带变号格林函数的三阶三点边值问题的正解的存在性

雷策宇; 韩晓玲

doi:10.6054/j.jscnun.2021032

带变号格林函数的三阶三点边值问题的正解的存在性

雷策宇,
韩晓玲^,

西北师范大学数学与统计学院, 兰州 730070

基金项目:

国家自然科学基金项目 11561063

详细信息

通讯作者:
韩晓玲, Email: hanxiaoling9@163.com

中图分类号: O175.8
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出版历程
- 收稿日期: 2020-08-01
- 网络出版日期: 2021-04-28
- 刊出日期: 2021-04-24

The Existence of Positive Solutions to A Third-order Three-point Boundary Value Problem with Sign-changing Green's Function

LEI Ceyu,
HAN Xiaoling^,

College of Mathematics and Statistics, Northwest Normal University, Lanzhou 730070, China

摘要

摘要: 应用格林函数的性质和迭代法, 研究了一类具有变号格林函数的三阶三点边值问题 $\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} u'''\left( t \right) = f\left( {t,u\left( t \right)} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {t \in \left[ {0,1} \right]} \right),\\ u\left( 1 \right) = 0,u'\left( 0 \right) = u''\left( 0 \right),\alpha u''\left( \eta \right) + \beta u\left( 0 \right) = 0 \end{array} \end{array}} \right.$ 正解的存在性, 其中, f∈C([0, 1]×[0, ∞), [0, ∞)), α∈[0, 1], $\frac{2}{7}$ α < β < $\frac{2}{3}$ α, η∈[ $\frac{2}{3}$ , 1). 得到了该边值问题正解存在性的条件.
- 迭代法 /
- 三阶三点边值问题 /
- 正解
Abstract: Using the properties of Green's function and the iterative method, the existence of positive solutions to a class of third-order three-point boundary value problems with sign-changing Green's function is studied: $\left\{ {\begin{array}{*{20}{c}} \begin{array}{l} u'''\left( t \right) = f\left( {t,u\left( t \right)} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {t \in \left[ {0,1} \right]} \right),\\ u\left( 1 \right) = 0,u'\left( 0 \right) = u''\left( 0 \right),\alpha u''\left( \eta \right) + \beta u\left( 0 \right) = 0, \end{array} \end{array}} \right.$ where f∈C([0, 1]×[0, ∞), [0, ∞)), α∈[0, 1], $\frac{2}{7}$ α < β < $\frac{2}{3}$ α, η∈[ $\frac{2}{3}$ , 1). The conditions for the existence of positive solutions to the boundary value problem are obtained.
- iteration method /
- third-order three-point boundary value problem /
- positive solutions

HTML全文

三阶常微分方程边值问题由于在工程、物理和流体力学等领域的显著应用而受到广泛关注. 学者们运用单调迭代法、上下解方法、Guo-Krasnosel'skii不动点定理和Leray-Schauder非线性抉择等, 研究了三阶三点边值问题在格林函数非负的情况下的单个或多个正解的存在性^[1-7]. 近年来, 学者们在格林函数变号的情况下也得到了很多结果^[8-17]. 如: LI等^[9]使用Guo-Krasnosel'skii不动点定理讨论了变号格林函数的三阶三点边值问题

$\left\{\begin{array} {l} { u ^ { \prime \prime \prime } ( t ) = f ( t , u ( t ) ) \quad ( t \in [ 0 , 1 ] ) , } \\ { u ( 1 ) = u ^ { \prime } ( 0 ) = 0 , u ^ { \prime \prime } ( \eta ) + \alpha u ( 0 ) = 0 } \end{array}\right.$

正解的存在性, 其中, α ∈[0, 2), η ∈ $[\frac{{\sqrt {121 + 24\alpha } - 5}}{{3\left( {4 + \alpha } \right)}}, 1)$ .

GAO和SUN^[10]运用Avery-Henderson不动点定理, 在格林函数变号时讨论了问题

$\left\{\begin{array}{l} u^{\prime \prime \prime}(t)=f(t, u(t)) \quad(t \in[0,1]), \\ u(1)=u^{\prime}(0)=0, u^{\prime \prime}(\eta)-\alpha u^{\prime}(1)=0 \end{array}\right.$

正解的存在性, 其中, α ∈[0, 2), η ∈ $[\frac{{4 + \alpha }}{{24 - 3\alpha }}, 1)$ .

ZHAO和LI^[11]运用迭代法讨论了变号格林函数的三阶三点边值问题

$\left\{\begin{array}{l} u^{\prime \prime \prime}(t)=f(t, u(t)) \quad(t \in[0,1]), \\ u(1)=u^{\prime}(0)=0, u^{\prime \prime}(\eta)+\alpha u(0)=0 \end{array}\right.$

正解的存在性, 其中, α∈[0, 2), η∈[2/3, 1).

受文献[11]的启发, 本文将运用迭代法, 研究如下具有变号格林函数的三阶三点边值问题

$\left\{\begin{array}{l} u^{\prime \prime \prime}(t)=f(t, u(t)) \quad(t \in[0,1]), \\ u(1)=0, u^{\prime}(0)=u^{\prime \prime}(0), \alpha u^{\prime \prime}(\eta)+\beta u(0)=0 \end{array}\right.$

(1)

正解的存在性, 其中, α ∈[0, 1], $\frac{2}{7}$ α < β < $\frac{2}{3}$ α, η ∈[2/3, 1). 在本文中, 总是假设f∈C([0, 1]×[0, ∞), [0, ∞)), 且f满足下列2个条件:

(H₁) 对于每一个u∈[0, +∞), 映射t↦f(t, u)是递减的；

(H₂) 对于每一个t∈[0, 1], 映射u↦f(t, u)是递增的.

1. 预备知识

本文所用到的空间是C[0, 1], 记E=C[0, 1], ‖u‖= $\mathop {\max }\limits_{t \in \left[ {0, 1} \right]} \left| {u\left( t \right)} \right|$ .

引理1 对任意的y ∈C[0, 1], 边值问题

$\left\{\begin{array}{l} u^{\prime \prime \prime}(t)=y(s) \quad(t \in[0,1]), \\ u(1)=0, u^{\prime}(0)=u^{\prime \prime}(0), \alpha u^{\prime \prime}(\eta)+\beta u(0)=0 \end{array}\right.$

(2)

有唯一解

$u(t)=\int_{0}^{1} G(t, s) f(s, u(s)) \mathrm{d} s,$

其中:

$\begin{array}{c} G(t, s)=g_{1}(t, s)+g_{2}(t, s)+g_{3}(\eta, t, s) ;\\ g_{1}(t, s)=\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}; \\ g_{2}(t, s)=\left\{\begin{array}{l} 0 \quad(0 \leqslant t \leqslant s \leqslant 1), \\ (t-s)^{2} / 2 \quad(0 \leqslant s \leqslant t \leqslant 1) ; \end{array}\right. \\ g_{3}(\eta, t, s)=\left\{\begin{array}{l} 0 \quad(0 \leqslant \eta \leqslant s \leqslant 1), \\ \frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta} \quad(0 \leqslant s \leqslant \eta \leqslant 1); \end{array}\right. \end{array}$

证明对u'''(t)=y(t)在[0, t]上两边积分, 得

$\begin{array}{c} u^{\prime \prime}(t)=u^{\prime \prime}(0)-\int_{0}^{t} y(s) \mathrm{d} s ,\\ u^{\prime}(t)=u^{\prime}(0)+u^{\prime \prime}(0) t-\int_{0}^{t}(t-s) y(s) \mathrm{d} s ,\\ u(t)=u(0)+u^{\prime}(0) t+\frac{1}{2} u^{\prime \prime}(t) t^{2}+\frac{1}{2} \int_{0}^{t}(t-s)^{2} y(s) \mathrm{d} s, \end{array}$

由边界条件可得

$\begin{array}{c} u(0)+\frac{3}{2} u^{\prime}(0)+\frac{1}{2} \int_{0}^{1}(1-s)^{2} y(s) \mathrm{d} s=0,\\ \alpha u^{\prime}(0)+\alpha \int_{0}^{\eta} y(s) \mathrm{d} s+\beta u(0)=0. \end{array}$

从而有

$\begin{array}{l} u(0)=\frac{3 \alpha}{2 \alpha-3 \beta} \int_{0}^{\eta} y(s) \mathrm{d} s-\frac{\alpha}{2 \alpha-3 \beta} \int_{0}^{1}(1-s)^{2} y(s) \mathrm{d} s ,\\ u^{\prime}(0)=\frac{\beta}{2 \alpha-3 \beta} \int_{0}^{1}(1-s)^{2} y(s) \mathrm{d} s-\frac{2 \alpha}{2 \alpha-3 \beta} \int_{0}^{\eta} y(s) \mathrm{d} s. \end{array}$

故边值问题(2)的解为

$\begin{aligned} u(t)=& \frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta} \int_{0}^{\eta} y(s) \mathrm{d} s+\\ & \frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta} \int_{0}^{1}(1-s)^{2} y(s) \mathrm{d} s+\\ & \frac{1}{2} \int_{0}^{t}(t-s)^{2} y(s) \mathrm{d} s. \end{aligned}$

证毕.

引理2 G(t, s)具有如下性质:

(1) 当0≤s≤η时, G(t, s)≥0;

(2) 当0≤η≤s时, G(t, s)≤0.

证明当0≤s≤η时, 有

$\begin{array}{l} G(t, s)= \\ \left\{\begin{array}{l} \frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta} \quad(0 \leqslant t \leqslant s \leqslant 1), \\ \frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta}+\frac{1}{2}(t-s)^{2} \\ \quad(0 \leqslant s \leqslant t \leqslant 1) . \end{array}\right. \end{array}$

分以下2种情况讨论:

(1) 当0≤t≤s≤1时, 有

$\begin{array}{c} G_{t}(t, s)=\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}= \\ \frac{(1+t)\left[\beta(1-s)^{2}-2 \alpha\right]}{2 \alpha-3 \beta}<0; \end{array}$

(2) 当0≤s≤t≤1时, 有

$\begin{array}{l} G_{t}(t, s)=\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)= \\ \ \ \ \ \frac{(\beta-\alpha)+\left(\beta s^{2}-\alpha\right)+\beta t\left(s^{2}-2\right)+s(\beta-2 \alpha)-2 \beta t s}{2 \alpha-3 \beta}<0. \end{array}$

综上可知, G(t, s)关于变量t单调递减. 从而有

$\begin{array}{c} \min \{G(t, s), t \in[0,1]\}=G(1, s)=0, \\ \max \{G(t, s), t \in[0,1]\}=G(0, s)=\frac{3 \alpha-\alpha(1-s)^{2}}{2 \alpha-3 \beta}>0 . \end{array}$

故当0≤s≤η时, G(t, s)≥0.

同理可证得: 当0≤η≤s时, G(t, s)关于变量t单调递增. 则有

$\begin{array}{c} \max \{G(t, s), t \in[0,1]\}=G(1, s)=0,\\ \min \{G(t, s), t \in[0,1]\}=G(0, s)=-\frac{\alpha(1-s)}{2 \alpha-3 \beta}<0 . \end{array}$

从而G(t, s)≤0. 证毕.

设M=max{|G(t, s)| |t, s ∈[0, 1]}, 则

$M=\max \left\{\frac{3 \alpha-\alpha(1-s)^{2}}{2 \alpha-3 \beta}, \frac{\alpha(1-s)}{2 \alpha-3 \beta}\right\}<\frac{3 \alpha}{2 \alpha-3 \beta}.$

设K={y∈E|y(t)在[0, 1]上非负且递减}, 则K是E中的一个锥, 且在E中定义一个序关系“≤”, u≤ν当且仅当ν-u∈K.

定义算子T: K→E

$(T u)(t)=\int_{0}^{1} G(t, s) f(s, u(s)) \mathrm{d} s \quad(u \in K, t \in[0,1]).$

显然, 若u是T中的不动点, 则u是边值问题(1)的递减非负解.

引理3 算子T: K→K是全连续的.

证明设u∈K, 当t∈[0, η]时, 有

$\begin{array}{l} (T u)(t)=\int_{0}^{t}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\right. \\ \ \ \ \ \ \ \ \ \left.\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta}+\frac{1}{2}(t-s)^{2}\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \ \ \ \ \int_{t}^{\eta}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta}\right] \times \\ \ \ \ \ \ \ \ \ f(s, u(s)) \mathrm{d} s+\int_{\eta}^{1}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}\right] \times \\ \ \ \ \ \ \ \ \ f(s, u(s)) \mathrm{d} s . \end{array}$

由条件(H₁)、(H₂), 可得

$\begin{array}{l} (T u)^{\prime}(t)= \\ \ \ \ \ \int_{0}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \int_{t}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \int_{\eta}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] f(s, u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ f(\eta, u(\eta))\left\{\int_{0}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right] \mathrm{d} s+\right. \\ \ \ \ \ \int_{t}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}\right] \mathrm{d} s+\\ \ \ \ \ \left.\int_{\eta}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] \mathrm{d} s\right\} \leqslant f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta} \times\right. \\ \ \ \ \ \int_{0}^{\eta}\left(-2 s+s^{2}\right) \mathrm{d} s+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \int_{0}^{\eta} \mathrm{d} s+\int_{0}^{t} t \mathrm{~d} s- \\ \ \ \ \ \left.\int_{0}^{t} s \mathrm{~d} s+\frac{\beta(1+t)}{2 \alpha-3 \beta} \int_{\eta}^{1}(1-s)^{2} \mathrm{~d} s\right] \leqslant\\ \ \ \ \ f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\frac{1}{2} t^{2}+\right. \\ \ \ \ \ \left.\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta\right] \leqslant f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\right. \\ \ \ \ \ \left.\frac{2(\beta-2 \alpha)(1+t) \eta+(2 \alpha-3 \beta)}{2(2 \alpha-3 \beta)}\right] \leqslant f(\eta, u(\eta)) \times \\ \ \ \ \ {\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\frac{2 \eta(\beta-2 \alpha)+(2 \alpha-3 \beta)}{2 \alpha-3 \beta}\right] \leqslant 0}. \end{array}$

当t ∈[η, 1]时, 有

$\begin{array}{l} (T u)(t)=\int_{0}^{\eta}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\frac{3 \alpha-2 \alpha\left(t+t^{2} / 2\right)}{2 \alpha-3 \beta}+\right. \\ \ \ \ \ \left.\frac{1}{2}(t-s)^{2}\right] f(s, u(s)) \mathrm{d} s+\int_{\eta}^{t}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}+\right. \\ \ \ \ \ \left.\frac{1}{2}(t-s)^{2}\right] f(s, u(s)) \mathrm{d} s+\int_{t}^{1}\left[\frac{\beta\left(t+t^{2} / 2\right)-\alpha}{2 \alpha-3 \beta}(1-s)^{2}\right] \times \\ \ \ \ \ f(s, u(s)) \mathrm{d} s, \end{array}$

则

$\begin{array}{l} (T u)^{\prime}(t)= \\ \ \ \ \ \int_{0}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \int_{\eta}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-(t-s)\right] f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \int_{t}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] f(s, u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ f(\eta, u(\eta))\left\{\int_{0}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha}{2 \alpha-3 \beta}+(t-s)\right] \mathrm{d} s+\right. \\ \ \ \ \ \left.\int_{\eta}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}+(t-s)\right] \mathrm{d} s+\int_{t}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] \mathrm{d} s\right\} \leqslant \\ \ \ \ \ f(\eta, u(\eta))\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta} \int_{0}^{\eta}\left(s^{2}-2 s\right) \mathrm{d} s-\int_{0}^{\eta} s \mathrm{~d} s+\right.\\ \ \ \ \ \frac{\beta-2 \beta t-2 \alpha}{2 \alpha-3 \beta} \int_{0}^{\eta} \mathrm{d} s+\frac{\beta(1+t)}{2 \alpha-3 \beta} \int_{\eta}^{1}(1-s)^{2} \mathrm{~d} s+ \\ \ \ \ \ \left.\int_{\eta}^{t}(t-s) \mathrm{d} s\right\}=f(\eta, u(\eta))\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\right. \\ \ \ \ \ \left.\frac{1}{2} t^{2}+\frac{\beta+\beta t-2 \alpha}{2 \alpha-3 \beta} \eta-t \eta\right\} \leqslant f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\right. \\ \ \ \ \ \left.\frac{1}{2}+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta\right] \leqslant f(\eta, u(\eta))\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(\frac{1}{3}-\eta\right)+\right. \\ \ \ \ \ \left.\frac{1}{2}-\frac{(2 \alpha-\beta) \eta}{2 \alpha-3 \beta}\right] \leqslant 0. \end{array}$

综上可知, (Tu)(t)在[0, 1]上单调递减. 又由于(Tu)(1)=0, 故(Tu)(t)在[0, 1]上非负, 从而Tu∈K. 假设D⊂K是有界集, 则存在一个常数C₁>0, 使得‖ u ‖≤C₁ (u∈D). 下证T(D)是相对紧的.

设C₂=sup{f(t, u)|(t, u)∈[0, 1]×[0, C₁]}. 对∀y∈T(D), ∃u∈D, 使得y=Tu, 则对∀t ∈[0, 1], 有

$\begin{array}{l} |y(t)|=|(T u)(t)|=\left|\int_{0}^{1} G(t, s) f(s, u(s)) \mathrm{d} s\right| \leqslant \\ \ \ \ \ \ \ \int_{0}^{1}|G(t, s)| f(s, u(s)) \mathrm{d} s \leqslant M \int_{0}^{1} f(s, u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ \ \ M C_{2}. \end{array}$

从而知T(D)是一致有界的. 另一方面, 当ε>0, 0 < τ < min{1-η, $\frac{\varepsilon }{{12{C_2}\left( {M + 1} \right)}}$ }时, 对∀u ∈D, 有

$\int_{\eta-\tau}^{\eta+\tau} f(s, u(s)) \mathrm{d} s \leqslant 2 C_{2} \tau<\frac{\varepsilon}{6(M+1)}.$

(3)

因为G(t, s)在[0, 1]×[0, η-τ]和[0, 1]×[η+ τ, 1]上一致连续, 故∃δ>0, 使得对∀t₁, t₂ ∈[0, 1], 当|t₁-t₂| < δ时, 有

$\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right|<\frac{\varepsilon}{3\left(C_{2}+1\right)(\eta-\tau)}\ \ \ \ (s \in[0, \eta-\tau]),$

(4)

$\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right|<\frac{\varepsilon}{3\left(C_{2}+1\right)\left(1-\eta^{-} \tau\right)}\ \ \ \ (s \in[\eta+\tau, 1]) .$

(5)

由式(3)~(5)及对∀y∈T(D), ∀t₁, t₂∈[0, 1]和|t₁-t₂| < δ, 有

$\begin{array}{l} \left|y\left(t_{1}\right)-y\left(t_{2}\right)\right|=\left|T\left(t_{1}\right)-T\left(t_{1}\right)\right|= \\ \ \ \ \ \ \ \ \ \left|\int_{0}^{1}\left(G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right) f(s, u(s)) \mathrm{d} s\right|= \\ \ \ \ \ \ \ \ \ \int_{0}^{\eta-\tau}\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right| f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \ \ \ \ \int_{\eta-\tau}^{\eta+\tau}\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right| f(s, u(s)) \mathrm{d} s+ \\ \ \ \ \ \ \ \ \ \int_{\eta+\tau}^{1}\left|G\left(t_{1}, s\right)-G\left(t_{2}, s\right)\right| f(s, u(s)) \mathrm{d} s \leqslant \\ \ \ \ \ \ \ \ \ C_{2} \frac{\varepsilon}{3\left(C_{2}+1\right)(\eta-\tau)}(\eta-\tau)+\frac{\varepsilon}{3(M+1)} M+ \\ \ \ \ \ \ \ \ \ C_{2} \frac{\varepsilon}{3\left(C_{2}+1\right)(1-\eta-\tau)}(1-\eta-\tau)= \\ \ \ \ \ \ \ \ \ \frac{C_{2} \varepsilon}{3\left(C_{2}+1\right)}+\frac{M \varepsilon}{3(M+1)}+\frac{C_{2} \varepsilon}{3\left(C_{2}+1\right)} \leqslant \varepsilon. \end{array}$

故T(D)是等度连续的, 从而由Arzela-Ascoli定理^[18]知T(D)是相对紧的. 因此, T是紧算子.

最后, 证明T是连续的. 设u_n (n=1, 2, …), u₀ ∈K, ‖u_n-u₀ ‖→0 (n→0), 则∃C₃>0, 使得对∀n, ‖ u ‖ ≤C₃.

设C₄=sup{f(t, u)|(t, u)∈[0, 1]×[0, C₃]}. 对∀n, t∈[0, 1], ∀s∈[0, 1], 有

$G(t, s) f\left(s, u_{n}(s)\right) \leqslant M C_{4}.$

由Lebesgue控制收敛定理^[19], 对∀t∈[0, 1], 有

$\begin{array}{l} \lim \limits_{n \rightarrow \infty}\left(T u_{n}\right)(t)=\lim \limits_{n \rightarrow \infty} \int_{0}^{1} G(t, s) f\left(s, u_{n}(s)\right) \mathrm{d} s= \\ \ \ \ \ \int_{0}^{1} G(t, s) \lim \limits_{n \rightarrow \infty} f\left(s, u_{n}(s)\right) \mathrm{d} s=\int_{0}^{1} G(t, s) f\left(s, u_{0}(s)\right) \mathrm{d} s= \\ \ \ \ \ \left(T u_{0}\right)(t), \end{array}$

这表明T是连续的. 因此, T: K→K是全连续的. 证毕.

2. 主要结果及证明

定理1 假设条件(H₁)、(H₂)成立, 且对于f(t, 0)≠0, ∀t∈[0, 1], 存在2个正实数a和b, 满足以下条件：

$\begin{array}{l} \ \ \ \ \left(\mathrm{H}_{3}\right) f(0, a) \leqslant \frac{2 \alpha-3 \beta}{3 \alpha} a ;\\ \ \ \ \ \left(\mathrm{H}_{4}\right) b\left(u_{2}-u_{1}\right) \leqslant f\left(t, u_{2}\right)-f\left(t, u_{1}\right) \leqslant 2 b\left(u_{2}-u_{1}\right) \\ \left(0 \leqslant t \leqslant 1,0 \leqslant u_{1} \leqslant u_{2} \leqslant a\right). \end{array}$

构造一个迭代序列ν_n+1=Tν_n (n=0, 1, 2, …), 其中ν₀(t)=0, 则{ν_n}_n=1^∞在E中收敛到ν^*, 且ν^*是边值问题(1)的一个递减正解.

证明设K_a= {u ∈K| | |u| | ≤a}. 由引理3可知Tu∈K, 且由条件(H₃)得到

$\begin{aligned} 0 \leqslant&(T u)(t)=\int_{0}^{1} G(t, s) f(s, u(s)) \mathrm{d} s \leqslant & \\ & \int_{0}^{1}|G(t, s)| f(0, a) \mathrm{d} s \leqslant \frac{2 \alpha-3 \beta}{3 \alpha} a M \leqslant a \quad(t \in[0,1]), \end{aligned}$

从而有‖ Tu ‖ ≤a, 故T: K_a→K_a.

下证{ν_n}_n=1^∞在E中收敛到ν^*, 且ν^*是边值问题(1)的一个递减正解.

事实上, 对于ν₀ ∈K_a和T: K_a→K_a, 有ν_n ∈K_a, n=0, 1, 2, …. 由于集合{ν_n}_n=0^∞是有界的且T是全连续算子, 所以集合{ν_n}_n=0^∞是相对紧的. 接下来, 通过归纳法证明{ν_n}_n=0^∞是单调的. 首先, 很明显有ν₁-ν₀=ν₁ ∈K, 这表明ν₀≤ν₁. 接下来, 假设ν_k-1≤ν_k. 从而由条件(H₄)可得

$\begin{array}{l} \nu_{k+1}^{\prime}-\nu_{k}^{\prime}=\left(T \nu_{k}\right)^{\prime}(t)-\left(T \nu_{k-1}\right)^{\prime}(t)= \\ \ \ \ \ \int_{0}^{1} G_{t}(t, s)\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s= \\ \ \ \ \ \int_{0}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right]\left[f\left(s, \nu_{k}(s)\right)-\right.\\ \ \ \ \ \left.f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s+\int_{t}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}\right] \times \\ \ \ \ \ {\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s+\int_{\eta}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] \times} \\ \ \ \ \ {\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s \leqslant b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right] \times} \\ \ \ \ \ \left\{\int_{0}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right] \mathrm{d} s+\right. \\ \ \ \ \ \int_{t}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}\right] \mathrm{d} s+\\ \ \ \ \ \left.\int_{\eta}^{1}\left[\frac{2 \beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right] \mathrm{d} s\right\}=b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right] \times \\ \ \ \ \ \left\{\int_{0}^{\eta} \frac{\beta(1+t)}{2 \alpha-3 \beta}\left(s^{2}-2 s\right) \mathrm{d} s+\int_{0}^{\eta} \frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \mathrm{d} s+\right. \\ \ \ \ \ \left.\int_{0}^{t}(t-s) \mathrm{d} s+\frac{2 \beta(1+t)}{2 \alpha-3 \beta} \int_{\eta}^{1}(1-s)^{2} \mathrm{~d} s\right\}= \\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(-\eta^{3}+3 \eta^{2}-6 \eta+2\right)+\right. \\ \ \ \ \ \left.\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta+\frac{1}{2} t^{2}\right\} \leqslant b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right] \times\\ \ \ \ \ \left\{\frac{\beta(1+t)}{2 \alpha-3 \beta}\left(-\eta^{3}+3 \eta^{2}-6 \eta+2\right)+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta+\frac{1}{2}\right\} \leqslant \\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta}(2-3 \eta)+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta {+}\right. \\ \ \ \ \ \left.\frac{1}{2}\right\} \leqslant b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left\{\frac{\beta(1+t)(2-3 \eta)}{2 \alpha-3 \beta}+\right. \\ \ \ \ \ \left.\frac{(2 \alpha-3 \beta)-2 \eta(2 \alpha-\beta)}{2 \alpha-3 \beta}\right\} \leqslant 0 \quad(t \in[0, \eta]);\\ \nu_{k+1}^{\prime}-\nu_{k}^{\prime}=\left(T \nu_{k}\right)^{\prime}(t)-\left(T \nu_{k-1}\right)^{\prime}(t)= \\ \ \ \ \ \int_{0}^{1} G_{t}(t, s)\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s= \\ \ \ \ \ \int_{0}^{\eta}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-\frac{2 \alpha(1+t)}{2 \alpha-3 \beta}+(t-s)\right]\left[f\left(s, \nu_{k}(s)\right)-\right. \\ \ \ \ \ \left.f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s+\int_{\eta}^{t}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}-(t-s)\right] \times \\ \ \ \ \ {\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s+}\\ \ \ \ \ \int_{t}^{1}\left[\frac{\beta(1+t)}{2 \alpha-3 \beta}(1-s)^{2}\right]\left[f\left(s, \nu_{k}(s)\right)-f\left(s, \nu_{k-1}(s)\right)\right] \mathrm{d} s \leqslant \\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left\{\frac{\beta(1+t)}{2 \alpha-3 \beta} \int_{0}^{\eta}\left(-2 s+s^{2}\right) \mathrm{d} s+\right. \\ \ \ \ \ \frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \int_{0}^{\eta} \mathrm{d} s+\int_{0}^{\eta}(t-s) \mathrm{d} s+\frac{2 \beta(1+t)}{2 \alpha-3 \beta} \int_{\eta}^{1}(1-s)^{2} \mathrm{~d} s+ \\ \ \ \ \ \left.2 \int_{\eta}^{t}(t-s) \mathrm{d} s\right\}=b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left[\frac{\beta(1+t)}{3(2 \alpha-3 \beta)} \times\right. \\ \ \ \ \ \left.\left(-\eta^{3}+3 \eta^{2}-6 \eta+2\right)+t^{2}+\frac{(\beta-2 \alpha)(1+t)}{2 \alpha-3 \beta} \eta-t \eta+\frac{1}{2} \eta^{2}\right] \leqslant\\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left[\frac{\beta(1+t)}{3(2 \alpha-3 \beta)}(-5 \eta+2)+\right. \\ \ \ \ \ \left.\frac{t[(2 \alpha-3 \beta)-\eta(2 \alpha-\beta)]-\eta(2 \alpha-\beta)}{2 \alpha-3 \beta}\right] \leqslant \\ \ \ \ \ b\left[\nu_{k}(\eta)-\nu_{k-1}(\eta)\right]\left[\frac{\beta(1+t)}{3(2 \alpha-3 \beta)}(-5 \eta+2)-\right. \\ \ \ \ \ \left.\frac{\eta(2 \alpha-\beta)}{2 \alpha-3 \beta}\right] \leqslant 0 . \end{array}$

因此,

$\nu_{k+1}^{\prime}(t)-\nu_{k}^{\prime}(t) \leqslant 0 \quad(t \in[0,1]),$

(6)

即ν_k+1(t)-ν_k(t)在[0, 1]上单调递减. 与此同时, 易得

$\nu_{k+1}(1)-\nu_{k}(1)=\int_{0}^{1} G(1, s) f\left(s, \nu_{k+1}(s)-\nu_{k}(s)\right) \mathrm{d} s=0,$

(7)

从而ν_k+1(t)-ν_k(t)≥0 (t∈[0, 1]).

由式(6)和式(7)可知ν_k+1-ν_k ∈K, 则ν_k+1≤ν_k (n=0, 1, 2, …). 由于{ν_n}_n=1^∞是相对紧集, 并且是单调的, 因此, 必存在一个ν^*∈K_a, 使得 $\mathop {\lim }\limits_{n \to \infty }$ ν_n=ν^*. 再由T的连续性以及ν_n+1=Tν_n可知ν^*=Tν^*. 这表明ν^*是边值问题(1)单调递减的非负解. 此外, 根据f(t, 0)≠0 (t ∈[0, 1])知, 0不是边值问题(1)的解, 从而ν^*是边值问题(1)的正解. 证毕.

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