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  浙江大学学报(理学版)  2017, Vol. 44 Issue (3): 281-286, 338  DOI:10.3785/j.issn.1008-9497.2017.03.006
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闫东明. 一类奇异边值问题正解的存在性及多解性[J]. 浙江大学学报(理学版), 2017, 44(3): 281-286, 338. DOI: 10.3785/j.issn.1008-9497.2017.03.006.
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YAN Dongming. Existence of single and multiple positive solutions of singular boundary value problem[J]. Journal of Zhejiang University(Science Edition), 2017, 44(3): 281-286, 338. DOI: 10.3785/j.issn.1008-9497.2017.03.006.
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基金项目

浙江省教育厅一般项目(Y201534471)

作者简介

闫东明(1982-), ORCID:http://orcid.org/0000-0002-2893-4871, 男, 博士, 讲师, 主要从事微分方程研究, E-mail:13547895541@126.com

文章历史

收稿日期:2015-01-04
一类奇异边值问题正解的存在性及多解性
闫东明     
浙江财经大学 数据科学学院, 浙江 杭州 310018
摘要: 应用Dancer全局分歧理论,研究奇异边值问题$\left\{ \begin{gathered} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + f\left( {t,u\left( t \right)} \right) = 0,\;\;t \in \left( {0,1} \right), \hfill \\ u\left( 0 \right) = u\left( 1 \right) = 0 \hfill \\ \end{gathered} \right. $正解的存在性和多解性,其中f:[0,1]×[0,∞)→[0,∞)连续.给出了关于此类问题正解存在的充分条件,该充分条件与相应线性问题的第1个特征值有关,且所涉及的值是最优的.
关键词: 奇异边值问题    全局分歧    正解    多解性    第1特征值    
Existence of single and multiple positive solutions of singular boundary value problem
YAN Dongming     
School of Mathematics and Statistics, Zhejiang University of Finance and Economics, Hangzhou 310018, China
Abstract: By using Dancer's global bifurcation theorem, we studied the existence of single and multiple positive solutions of the singular boundary value problem $\left\{ \begin{gathered} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + f\left( {t,u\left( t \right)} \right) = 0,\;\;t \in \left( {0,1} \right), \hfill \\ u\left( 0 \right) = u\left( 1 \right) = 0 \hfill \\ \end{gathered} \right. $ And the optimal sufficient conditions for the existence of single and multiple positive solutions of the mentioned problem are obtained, which is relate to the first eigenvalue of the relevant linear problem.
Key words: singular boundary value problem    global bifurcation    positive solutions    multiple positive solutions    first eigenvalue    
0 引言

奇异边值问题因其具有的物理意义引起了人们的广泛关注, 并且此类问题的相关研究也取得了许多深刻的结果[1-8].同时,鉴于正解及多个正解的实际意义, 奇异边值问题正解及多个正解存在性的研究显得更为活跃.文献[6]研究了奇异边值问题

$ \left\{ \begin{array}{l} u''\left( t \right) + \lambda f\left( {t,u\left( t \right)} \right) = 0,t \in \left( {0,1} \right),\\ u\left( 0 \right) = u\left( 1 \right) = 0 \end{array} \right. $ (1)

正解的存在性.在非线性项f满足非负性及一些局部增长性条件下, 获得了奇异边值问题(1) 的一个正解的存在性结果, 并给出了问题(1) 有正解存在和无正解存在时非负参数λ的取值范围.文献[8]研究了带一般微分算子的奇异边值问题:

$ \left\{ \begin{array}{l} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + \\ \;\;\;\;\;\;\;\;\;f\left( {t,u\left( t \right)} \right) = 0,t \in \left( {0,1} \right),\\ u\left( 0 \right) = u\left( 1 \right) = 0 \end{array} \right. $ (2)

多个正解的存在性.通过构造奇异边值问题(2) 的格林函数, 并运用Leggett-Williams不动点定理, 在非线性项f满足非负性条件及一些局部条件下, 获得了奇异边值问题(2) 多个正解的存在性结果.

另外, 文献[9-10]运用Rabinowitz全局分歧理论以及区间分歧方法,对Dirichlet边值问题正解集的结构和全局分歧进行了讨论, 获得了大量有价值的结果.

上述文献从不同的角度丰富了常微分方程奇异边值问题正解的存在性理论.然而, 这些研究所得到的奇异边值问题有正解及多个正解存在的充分性条件中所涉及的数值均非最优, 且所得充分性条件与奇异边值问题对应的线性问题的第1个特征值之间没有建立联系,而边值问题对应的线性问题的特征值在边值问题正解及多个正解存在性的研究中是一个很本质的量.本文运用Dancer全局分歧定理[11]对二阶常微分方程奇异边值问题(2) 进行研究,试图给出此类问题有正解存在且与其相应线性问题第1个特征值有关的充分条件, 以及说明这些充分条件中所涉及的值是最优的.无论是方法还是所得结果, 本研究都将进一步丰富二阶常微分方程奇异边值问题正解的存在性理论.

本文总假定

$ \begin{array}{l} \left( {{A_1}} \right)a \in C\left( {0,1} \right) \cap {L^1}\left( {0,1} \right);b \in \left( {\left( {0,1} \right),} \right.\\ \left. {\left( { - \infty ,0} \right)} \right)且\int_0^1 {s\left( {1 - s} \right)\left| {b\left( s \right)} \right|{\rm{d}}s < + \infty } .\\ \left( {{A_2}} \right)f:\left[ {0,1} \right] \times \left[ {0, + \infty } \right) \to \left[ {0, + \infty } \right)连续. \end{array} $
1 预备知识及引理

C[0, 1]为定义在[0, 1]上的连续实值函数构成的集合, 定义范数

$ {\left\| x \right\|_\infty } = \sup \left\{ {\left| {x\left( t \right)} \right|\left| {t \in \left[ {0,1} \right]} \right|} \right\}, $

C[0, 1]为Banach空间.记[0, 1]上的所有绝对连续函数全体为AC[0, 1].令

$ \begin{array}{l} A{C_{{\rm{loc}}}}\left[ {0,1} \right) = \left\{ {u{{\left| u \right|}_{\left[ {0,d} \right]}} \in AC\left[ {0,d} \right],\left[ {0,d} \right] \subseteq \left[ {0,1} \right)} \right\},\\ A{C_{{\rm{loc}}}}\left( {0,1} \right] = \left\{ {u{{\left| u \right|}_{\left[ {d,1} \right]}} \in AC\left[ {d,1} \right],\left[ {d,1} \right] \subseteq \left( {0,1} \right]} \right\}, \end{array} $

方便起见,记

$ {f_0}: = \mathop {\lim \inf }\limits_{u \to {0^ + }} \mathop {\min }\limits_{t \in \left[ {0,1} \right]} \frac{{f\left( {t,u} \right)}}{u}, $
$ {f^0}: = \mathop {\lim \sup }\limits_{u \to {0^ + }} \mathop {\max }\limits_{t \in \left[ {0,1} \right]} \frac{{f\left( {t,u} \right)}}{u}, $
$ {f_\infty }: = \mathop {\lim \inf }\limits_{u \to + \infty } \mathop {\min }\limits_{t \in \left[ {0,1} \right]} \frac{{f\left( {t,u} \right)}}{u}, $
$ {f^\infty }: = \mathop {\lim \sup }\limits_{u \to + \infty } \mathop {\max }\limits_{t \in \left[ {0,1} \right]} \frac{{f\left( {t,u} \right)}}{u}. $

引理1[8] 假设(A1)成立.则

(ⅰ)初值问题

$ \left\{ \begin{array}{l} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) = 0,t \in \left( {0,1} \right),\\ u\left( 0 \right) = 0,u'\left( 0 \right) = 1 \end{array} \right. $

有唯一解αAC[0, 1]∩C1[0, 1) 且α′∈ACloc[0, 1);

(ⅱ)初值问题

$ \left\{ \begin{array}{l} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) = 0,t \in \left( {0,1} \right),\\ u\left( 1 \right) = 0,u'\left( 1 \right) = - 1 \end{array} \right. $

有唯一解βAC[0, 1]∩C1(0, 1]且β′∈ACloc(0, 1];

(ⅲ) α在[0, 1]非减, β在[0, 1]非增.

$ G\left( {t,s} \right) = \frac{1}{\rho }\left\{ {\begin{array}{*{20}{l}} {\alpha \left( s \right)\beta \left( t \right),\;\;\;0 \leqslant s \leqslant t \leqslant 1,} \\ {\alpha \left( t \right)\beta \left( s \right),\;\;\;0 \leqslant t \leqslant s \leqslant 1,} \end{array}} \right. $

其中$\rho {\text{ = }}\alpha '\left( {\frac{1}{2}} \right)\beta \left( {\frac{1}{2}} \right)-\alpha \left( {\frac{1}{2}} \right)\beta '\left( {\frac{1}{2}} \right) $是一个正数, 则对G有下述引理.

引理2 设(A1)成立.则G具有以下性质:

(ⅰ) G(t, s)>0, (t, s)∈(0, 1)×(0, 1);

(ⅱ) G(t, s)≤G(t, t), (t, s)∈[0, 1]×[0, 1];

(ⅲ) G(t, s)≥δt(1-t)G(s, s), (t, s)∈[0, 1]×[0, 1], 其中δ>0为常数.

证明 由引理1易证(ⅰ)、(ⅱ).以下证(ⅲ)成立.

αAC[0, 1]∩C1[0, 1) 且α′(0)=1可知, 存在$ {\tilde c_3} > {\tilde c_2} > 0$, 使得

$ {{\tilde c}_2}t \le \alpha \left( t \right) \le {{\tilde c}_3}t,\;\;\;\;t \in \left[ {0,1} \right]. $ (3)

βAC[0, 1]∩C1(0, 1]且β′(1)=-1可知, 存在${\tilde c_4} > {\tilde c_1} > 0 $, 使得

$ {{\tilde c}_1}\left( {1 - t} \right) \le \beta \left( t \right) \le {{\tilde c}_4}\left( {1 - t} \right),\;\;\;\;t \in \left[ {0,1} \right]. $ (4)

G的定义、引理1以及式(3) 和(4),有

$ \begin{array}{l} \frac{{G\left( {t,s} \right)}}{{G\left( {s,s} \right)}} = \left\{ \begin{array}{l} \frac{{\beta \left( t \right)}}{{\beta \left( s \right)}}\left( {0 \le s \le t \le 1} \right) \ge \\ \frac{{\alpha \left( t \right)}}{{\alpha \left( s \right)}}\left( {0 \le s \le t \le 1} \right) \ge \end{array} \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left\{ \begin{array}{l} \frac{{\beta \left( t \right)}}{{\beta \left( 0 \right)}} \ge \\ \frac{{\alpha \left( t \right)}}{{\alpha \left( 1 \right)}} \ge \end{array} \right.\left\{ \begin{array}{l} \frac{{{{\tilde c}_1}\left( {1 - t} \right)}}{{\beta \left( 0 \right)}} \ge \delta t\left( {1 - t} \right),\\ \frac{{{{\tilde c}_2}t}}{{\alpha \left( 1 \right)}} \ge \delta t\left( {1 - t} \right), \end{array} \right. \end{array} $

$ G\left( {t,s} \right) \ge \delta t\left( {1 - t} \right)G\left( {s,s} \right),\;\left( {t,s} \right) \in \left[ {0,1} \right] \times \left[ {0,1} \right], $

其中$\delta = \min \left\{ {\frac{{{{\tilde c}_1}}}{{\beta \left( 0 \right)}}, \frac{{{{\tilde c}_2}}}{{\alpha \left( 1 \right)}}} \right\} > 0 $.引理证毕.

定义P={uC[0, 1]:u(t)≥0, t∈[0, 1], u(0)=u(1)=0}.则PC[0, 1]中的一个锥.设

$ \left\{ \begin{array}{l} \left( {Au} \right)\left( t \right) = \int_0^1 {G\left( {t,s} \right)q\left( s \right)f\left( {s,u\left( s \right)} \right){\rm{d}}s} ,t \in \left[ {0,1} \right],u \in P,\\ \left( {Tu} \right)\left( t \right) = \int_0^1 {G\left( {t,s} \right)q\left( s \right)u\left( s \right){\rm{d}}s} ,t \in \left[ {0,1} \right],u \in C\left[ {0,1} \right], \end{array} \right. $ (5)

其中$q\left( s \right) = {{\text{e}}^{\int_{\frac{1}{2}}^s {a\left( \tau \right){\text{d}}\tau } }} $.则可以验证算子A在锥P中的非零不动点是奇异问题(2) 的一个正解.

根据Krein-Rutmann定理[11-12], 有下述引理.

引理3 设E为一个Banach空间, K$\subset $E是一个锥且满足$\mathop K\limits^ \circ \ne \emptyset $.若有界线性算子T是一个紧的强正算子, 则T的谱半径r(T)≠0, 且r-1(T)是T的一个具有正特征函数φ1$\mathop K\limits^ \circ $的简单特征值.

Eu0={u|uC[0, 1]且存在λ>0使得

$ \left. { - \lambda t\left( {1 - t} \right) \le u\left( t \right) \le \lambda t\left( {1 - t} \right),t \in \left[ {0,1} \right]} \right\}. $

uEu0, 则令u的范数为

$ \begin{array}{l} {\left\| u \right\|_{{u_0}}} = \inf \left\{ {\lambda \left| {\lambda > 0,} \right. - \lambda t\left( {1 - t} \right) \le } \right.\\ \;\;\;\;\;\;\;\;\;\;\left. {u\left( t \right) \le \lambda t\left( {1 - t} \right),t \in \left[ {0,1} \right]} \right\}. \end{array} $ (6)

此处‖uu0uu0-范数, 相关内容可参见文献[14].

引理4 (Eu0, ‖·‖u0)是Banach空间.

证明 设{un}是(Eu0, ‖·‖u0)中的Cauchy列.则有

$ \left\| {{u_n} - {u_m}} \right\| \to 0,\;\;\;n,m \to \infty . $ (7)

由式(6) 和(7), 对$\forall $ε>0,存在N>0使得

$ \begin{array}{l} - \varepsilon t\left( {1 - t} \right) \le {u_n}\left( t \right) - {u_m}\left( t \right) \le \varepsilon t\left( {1 - t} \right),\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;t \in \left[ {0,1} \right],\;\;\;\;\forall n,m > N. \end{array} $ (8)

从而,

$ {\left\| {{u_n} - {u_m}} \right\|_\infty } \le \frac{\varepsilon }{4} \to 0,\;\;\;\;n,m \to \infty , $

即{un}是(C[0, 1], ‖·‖)中的Cauchy列.根据(C[0, 1], ‖·‖)的完备性, 存在u0C[0, 1]使得

$ {\left\| {{u_n} - {u_0}} \right\|_\infty } \to 0,\;\;\;\;n \to \infty . $

在式(8) 中令m→∞,则有

$ \begin{array}{l} - \varepsilon t\left( {1 - t} \right) \le {u_n}\left( t \right) - {u_0}\left( t \right) \le \varepsilon t\left( {1 - t} \right),\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;t \in \left[ {0,1} \right],\left( {n > N} \right),\\ 即\;\;\;\;\;\;{\left\| {{u_n} - {u_0}} \right\|_{{u_0}}} \le \varepsilon ,\;\;\;n > N. \end{array} $

因此, 对Cauchy列{un}$ \subset $(Eu0, ‖·‖u0), 存在u0Eu0使得

$ {\left\| {{u_n} - {u_0}} \right\|_{{u_0}}} \to 0,\;\;\;\;n \to \infty . $

所以, (Eu0, ‖·‖u0)是Banach空间.引理证毕.

引理5 设(A1)成立, 算子T如式(5) 所定义.则T的谱半径r(T)>0, 且T有一个正的特征函数φ1对应于其第1个特征值λ1=(r(T))-1, 即φ1=λ11.

证明 令K=Eu0P.则K$\subset $Eu0是一个锥, 且满足$\mathop K\limits^ \circ \ne \emptyset $.事实上, 若令u0(t)=t(1-t), t∈[0, 1], 则u0$\mathop K\limits^ \circ $.易证T:Eu0Eu0是全连续算子, 且T(K)$\subset $K.

下证T是一个强正算子.设uK\{θ}.则u在[0, 1]上不恒为0, 并且由K的定义知, 存在λ0>0, 使得

$ 0 \le u\left( t \right) \le {\lambda _0}t\left( {1 - t} \right),\;\;\;\;t \in \left[ {0,1} \right]. $ (9)

由引理1以及式(3)、(4) 和(9),有

$ \begin{array}{l} \left( {Tu} \right)\left( t \right) = \int_0^1 {G\left( {t,s} \right)q\left( s \right)u\left( s \right){\rm{d}}s} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^1 {G\left( {t,t} \right)q\left( s \right)u\left( s \right){\rm{d}}s} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{{{\lambda _0}}}{\rho }\alpha \left( t \right)\beta \left( t \right)\int_0^1 {q\left( s \right)s\left( {1 - s} \right){\rm{d}}s} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\frac{{{\lambda _0}{{\tilde c}_3}{{\tilde c}_4}}}{\rho }\int_0^1 {q\left( s \right)s\left( {1 - s} \right){\rm{d}}s} } \right]\left[ {t\left( {1 - t} \right)} \right] = :\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;{c_{2u}}\left[ {t\left( {1 - t} \right)} \right],t \in \left[ {0,1} \right] \end{array} $

$ \begin{array}{l} \left( {Tu} \right)\left( t \right) = \int_0^1 {G\left( {t,s} \right)q\left( s \right)u\left( s \right){\rm{d}}s} \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_0^1 {\delta t\left( {1 - t} \right)G\left( {s,s} \right)q\left( s \right)u\left( s \right){\rm{d}}s} \ge \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\left[ {\frac{{\delta {{\tilde c}_1}{{\tilde c}_2}}}{\rho }\int_0^1 {q\left( s \right)u\left( s \right)s\left( {1 - s} \right){\rm{d}}s} } \right]\left[ {t\left( {1 - t} \right)} \right] = :\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;{c_{1u}}\left[ {t\left( {1 - t} \right)} \right],t \in \left[ {0,1} \right], \end{array} $

其中,

$ \begin{array}{l} {c_{1u}} = \frac{{\delta {{\tilde c}_1}{{\tilde c}_2}}}{\rho }\int_0^1 {q\left( s \right)u\left( s \right)s\left( {1 - s} \right){\rm{d}}s} > 0,\\ {c_{2u}} = \frac{{{\lambda _0}{{\tilde c}_3}{{\tilde c}_4}}}{\rho }\int_0^1 {q\left( s \right)s\left( {1 - s} \right){\rm{d}}s} > 0, \end{array} $

ρ, δ见引理2.从而, 对uK\{θ}存在常数c1u, c2u>0, 使得

$ {c_{1u}}\left[ {t\left( {1 - t} \right)} \right] \le \left( {Tu} \right)\left( t \right) \le {c_{2u}}\left[ {t\left( {1 - t} \right)} \right],t \in \left[ {0,1} \right], $

$ T\left( {K\backslash \left\{ \theta \right\}} \right) \subset \mathop K\limits^ \circ . $

所以T是一个强正算子.

由引理3可知, T的谱半径r(T)≠0, 且T有一个正的特征函数φ1对应于其第1个特征值λ1=(r(T))-1, 并且满足φ1=λ11.引理证毕.

注1 由引理5可知, 当条件(A1)成立时, 线性特征值问题

$ \left\{ \begin{array}{l} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + \lambda u\left( t \right) = 0,t \in \left( {0,1} \right),\\ u\left( 0 \right) = u\left( 1 \right) = 0 \end{array} \right. $

的第1个特征值λ1是大于0的, 并且λ1所对应的特征函数φ1为正.

下面叙述一个关于参数化非线性算子方程正解集合的全局结构.此结果为Dancer全局分歧定理[11]的推论.

假设E为一个实Banach空间, 其上范数为‖·‖.令K$\subset $E为一个锥, 若A([0, ∞)×K)$\subset $K,则称非线性映射A:[0, ∞)×KE是正的.若算子A连续, 并且把[0, ∞)×K中的有界子集映为E中的相对紧集,则称上述非线性映射AK-全连续的.最后, 对于定义在E上的连续线性算子B, 用r(B)表示算子B的谱半径.

引理6[11] 假设

(ⅰ) K有非空内部且E= $\overline {K-K} $.

(ⅱ) A:[0, ∞)×KEK-全连续的正算子.对任意的λR, 有A(λ, 0)=0;对任意的uK, 有A(0, u)=0, 且

$ A\left( {\lambda ,u} \right) = \lambda Bu + F\left( {\lambda ,u} \right), $

其中B:EE为定义在E上的强正线性紧算子且满足r(B)>0, F:[0, ∞)×KE满足:当‖u‖→0时, ‖F(λ, u)‖=o(‖u‖)对λ局部一致成立.

则存在

$ \begin{array}{l} {\boldsymbol{{\cal D}}_K}\left( A \right) = \left\{ {\left( {\lambda ,u} \right) \in \left[ {0,\infty } \right) \times K:u = A\left( {\lambda ,u} \right),} \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. {u \ne 0} \right\} \cup \left\{ {\left( {r{{\left( B \right)}^{ - 1}},0} \right)} \right\} \end{array} $

的一个无界连通子集$\mathscr{C}$, 使得(r(B)-1, 0)∈$\mathscr{C}$.

2 主要结果及其证明

定理1 设(A1)和(A2)成立, λ1=(r(T))-1.若

$ {f^\infty } < {\lambda _1} < {f_0} $ (10)

$ {f^0} < {\lambda _1} < {f_\infty } $ (11)

成立, 则奇异边值问题(2) 至少有1个正解.

注2 定理1中充分性条件(10)、(11) 与奇异边值问题(2) 对应的线性问题的第1个特征值建立起了联系.

注3 条件(10)、(11) 中涉及的值λ1是最优的.若f=λ1=f0f0=λ1=f, 则奇异边值问题(2) 正解的存在性不能保证.考虑如下边值问题:

$ \begin{array}{*{20}{c}} {u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + {\lambda _1}u\left( t \right) + }\\ {\rho \left( {u\left( t \right)} \right) = 0,\;\;\;\;t \in \left( {0,1} \right),} \end{array} $ (12)
$ u\left( 0 \right) = u\left( 1 \right) = 0, $ (13)

其中,

$ \rho \left( x \right) = \left\{ \begin{array}{l} {x^2},\;\;\;\;x \in \left[ {0,1} \right],\\ \sqrt x ,\;\;\;x \in \left[ {1,\infty } \right). \end{array} \right. $

显然, f(t, u)=λ1u+ρ(u)满足条件(A2), 且f=λ1=f0, f0=λ1=f, 但奇异边值问题(12) 和(13) 无正解.事实上, 反设u是式(12) 和(13) 的一个正解, 并设λ1所对应的正的特征函数为φ1(t), 将式(12) 两端同乘以q(t)φ1(t),再从0到1积分, 可得$\int_0^1 {\rho \left( {u\left( t \right)} \right)} q\left( t \right){\varphi _1}\left( t \right){\text{d}}t = 0 $, 矛盾!因此, 边值问题(12) 和(13) 无正解.

证明 仅给出式(10) 成立时的证明,式(11) 可类似证得.

为了研究问题(2) 正解的存在性, 考虑以下参数化问题:

$ \left\{ \begin{array}{l} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + \\ \;\;\;\;\;\;\mu f\left( {t,u\left( t \right)} \right) = 0,\;\;\;t \in \left( {0,1} \right),\\ u\left( 0 \right) = u\left( 1 \right) = 0. \end{array} \right. $ (14)

$ f\left( {t,s} \right) = {f_0}s + \xi \left( {t,s} \right),f\left( {t,s} \right) = {f^\infty }s + \zeta \left( {t,s} \right). $ (15)

$ \mathop {\lim \inf }\limits_{s \to 0 + } \mathop {\min }\limits_{t \in \left[ {0,1} \right]} \frac{{\xi \left( {t,s} \right)}}{s} = 0,\;\;\;\;\;\;\mathop {\lim \sup }\limits_{s \to + \infty } \mathop {\max }\limits_{t \in \left[ {0,1} \right]} \frac{{\zeta \left( {t,s} \right)}}{s} = 0. $ (16)

于是, 参数化问题(14) 可重写为

$ \left\{ \begin{array}{l} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + \mu {f_0}u\left( t \right) + \\ \;\;\;\;\;\;\;\;\mu \xi \left( {t,u\left( t \right)} \right) = 0,\;t \in \left( {0,1} \right),\\ u\left( 0 \right) = u\left( 1 \right) = 0. \end{array} \right. $ (17)

对空间Eu0以及其上的锥K,有

$ {{\bf{E}}_{{u_0}}} = \overline {K - K} , $

此闭包在u0-范数‖·‖u0下取得.事实上, 设uEu0,则存在常数l>0,使得

$ - l\left[ {t\left( {1 - t} \right)} \right] \le u\left( t \right) \le l\left[ {t\left( {1 - t} \right)} \right],\;t \in \left[ {0,1} \right]. $

若取$0 < \rho < \frac{1}{l} $, 则有

$ \begin{array}{l} - \left( {1 + \rho l} \right)\left[ {t\left( {1 - t} \right)} \right] \le \left[ {t\left( {1 - t} \right) \pm \rho u\left( t \right)} \right] \le \\ \;\;\;\;\;\;\;\left( {1 + \rho l} \right)\left[ {t\left( {1 - t} \right)} \right],t \in \left[ {0,1} \right], \end{array} $ (18)
$ \begin{array}{l} \left[ {t\left( {1 - t} \right) \pm \rho u\left( t \right)} \right] \ge \left( {1 - \rho l} \right)\left[ {t\left( {1 - t} \right)} \right] \ge 0,\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;t \in \left[ {0,1} \right], \end{array} $ (19)

u0(t)=t(1-t), t∈[0, 1].则由式(18) 和(19), 有

$ {u_0} \pm \rho u \in K. $

$ {u_1} = \frac{1}{{2\rho }}\left( {{u_0} + \rho u} \right),\;\;\;\;{u_2} = \frac{1}{{2\rho }}\left( {{u_0} - \rho u} \right). $

u1, u2K, 且满足

$ u = {u_1} - {u_2}, $

uK-K, 从而有

$ {{\bf{E}}_{{u_0}}} \subseteq \overline {K - K} . $

另一方面, 由空间(Eu0, ‖·‖u0)的完备性(见引理4), 可推得

$ \overline {K - K} \subseteq {{\bf{E}}_{{u_0}}}. $

因此, Eu0=$ \overline {K-K} $, 即满足引理6的条件(ⅰ).方程(17) 等价于如下算子方程:

$ u = A\left( {\mu ,u} \right), $ (20)

其中,

$ A\left( {\mu ,u} \right) = \mu Bu + F\left( {\mu ,u} \right), $

$ \begin{array}{l} \;\;\;\left( {Bu} \right)\left( t \right): = \int_0^1 {G\left( {t,s} \right)q\left( s \right){f_0}u\left( s \right){\rm{d}}s} ,\\ F\left( {\mu ,u} \right)\left( t \right): = \mu \int_0^1 {G\left( {t,s} \right)q\left( s \right)\xi \left( {s,u\left( s \right)} \right){\rm{d}}s} . \end{array} $

容易验证A:[0, ∞)×KEu0K-全连续的正算子, 对任意μR, 有A(μ, 0)=0;对任意的uK, 有A(0, u)=0.另外, 由引理5的证明过程易知B是紧的强正算子, 并且由引理5可推知$r\left( B \right) = \frac{{{\lambda _1}}}{{{f_0}}} > 0 $.由式(16) 可推得

$ {\left\| {F\left( {\mu ,u} \right)} \right\|_{{u_0}}} = o\left( {{{\left\| u \right\|}_{{u_0}}}} \right) $

μ局部一致成立.从而, 也满足了引理6的条件(ⅱ).因此, 由引理6可知:存在问题(14) 正解集的一个无界连通分支C+, 且满足$\left( {\frac{{{\lambda _1}}}{{{f_0}}}, 0} \right) \in {C^ + } $.

$ \begin{array}{l} {\mathit{\Phi }^ + }: = \left\{ {u \in C\left[ {0,1} \right]:u\left( t \right) > 0,t \in \left[ {0,1} \right],} \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. {u\left( 0 \right) = u\left( 1 \right) = 0} \right\}. \end{array} $ (21)

下面只需证明连通分支C+Φ+中连接$\left( {\frac{{{\lambda _1}}}{{{f_0}}}, 0} \right) $$\left( {\frac{{{\lambda _1}}}{{{f^\infty }}}, \infty } \right) $.从而由

$ {f^\infty } <{\lambda _1} < {f_0}, $

$ \frac{{{\lambda _1}}}{{{f_0}}} < 1 < \frac{{{\lambda _1}}}{{{f^\infty }}}, $

可推得当μ=1时, 问题(14) 即原问题(2) 至少有1个正解u存在.

假设(ηk, yk)∈C+, 并且满足|ηk|+‖yk→∞, k→∞.下面分2步来证明.

第1步 证明{ηk}有界.

反设ηk→∞, k→∞.则易证yk在区间[0, 1]上变号.事实上, 由(ηk, yk)∈C+

$ \begin{array}{l} {{y''}_k}\left( t \right) + a\left( t \right){{y'}_k}\left( t \right) + b\left( t \right){y_k}\left( t \right) + \\ \;\;\;\;\;\;\;\;\;\;\;{\eta _k}\frac{{f\left( {t,{y_k}\left( t \right)} \right)}}{{{y_k}\left( t \right)}}{y_k}\left( t \right) = 0. \end{array} $

定义线性算子(Lyk)(t):=yk(t)+a(t)yk(t)+b(t)yk(t), 则有

$ L{y_k} + {\eta _k}\frac{{f\left( {t,{y_k}} \right)}}{{{y_k}}}{y_k} = 0. $ (22)

另一方面, 由假设(A2)可知,存在eC[0, 1], e(t)>0, t∈[0, 1], 使得

$ \frac{{f\left( {t,s} \right)}}{s} \ge e\left( t \right),\left( {t,s} \right) \in \left[ {0,1} \right] \times \left( {0,\infty } \right). $

由上述事实以及ηk→∞, k→∞的假设可得

$ {\eta _k}\frac{{f\left( {t,{y_k}} \right)}}{{{y_k}}} \to \infty ,\;\;\;\;k \to \infty . $

于是, 由Sturm比较定理可知,对充分大的k, yk在区间[0, 1]中至少有1个零点, 这与yk>0矛盾!因此, {ηk}是有界的.进一步, 由连通分支C+的无界性,可知‖yk→∞, k→∞.

第2步 证明C+Φ+中连接$\left( {\frac{{{\lambda _1}}}{{{f_0}}}, 0} \right) $$\left( {\frac{{{\lambda _1}}}{{{f^\infty }}}, \infty } \right) $.

由式(15) 以及(ηk, yk)∈C+,知(ηk, yk)(kN)也满足

$ \left\{ \begin{array}{l} {{y''}_k}\left( t \right) + a\left( t \right){y_k}\left( t \right) + b\left( t \right){y_k}\left( t \right) + {\eta _k}{f^\infty }{y_k}\left( t \right) + \\ \;\;\;\;\;\;\;\;\;\;\;{\eta _k}\zeta \left( {t,{y_k}\left( t \right)} \right) = 0,\;\;\;\;\;t \in \left( {0,1} \right),\\ {y_k}\left( 0 \right) = {y_k}\left( 1 \right) = 0. \end{array} \right. $ (23)

$ {v_k}: = \frac{{{y_k}}}{{{{\left\| {{y_k}} \right\|}_\infty }}}. $

则式(23) 可变为

$ \left\{ \begin{array}{l} {{v''}_k}\left( t \right) + a\left( t \right){v_k}\left( t \right) + b\left( t \right){v_k}\left( t \right) + {\eta _k}{f^\infty }{v_k}\left( t \right) + \\ \;\;\;\;\;\;\;\;\;\;\;{\eta _k}\frac{{\zeta \left( {t,{y_k}\left( t \right)} \right)}}{{{y_k}\left( t \right)}}{v_k}\left( t \right) = 0,\;t \in \left( {0,1} \right),\\ {v_k}\left( 0 \right) = {v_k} = \left( 1 \right) = 0. \end{array} \right. $ (24)

易知式(24) 等价于

$ \begin{array}{l} {v_k}\left( t \right) = {\eta _k}\int_0^1 {G\left( {t,s} \right)q\left( s \right)\left( {{f^\infty }{v_k}\left( s \right) + } \right.} \\ \;\;\;\;\;\;\;\;\;\;\;\left. {\frac{{\zeta \left( {s,{y_k}\left( s \right)} \right)}}{{{y_k}\left( s \right)}}{v_k}\left( s \right)} \right){\rm{d}}s. \end{array} $ (25)

$ {w_k}\left( t \right): = {f^\infty }{v_k}\left( t \right) + \frac{{\zeta \left( {t,{y_k}\left( t \right)} \right)}}{{{y_k}\left( t \right)}}{v_k}\left( t \right). $

因为f(t, yk(t))=fyk(t)+ζ(t, yk(t)), 所以由式(16) 以及事实yk(t)>0, t∈[0, 1]可推知:

$ \begin{array}{l} \left| {\frac{{\zeta \left( {t,{y_k}\left( t \right)} \right)}}{{{y_k}\left( t \right)}}} \right| \le {f^\infty } + \left| {\frac{{f\left( {t,{y_k}\left( t \right)} \right)}}{{{y_k}\left( t \right)}}} \right| \le \\ \;\;\;\;\;2{f^\infty } + \max \left\{ {\left| {\frac{{f\left( {t,{y_k}\left( t \right)} \right)}}{{{y_k}\left( t \right)}}} \right|:0 \le {y_k}\left( t \right) \le L} \right\} \le \\ \;\;\;\;\;2{f^\infty } + \max \left\{ {\left| {\frac{{f\left( {t,\tau } \right)}}{\tau }} \right|:0 \le \tau \le L} \right\}, \end{array} $

这意味着存在一个不依赖于k的函数σC[0, 1], 使得

$ \left| {\frac{{\zeta \left( {t,{y_k}\left( t \right)} \right)}}{{{y_k}\left( t \right)}}} \right| \le \sigma \left( t \right),\;\;\;t \in \left[ {0,1} \right]. $

于是{wk(t)}(kN)在C[0, 1]中一致有界.并且, 不难验证

$ \left\{ {{\eta _k}\int_0^1 {G\left( {t,s} \right)q\left( s \right){w_k}\left( s \right){\rm{d}}s} } \right\} \subset {C^1}\left[ {0,1} \right], $

C1[0, 1]紧嵌入C[0, 1],说明存在vk的子列(不妨仍记为vk), 使得对某个v*C[0, 1]及η*∈[0, ∞), 有vkv*ηkη*(k→∞).对式(25) 运用Lebesgue控制收敛定理可得

$ {v^ * }\left( t \right) = {\eta ^ * }\int_0^1 {G\left( {t,s} \right)q\left( s \right){f^\infty }{v^ * }\left( s \right){\rm{d}}s} . $

上式蕴含了v*C2[0, 1], 且满足

$ \left\{ \begin{array}{l} {v^{ * ''}} + a\left( t \right){v^{ * '}} + b\left( t \right){v^ * } + {\eta ^ * }{f^\infty }{v^ * } = 0,t \in \left( {0,1} \right),\\ {v^ * }\left( 0 \right) = {v^ * }\left( 1 \right) = 0. \end{array} \right. $

于是,

$ {\eta ^ * } = \frac{{{\lambda ^1}}}{{{f^\infty }}}. $

因此, 连通分支C+Φ+中连接$\left( {\frac{{{\lambda _1}}}{{{f_0}}}, 0} \right) $$\left( {\frac{{{\lambda _1}}}{{{f^\infty }}}, \infty } \right) $.定理证毕.

由定理1易得如下推论.

推论1 设(A1)和(A2)成立, λ1=(r(T))-1.若以下两条件之一成立:

$ \left( {\rm{i}} \right)\frac{{{\lambda _1}}}{{{f_\infty }}} < \lambda < \frac{{{\lambda _1}}}{{{f^0}}}; $
$ \left( {{\rm{ii}}} \right)\frac{{{\lambda _1}}}{{{f_0}}} < \lambda < \frac{{{\lambda _1}}}{{{f^\infty }}}. $

则奇异特征值问题

$ \left\{ \begin{array}{l} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + \lambda f\left( {t,u\left( t \right)} \right) = 0,\\ \;\;\;\;\;\;t \in \left( {0,1} \right),\\ u\left( 0 \right) = u\left( 1 \right) = 0 \end{array} \right. $

至少有1个正解.

定理2 设(A1)和(A2)成立, λ1=(r(T))-1.若以下两条件之一成立:

(ⅰ) f0>λ1, f>λ1, 且存在η*>0, 使得f(t, u) < A*η*, (t, u)∈[0, 1]×[0, η*];

(ⅱ) f0 < λ1, f < λ1, 且存在λ*>0, 使得f(t, u)>B*λ*, (t, u)∈[0, 1]×[0, λ*],

其中,

$ \begin{array}{l} {A^ * } = {\left[ {\int_0^1 {G\left( {s,s} \right)q\left( s \right){\rm{d}}s} } \right]^{ - 1}},\\ {B^ * } = {\left[ {\int_0^1 {G\left( {\frac{1}{2},s} \right)q\left( s \right){\rm{d}}s} } \right]^{ - 1}}, \end{array} $

则奇异问题(2) 至少有2个正解u1, u2.

证明 仅给出(ⅰ)成立时的证明.(ⅱ)可类似证得.

由定理1的证明知, 参数化问题

$ \left\{ \begin{array}{l} u''\left( t \right) + a\left( t \right)u'\left( t \right) + b\left( t \right)u\left( t \right) + \mu f\left( {t,u\left( t \right)} \right) = 0,\\ \;\;\;\;\;\;t \in \left( {0,1} \right),\\ u\left( 0 \right) = u\left( 1 \right) = 0 \end{array} \right. $

的解集中存在一条无界的连通分支C+,在Φ+中连接$\left( {\frac{{{\lambda _1}}}{{{f_0}}}, 0} \right) $$\left( {\frac{{{\lambda _1}}}{{{f^\infty }}}, \infty } \right) $.以下只需证明C+与{1}×E有2个互异的交点.

设(μ, u)∈C+, 且满足‖u=η*.则有

$ 0 < u\left( t \right) \le {\left\| u \right\|_\infty } = {\eta ^ * },\;\;\;t \in \left[ {0,1} \right]. $ (26)

因此, 由式(26) 和条件(ⅰ)可得

$ f\left( {t,u\left( t \right)} \right) < {A^ * }{\eta ^ * },t \in \left[ {0,1} \right],{\left\| u \right\|_\infty } = {\eta ^ * }. $ (27)

由引理2的(ⅱ)以及式(27), 对满足‖u=η*的(μ, u)有

$ \begin{array}{*{20}{c}} {u\left( t \right) = \mu \left( {Au} \right)\left( t \right) \le \mu \int_0^1 {G\left( {s,s} \right)q\left( s \right)f\left( {s,u\left( s \right)} \right){\rm{d}}s} < }\\ {\mu {A^ * }{\eta ^ * }\int_0^1 {G\left( {s,s} \right)q\left( s \right){\rm{d}}s} = \mu {{\left\| u \right\|}_\infty }.} \end{array} $

从而, 对满足‖u=η*的(μ, u)一定有

$ \mu > 1. $

又由假设条件(ⅰ)可得

$ \frac{{{\lambda _1}}}{{{f_0}}} < 1,\;\;\;\frac{{{\lambda _1}}}{{{f^\infty }}} \le \frac{{{\lambda _1}}}{{{f_\infty }}} < 1. $

从而由C+的无界连通性以及μ>1, $ \frac{{{\lambda _1}}}{{{f_0}}} < 1, \frac{{{\lambda _1}}}{{{f^\infty }}} < 1$的事实, 可推知存在

$ \left( {1,{u_1}} \right),\left( {1,{u_2}} \right) \in {C^ + }, $

且满足‖u1 < η*, ‖u2>η*, 即问题(2) 有2个正解.定理证毕.

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