0 引言

非局部边值问题在物理模型、生物学、数学等领域得到了广泛研究.在过去的几十年里, 大量文献研究了此类边值问题正解的存在性^[1-5].近几年, 出现了对此类边值问题对称正解的存在性的研究^[6-9].最近, 文献[10]讨论了二阶三点边值问题:

$ \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. $

对称正解的存在性, 其中a: (0, 1)→[0, +∞)在(0, 1)上是对称的, f: [0, 1]×[0, +∞)×R→[0, +∞)是连续的, 并且f(·, u)在[0, 1]上对称, 通过Krasnosel'skii不动点定理获得了上述边值问题对称正解的存在性.

受以上结果的启发, 本文主要研究二阶三点边值问题

$ 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)

对称正迭代解的存在性.本文假设

(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

定义1  设E是Banach空间, 且P是E上的非空凸闭集.满足:

(1) 对任意的u∈P, λ≥0, 有λu∈P;

(2) u, -u∈P, 有u=θ,

则P是一个锥.

定义2  设(E, ≤)是序Banach空间, 若对任意u, v∈E且u≤v有φ(u)≤φ(v), 则算子φ: E→E非减.若不等号是严格的, 则称φ是严格非减的.

定义3  设E=C¹[0, 1], u∈E, 若对任意t₁, t₂∈[0, 1]且λ∈[0, 1]有u(λt₁+(1-λ)t₂)≥λu(t₁)+(1-λ)u(t₂), 则称u在[0, 1]上是凹的.

定义4  若函数u(t)满足u(t)=u(1-t), 则称u(t)在[0, 1]上对称.

定义5  若函数u^*在[0, 1]上是正对称的, 且满足式(1)和(2), 则称函数u^*是边值问题(1)和(2)的一个对称正解.

记空间E=C¹[0, 1], 令其范数为

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

这里‖u‖_∞=$\mathop {\max }\limits_{t \in \left[ {0, 1} \right]} $|u(t)|.

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

定义锥P={u∈E: u(t)≥0, u是[0, 1]上的凹函数, 且u(t)=u(1-t)}.

引理1  设y(t)∈C[0, 1]且在[0, 1]上对称, 则三点边值问题

$ 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)

有唯一对称解

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

其中,

$ \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. $

对于任意u(t)∈C⁺[0, 1], 定义算子T:

$ \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)

则算子T的不动点是问题(1)和(2)的解.

注1  由算子的定义, 易得

$ \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} $

引理2  若(H2)成立, 则算子T: P→P全连续, 且T是非减的.

证明  对任意y∈P, 根据算子T的定义, 有

$ \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} $

即(Tu)(t)在[0, 1]上是凹的, 且(Tu)(t)=(Tu)(1-t), 故T(P)⊂P.

下证算子T: P→P是全连续的.算子T显然是连续的, 只须证T是紧的.

设有界集Ω₀⊂P, 则存在一个R使得Ω₀⊂{u∈P: ‖u‖≤R}.对任意u∈P, 有

$ \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} $

因此, 由式(5)有

$ {\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} $

上式显然有界, 故(Tu)(t)一致有界.

下证(Tu)(t)等度连续.对任意的t₁, t₂∈[0, 1], t₁≤t₂, 有

$ \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} $

更进一步

$ \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} $

故(Tu)(t)是等度连续的, 由Ascoli-Arzela定理得T(Ω₀)是紧集.因此T: P→P是全连续的.

最后证明Tu关于u是非减的.对于任意的u_i∈P(i=1, 2), u₁(t)≤u₂(t), 由锥的性质, 有u₂(t)-u₁(t)∈P.因此, 有

$ \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. $

当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} $

为证明(Tu₂)(t)-(Tu₁)(t)是凹的, 只须证(Tu₂)′(t)-(Tu₁)′(t)是非减的.令0≤t₁≤t₂≤$\frac{1}{2}$, 则

$ \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} $

同理可证当t∈$\left[ {\frac{1}{2}, 1} \right]$时, 上述结论也成立.且易证(Tu₂)(t)-(Tu₁)(t)是对称的, 故(Tu₂-Tu₁)∈P, 则T是非减的.

2 主要结果及证明

定理1  假设(H1)~(H3)成立, 若存在2个正数a₁≤a, 使得

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

(6)

其中a和a₁满足

$ \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)

则式(1)和式(2)有2个正的凹解u^*和w^*, 使得

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

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

并且

$ \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 }. $

其中, u₀(t)=$\frac{4}{5}$at(1-t)+$\frac{4}{5}$a, 0≤t≤1;

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

迭代方法为u_n+1=Tu_n=Tⁿu₀, w_n+1=Tw_n=Tⁿw₀.

证明  令$\overline {{P_a}} $={u∈P: 0≤‖u‖≤a}. 先证T$\overline {{P_a}} $⊂$\overline {{P_a}} $.

对任意的u∈$\overline {{P_a}} $, 有

$ 0 \le u\left( t \right) \le \mathop {\max }\limits_{t \in \left[ {0,1} \right]} \left| {u\left( t \right)} \right| \le \left\| u \right\| \le a, $

$ \left| {u'\left( t \right)} \right| \le \mathop {\max }\limits_{t \in \left[ {0,1} \right]} \left| {u'\left( t \right)} \right| \le \left\| u \right\| \le a. $

由(H2)及式(6)得

$ 0 \le f\left( {t,u\left( t \right),u'\left( t \right)} \right) \le \mathop {\sup }\limits_{t \in \left[ {0,1} \right]} f\left( {t,a,a} \right) \le {a_1}. $

(8)

对任意u(t)∈$\overline {{P_a}} $, 由引理2有Tu∈P, 且

$ \begin{array}{l} {\left\| {Tu} \right\|_\infty }\left( {Tu} \right)\left( {\frac{1}{2}} \right) = \\ \;\;\;\;\int_0^1 {G\left( {\frac{1}{2},s} \right)q\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} = \\ \;\;\;\;\int_0^{\frac{1}{2}} {\frac{1}{2}sq\left( s \right)f\left( {s,u\left( s \right),u'\left( s \right)} \right){\rm{d}}s} + \\ \;\;\;\;\int_0^{\frac{1}{2}} {\frac{1}{2}\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} \le \\ \;\;\;\;{a_1}\int_0^{\frac{1}{2}} {q\left( s \right){\rm{d}}s} + {a_1}\int_{\frac{1}{2}}^1 {q\left( s \right){\rm{d}}s} \le \\ \;\;\;\;{a_1}\int_0^1 {q\left( s \right){\rm{d}}s} \le a; \end{array} $

$ \begin{array}{l} {\left\| {{{\left( {Tu} \right)}^\prime }} \right\|_\infty } = {\left( {Tu} \right)^\prime }\left( 0 \right) = \\ \;\;\;\;\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 \\ \;\;\;\;{a_1}\int_0^1 {\left( {1 - s} \right)q\left( s \right){\rm{d}}s} \le a. \end{array} $

得到‖Tu‖≤a, 故T$\overline {{P_a}} $⊂$\overline {{P_a}} $.

一方面, 令

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

则

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

从而u₀∈$\overline {{P_a}} $.

令u₁=Tu₀, 同理可证u₁⊂$\overline {{P_a}} $.

定义

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

(9)

由于T是全连续的, 故{u_n}是紧集.因为

$ \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} $

而且

$ \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} $

从而u₁(t)≤u₀(t), |u₁′(t)|≤|u₀′(t)|, 0＜t＜1.

由引理2及式(9), 有

$ {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 , $

因此存在u^*∈$\overline {{P_a}} $, 使得u_n→u^*.根据T的连续性和u_n+1(t)=Tu_n(t), 可得Tu^*(t)=u^*(t), 则u^*是问题(1)和(2)凹的对称正解.

另一方面, 令

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

由于T$\overline {{p_a}} $⊂$\overline {{P_a}} $, 则ω_n∈$\overline {{p_a}} $.因为T是全连续的, 故{ω_n}是紧集.

因为

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

则有

$ {\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, $

由引理2及式(9), 有

$ \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), $

因此存在ω^*∈$\overline {{p_a}} $, 使ω_n→ω^*.由T的连续性和ω_n+1(t)=Tω_n(t), 可得Tω^*(t)=ω^*(t), 且ω^*＞0, 则ω^*是问题(1)和(2)凹的对称正解.证毕!

3 举例

考虑边值问题

$ 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)

其中,     f(t, u, v)=$\sqrt u $+(u′)²+t(1-t),

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

则(H1)~(H3)成立.令a=1, a₁=3满足式(6), 则由定理1可得边值问题(10)和(11)有2个凹的对称解u^*, ω^*, 使得

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

并且

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

其中,

$ {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. $