利用锥上不动点理论, 文献[1-6]研究了边值问题正解的存在性; 文献[7-8]研究了带Riemann-Stieltjes积分边值条件的高阶问题, 其中2012年, 当$r a(t)f(x)=\lambda f(t, x)$时, 文献[8]研究了下列奇异高阶问题

$\left\{ \begin{array}{l} {x^{(n)}}(t) + ra(t)f(x) = 0,{\mkern 1mu} {\mkern 1mu} 0 ＜ t ＜ 1,\\ x(0) = \int_0^1 {x(t)d\alpha (t)} ,{\mkern 1mu} {\mkern 1mu} x'(0) = \cdots = {x^{(n - 2)}}(0) = 0,{\mkern 1mu} {\mkern 1mu} x(1) = \int_0^1 {x(t)d\beta (t),} \end{array} \right.$

(1.1)

其中$f(t, x)$在$t=0, t=1$处奇异, $\alpha, \, \, \beta:[0,1]\rightarrow \Re$分别是有界变差函数.

应用分歧方法, 文献[9-11]研究了二阶边值问题; 文献[12-14]研究了四阶边值问题; 文献[15]研究了高维问题; 文献[16-17]研究了带Riemann-Stieltjes积分边值条件问题.

受上述文献的启发, 本文研究奇异高阶含Riemann-Stieltjes积分边值条件的问题(1.1) 正解的存在性问题.本文做如下假设

(H1) 假设$\alpha, \, \, \beta:[0,1]\rightarrow \Re$分别是非减函数且在[0, 1]上不恒为常数. ${g_\alpha }(s) = \int_0^1 {k(t,s)d\alpha (t)} $, ${g_\beta }(s) = \int_0^1 {k(t,s)d\beta (t)} $ $g_{\alpha}(s)\geq0$且$g_{\alpha}(s)\geq0, g_{\beta}(s)\geq0, \forall s\in[0,1]$;

(H2)

$\begin{array}{l} 0 ＜ \int_0^1 {(1 - {t^{n - 1}})d\alpha (t) ＜ 1} ,0 ＜ \int_0^1 {{t^{n - 1}}d\beta (t) ＜ 1,} \\ 0 ＜ \int_0^1 {(1 - {t^{n - 1}})d\beta (t) ＜ \infty } ,0 ＜ \int_0^1 {{t^{n - 1}}d\alpha (t) ＜ \infty ,} \end{array}$

$D = \int_0^1 {{t^{n - 1}}d\alpha (t)} \left( {1 - \int_0^1 {d\beta (t)} } \right) + \left( {1 - \int_0^0 {d\alpha (t)} } \right)\left( {1 - \int_0^1 {{t^{n - 1}}d\beta (t)} } \right) ＞ 0;$

(H3) $a(\cdot)\in C((0, 1), [0, \infty))$, 在$(0, 1)$的任何子区间上$a(t)\not\equiv0$, 且

$0 ＜ \int_0^1 k (\tau (s),s)a(s)ds ＜ \infty ,0 ＜ \int_0^1 {{k_i}({\tau _i}(s),s)a(s)ds ＜ \infty } ,$

$k(\tau(s), s), k_{i}(\tau_{i}(s), s)$分别由引理2.2与引理2.3给出;

(H4) $f(\cdot)\in C([0, \infty), [0, \infty))$, 对任何$s＞0$, 都有$f(s)＞0$成立;

(H5) $f_{0}, f_{\infty}\in (0, +\infty)$;

(H6) $f_{0}\in (0, +\infty)$且$f_{\infty}=\infty;$

(H7) $f_{0}=0$且$f_{\infty}=\infty;$

(H8) $f_{0}=\infty$且$f_{\infty}=\infty, $

其中

${f_0} = \mathop {\lim }\limits_{s \to {0^ + }} \frac{{f(s)}}{s},{f_\infty } = \mathop {\lim }\limits_{s \to + \infty } \frac{{f(s)}}{s},{f_0} = \mathop {\lim }\limits_{s \to {0^ + }} \frac{{f(s)}}{s},{f_\infty } = \mathop {\lim }\limits_{s \to + \infty } \frac{{f(s)}}{s}.$

本章安排如下:在第二部分给出格林函数及其性质; 第三部分给出预备知识; 第四部分给出问题(1.1) 至少存在一个正解的主要定理及证明.

考虑如下边值问题

$\left\{ \begin{array}{l} {x^{(n)}}(t) + y(t) = 0,0 ＜ t ＜ 1,\\ x(0) = \int_0^1 {x(t)d\alpha (t)} ,{\mkern 1mu} {\mkern 1mu} x'(0) = \cdots = {x^{(n - 2)}}(0) = 0,{\mkern 1mu} {\mkern 1mu} x(1) = \int_0^1 {x(t)d\beta (t)} . \end{array} \right.$

(2.1)

引理2.1  (见文献[8, 引理1])假设条件(H1) 和(H2) 成立.对于任何$y\in C[0,1]$, 则问题(2.1) 存在唯一解

$x(t) = \int_0^1 {K(t,s)y(s)ds} ,$

(2.2)

其中

$K(t,s) = k(t,s) + \frac{{A(t)}}{D}{g_\alpha }(s) + \frac{{B(t)}}{D}{g_\beta }(s),$

(2.3)

$k(t,s) = \frac{1}{{(n - 1)!}}\left\{ {\begin{array}{*{20}{l}} {{t^{n - 1}}{{(1 - s)}^{n - 1}} - {{(t - s)}^{n - 1}},}&{{\rm{0}} \le s \le t \le {\rm{1}},}\\ {{t^{n - 1}}{{(1 - s)}^{n - 1}},}&{{\rm{0}} \le t \le s \le {\rm{1}}} \end{array}} \right.$

(2.4)

$\begin{array}{l} A(t) = \left( {1 - \int_0^1 {{t^{n - 1}}d\beta (t)} } \right) - \left( {1 - \int_0^1 {d\beta (t)} } \right){t^{n - 1}},\\ B(t) = \int_0^1 {{t^{n - 1}}d\alpha (t)} + \left( {1 - \int_0^1 {d\alpha (t)} } \right){t^{n - 1}},\\ D = \int_0^1 {{t^{n - 1}}d\alpha (t)} \left( {1 - \int_0^1 {d\beta (t)} } \right) + \left( {1 - \int_0^1 {d\alpha (t)} } \right)\left( {1 - \int_0^1 {{t^{n - 1}}d\beta (t)} } \right),\\ {g_\alpha }(s) = \int_0^1 {k(t,s)d\alpha (t)} ,{\mkern 1mu} {\mkern 1mu} {g_\beta }(s) = \int_0^1 {k(t,s)d\alpha (t)} . \end{array}$

引理2.2 (见文献[8, 引理2])由$(2.4)$式定义的$k(t, s)$满足下列性质

$c(t)k(\tau(s), s)\leq k(t, s)\leq k(\tau(s), s), \, \, \forall t, \, \, s\in[0,1],$

(2.5)

其中

$\begin{array}{l} \tau (s) = \frac{s}{{1 - {{(1 - s)}^{\frac{{n - 1}}{{n - 2}}}}}},{\mkern 1mu} {\mkern 1mu} k(\tau (s),s) = \frac{{\tau {{(s)}^{n - 2}}s{{(1 - s)}^{n - 1}}}}{{(n - 1)!}},\\ c(t) = \min \left\{ {\frac{{{{(n - 1)}^{n - 1}}{t^{n - 2}}(1 - t)}}{{{{(n - 2)}^{n - 2}}}},{\mkern 1mu} {\mkern 1mu} {t^{n - 1}}} \right\}. \end{array}$

(2.6)

引理2.3  $k(t, s)$由(2.4) 式定义, $i=2, \cdots, n$, 下式成立

$\frac{{{\partial ^{(i - 1)}}}}{{\partial {t^{(i - 1)}}}}k(t,s) = {k_i}(t,s) = \frac{1}{{(n - i)!}}\left\{ {\begin{array}{*{20}{l}} {{t^{n - i}}{{(1 - s)}^{n - 1}} - {{(t - s)}^{n - i}},}&{0 \le s \le t \le 1,}\\ {{t^{n - i}}{{(1 - s)}^{n - i}},}&{0 \le t \le s \le 1} \end{array}} \right.$

(2.7)

并且$k_{i}(t, s)$满足

$c_{i}(t)k_{i}(\tau_{i}(s), s)\leq k_{i}(t, s)\leq k_{i}(\tau_{i}(s), s), \, \, \forall t, \, \, s\in[0,1],$

(2.8)

其中

$\begin{array}{l} {\tau _i}(s) = \frac{s}{{1 - {{(1 - s)}^{\frac{{n - 1}}{{n - i - 2}}}}}},{\mkern 1mu} {\mkern 1mu} {k_i}({\tau _i}(s),s) = \frac{{{\tau _i}{{(s)}^{n - i - 2}}s{{(1 - s)}^{n - 1}}}}{{(n - i - 1)!}},\\ {c_i}(t) = \min \left\{ {\frac{{{{(n - i - 1)}^{n - i - 1}}{t^{n - i - 2}}(1 - t)}}{{{{(n - i - 2)}^{n - i - 2}}}},{\mkern 1mu} {\mkern 1mu} {t^{n - i - 1}}} \right\}. \end{array}$

(2.9)

证    相似于文献[7]第1937-1938页中定理3.1的证明方法, 易得引理2.3, 故证明略.

引理2.4 (见文献[8, 引理3])假设条件(H1) 和(H2) 成立.由$(2.3)$式定义的$K(t, s)$满足下列性质

(ⅰ) $K(t, s)$在$[0,1]\times[0,1]$上连续且$K(t, s)\geq0$;

(ⅱ) 对于任意$t, \, \, s\in[0,1]$都有$K(t, s)\leq K(s)$成立, 对于任意$t, \, \, s\in[0,1]$, 下式成立

$\mathop {\min }\limits_{t \in [0,1]} K(t,s) \ge q(t)K(s),{\mkern 1mu} {\mkern 1mu} \forall s \in [0,1],$

(2.10)

其中

$\begin{array}{l} K(s) = k(\tau (s),s) + \frac{{{M_1}}}{D}{g_\alpha }(s) + \frac{{{M_2}}}{D}{g_\beta }(s),\\ q(t) = \min \left\{ {c(t),{\mkern 1mu} {\mkern 1mu} \frac{{A(t)}}{{{M_1}}},\frac{{B(t)}}{{{M_2}}}} \right\}, \end{array}$

(2.11)

$\begin{array}{l} {M_1} = \max \left\{ {1 - \int_0^1 {{t^{n - 1}}d\beta (t)} ,\int_0^1 {(1 - {t^{n - 1}})d\beta (t)} } \right\},{\mkern 1mu} {\mkern 1mu} \\ {M_2} = \max \left\{ {\int_0^1 {{t^{n - 1}}d\alpha (t)} ,1 - \int_0^1 {(1 - {t^{n - 1}})d\alpha (t)} } \right\}. \end{array}$

(2.12)

引理2.5 (见文献[8, 引理4])假设条件(H1) 和(H2) 成立.则对于$y \in C[0,1]$且$y\geq0$, (2.1) 式的唯一解满足

(ⅰ) $x(t)\geq0, $ $\forall t\in[0,1]$;

(ⅱ) $\mathop {\min }\limits_{t \in [0,1]} x(t) \ge q(t)\|x\|$,

其中$q(t)$由引理2.3 (ⅱ)给出.

记$Y=C[0,1], $其上范数为$\|x\|_{\infty}=\max\limits_{t\in[0,1]}|x(t)|.$

记$E=\{x\in C^{n-1}[0,1]|x(0)=\alpha[x], x'(0)=\cdots=x^{(n-2)}(0)=0, x(1)=\beta[x]\}, $其上范数为

$\|x\|_{E}=\max\{\|x\|_{\infty}, \, \, \|x'\|_{\infty}, \cdots\|x^{(n-1)}\|_{\infty}\}.$

定义算子$L:D(L)\subset E\rightarrow E, $ $Lx=x^{(n)}, x\in D(L), $其中

$D(L)=\{x\in C^{n}[0,1]|x(0)=\alpha[x], x'(0)=\cdots=x^{(n-2)}(0)=0, x(1)=\beta[x]\}.$

容易验证$L$为闭算子且$L^{-1}:Y\rightarrow D(L)$是全连续算子.

令$\Sigma$为(1.1) 在$[0, \infty)\times E$上正解集合的闭包.

定义锥

$P=\{x\in C[0,1]|x(t)\geq0, t\in[0,1], \min\limits_{t\in[\delta, 1-\delta]}x(t)\geq q(t)\|x\|_{\infty}\},$

其中$q(t)$由引理2.3 (ⅱ)给出, 且对于$r ＞ 0$, 令$\Omega_{r}=\{x\in P | \|x\|_{E} ＜ r\}.$

首先考虑线性问题

$\begin{array}{l} {x^{(n)}} = \lambda a(t)x(t),{\mkern 1mu} {\mkern 1mu} t \in (0,1),\\ x(0) = \alpha [x],x'(0) = \cdots = {x^{(n - 2)}}(0) = 0,x(1) = \beta [x]. \end{array}$

令

$\begin{array}{l} {L_\lambda }x(t) = \lambda \int_0^1 {K(t,s)a(s)x(s)ds} ,t \in [0,1].\\ {T_\lambda }x(t) = \lambda \int_0^1 {K(t,s)a(s)f(x(s))ds} ,t \in [0,1]. \end{array}$

由Krein-Rutmann定理(见文献[18, 定理2.5], 亦可参考文献[19]或[20]), 可得下列引理.

引理3.1 设(H1)-(H3) 成立, $r(L_{\lambda})$是$L_{\lambda}$的谱半径.则$r(L_{\lambda})\neq0$且$L_{\lambda}$有一个对应于第一特征值$\lambda_{1}=\frac{1}{r(L_{\lambda})}$的正的特征函数$\phi_{1}\in\mathrm{int}P$, 它是简单的并且再没有别的特征值对应正的特征函数.

引理3.2 设(H1)-(H4) 成立, 则问题$(1.1)$的解$x(t)$满足

$\|x(t)\|_{\infty}\leq\frac{1}{1-\alpha[1]}\|x'\|_{\infty}, \|x'\|_{\infty}\leq\|x''\|_{\infty}, \cdots, \|x^{(n-2)}\|_{\infty}\leq\|x^{(n-1)}\|_{\infty}.$

证    由$|\int_0^1 {x(s)d\alpha (s)} | \le \mathop {\max }\limits_{s \in [0,1]} |x(s)| \cdot \alpha [1]$和$x(0) = \int_0^1 {x(s)d\alpha (s)} ,$可得

$\mathop {\max }\limits_{t \in [0,1]} |x(t)| = \mathop {\max }\limits_{t \in [0,1]} \left| {x(0) + \int_0^t {x'(s)ds} } \right| \le \mathop {\max }\limits_{t \in [0,1]} |x(t)|\alpha [1] + \int_0^1 {\left| {x'(s)} \right|ds} ,$

即

$\mathop {\max }\limits_{t \in [0,1]} |x(t)| \le \frac{1}{{1 - \alpha [1]}}\int_0^1 {\left| {x'(s)} \right|ds} .$

进而$\|x(t)\|_{\infty}\leq\frac{1}{1-\alpha[1]}\|x'\|_{\infty}.$由$x'(0)=0, $可得

$|x'(t)| = \left| {x'(0) + \int_0^t {x''(s)ds} } \right| \le \left| {\int_0^1 {x''(s)ds} } \right| \le \int_0^1 {\left| {x''(s)} \right|ds} ,$

进而$ \|x'\|_{\infty}\leq\|x''\|_{\infty}. $继续这样一个过程, 由条件$x'(0)=x''(0)=\cdots=x^{(n-2)}(0)=0$, 可得

$\|{x}''{{\|}_{\infty }}\le \|{x}'''{{\|}_{\infty }},\cdots \|{{x}^{(n-2)}}{{\|}_{\infty }}\le \|{{x}^{(n-1)}}{{\|}_{\infty }}.$

结论获证.

引理3.3    设(H1)-(H4) 成立.假设$\{(\mu_{k}, x_{k})\}\subset(0, \infty)\times P$是问题(1.1) 的一个正解序列, 存在常数$c_{0} ＞ 0$, 使得$\|\mu_{k}\|\leq c_{0}$, 且

$\mathop {\lim }\limits_{k \to \infty } \|{x_k}\|{_E} = \infty ,$

(3.1)

则$\mathop {\lim }\limits_{k \to \infty } \|{x_k}\|{_\infty } = \infty .$

证  反设存在常数$M_{0} ＞ 0$, 使得$\|x_{k}\|_{\infty}\leq M_{0}.$由于$(\mu_{k}, x_{k})$是问题(1.1) 的正解, 则

${x_k}(t) = {\mu _k}\int_0^1 {K(t,s)a(s)f({x_k}(s))ds} ,t \in [0,1].$

进而

$x_k^{(n - 1)}(t) = {\mu _k}\int_0^1 {\frac{{{\partial ^{(n - 1)}}}}{{\partial {t^{(n - 1)}}}}K(t,s) \cdot a(s)f({x_k}(s))ds} .$

(3.2)

又

$\begin{array}{l} \frac{{{\partial ^{(n - 1)}}}}{{\partial {t^{(n - 1)}}}}K(t,s) = \left| {\frac{{{\partial ^{(n - 1)}}}}{{\partial {t^{(n - 1)}}}}k(t,s) + \frac{{{g_\alpha }(s)}}{D}\frac{{{d^{(n - 1)}}A(t)}}{{{d^{(n - 1)}}t}} + \frac{{{g_\beta }(s)}}{D}\frac{{{d^{(n - 1)}}B(t)}}{{{d^{(n - 1)}}t}}} \right|\\ \quad \quad \quad \quad \quad \quad \le {\Phi _1}(s) + {\Phi _2}(s), \end{array}$

(3.3)

其中

$ \Phi_{1}(s):=\frac{\partial^{(n-1)}}{\partial t^{(n-1)}}k(t, s)=(1-s)^{n-1}, \\ \Phi_{2}(s):=\frac{(n-1)![(1-\alpha[1])\alpha[1]+(1-\beta[1])\beta[1]]}{D}k(\tau(s), s).$

由(H3) 可得

$0 ＜ \int_0^1 {{\Phi _1}a(s)ds ＜ \infty } ,0 ＜ \int_0^1 {{\Phi _2}a(s)ds ＜ \infty } .$

(3.4)

假设$b = \mathop {\max }\limits_{s \in [0,{M_0}]} \{ f(s)\} $, 结合(3.2), (3.3) 和(3.4) 式, 可得

$\|x_{k}^{(n-1)}(t){{\|}_{\infty }}\le b\cdot \left[ \int_{0}^{1}{{{\Phi }_{1}}(s)a(s)ds}+\int_{0}^{1}{{{\Phi }_{1}}(s)a(s)ds.} \right]$

即由$\|x_{k}(t)\|_{\infty}$有界可推出$\|x_{k}^{(n-1)}(t)\|_{\infty}$有界.

结合引理3.2, 存在常数$M_{2} ＞ 0$满足$\|x_{k}(t)\|_{E}\leq M_{2}.$与已知条件矛盾, 结论获证.

引理3.4 (见文献[17]) 设$X$是一个Banach空间且令$\{C_{n}|n=1, 2, \cdots\}$是$X$中的闭连通分支序列.假设

(ⅰ) 存在$z_{n}\in C_{n}, \, \, n=1, 2, \cdots$和$z^{\ast}\in X$, 使得$z_{n}\rightarrow z^{\ast}$;

(ⅱ) $r_{n}=\sup\{\|x\||x\in C_{n}\}=\infty;$

(ⅲ) 对所有$R＞0$, $\left(\cup_{n=1}^{\infty}C_{n}\right)\cap B_{R}$是$X$中的相对紧子集, 其中$ B_{R}=\{x\in X|\|x\|\leq R\}. $

则在$\mathbb{D}$中存在一个无界连通分支$C$使得$z^{\ast}\in C, $其中$\mathbb{D}:=\underset{n\to \infty }{\mathop{\lim \sup }}\,{{C}_{n}}=\{x\in X|\exists \{{{n}_{i}}\}\subset \mathbb{N}$和$x_{n_{i}}\in C_{n_{i}}$, 使得$x_{n_{i}}\rightarrow x\}$ (见文献[21]).

首先考虑下列特征值问题

$\left\{ \begin{align} &{{x}^{(n)}}(t)+\lambda ra(t)f(x)=0,0＜t＜1, \\ &x(0)=\int_{0}^{1}{x(t)d\alpha (t)},{x}'(0)=\cdots ={{x}^{(n-2)}}(0)=0,x(1)=\int_{0}^{1}{x}(t)d\beta (t), \\ \end{align} \right.$

(4.1)

其中$\lambda＞0$是一个参数.

设$\zeta\in C(\mathbb{R})$使得$f(x)=f_{0}x+\zeta(x)$且满足$\mathop {\lim }\limits_{|s| \to 0} \frac{{\zeta (s)}}{s} = 0.$

考虑

$\left\{ \begin{array}{l} {x^{(n)}}(t) + \lambda ra(t){f_0}x + \lambda ra(t)\zeta (x) = 0,0 ＜ t ＜ 1,\\ x(0) = \int_0^1 {x(t)d\alpha (t)} ,{\mkern 1mu} {\mkern 1mu} x'(0) = \cdots = {x^{(n - 2)}}(0) = 0,{\mkern 1mu} {\mkern 1mu} x(1) = \int_0^1 {x(t)d\beta (t)} \end{array} \right.$

(4.2)

作为从平凡解$x\equiv0$发出的一个分岐问题.方程(4.2) 等价于

$\begin{array}{l} x(t) = \int_0^1 {G(t,s)\left[ {\lambda ra(s){f_0}x(s) + \lambda ra(s)\zeta (x(s))} \right]ds} \\ \quad \quad : = \lambda {L^{ - 1}}\left[ {ra( \cdot ){f_0}x( \cdot )} \right](t) + \lambda {L^{ - 1}}\left[ {ra( \cdot )\zeta (x( \cdot ))} \right](t). \end{array}$

进而可以证明

$\|L^{-1}[a(\cdot)\zeta(x(\cdot))]\|_{E}=o(\|x\|_{E}), |x\|_{E}\rightarrow0\, .$

(4.3)

事实上, 对所有$(t, s)\in[0,1]\times[0,1], $由引理2.1-2.3可得

$K_{i}(t, s)=k_{i}(t, s)+\frac{g_{\alpha}(s)}{D}A_{i}(t)+\frac{g_{\beta}(s)}{D}B_{i}(t),$

其中

$\begin{array}{l} \frac{{{\partial ^{(i - 1)}}}}{{\partial {t^{(i - 1)}}}}K(t,s) = {K_i}(t,s)({K_i}(t,s)由{\rm{(2}}{\rm{.7)}} 式给出),\\ {A_i}(t) = \frac{{{d^{(i - 1)}}A}}{{{d^{(i - 1)}}t}} = \left\{ {\begin{array}{*{20}{l}} {A(t),}&{i = 1,}\\ { - (1 - \int_0^1 {d\beta (t))P_{n - 1}^{i - 1}{t^{n - i}}} ,}&{i = 2, \cdots ,n,} \end{array}} \right.\\ {B_i}(t) = \frac{{{d^{(i - 1)}}B}}{{{d^{(i - 1)}}t}} = \left\{ {\begin{array}{*{20}{l}} {B(t),}&{i = 1,}\\ {(1 - \int_0^1 {d\alpha (t))P_{n - 1}^{i - 1}{t^{n - i}}} ,}&{i = 2, \cdots ,n.} \end{array}} \right. \end{array}$

则对于任意$t, \, \, s\in[0,1]$, $i=1, \cdots, n$, 都有

$\begin{array}{l} |{K_i}(t,s)| \le {k_i}(t,s) + \frac{{|{A_i}(t)|}}{D}{g_\alpha }(s) + \frac{{|{B_i}(t)|}}{D}{g_\beta }(s),{\mkern 1mu} {\mkern 1mu} \\ \quad \quad \quad \quad \le {k_i}(t,s) + \frac{{{M_{i1}}}}{D}\int_0^1 {k(t,s)d\alpha (t)} + \frac{{{M_{i2}}}}{D}\int_0^1 {k(t,s)d\beta (t)} \\ \quad \quad \quad \quad \le {k_i}({\tau _i}(s),s) + \frac{{{M_{i1}}\int_0^1 {d\alpha (t) + {M_{i2}}} \int_0^1 {d\beta (t)} }}{D}k(\tau (s),s), \end{array}$

(4.4)

其中

$\begin{array}{l} {M_{i1}} = \left\{ {\begin{array}{*{20}{l}} {{M_1},}&{i = 1,}\\ {\left| {1 - \int_0^1 {d\beta (t)} } \right|P_{n - 1}^{i - 1},}&{i = 2, \cdots ,n,} \end{array}} \right.\\ {M_{i2}} = \left\{ {\begin{array}{*{20}{l}} {{M_2},}&{i = 1,}\\ {\left| {1 - \int_0^1 {d\alpha (t)} } \right|P_{n - 1}^{i - 1},}&{i = 2, \cdots ,n.} \end{array}} \right. \end{array}$

由(4.4) 式, $i=1, \cdots, n$, 可得

$\begin{array}{l} \left| {{{\left( {{L^{ - 1}}\left[ {a( \cdot )\zeta (x( \cdot ))} \right](t)} \right)}^{(i - 1)}}} \right| = \left| {\int_0^1 {\frac{{{\partial ^{i - 1}}}}{{\partial {t^{i - 1}}}}K(t,s)a(s)\zeta (x(s))ds} } \right|\\ \le \int_0^1 {\left[ {{k_i}({\tau _i}(s),s)a(s) + \frac{{{M_{i1}}\int_0^1 {\alpha (t)} + {M_{i2}}\int_0^1 {\beta (t)} }}{D}k(\tau (s),s)a(s)} \right]} \zeta (x(s))ds. \end{array}$

由$L^{-1}$的紧性结合(H3), $i=1, \cdots, n$, 可得

$\|(L^{-1}[a(\cdot)\zeta(x(\cdot))])^{(i-1)}\|_{\infty}=o(\|x\|_{\infty}), $

进而$i=1, \cdots, n$, $\|(L^{-1}[a(\cdot)\zeta(x(\cdot))])^{(i-1)}\|_{\infty}=o(\|x\|_{E}).$即(4.3) 式得证.

由引理3.1和全局分岐定理(可参考Dancer^[22]和Zeidler^[23]推论15.12), 对于问题(4.2), 可得如下结论.

引理4.1    令(H1)-(H5) 成立, $(\frac{\lambda_{1}}{rf_{0}}, 0)$是问题(4.2) 的一个分岐点.进而, 存在(4.2) 式正解的一个连通分支$\mathscr{C}$, 满足$\mathscr{C}(\subset[0, \infty)\times E)$, 并且$\mathscr{C}$在$[0, \infty)\times P$中连接$\left(\frac{\lambda_{1}}{rf_{0}}, 0\right)$和$\left(\frac{\lambda_{1}}{rf_{\infty}}, \infty\right)$.

注4.1    问题(4.1) 的形如$(1, x)$的任何解将产生问题$(1.1)$的一个解$x$.为了获得结论, 仅仅证明$\mathscr{C}$在$[0, \infty)\times P$中穿过超平面$\{1\}\times E$即可.

下面是本文主要结果.

定理4.1    令(H1)-(H5) 成立.要么$\lambda_{1}/ f_{\infty}＜r＜\lambda_{1}/ f_{0}$成立, 要么$\lambda_{1}/ f_{0}＜r＜\lambda_{1}/ f_{\infty}$成立.则问题(1.1) 至少有一个正解.

证    由引理4.1易得结论, 故证明略.

定理4.2    令(H1)-(H4) 和(H6) 成立.假设$r\in(0, \frac{\lambda_{1}}{f_{0}})$.则问题(1.1) 至少有一个正解.

证    受文献[24]的启发, 可以定义截断函数$f$如下

${f^{[n]}}(s): = \left\{ {\begin{array}{*{20}{l}} {f(s),}&{s \in [ - n,n],}\\ {\frac{{2{n^2} - f(n)}}{n}(s - n) + f(n),}&{s \in (n,2n),}\\ {\frac{{2{n^2} + f( - n)}}{n}(s + n) + f( - n),}&{s \in ( - 2n, - n),}\\ {ns,}&{s \in ( - \infty , - 2n] \cup [2n, + \infty ).} \end{array}} \right.$

考虑

$\left\{ \begin{array}{l} {x^{(n)}}(t) + \lambda ra(t){f^{[n]}}(x) = 0,0 ＜ t ＜ 1,\\ x(0) = \int_0^1 {x(t)d\alpha (t)} ,{\mkern 1mu} {\mkern 1mu} x'(0) = \cdots = {x^{(n - 2)}}(0) = 0,{\mkern 1mu} {\mkern 1mu} x(1) = \int_0^1 {x(t)d\beta (t)} . \end{array} \right.$

(4.5)

显然, 可得

$\mathop {\lim }\limits_{n \to + \infty } {f^{[n]}}(s) = f(s),{({f^{[n]}})_0} = {f_0},{({f^{[n]}})_\infty } = n.$

相似于定理4.1, 由引理4.1可知, 存在问题(4.5) 从$(\frac{\lambda_{1}}{rf_{0}}, 0)$发出的一个解的无界连通分支$\mathscr{C}^{[n]}$, 满足$\mathscr{C}^{[n]}\subset([0, \infty)\times E)$, 并且$\mathscr{C}^{[n]}$在$[0, \infty)\times P$中连接$(\frac{\lambda_{1}}{rf_{0}}, 0)$到$(\frac{\lambda_{1}}{nr}, \infty)$.

令$z_{n}=(\frac{\lambda_{1}}{nr}, \infty)$且$z^{\ast}=(0, \infty), $则$z_{n}\rightarrow z^{\ast}.$因此引理3.4 (ⅰ)满足且$z^{\ast}=(0, \infty).$显然

$r_{n}=\sup\left\{\lambda+\|x\||(\lambda, x)\in \mathscr{C}^{[n]}\right\}=\infty,$

相应的, 引理3.4 (ⅱ)成立.由Arezela-Ascoli定理和$f^{[n]}$直接可得引理3.4 (ⅲ).因此由引理3.4可知$\mathop {\lim \sup }\limits_{n \to \infty } {{\mathscr {C}}^{[n]}}$包括一个无界连通分支$\mathscr{C}$满足$(\infty, 0)\in \mathscr{C}$并且$(\frac{\lambda_{1}}{rf_{0}}, 0)\in \mathscr{C}$.

定理4.3    令(H1)-(H4) 和(H7) 成立.假设$r\in(0, +\infty)$.则问题(1.1) 至少有一个正解.

证    定义

${f^{[n]}}(s): = \left\{ {\begin{array}{*{20}{l}} {f(s),}&{s \in [ - \infty , - \frac{2}{n}] \cup [\frac{2}{n},\infty ],}\\ { - \left[ {f( - \frac{2}{n}) + \frac{1}{{{n^2}}}} \right](ns + 2) + f( - \frac{2}{n}),}&{s \in ( - \frac{2}{n}, - \frac{1}{n}),}\\ {\left[ {f(\frac{2}{n}) - \frac{1}{{{n^2}}}} \right](ns - 2) + f(\frac{2}{n}),}&{s \in (\frac{1}{n},\frac{2}{n}),}\\ {\frac{1}{n}s,}&{s \in \left[ { - \frac{1}{n},\frac{1}{n}} \right].} \end{array}} \right.$

考虑问题

$\left\{ \begin{array}{l} {x^{(n)}}(t) + \lambda ra(t){f^{[n]}}(x) = 0,0 ＜ t ＜ 1,\\ x(0) = \int_0^1 {x(t)d\alpha (t)} ,{\mkern 1mu} {\mkern 1mu} x'(0) = \cdots = {x^{(n - 2)}}(0) = 0,{\mkern 1mu} {\mkern 1mu} x(1) = \int_0^1 {x(t)d\beta (t)} . \end{array} \right.$

(4.6)

显然, 可得

$\mathop {\lim }\limits_{n \to + \infty } {f^{[n]}}(s) = f(s),{({f^{[n]}})_0} = \frac{1}{n},{({f^{[n]}})_\infty } = {f_\infty } = \infty .$

相似于定理4.2的证明方法, 由引理4.1可知, 问题(4.6) 存在从$(\frac{n\lambda_{1}}{r}, 0)$发出的一个解的无界连通分支$\mathscr{C}^{[n]}$, 满足$\mathscr{C}^{[n]}\subset([0, \infty)\times E)$, 并且$\mathscr{C}^{[n]}$在$[0, \infty)\times P$中连接$(\frac{n\lambda_{1}}{r}, 0)$到$(0, \infty)$.

令$z_{n}=(\frac{n\lambda_{1}}{r}, 0)$且$z^{\ast}=(\infty, 0), $则$z_{n}\rightarrow z^{\ast}.$因此引理3.4 (ⅰ)满足且$z^{\ast}=(\infty, 0).$显然

$\begin{equation} r_{n}=\sup\left\{\lambda+\|x\||(\lambda, x)\in \mathscr{C}^{[n]}\right\}=\infty. \end{equation}$

相应的, 引理3.4 (ⅱ)成立.由Arezela-Ascoli定理和$f^{[n]}$直接可得引理3.4 (ⅲ).因此由引理3.4可知$\limsup\limits_{n\rightarrow\infty}\mathscr{C}^{[n]}$包括一个无界连通分支$\mathscr{C}$满足$(\infty, 0)\in \mathscr{C}$并且$(0, \infty)\in \mathscr{C}$.

定理4.4    令(H1)-(H4) 和(H8) 成立.存在一个$\lambda^{+}＞0$使得$r\in(0, \lambda^{+})$成立.则问题(1.1) 至少有一个正解.

证    定义

${f^{[n]}}(s): = \left\{ {\begin{array}{*{20}{l}} {f(s),}&{s \in \left[ { - \infty , - \frac{2}{n}} \right] \cup \left[ {\frac{2}{n},\infty } \right],}\\ { - \left[ {f( - \frac{2}{n}) + 1} \right](ns + 2) + f\left( { - \frac{2}{n}} \right),}&{s \in \left( { - \frac{2}{n}, - \frac{1}{n}} \right),}\\ {\left[ {f\left( {\frac{2}{n}} \right) - 1} \right](ns - 2) + f\left( {\frac{2}{n}} \right),}&{s \in \left( {\frac{1}{n},\frac{2}{n}} \right),}\\ {ns,}&{s \in \left[ { - \frac{1}{n},\frac{1}{n}} \right].} \end{array}} \right.$

考虑问题

$\left\{ \begin{array}{l} {x^{(n)}}(t) + \lambda ra(t){f^{[n]}}(x) = 0,0 ＜ t ＜ 1,\\ x(0) = \int_0^1 {x(t)d\alpha (t)} ,{\mkern 1mu} {\mkern 1mu} x'(0) = \cdots = {x^{(n - 2)}}(0) = 0,{\mkern 1mu} {\mkern 1mu} x(1) = \int_0^1 {x(t)d\beta (t)} . \end{array} \right.$

(4.7)

显然,

$\mathop {\lim }\limits_{n \to + \infty } {f^{[n]}}(s) = f(s),{({f^{[n]}})_0} = n,{({f^{[n]}})_\infty } = {f_\infty } = \infty .$

相似于定理4.2的证明方法, 由引理4.1可知, 问题(4.7) 存在从$(\frac{\lambda_{1}}{nr},0)$发出的一个解的无界连通分支$\mathscr{C}^{[n]}$, 满足$\mathscr{C}^{[n]}\subset([0,\infty)\times E)$, 并且$\mathscr{C}^{[n]}$在$[0,\infty)\times P$中连接$(\frac{\lambda_{1}}{nr},0)$到$(\infty,\infty)$.

令$z_{n}=(\frac{\lambda_{1}}{nr},0)$且$z^{\ast}=(0,0),$则$z_{n}\rightarrow z^{\ast}.$因此引理3.4(ⅰ)满足且$z^{\ast}=(0,0).$显然

$\begin{equation} r_{n}=\sup\left\{\lambda+\|x\||(\lambda,x)\in \mathscr{C}^{[n]}\right\}=\infty,\nonumber \end{equation}$

相应的, 引理3.4 (ⅱ)成立.由Arezela-Ascoli定理和$f^{[n]}$直接可得引理3.4 (ⅲ).因此由引理3.4可知$\limsup\limits_{n\rightarrow\infty}\mathscr{C}^{[n]}$包括一个无界连通分支$\mathscr{C}$满足$(0,0)\in \mathscr{C}$并且$(0,\infty)\in \mathscr{C}$.