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 浙江大学学报(理学版)  2018, Vol. 45 Issue (5): 540-544  DOI:10.3785/j.issn.1008-9497.2018.05.004 0

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ZHAO Ruhui, HAN Xiaoling. Symmetric positive solutions for the second-order three-point boundary value problems[J]. Journal of Zhejiang University(Science Edition), 2018, 45(5): 540-544. DOI: 10.3785/j.issn.1008-9497.2018.05.004.
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### 文章历史

Symmetric positive solutions for the second-order three-point boundary value problems
ZHAO Ruhui, HAN Xiaoling
College of Mathematics and Statistics, Northwest Normal University, Lanzhou 730070, China
Abstract: In this paper, we investigate the existence of symmetric positive solutions for the following second-order three-point boundary value problems:$\left\{ \begin{array}{l} u''\left( t \right) + q\left( t \right)f\left( {t,u\left( t \right),u'\left( t \right)} \right) = 0,t \in \left( {0,1} \right),\\ u\left( t \right) = u\left( {1 - t} \right),u'\left( 0 \right) - u'\left( 1 \right) = u\left( {\frac{1}{2}} \right) \end{array} \right.$, where f:[0, 1]×[0, +∞)×R→[0, +∞) is continuous; q(t) ≥ 0, t∈(0, 1).
Key words: second-order three-point boundary value problem    symmetric positive solutions    monotone iterative
0 引言

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

 $u''\left( t \right) + q\left( t \right)f\left( {t,u\left( t \right),u'\left( t \right)} \right) = 0,t \in \left( {0,1} \right),$ (1)
 $u\left( t \right) = u\left( {1 - t} \right),u'\left( 0 \right) - u'\left( 1 \right) = u\left( {\frac{1}{2}} \right)$ (2)

(H1)  f: [0, 1]×[0, +∞)×R→[0, +∞)连续, f(t, u, v)＞0并且在t∈[0, 1]上f(t, u, v)=f(1-t, u, -v);

(H2)  对任意(t, v)∈$\left[ {0, \frac{1}{2}} \right]$×R, f(t, ·, v)非减; 对任意(t, u)∈$\left[ {0, \frac{1}{2}} \right]$×[0, +∞), f(t, u, ·)非减;

(H3)  q(t)是定义在(0, 1)上的非负连续函数, 满足q(t)=q(1-t), 并且在(0, 1)的任何子区间上不恒为零, 且

 $\int_0^1 {q\left( t \right){\rm{d}}t} < + \infty .$
1 预备知识

(1) 对任意的uP, λ≥0, 有λuP;

(2) u, -uP, 有u=θ,

P是一个锥.

 $\left\| u \right\| = \max \left\{ {{{\left\| u \right\|}_\infty },{{\left\| {u'} \right\|}_\infty }} \right\},$

 ${C^ + }\left[ {0,1} \right] = \left\{ {u \in E:u\left( t \right) \ge 0,t \in \left[ {0,1} \right]} \right\}$

 $u''\left( t \right) + y\left( t \right) = 0,t \in \left( {0,1} \right),$ (3)
 $u\left( t \right) = u\left( {1 - t} \right),u'\left( 0 \right) - u'\left( 1 \right) = u\left( {\frac{1}{2}} \right)$ (4)

 $u\left( t \right) = \int_0^1 {G\left( {t,s} \right)y\left( s \right){\rm{d}}s} ,$

 $\begin{array}{l} \\ G\left( {t,s} \right) = {G_1}\left( {t,s} \right) + {G_2}\left( s \right), \end{array}$
 ${G_1}\left( {t,s} \right) = \left\{ \begin{array}{l} t\left( {1 - s} \right),0 \le t \le s \le 1,\\ s\left( {1 - t} \right),0 \le s \le t \le 1, \end{array} \right.$
 ${G_2}\left( {t,s} \right) = \left\{ \begin{array}{l} 1 - \frac{s}{2},0 \le s \le \frac{1}{2},\\ \frac{{1 + s}}{2},\frac{1}{2} \le s \le 1. \end{array} \right.$

 $\left( {Tu} \right)\left( t \right) = \int_0^1 {G\left( {t,s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} ,t \in \left[ {0,1} \right],$ (5)

 $\begin{array}{l} {\left( {Tu} \right)^\prime }\left( t \right) = - \int_0^t {sq\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\int_t^1 {\left( {1 - s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} . \end{array}$

 $\left\{ \begin{array}{l} {\left( {Tu} \right)^{\prime \prime }}\left( t \right) + y\left( t \right) = 0,t \in \left( {0,1} \right),\\ \left( {Tu} \right)\left( t \right) = \left( {Tu} \right)\left( {1 - t} \right),\\ {\left( {Tu} \right)^\prime }\left( 0 \right) - {\left( {Tu} \right)^\prime }\left( 1 \right) = \left( {Tu} \right)\left( {\frac{1}{2}} \right). \end{array} \right.$
 $\begin{array}{l} \left( {Tu} \right)\left( {1 - t} \right) = \int_0^{1 - t} {stq\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\;\;\int_{1 - t}^1 {\left( {1 - t} \right)\left( {1 - s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\;\;\int_0^1 {{G_2}\left( s \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} = \\ \;\;\;\;\;\;\int_1^t {\left( {1 - s} \right)sq\left( {1 - s} \right)f\left( {1 - s,u'\left( {1 - s} \right)} \right){\rm{d}}\left( {1 - s} \right)} + \\ \;\;\;\;\;\;\int_t^0 {\left( {1 - t} \right)sq\left( {1 - s} \right)f\left( {1 - s,u'\left( {1 - s} \right)} \right){\rm{d}}\left( {1 - s} \right)} + \\ \;\;\;\;\;\;\int_0^1 {{G_2}\left( s \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} = \\ \;\;\;\;\;\;\int_0^t {\left( {1 - t} \right)sq\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\;\;\int_t^1 {\left( {1 - s} \right)tq\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\;\;\int_0^1 {{G_2}\left( s \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} = \left( {Tu} \right)\left( t \right), \end{array}$

 $\begin{array}{l} 0 \le f\left( {s,u\left( s \right),u'\left( s \right)} \right) \le \max \left\{ {f\left( {s,u\left( s \right),u'\left( s \right)} \right)\left| {s \in } \right.} \right.\\ \;\;\;\left. {\left[ {0,1} \right],u \in \left[ {0,R} \right],u' \in \left[ { - R,R} \right]} \right\} = M. \end{array}$

 ${\left\| {Tu} \right\|_\infty } \le M\int_0^1 {q\left( s \right){\rm{d}}s} ,$
 $\begin{array}{l} {\left\| {{{\left( {Tu} \right)}^\prime }} \right\|_\infty } \le \int_0^1 {\left( {1 - s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} \le \\ \;\;\;\;M\int_0^1 {\left( {1 - s} \right)q\left( s \right){\rm{d}}s} \le M\int_0^1 {q\left( s \right){\rm{d}}s} , \end{array}$

 $\begin{array}{l} \left| {\left( {Tu} \right)\left( {{t_2}} \right) - \left( {Tu} \right)\left( {{t_1}} \right)} \right| = \\ \;\;\;\;\left| {\int_0^1 {G\left( {{t_2},s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} - } \right.\\ \;\;\;\;\left. {\int_0^1 {G\left( {{t_1},s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} } \right| = \\ \;\;\;\;\left| {\int_0^1 {\left[ {G\left( {{t_2},s} \right) - G\left( {{t_1},s} \right)} \right]q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} } \right| \le \\ \;\;\;\;M\int_0^1 {q\left( s \right){\rm{d}}s\left| {{t_2} - {t_1}} \right|} , \end{array}$

 $\begin{array}{l} \left| {{{\left( {Tu} \right)}^\prime }\left( {{t_2}} \right) - {{\left( {Tu} \right)}^\prime }\left( {{t_1}} \right)} \right| = \\ \;\;\;\;\;\left| { - \int_0^{{t_2}} {sq\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + } \right.\\ \;\;\;\;\;\int_{{t_2}}^1 {\left( {1 - s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\;\int_0^{{t_1}} {sq\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} - \\ \;\;\;\;\;\left. {\int_{{t_1}}^1 {\left( {1 - s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} } \right| = \\ \;\;\;\;\;\left| {\int_{{t_1}}^{{t_2}} {sq\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + } \right.\\ \;\;\;\;\;\left. {\int_{{t_1}}^{{t_2}} {\left( {1 - s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} } \right| = \\ \;\;\;\;\;\left| {\int_{{t_1}}^{{t_2}} {q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} } \right| \le M\int_{{t_1}}^{{t_2}} {q\left( s \right){\rm{d}}s} . \end{array}$

 $\left\{ \begin{array}{l} {{u'}_2}\left( t \right) \ge {{u'}_1}\left( t \right),t \in \left[ {0,\frac{1}{2}} \right],\\ {{u'}_2}\left( t \right) \le {{u'}_1}\left( t \right),t \in \left[ {\frac{1}{2},1} \right]. \end{array} \right.$

t$\left[ {0, \frac{1}{2}} \right]$时, 根据(H2)和Tu的定义, 有

 $\begin{array}{l} \left( {T{u_1}} \right)\left( t \right) - \left( {T{u_2}} \right)\left( t \right) = \\ \;\;\;\;\;\int_0^1 {G\left( {t,s} \right)q\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} - \\ \;\;\;\;\;\int_0^1 {G\left( {t,s} \right)q\left( s \right)f\left( {s,{u_2}\left( s \right),{{u'}_2}\left( s \right)} \right){\rm{d}}s} , \end{array}$
 $\begin{array}{l} {\left( {T{u_2}} \right)^\prime }\left( t \right) - {\left( {T{u_1}} \right)^\prime }\left( t \right) = \\ \;\;\;\; - \int_0^t {sq\left( s \right)f\left( {s,{u_2}\left( s \right),{{u'}_2}\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\int_t^1 {\left( {1 - s} \right)q\left( s \right)f\left( {s,{u_2}\left( s \right),{{u'}_2}\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\int_0^t {sq\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\int_t^1 {\left( {1 - s} \right)q\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} . \end{array}$

 $\begin{array}{l} {\left( {T{u_2}} \right)^\prime }\left( {{t_2}} \right) - {\left( {T{u_1}} \right)^\prime }\left( {{t_2}} \right) - {\left( {T{u_2}} \right)^\prime }\left( {{t_1}} \right) + {\left( {T{u_1}} \right)^\prime }\left( {{t_1}} \right) = \\ \;\;\;\;\;\;\;\int_{{t_1}}^{{t_2}} {q\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} - \\ \;\;\;\;\;\;\;\int_{{t_1}}^{{t_2}} {q\left( s \right)f\left( {s,{u_2}\left( s \right),{{u'}_2}\left( s \right)} \right){\rm{d}}s} \le 0. \end{array}$

2 主要结果及证明

 ${\sup _{t \in \left[ {0,1} \right]}}f\left( {t,a,a} \right) \le {a_1},$ (6)

 $\frac{{4a}}{5} \ge \max \left\{ {\int_0^1 {q\left( s \right){\rm{d}}s} ,\int_0^1 {\left( {1 - s} \right)q\left( s \right){\rm{d}}s} ,\int_0^1 {sq\left( s \right){\rm{d}}s} } \right\}{a_1},$ (7)

 $0 < \left\| {{u^ * }} \right\| \le a,$
 $0 < \left\| {{w^ * }} \right\| \le a,$

 $\mathop {\lim }\limits_{n \to \infty } {u_n} = \mathop {\lim }\limits_{n \to \infty } {T^n}{u_0} = {u^ * },$
 $\mathop {\lim }\limits_{n \to \infty } {\left( {{u_n}} \right)^\prime } = \mathop {\lim }\limits_{n \to \infty } {\left( {{T^n}{u_0}} \right)^\prime } = {\left( {{u^ * }} \right)^\prime },$
 $\mathop {\lim }\limits_{n \to \infty } {w_n} = \mathop {\lim }\limits_{n \to \infty } {T^n}{w_0} = {w^ * },$
 $\mathop {\lim }\limits_{n \to \infty } {\left( {{w_n}} \right)^\prime } = \mathop {\lim }\limits_{n \to \infty } {\left( {{T^n}{w_0}} \right)^\prime } = {\left( {{w^ * }} \right)^\prime }.$

 ${w_0}\left( t \right) = 0,0 \le t \le 1.$

 ${u_0}\left( t \right) = \frac{4}{5}at\left( {1 - t} \right) + \frac{4}{5}a \le a,$

 ${u_0}^\prime \left( t \right) = - \frac{8}{5}at + \frac{4}{5}a \le a,$

u1=Tu0, 同理可证u1$\overline {{P_a}}$.

 ${u_{n + 1}} = T{u_n} = {T^{n + 1}}{u_0},n = 0,1,2, \cdots ,$ (9)

 $\begin{array}{l} {u_1}\left( t \right) = \left( {T{u_0}} \right)\left( t \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\int_0^1 {G\left( {t,s} \right)q\left( s \right)f\left( {s,{u_0}\left( s \right),{{u'}_0}\left( s \right)} \right){\rm{d}}s} = \\ \;\;\;\;\;\;\;\;\;\;\;\int_0^1 {{G_1}\left( {t,s} \right)q\left( s \right)f\left( {s,{u_0}\left( s \right),{{u'}_0}\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\;\;\;\;\;\;\;\int_0^1 {{G_2}\left( s \right)q\left( s \right)f\left( {s,{u_0}\left( s \right),{{u'}_0}\left( s \right)} \right){\rm{d}}s} \le \\ \;\;\;\;\;\;\;\;\;\;\;{a^1}t\left( {1 - t} \right)\int_0^1 {q\left( s \right)ds} + {a_1}\int_0^1 {q\left( s \right){\rm d}s} \le \\ \;\;\;\;\;\;\;\;\;\;\;\frac{4}{5}at\left( {1 - t} \right) + \frac{4}{5}a = {u_0}, \end{array}$

 $\begin{array}{l} \left| {{{u'}_1}\left( t \right)} \right| = \left| {{{\left( {T{u_0}} \right)}^\prime }\left( t \right)} \right| = \\ \;\;\;\;\;\;\;\;\;\;\;\;\left| { - \int_0^1 {sq\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} + } \right.\\ \;\;\;\;\;\;\;\;\;\;\;\;\left. {\int_0^1 {\left( {1 - s} \right)q\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} } \right| \le \\ \;\;\;\;\;\;\;\;\;\;\;\;\int_0^1 {sq\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\;\;\;\;\;\;\;\;\int_0^1 {\left( {1 - s} \right)q\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} = \\ \;\;\;\;\;\;\;\;\;\;\;\;\int_0^1 {q\left( s \right)f\left( {s,{u_1}\left( s \right),{{u'}_1}\left( s \right)} \right){\rm{d}}s} \le \\ \;\;\;\;\;\;\;\;\;\;\;\;{a_1}\int_0^1 {q\left( s \right){\rm d}s} \le \frac{4}{5}a = \left| {{{u'}_0}\left( t \right)} \right|, \end{array}$

 ${u_{n + 1}}\left( t \right) \le {u_n}\left( t \right),{{u'}_{n + 1}}\left( t \right) \le {{u'}_n}\left( t \right),$
 $0 < t < 1,n = 1,2, \cdots ,$

 ${\omega _{n + 1}} = T{\omega _n} = {T^{n + 1}}{\omega _0},n = 1,2, \cdots ,$

 ${\omega _1} = T{\omega _0} = T\left( 0 \right) \in \overline {{p_a}} ,$

 ${\omega _1} = T{\omega _0} = T\left( 0 \right)\left( t \right) \ge 0,0 < t < 1,$
 $\left| {{{\omega '}_1}\left( t \right)} \right| = \left| {{{\left( {T{\omega _0}} \right)}^\prime }\left( t \right)} \right| = \left| {T'\left( 0 \right)\left( t \right)} \right| \ge 0,0 < t < 1,$

 $\begin{array}{l} {\omega _{n + 1}}\left( t \right) \ge {\omega _n}\left( t \right),\left| {{{\omega '}_{n + 1}}\left( t \right)} \right| \ge \left| {{{\omega '}_n}\left( t \right)} \right|0 < t < 1,\\ \;\;\;\;\;n = 0,1, \cdots , \end{array}$
 $f\left( {t,u,v} \right) = \sqrt u + {\left( {u'} \right)^2} + t\left( {1 - t} \right),$

3 举例

 $u''\left( t \right) + \frac{1}{{20}}\left( {\frac{1}{{\sqrt t }} + \frac{1}{{\sqrt {1 - t} }}} \right)\left[ {\sqrt u + {{\left( {u'} \right)}^2} + t\left( {1 - t} \right)} \right] = 0,$ (10)
 $u\left( t \right) = u\left( {1 - t} \right),u'\left( 0 \right) - u'\left( 1 \right) = u\left( {\frac{1}{2}} \right),t \in \left( {0,1} \right),$ (11)

 $q\left( t \right) = \frac{1}{{20}}\left( {\frac{1}{{\sqrt t }} + \frac{1}{{\sqrt {1 - t} }}} \right).$

 ${\left\| {{u^ * }} \right\|_\infty } \le 1,{\left\| {{\omega ^ * }} \right\|_\infty } \le 1,$

 $\mathop {\lim }\limits_{n \to \infty } {T^n}{u_0} = {u^ * },\mathop {\lim }\limits_{n \to \infty } {T^n}{\omega _0} = {\omega ^ * }.$

 ${u_0}\left( t \right) = \frac{4}{5}t\left( {1 - t} \right) + \frac{4}{5},{\omega _0}\left( t \right) = 0,0 \le t \le 1.$

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