﻿ 一类奇异边值问题正解的存在性及多解性
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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 0

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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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### 文章历史

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

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

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

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

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

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

(ⅰ)初值问题

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

(ⅱ)初值问题

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

(ⅲ) α在[0, 1]非减, β在[0, 1]非增.

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

(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.

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

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

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

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

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(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

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

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

2 主要结果及其证明

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

 $\mu > 1.$

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

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

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