w(s){\rm d}s +\sum_{k=1}^m t_k^{1-\alpha}l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t w(t_k), \end{eqnarray*}$

解此不等式得

$\|u\|\le \frac{\frac{\sum\limits_{k=1}^m t_k^{1-\alpha}c_k+c\kappa_2}{\alpha(\alpha-1)} +b \kappa_2\int_{0}^{1} \varphi( s) w(s){\rm d}s +\sum\limits_{k=1}^m t_k^{1-\alpha}l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t w(t_k)}{\omega_{\kappa_1,\kappa_2}\left[\frac{(b\kappa_1 -1) \Gamma(\alpha+1)\Gamma(\alpha-1)}{\Gamma(2\alpha)}+ \sum\limits_{k=1}^m l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t\right] }:=\Lambda_1.$

2) $b\kappa_1< 1$. 仍由(3.3)式知

$\begin{eqnarray*} && (b\kappa_1-1) \int_{0}^{1} \|u\|\varphi(t){\rm d}t+ \omega_{\kappa_1,\kappa_2} \sum_{k=1}^m l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t \|u\|\\ & \le &\frac{\sum\limits_{k=1}^m t_k^{1-\alpha}c_k+c\kappa_2}{\alpha(\alpha-1)}+b \kappa_2\int_{0}^{1} \varphi( s) w(s){\rm d}s +\sum_{k=1}^m t_k^{1-\alpha}l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t w(t_k), \end{eqnarray*}$

解之得

$\|u\|\le \frac{\frac{\sum\limits_{k=1}^m t_k^{1-\alpha}c_k+c\kappa_2}{\alpha(\alpha-1)}+b \kappa_2\int_{0}^{1} \varphi( s) w(s){\rm d}s +\sum\limits_{k=1}^m t_k^{1-\alpha}l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t w(t_k)}{\frac{b\kappa_1-1}{\alpha(\alpha-1)} + \omega_{\kappa_1,\kappa_2} \sum\limits_{k=1}^m l_k\int_0^{t_k} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t }:=\Lambda_2.$

综合上述两种情形, 取$R_1>\max\{\Theta_{\kappa_1,\kappa_2,M,M_k},\Lambda_1,\Lambda_2\}$, 则当$u\in \partial B_{R_1}\cap P $时(3.2)式不可能成立, 即(3.1)式成立, 反证法奏效. 结合引理2.7知

$\begin{equation}\label{idex0}i(A, B_{R_1}\cap P, P)=0.\end{equation}$

第二步:证明

$\begin{equation}\label{tonglun}u\not = \lambda Au, \lambda\in [0,1], u\in \partial B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P.\end{equation}$

事实上, 若上式不成立, 则存在$\lambda\in [0,1]$, $u\in \partial B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P$$(\|u\|=\Theta_{\kappa_1,\kappa_2,M,M_k})$ 使得

$u=\lambda Au,$

由此结合(H2), 我们有

$\begin{eqnarray*} u(t)& \le &(Au)(t)\\ & \le &\int_{0}^{1} \frac{\varphi( s)}{\Gamma(\alpha)} \left[1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right] \frac{\Gamma(\alpha)}{1+\frac{\int_0^1 {\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}}\frac{\Theta_{\kappa_1,\kappa_2,M,M_k} Q(s)}{3\int_0^1 Q(t)\varphi(t){\rm d}t}{\rm d}s+ \sum_{k=1}^m t_k^{1-\alpha} \frac{\Theta_{\kappa_1,\kappa_2,M,M_k} N_k}{2\sum\limits_{k=1}^m t_k^{1-\alpha}N_k} \\ & =&\frac{5}{6} \Theta_{\kappa_1,\kappa_2,M,M_k}, \end{eqnarray*}$

该式右侧是常数, 则 $\|u\|\le \frac{5}{6} \Theta_{\kappa_1,\kappa_2,M,M_k}$, 这与$u\in \partial B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P$矛盾. 矛盾表明(3.5)式成立. 根据引理2.8得

$\begin{equation}\label{idex1}i(A, B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P, P)=1.\end{equation}$

结合(3.4)和(3.6)式可得

$ \begin{eqnarray*} i\left(A,\left(B_{R_1} \backslash \overline{B}_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\right) \cap P, P\right) &=&i(A, B_{R_1}\cap P, P)- i(A, B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P, P)\\ &=&0-1=-1. \end{eqnarray*}$

由此可得算子$A$$\left(B_{R_1} \backslash \overline{B}_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\right) \cap P$中至少有一个不动点$u^*$, 且$\|u^*\|\ge \Theta_{\kappa_1,\kappa_2,M,M_k}$, 故而知$u^*-w$是问题(1.1)的正解. 证毕.

定理3.2 若(H$f$), (H$I_k$), (H$\eta$), (H3)-(H4)成立, 则问题(1.1)至少有一个正解.

第一步: 证明存在足够大的$R_2>\Theta_{\kappa_1,\kappa_2,M,M_k}$使得

$\begin{equation}\label{tonglun1}u\not = \lambda Au, \lambda\in [0,1], u\in \partial B_{R_2}\cap P.\end{equation}$

反证法. 若上式不成立, 则存在$\lambda\in [0,1]$, $u\in \partial B_{R_2}\cap P$使得

$\begin{equation}\label{antitonglun1}u = \lambda Au.\end{equation}$

注意到引理2.6, $u\in P_0$. 根据(H3)我们可得

$\limsup_{u\to +\infty}\frac{I_k(u)+M_k}{u}\le e_k, k=1,2,\cdots,m, \ \limsup_{u\to +\infty}\frac{f(t,u)+M}{u}\le d,$

对任意的$t\in [0,1]$一致成立, 从而存在$\widetilde{c}>0,\widetilde{c}_k>0,k=1,2,\cdots,m$使得

$\widetilde{f}(t,u)\le du+\widetilde{c}, \ \widetilde{I}_k(u)\le e_k u+\widetilde{c}_k, \forall u\in \Bbb R ^+, t\in [0,1],k=1,2,\cdots,m.$

根据(3.8)式有

$\begin{eqnarray*} u(t)&\le& (Au)(t)\\ & \le& \int_{0}^{1} G(t, s) [d(u(s)-w(s))+\widetilde{c}]{\rm d}s+ \sum_{k=1}^m H(t,t_k) [e_k\left(u\left(t_{k}\right)-w(t_k)\right)+\widetilde{c}_k ] \\ & \le& d\int_{0}^{1} G(t, s) u(s){\rm d}s+ \sum_{k=1}^m H(t,t_k) e_k u\left(t_{k}\right) + \sum_{k=1}^m t_k^{1-\alpha} \widetilde{c}_k+\widetilde{c}\kappa_2. \end{eqnarray*}$

在上式两端乘上$\varphi(t)$并在[0,1]上积分, 运用引理2.4并经整理可得

$\int_0^1 u(t)\varphi(t){\rm d}t \le d\kappa_2 \int_0^1 u(t)\varphi(t){\rm d}t+\sum_{k=1}^m t_k^{1-\alpha} e_k \int_0^{t_k}t^\alpha (1-t)^{\alpha-2}{\rm d}t u\left(t_{k}\right)+ \frac{\sum\limits_{k=1}^m t_k^{1-\alpha} \widetilde{c}_k+\widetilde{c}\kappa_2}{\alpha(\alpha-1)}. $

注意到$u\in P_0$, 由(H3)得

$\begin{eqnarray*} && \omega_{\kappa_1,\kappa_2}(1-d\kappa_2) \int_0^1 t^{\alpha-1}\|u\|\varphi(t){\rm d}t\le (1-d\kappa_2)\int_0^1 u(t)\varphi(t){\rm d}t\\ & \le& \sum_{k=1}^m t_k^{1-\alpha} e_k \int_0^{t_k}t^\alpha (1-t)^{\alpha-2}{\rm d}t \|u\|+ \frac{\sum\limits_{k=1}^m t_k^{1-\alpha} \widetilde{c}_k+\widetilde{c}\kappa_2}{\alpha(\alpha-1)}, \end{eqnarray*}$

解之得

$\|u\|\le \frac{\frac{\sum\limits_{k=1}^m t_k^{1-\alpha} \widetilde{c}_k+\widetilde{c}\kappa_2}{\alpha(\alpha-1)}}{\frac{\omega_{\kappa_1,\kappa_2}(1-d\kappa_2)\Gamma(\alpha+1)\Gamma(\alpha-1)}{\Gamma(2\alpha)}- \sum\limits_{k=1}^m t_k^{1-\alpha} e_k \int_0^{t_k}t^\alpha (1-t)^{\alpha-2}{\rm d}t}:=\Lambda_3.$

这时若取$R_2>\max\{\Theta_{\kappa_1,\kappa_2,M,M_k},\Lambda_3\}$, 则(3.8)式不成立, 从而反证法奏效,(3.7)式成立. 根据引理2.8得

$\begin{equation}\label{idex2}i(A, B_{R_2}\cap P, P)=1.\end{equation}$

第二步: 证明

$\begin{equation}\label{que2}u\not = Au +\lambda \widetilde{\phi}, \lambda \ge 0, u\in \partial B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P,\end{equation}$

其中$\widetilde{\phi}\in P$是一固定元素. 若上式不成立, 则存在$\lambda \ge 0, u\in \partial B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P$使得

$u = Au +\lambda \widetilde{\phi},$

此式表明

$\Theta_{\kappa_1,\kappa_2,M,M_k}=\|u\|\ge u(t)\ge (Au)(t), \forall t\in [0,1].$

该不等式左端是常数, 从而有

$\begin{equation}\label{normbudengshi}\|u\|\ge \|Au\|.\end{equation}$

另一方面, 注意到$t_0\in (0,1]$, 从而我们有

$\begin{eqnarray*} \|Au\|& \ge& (Au)(t_0)= \int_{0}^{1} G(t_0, s) \widetilde{f}(s,u(s)-w(s)){\rm d}s+ \sum_{k=1}^m H(t_0,t_k) \widetilde{I}_{k}\left(u\left(t_{k}\right)-w(t_k)\right)\\ & \ge& \int_{0}^{1} t_0^{\alpha-1} \frac{\varphi(s)}{\Gamma(\alpha)} \left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right] \frac{\widetilde{Q}(s) \Theta_{\kappa_1,\kappa_2,M,M_k}\Gamma(\alpha)}{t_0^{\alpha-1}\left[1+\frac{\int_0^1 \tau^{\alpha-1}{\rm d}\eta(\tau)}{\delta_{\alpha,\eta}}\right]\int_0^1 \varphi(s)\widetilde{Q}(s){\rm d}s }{\rm d}s\\ && + \sum_{t \leq t_{k}<1} t_0^{\alpha-1}t_k^{1-\alpha} \frac{\Theta_{\kappa_1,\kappa_2,M,M_k} \widetilde{N}_k}{t_0^{\alpha-1}\sum\limits_{t \leq t_{k}<1} t_k^{1-\alpha} \widetilde{N}_k }\\ & = &2\Theta_{\kappa_1,\kappa_2,M,M_k}, \end{eqnarray*}$

这与(3.11)式矛盾. 矛盾表明(3.10)式成立. 根据引理2.7知

$\begin{equation}\label{idex20}i(A, B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P, P)=0.\end{equation}$

综合(3.9)和(3.12)式有

$\begin{eqnarray*} i\left(A,\left(B_{R_2} \backslash \overline{B}_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\right) \cap P, P\right) &=&i(A, B_{R_2}\cap P, P)- i(A, B_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\cap P, P)\\ &=&1-0=1. \end{eqnarray*}$

由此可得算子$A$$\left(B_{R_2} \backslash \overline{B}_{\Theta_{\kappa_1,\kappa_2,M,M_k}}\right) \cap P$中至少有一个不动点$u^*$, 且$\|u^*\|\ge \Theta_{\kappa_1,\kappa_2,M,M_k}$, 故而知$u^*-w$是问题(1.1)的正解. 证毕.

注 3.3 虽然条件(H1)-(H4)中有很多不等式, 看上去同时满足这些不等式较为困难, 然而我们可以通过调整各个参数使所有的不等式均成立. 另外我们注意到, 极限条件

$\liminf\limits_{u\to +\infty}\frac{f(t,u)}{u}\ge b $

中的数$b$仅需大于或者等于0, 是一个很弱的条件.

$n=3,\alpha=2.5,\eta(t)=t$, 则

$\delta_{\alpha,\eta}=1.5-\int_0^1 t^{1.5}{\rm d}t=1.1, \kappa_1=\frac{2.5 \Gamma(1.5)}{\Gamma(5)}\left[1+\frac{\int_0^1 \tau^{1.5}{\rm d}\tau}{1.1}\right]=0.126, $
$ \kappa_2=\frac{1+\frac{\int_0^1 {\rm d}\tau}{1.1}}{2.5\times 1.5 \Gamma(2.5)}=0.383, \omega_{\kappa_1,\kappa_2}=\frac{0.126\Gamma(5)}{2.5^2\times 1.5^2\times 0.383 \Gamma^2(1.5)}=0.715. $

为了便于计算, 仅取两个脉冲点, 即令$m=2$, 脉冲点$t_1=0.5,t_2=0.6$, 则

$\int_0^{t_1} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t=\int_0^{0.5} t^{2.5}(1-t)^{0.5}{\rm d}t=0.02, $
$ \int_0^{t_2} t^{\alpha}(1-t)^{\alpha-2}{\rm d}t=\int_0^{0.6} t^{2.5}(1-t)^{0.5}{\rm d}t=0.035, $
$ \Theta_{\kappa_1,\kappa_2,M,M_k}=0.715^{-1}\left[\frac{M}{1.5\Gamma(2.5)}+ t_1^{-1.5} M_1+t_2^{-1.5} M_2\right]=0.701 M+3.956M_1+3.01M_2. $

例3.4 重写(H1)-(H2)

(H1) 存在$b\ge 0$, $l_1,l_2\ge 0(\not \equiv 0)$使得

$\mbox{当$b<\kappa_1^{-1}$时:}\ \frac{0.126 b-1}{3.75} + 0.0143l_1+0.025l_2 >0, $

$\liminf_{u\to +\infty}\frac{I_1(u)}{u}\ge l_1,\ \liminf_{u\to +\infty}\frac{I_2(u)}{u}\ge l_2, \ \liminf_{u\to +\infty}\frac{f(t,u)}{u}\ge b,$

对任意的$t\in [0,1]$一致成立;

(H2) 取$M=5,M_1=6,M_2=7$, $N_1=2.5,N_2=3.4$, $Q(t)\equiv 10, \forall t\in [0,1]$, 则$0.701 M+3.956M_1+3.01M_2=48.311,$$6.657N_1+4.303N_2=31.27$,

$ \widetilde{f}(t,u)\le \frac{\Gamma(2.5)}{1+\frac{\int_0^1 {\rm d}\tau}{1.1}}\frac{(0.701 M+3.956M_1+3.01M_2) Q(t)}{3\int_0^1 Q(t)\varphi(t){\rm d}t}=42.03,$
$ \widetilde{I}_1(u)\le \frac{(0.701 M+3.956M_1+3.01M_2)N_1 }{6.657N_1+4.303N_2}=3.862,$
$ \widetilde{I}_2(u)\le \frac{(0.701 M+3.956M_1+3.01M_2)N_2 }{6.657N_1+4.303N_2}=5.253, \forall t\in [0,1], u\in [0,48.311].$

从而令

${f}(t,u)=\sin \pi t+\frac{1}{25}u-5,I_1(u)=\frac{1}{650}u^2-6, I_2(u)=\frac{1}{2}\sqrt{u}-7, t\in [0,1],u\in \Bbb R ^+,$

$\widetilde{f}(t,u)=\sin \pi t+\frac{1}{25}u\le 2.932$, $\widetilde{I}_1(u)=\frac{1}{650}u^2\le 3.59$, $\widetilde{I}_2(u)=\frac{1}{2}\sqrt{u}\le 3.475,t\in [0,1],$$u\in [0,48.311]$.

另外可计算极限

$\liminf_{u\to +\infty}\frac{f(t,u)}{u}=\liminf_{u\to +\infty}\frac{\sin \pi t+\frac{1}{25}u-5}{u}=\frac{1}{25}<\kappa_1^{-1},$
$\liminf_{u\to +\infty}\frac{I_1(u)}{u}=\liminf_{u\to +\infty}\frac{\frac{1}{650}u^2-6}{u}=+\infty,$
$\liminf_{u\to +\infty}\frac{I_2(u)}{u}=\liminf_{u\to +\infty}\frac{\frac{1}{2}\sqrt{u}-7}{u}=0, $

从而可取$b=\frac{1}{25}$, $l_1=20,l_2=0$使得$\frac{0.126 b-1}{3.75} + 0.0143l_1+0.025l_2 >0$.

综上所得(H1)-(H2)满足, 另(H$f$), (H$I_k$), (H$\eta$)显然成立, 则定理3.1所有的条件均满足.

例3.5 重写(H3)-(H4)

(H3) 存在$d\ge 0,$$e_k\ge 0(\not \equiv 0)$, $k=1,2$使得

$d<\kappa_2^{-1}\Longrightarrow 0.088(1-0.383d)-0.057e_1-0.075e_2>0,$
$\limsup_{u\to +\infty}\frac{I_k(u)}{u}\le e_k, k=1,2, \ \limsup_{u\to +\infty}\frac{f(t,u)}{u}\le d,$

对任意的$t\in [0,1]$一致成立;

(H4) 取$t_0=0.5,M=7,M_1=8,M_2=9$, $\widetilde{N}_1=1.5,\widetilde{N}_2=1.7$, $\widetilde{Q}(t)\equiv 20,t\in [0,1]$, 则

$\Theta_{\kappa_1,\kappa_2,M,M_k}=0.701 M+3.956M_1+3.01M_2=63.645,$
$ \widetilde{f}(t,u)\ge 642,\widetilde{I}_1(u)\ge 34.178, \widetilde{I}_2(u)\ge 38.735, \forall t\in [0,1], u\in [0,63.645], k=1,2.$

$f(t,u)=\ln(e^{650}+u)+|\cos\pi t|-7,$
$ I_1(u)=\ln (e^{35}+u)-8,I_2(u)=\ln (e^{39}+u)-9,t\in [0,1], u\in \Bbb R ^+,$

$\widetilde{f}(t,u)=\ln(e^{650}+u)+|\cos\pi t|\ge 650, \widetilde{I}_1(u)=\ln (e^{35}+u)\ge 35, $
$ \widetilde{I}_2(u)=\ln (e^{39}+u)\ge 39, \forall t\in [0,1],u\in [0,63.645]. $

另外可计算极限

$\limsup_{u\to +\infty}\frac{f(t,u)}{u}=\limsup_{u\to +\infty}\frac{\ln(e^{650}+u)+|\cos\pi t|-7}{u}=0,$
$\limsup_{u\to +\infty}\frac{I_1(u)}{u}=\limsup_{u\to +\infty}\frac{\ln (e^{35}+u)-8}{u}=0, $$\limsup_{u\to +\infty}\frac{I_2(u)}{u}=\limsup_{u\to +\infty}\frac{\ln (e^{39}+u)-9}{u}=0, $

则可取$d=0.001,e_1=0.2,e_2=0.3$使得$0.088(1-0.383d)-0.057e_1-0.075e_2>0$.

综上所得(H3)-(H4)满足, 另(H$f$), (H$I_k$), (H$\eta$)显然成立, 则定理3.2所有的条件均满足.

参考文献

Wang Y, Liu L, Zhang X, Wu Y.

Positive solutions of an abstract fractional semipositone differential system model for bioprocesses of HIV infection

Applied Mathematics and Computation, 2015, 258: 312-324

DOI:10.1016/j.amc.2015.01.080      URL     [本文引用: 3]

Zhong Q, Zhang X, Gu L, Lei L, Zhao Z.

Multiple positive solutions for singular higher-order semipositone fractional differential equations with $p$-Laplacian

Nonlinear Analysis: Modelling and Control, 2020, 25(5): 806-826

[本文引用: 3]

Hao X, Sun H, Liu L, Wang D.

Positive solutions for semipositone fractional integral boundary value problem on the half-line

RACSAM, 2019, 113(4): 3055-3067

DOI:10.1007/s13398-019-00673-w      URL     [本文引用: 2]

Ding Y, Jiang J, O'Regan D, Xu J.

Positive solutions for a system of Hadamard-type fractional differential equations with semipositone nonlinearities

Complexity, 2020, Aarticle ID 9742418

[本文引用: 2]

Xu J, Goodrich C, Cui Y.

Positive solutions for a system of first-order discrete fractional boundary value problems with semipositone nonlinearities

RACSAM, 2019, 113(2): 1343-1358

DOI:10.1007/s13398-018-0551-7      URL     [本文引用: 2]

Yang W.

Positive solutions for nonlinear semipositone fractional $q$-difference system with coupled integral boundary conditions

Applied Mathematics and Computation, 2014, 244: 702-725

DOI:10.1016/j.amc.2014.07.039      URL     [本文引用: 2]

Yuan C.

Two positive solutions for $(n-1,1)$-type semipositone integral boundary value problems for coupled systems of nonlinear fractional differential equations

Communications in Nonlinear Science and Numerical Simulation, 2012, 17(2): 930-942

DOI:10.1016/j.cnsns.2011.06.008      URL     [本文引用: 2]

Xu X, Jiang D, Yuan C.

Multiple positive solutions to singular positone and semipositone Dirichlet-type boundary value problems of nonlinear fractional differential equations

Nonlinear Analysis: Theory Methods & Applications, 2011, 74(16): 5685-5696

[本文引用: 2]

Henderson J, Luca R.

Existence of positive solutions for a system of semipositone fractional boundary value problems

Electronic Journal of Qualitative Theory of Differential Equations, 2016, 22: 1-28

[本文引用: 2]

Ege S, Topal F.

Existence of multiple positive solutions for semipositone fractional boundary value problems

Filomat, 2019, 33(3): 749-759

DOI:10.2298/FIL1903749E      URL     [本文引用: 2]

Xu J, Wei Z, Ding Y.

Positive solutions for a boundary-value problem with Riemann-Liouville fractional derivative

Lithuanian Mathematical Journal, 2012, 52(4): 462-476

DOI:10.1007/s10986-012-9187-z      URL     [本文引用: 2]

Jie Z, Feng M.

Green's function for Sturm-Liouville-type boundary value problems of fractional order impulsive differential equations and its application

Boundary Value Problems, 2014, 2014: Article 69

DOI:10.1186/1687-2770-2014-69      URL     [本文引用: 3]

Wang G, Ahmad B, Zhang L.

Some existence results for impulsive nonlinear fractional differential equations with mixed boundary conditions

Computers and Mathematics with Applications, 2011, 62: 1389-1397

DOI:10.1016/j.camwa.2011.04.004      URL     [本文引用: 3]

Zhang K, Xu J.

Positive solutions for an impulsive boundary value problem with Caputo fractional derivative

Journal of Nonlinear Sciences and Applications, 2016, 9(6): 4628-4638

DOI:10.22436/jnsa.009.06.101      URL     [本文引用: 3]

Zhao K.

Impulsive integral boundary value problems of the higher-order fractional differential equation with eigenvalue arguments

Advances in Difference Equations, 2015, 2015: Article 382

DOI:10.1186/s13662-015-0725-y      URL     [本文引用: 3]

Ahmad B, Sivasundaram S.

Existence results for nonlinear impulsive hybrid boundary value problems involving fractional differential equations

Nonlinear Analysis: Hybrid Systems, 2009, 3(3): 251-258

DOI:10.1016/j.nahs.2009.01.008      URL     [本文引用: 2]

Ahmad B, Sivasundaram S.

Existence of solutions for impulsive integral boundary value problems of fractional order

Nonlinear Analysis: Hybrid Systems, 2010, 4(1): 134-141

DOI:10.1016/j.nahs.2009.09.002      URL     [本文引用: 2]

Wang G, Ahmad B, Zhang L, Nieto J.

Comments on the concept of existence of solution for impulsive fractional differential equations

Communications in Nonlinear Science and Numerical Simulation, 2014, 19(3): 401-403

DOI:10.1016/j.cnsns.2013.04.003      URL     [本文引用: 2]

Bouzaroura A, Mazouzi S.

Existence results for certain multi-orders impulsive fractional boundary value problem

Results in Mathematics, 2014, 66(1/2): 1-20

DOI:10.1007/s00025-014-0403-5      URL     [本文引用: 2]

Tian Y, Bai Z.

Existence results for the three-point impulsive boundary value problem involving fractional differential equations

Computers & Mathematics with Applications, 2010, 59(8): 2601-2609

DOI:10.1016/j.camwa.2010.01.028      URL     [本文引用: 2]

Fu X, Bao X.

Some existence results for nonlinear fractional differential equations with impulsive and fractional integral boundary conditions

Advances in Difference Equations, 2014, 2014: Article 129

DOI:10.1186/1687-1847-2014-129      URL     [本文引用: 2]

Liu Z, Lu L, Szanto I.

Existence of solutions for fractional impulsive differential equations with $p$-Laplacian operator

Acta Mathematica Hungarica, 2013, 141(3): 203-219

DOI:10.1007/s10474-013-0305-0      URL     [本文引用: 2]

Zhao X, Ge W.

Some results for fractional impulsive boundary value problems on infinite intervals

Applications of Mathematics, 2011, 56(4): 371-387

DOI:10.1007/s10492-011-0021-4      URL     [本文引用: 2]

Liu Y.

Solvability of impulsive periodic boundary value problems for higher order fractional differential equations

Arabian Journal of Mathematics, 2016, 5(4): 195-214

DOI:10.1007/s40065-016-0153-1      URL     [本文引用: 2]

Wang H, Lin X.

Anti-periodic BVP of fractional order with fractional impulsive conditions and variable parameter

Journal of Applied Mathematics and Computing, 2017, 53(1/2): 285-301

DOI:10.1007/s12190-015-0968-5      URL     [本文引用: 2]

El-shahed M.

Positive solutions for boundary-value problems of nonlinear fractional differential equation

Abstract and Applied Analysis, 2007, ID: 010368

[本文引用: 2]

Kilbas A, Srivastava H, Trujillo J.

Theory and Applications of Fractional Differential Equations

Amsterdam: NorthHolland, 2006

[本文引用: 2]

Podlubny I. Fractional Differential Equations, Mathematics in Science and Engineering, Vol 198. San Diego: Academic Press, 1999

[本文引用: 2]

Samko S, Kilbas A, Marichev O. Fractional Integrals and Derivatives: Theory and Applications. Yverdon: Gordon and Breach Science Publisher, 1993

[本文引用: 2]

Guo D, Lakshmikantham V. Nonlinear Problems in Abstract Cones. Orlando: Academic Press, 1988

[本文引用: 2]

/