分数阶模型作为一种重要的数学模型, 在许多领域中都得到了广泛的应用, 如空气动力学、流体力学、黏弹性力学、生物数学等, 参见文献[1-4]. 例如文献[1]中提到眼球瞬间运动的神经控制过程, 认为前庭视觉反射效应是分数阶的, 运动神经控制是整数阶的, 其模型为

$ \tau_{1} D r(t)+r(t) = \tau_{1} \tau_{2} D^{\nu+1} v(t)+\tau_{1} D^{\nu} v(t), $

其中$ D^\nu, D^{\nu+1} $是Riemann-Liouville型分数阶导数. 然而我们注意到大多数的研究成果仅限于分数阶微分方程, 对分数阶差分方程却鲜有问津, 正如郑祖庥教授在文献[1]中所阐述的那样: “对分数微分方程来说, 离散化或者问题提出时便是离散的差分方程是不可避免的, 迄今只作为近似解计算的出发点, 没有对差分分数方程的专门研究”. 值得喜悦的是, 近些年来已有诸多学者致力于这方面的研究, 参见专著[5-6]和文献[7-20]及其所附参考文献. 例如在专著[5]中程金发教授系统地介绍了分数阶差分、分数阶和分的概念和性质, 给出了类似于微积分中的莱布尼兹公式、分数阶差分及和分的$ Z $变换公式, 讨论了分数阶差分方程解的存在唯一性和连续依赖性, 归纳了求分数阶差分方程显示解方法, 并研究了分数阶差分方程边值问题的格林函数及其性质.

另外我们注意到美国学者Goodrich在过去几年里对分数阶差分方程边值问题进行了比较深入详细的研究, 其研究成果可参见专著[6]. 例如在文献[7]中Goodrich运用锥拉伸与压缩不动点定理研究了如下分数阶差分方程边值问题正解的存在性

$ \left\{\begin{array}{ll} \Delta^{\nu} y(t) = \lambda f(t+\nu-1, y(t+\nu-1)), {\quad} t \in[0, T]_{Z}, \\ { } y(\nu-1) = y(\nu+T)+\sum\limits_{i = 1}^{N} F\big(t_{i}, y(t_{i})\big), \end{array}\right. $

其中非线性项$ f $下方有界, 且关于未知函数$ y $在无穷远处次线性增长. 在文献[8]中Goodrich及其合作者运用锥上的不动点指数理论研究了如下分数阶差分方程组边值问题正解的存在性

$ \left\{\begin{array}{l} {\Delta^{\nu} x(t) = f_{1}(t+\nu-1, x(t+\nu-1), y(t+\nu-1)), t \in[0, T]_{{\Bbb Z}}}, \\ {\Delta^{\nu} y(t) = f_{2}(t+\nu-1, x(t+\nu-1), y(t+\nu-1)), t \in[0, T]_{{\Bbb Z}}}, \\ {x(\nu-1) = x(\nu+T), y(\nu-1) = y(\nu+T)}, \end{array}\right. $

其中非线性项$ f_i(i = 1, 2) $下方有界, 并采用合适的凹凸函数刻画它们之间的耦合行为.

另一方面我们注意到$ p $-Laplace方程边值问题来源于非牛顿流体力学、冰川学、燃烧理论、多孔介质的气体湍流、弹性理论、血浆问题、宇宙物理等应用学科, 然而具有$ p $-Laplacian算子的分数阶差分方程边值问题却很少出现在研究文献里, 参见文献[15-20]及其所附参考文献. 在文献[15]中王金华等人运用锥拉伸与压缩不动点定理研究了如下$ p $-Laplacian分数阶差分边值问题解的存在性

$ \left\{\begin{array}{l} {\Delta^{\beta}\left[\phi_{p}\big(\Delta^{\alpha} y\big)\right](t)+f(t+\alpha+\beta-1, y(t+\alpha+\beta-1)) = 0, t \in[0, b+1]_{\mathrm{N}_{0}}}, \\ {\Delta^{\alpha} y(\beta-1) = 0}, \\ {y(\alpha+\beta-3) = \Delta^{\gamma} y(\alpha+\beta+b+1-\gamma) = 0}, \end{array}\right. $

其中非线性项$ f $满足Lipschitz条件. 在文献[16]中作者运用Banach压缩原理和Brouwer不动点定理研究如下$ p $-Laplacian分数阶差分边值问题解的存在唯一性

$ \left\{\begin{array}{l} {\Delta^{\beta}\left[\phi_{p}\big(\Delta^{\alpha} y\big)\right](t)+f(\alpha+\beta+t-1, y(\alpha+\beta+t-1)) = 0, t \in[0, b]_{{\Bbb N}_{0}}}, \\ {\Delta^{\alpha} y(\beta-2) = \Delta^{\alpha} y(\beta+b) = 0} , \\ {y(\alpha+\beta-4) = y(\alpha+\beta+b) = 0}, \end{array}\right. $

其中非线性项$ f $满足Lipschitz条件和有界性条件.

受上述文献的启发, 本文运用锥上的不动点指数理论研究如下的$ p $-Laplacian分数阶差分边值问题正解的存在性

(1.1) $ \begin{equation} \left\{\begin{array}{l} {\Delta_{\nu-1}^{\nu}\big(\phi_{p}\big(\Delta_{\nu-1}^{\nu} y(t)\big)\big) = f(t+\nu-1, y(t+\nu-1)), t \in[0, T]_{{\Bbb Z}}}, \\[5pt] {y(\nu-1) = y(\nu+T), \Delta_{\nu-1}^{\nu} y(\nu-1) = \Delta_{\nu-1}^{\nu} y(\nu+T)}, \end{array}\right. \end{equation} $

其中$ \nu \in(0, 1) $是一实数, $ \Delta_{\nu-1}^{\nu} $是离散的分数阶差分, $ \phi_{p}(s) = |s|^{p-2} s $是$ p $-Laplacian算子, $ p>1, s\in {{\Bbb R}} $, 非线性项$ f $满足条件

(H0) $ f\in C([0, T]_{{\Bbb Z}}\times{{\Bbb R}} ^+, {{\Bbb R}} ^+) $, 其中$ {{\Bbb R}} ^+ = [0, +\infty) $, $ [0, T]_{{\Bbb Z}} = [0, T]\cap \mathbb Z $.

借助离散型Jensen不等式考虑问题(1.1)与相应的不具有$ p $-Laplacian算子的分数阶差分方程边值问题之间的关系, 在非线性项超(次)$ p-1 $次增长条件下获得问题(1.1)正解的存在性. 并且我们的方法也适合于不具有$ p $-Laplacian算子的情形($ p = 2 $).

以下我们先给出分数阶差分的定义.

定义 1.1^[5-6]  定义$ t^{\underline{\nu}}: = \frac{\Gamma(t+1)}{\Gamma(t+1-\nu)} $, 对任意$ t $和$ \nu $右边有定义. 如果$ t + 1 -\nu $是$ \rm{Gamma} $函数的极点, 而$ t + 1 $不是$ \rm{Gamma} $函数的极点, 则$ t^{\underline{\nu}} = 0 $.

定义 1.2^[5-6]  对$ \nu > 0 $, 函数$ f $的$ \nu $阶和分定义为

$ \Delta_{a}^{-\nu} f(t) = \frac{1}{\Gamma(\nu)} \sum\limits_{s = a}^{t-\nu}(t-s-1)^{\underline{\nu-1}} f(s), t \in {\Bbb N}_{a+N-\nu}, $

其中$ {\Bbb N}_{a+N-\nu} = \{a+N-\nu, a+N-\nu+1, \cdots\} $.

对$ \nu > 0 $, $ f $的$ \nu $阶分数阶差分定义为

$ \Delta_{a}^{\nu} f(t) = \Delta^{N} \Delta_{a}^{\nu-N} f(t), t \in {\Bbb N}_{a+N-\nu}, $

其中$ N \in {\Bbb N} $: $ 0 \leq N-1<\nu \leq N $.

令$ \phi_{q} = \phi_{p}^{-1} $, 其中$ p, q $是共轭数, 即$ \frac{1}{q}+ \frac{1}{p} = 1 $. 根据文献[17, 引理2.1], 我们可以得到与问题(1.1)等价的和方程

(1.2) $ \begin{eqnarray} y(t)& = &\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \phi_{q}\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(r+\nu-1, y(r+\nu-1))\bigg), {}\\ &&{\qquad}{\qquad}{\qquad} t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}, \end{eqnarray} $

其中

(1.3) $ \begin{equation} G(t, s) = \left\{\begin{array}{ll} { }\frac{(\nu+T-s-1)^{\underline{\nu-1}} t^{\underline{\nu-1}}}{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}}+(t-s-1)^{\underline {\nu-1}}, &0 \leq s \leq t-\nu \leq T, \\ { }\frac{(\nu+T-s-1)^{\underline{\nu-1}} t^{\underline{\nu-1}}}{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}}, &t-\nu<s \leq T. \end{array}\right. \end{equation} $

引理 1.1^{[7-8, 17]}  令$ C^{*} = 1+\frac{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}}{(\nu+T-1)^{\underline{\nu-1}}} $, 则对任意的$ (t, s)\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}\times [0, T]_{{\Bbb Z}} $, 我们有

(1.4) $ \begin{eqnarray} 0&<&\frac{(\nu+T)^{\underline{\nu-1}}}{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}}(\nu+T-s-1)^{\underline{\nu-1}} {}\\ &\leq& G(t, s) \leq \frac{C^{*} \Gamma(\nu)}{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}}(\nu+T-s-1) ^{\underline{\nu-1}}. \end{eqnarray} $

引理 1.2^[8]  令$ \varphi(t+\nu-1) = (\nu+T-t-1)^{\underline{\nu-1}}, t \in[0, T]_{{\Bbb Z}} $, $ \varphi^*(t) = (2\nu+T-t-2)^{\underline{\nu-1}}, $$ t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $, 则对任意的$ [0, T]_{{\Bbb Z}} $, 我们有

(1.5) $ \begin{equation} \sum\limits_{t = \nu-1}^{\nu+T-1} \frac{G(t, s)}{\Gamma(\nu)} \varphi^{*}(t) = \sum\limits_{t = 0}^{T} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} \varphi(t+\nu-1) \end{equation} $

和

(1.6) $ \begin{equation} \kappa_1 \varphi(s+\nu-1) \leq \sum\limits_{t = \nu-1}^{\nu+T-1} \frac{G(t, s)}{\Gamma(\nu)} \varphi^{*}(t)\le \kappa_2 \varphi(s+\nu-1), \end{equation} $

其中

(1.7) $ \begin{equation} \begin{array}{ll} { }\kappa_{1} = \sum\limits_{t = 0}^{T} \frac{(\nu+T)^{\underline{\nu-1}}}{\Gamma(\nu)\big(\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}\big)} \varphi(t+\nu-1), \\ [4mm] { }\kappa_{2} = \sum\limits_{t = 0}^{T} \frac{C^{*}}{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}} \varphi(t+\nu-1). \end{array} \end{equation} $

令$ E $是从$ [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $到$ {{\Bbb R}} $全体映射构成的集合, 并在其上赋予最大值范数, 即

$ { } \|y\| = \max\limits_{t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}}|y(t)|, y\in E. $

则$ E $是一Banach空间. 定义

$ P = \{y\in E: y(t)\ge 0, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}\}, $

则$ P $是$ E $上的锥. 根据(1.2)式, 我们可以在$ E $中定义如下的算子

$ \begin{eqnarray*} (Ay)(t)& = &\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \phi_{q}\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(r+\nu-1, y(r+\nu-1))\bigg), \\ &&{\qquad}{\qquad}{\qquad} \forall y\in E, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}, \end{eqnarray*} $

并且我们注意到如果存在$ \overline{y}\in P\backslash \{0\} $使得$ A\overline{y} = \overline{y} $, 则$ \overline{y} $是方程(1.1)的正解. 从而以下我们转而寻求算子$ A $不动点的存在性.

引理 1.3  令$ P_0 = \{y \in E: y(t) \geq \frac{(\nu+T)^{\underline{\nu-1}}}{C^{*} \Gamma(\nu)}\|y\|, t \in[\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}\} $, 则$ A(P)\subset P_0 $.

证  为方便书写, 令$ f(r+\nu-1, y(r+\nu-1)) = f(\cdot, y(\cdot)) $, $ y\in P $. 根据引理, 若$ y\in P $,

$ \begin{eqnarray*} (Ay)(t)& = &\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \phi_{q}\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(\cdot, y(\cdot))\bigg)\\ & \le& \sum\limits_{s = 0}^{T} \frac{C^{*}\varphi(s+\nu-1) }{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}} \phi_{q}\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(\cdot, y(\cdot))\bigg). \end{eqnarray*} $

从而$ y\in P $, 我们可得

$ \begin{eqnarray*} (Ay)(t)& \ge& \sum\limits_{s = 0}^{T}\frac{(\nu+T)^{\underline{\nu-1}}\varphi(s+\nu-1)}{\Gamma(\nu)(\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}})} \phi_{q}\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(\cdot, y(\cdot))\bigg)\\ & = &\frac{(\nu+T)^{\underline{\nu-1}}}{C^{*} \Gamma(\nu)} \sum\limits_{s = 0}^{T} \frac{C^{*}\varphi(s+\nu-1) }{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}} \phi_{q}\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(\cdot, y(\cdot))\bigg)\\ & \ge& \frac{(\nu+T)^{\underline{\nu-1}}}{C^{*} \Gamma(\nu)} \|Ay\|. \end{eqnarray*} $

证毕.

引理 1.4^[21]  设$ E $是实$ \rm Banach $空间, $ P $是$ E $上的锥, $ \Omega $是$ E $中的有界开集, $ A:\overline{\Omega}\cap P\to P $全连续. 若存在$ v_0\in P\backslash\{0\} $使得$ v\not = Av+\lambda v_0, \forall v\in \partial \Omega \cap P, \lambda\ge 0, $则$ i(A, \Omega\cap P, P) = 0, $其中$ i $为锥$ P $上的不动点指数.

引理 1.5^[21]  设$ E $是实$ \rm Banach $空间, $ P $是$ E $上的锥, $ \Omega $是$ E $中的有界开集, $ 0\in \Omega $. $ A:\overline{\Omega}\cap P\to P $全连续. 若$ v\not = \lambda Av, \forall v\in \partial \Omega \cap P, \lambda\in [0, 1], $则$ i(A, \Omega\cap P, P) = 1 $.

引理 1.6^[22]  令$ \theta>0, n \geq 1, a_{i} \geq 0\ (i = 1, 2, \cdots , n) $, 则以下离散型Jensen不等式成立

(1.8) $ \begin{equation} \bigg(\sum\limits_{i = 1}^{n} a_{i}\bigg)^{\theta} \leq n^{\theta-1} \sum\limits_{i = 1}^{n} a_{i}^{\theta}, \quad \forall \theta \geq 1 \end{equation} $

和

(1.9) $ \begin{equation} \bigg(\sum\limits_{i = 1}^{n} a_{i}\bigg)^{\theta} \geq n^{\theta-1} \sum\limits_{i = 1}^{n} a_{i}^{\theta}, \quad \forall 0<\theta \leq 1. \end{equation} $

本节我们首先给出关于非线性项$ f $的增长性条件

(H1) 存在$ a_1>\Big(\kappa_2^{\frac{pp_*}{p-1}-2}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} \Big)^{\frac{1-p}{p_*}}, c_1>0 $使得

$ f(t, y)\ge a_1y^{p-1}-c_1, y\in {{\Bbb R}} ^+, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}. $

(H2) 存在$ a_2\in \Big(0, \kappa_2^{-p}(T+1)^{\frac{2p-pp^*-2}{p^*}} \Big), r_1>0 $使得

$ f(t, y)\le a_2y^{p-1}, y\in [0, r_1], t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}. $

(H3) 存在$ a_3>\Big(\kappa_2^{\frac{pp_*}{p-1}-2}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} \Big)^{\frac{1-p}{p_*}}, r_2>0 $使得

$ f(t, y)\ge a_3y^{p-1}, y\in [0, r_2], t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}. $

(H4) 存在$ a_4\in \Big(0, \kappa_2^{-p}2^{\frac{p-p^*-1}{p^*}}(T+1)^{\frac{2p-pp^*-2}{p^*}} \Big), c_2>0 $使得

$ f(t, y)\le a_4y^{p-1}+c_2, y\in {{\Bbb R}} ^+, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}. $

(H5) 存在$ \rho_1>r_2 $和正数$ \delta_1<\left\{\begin{array}{ll}\kappa_2^{-p} (T+1)^{2-p}, p\ge 2, \\ \kappa_2^{-p} (T+1)^{p-2}, p\in (1, 2), \end{array}\right. $使得

$ f(t, y)\le \delta_1 \rho_1^{p-1}, y\in [0, \rho_1], t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}. $

令$ p^* = \max\{1, p-1\} $, $ p_* = \min\{1, p-1\} $, $ B_r = \{v\in E: \|v\|< r\}, r>0 $. 以下给出本文的主要结论及其证明.

定理 2.1  若条件(H0)–(H2)成立, 则方程(1.1)至少有一个正解.

证  定义集合$ S = \{y\in P: y = Ay+\lambda y_0, \lambda\ge 0\} $, 其中$ y_0 $是集合$ P_0 $中某一给定的元素. 下证集合$ S $在$ P $中有界. 若$ y\in P $, 则

$ y(t) = (Ay)(t)+\lambda y_0(t)\ge (Ay)(t), t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}. $

从而根据引理1.6可得

(2.1) $ \begin{eqnarray} y^{p_*}(t)&\ge & [(Ay)(t)]^{p_*}{}\\ & = & \left[ \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \phi_{q}\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(r+\nu-1, y(r+\nu-1))\bigg)\right]^{p_*}{}\\ & = & \left[ \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\kappa_2\Gamma(\nu)} \kappa_2\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\kappa_2\Gamma(\nu)}\kappa_2 f(r+\nu-1, y(r+\nu-1))\bigg)^{\frac{1}{p-1}}\right]^{p_*} {}\\ & \ge&\kappa_2^{\frac{pp_*}{p-1}}(T+1)^{p_*-1}\sum\limits_{s = 0}^{T} \bigg( \frac{G(t, s)}{\kappa_2\Gamma(\nu)}\bigg)^{p_*} {}\\ &&\times\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\kappa_2\Gamma(\nu)} f(r+\nu-1, y(r+\nu-1))\bigg)^{\frac{p_*}{p-1}}{}\\ & \ge &\kappa_2^{\frac{pp_*}{p-1}}(T+1)^{\frac{pp_*}{p-1}-2} {}\\ &&\times\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\kappa_2\Gamma(\nu)}\sum\limits_{r = 0}^{T} \bigg( \frac{G(s+\nu-1, r)}{\kappa_2\Gamma(\nu)}\bigg)^{\frac{p_*}{p-1}} f^{\frac{p_*}{p-1}}(r+\nu-1, y(r+\nu-1)){}\\ & \ge& \kappa_2^{\frac{pp_*}{p-1}-2}(T+1)^{\frac{pp_*}{p-1}-2} {}\\ &&\times\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)}\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f^{\frac{p_*}{p-1}}(r+\nu-1, y(r+\nu-1)). \end{eqnarray} $

注意到条件(H1), 我们可以得到

$ a_1^{\frac{p_*}{p-1}} y^{p_*}\le (f(t, y)+ c_1)^{\frac{p_*}{p-1}}\le f^{\frac{p_*}{p-1}}(t, y)+c^{\frac{p_*}{p-1}}_1, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}. $

将此不等式代入(2.1)式可得

$ y^{p_*}(t) \ge \kappa_2^{\frac{pp_*}{p-1}-2}(T+1)^{\frac{pp_*}{p-1}-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)}\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} \left[a_1^{\frac{p_*}{p-1}} y^{p_*}(r+\nu-1)-c^{\frac{p_*}{p-1}}_1\right]. $

在上式两端乘以$ \varphi^* $, 并在区间$ [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $上求和可得

$ \begin{eqnarray*} \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p_*}(t) \varphi^*(t) & \ge& \kappa_2^{\frac{pp_*}{p-1}-2}(T+1)^{\frac{pp_*}{p-1}-2} \sum\limits_{t = \nu-1}^{\nu+T-1} \varphi^*(t){\nonumber}\\ &&\times\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)}\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} \left[a_1^{\frac{p_*}{p-1}} y^{p_*}(r+\nu-1)-c^{\frac{p_*}{p-1}}_1\right]\\ & \ge& \kappa_2^{\frac{pp_*}{p-1}-2}\kappa_1 (T+1)^{\frac{pp_*}{p-1}-2} \sum\limits_{s = 0}^{T} \varphi(s+\nu-1) {\nonumber}\\ &&\times\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} \left[a_1^{\frac{p_*}{p-1}} y^{p_*}(r+\nu-1)-c^{\frac{p_*}{p-1}}_1\right]\\ & \ge &\kappa_2^{\frac{pp_*}{p-1}-2}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} {\nonumber}\\ &&\times \sum\limits_{r = 0}^{T} \varphi(r+\nu-1)\left[a_1^{\frac{p_*}{p-1}} y^{p_*}(r+\nu-1)-c^{\frac{p_*}{p-1}}_1\right]\\ & = &a_1^{\frac{p_*}{p-1}} \kappa_2^{\frac{pp_*}{p-1}-2}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} \sum\limits_{r = \nu-1}^{\nu+T-1} y^{p_*}(r) \varphi(r)\\ &&-c^{\frac{p_*}{p-1}}_1 \kappa_2^{\frac{pp_*}{p-1}-1}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} \frac{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}} }{C^{*}}. \end{eqnarray*} $

解上述的不等式我们有

(2.2) $ \begin{equation} \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p_*}(t) \varphi^*(t)\le \frac{c^{\frac{p_*}{p-1}}_1 \kappa_2^{\frac{pp_*}{p-1}-1}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} \frac{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}} }{C^{*}}}{a_1^{\frac{p_*}{p-1}} \kappa_2^{\frac{pp_*}{p-1}-2}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} -1}. \end{equation} $

注意到$ y\in P $, $ y_0\in P_0 $, 从而由引理1.3知$ y\in P_0 $. 则由(2.2)式我们可知

$ \bigg(\frac{(\nu+T)^{\underline{\nu-1}}}{C^{*} \Gamma(\nu)}\bigg)^{p_*}\|y\|^{p_*} \sum\limits_{t = \nu-1}^{\nu+T-1} \varphi^*(t)\le \frac{c^{\frac{p_*}{p-1}}_1 \kappa_2^{\frac{pp_*}{p-1}-1}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} \frac{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}} }{C^{*}}}{a_1^{\frac{p_*}{p-1}} \kappa_2^{\frac{pp_*}{p-1}-2}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} -1}, $

即

$ \|y\| \le \sqrt[p_*]{ \frac{c^{\frac{p_*}{p-1}}_1 \kappa_2^{\frac{pp_*}{p-1}-2}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} }{a_1^{\frac{p_*}{p-1}} \kappa_2^{\frac{pp_*}{p-1}-2}\kappa^2_1 (T+1)^{\frac{pp_*}{p-1}-2} -1}}\cdot\frac{C^{*} \Gamma(\nu)}{(\nu+T)^{\underline{\nu-1}}}: = M_1. $

这就证明了$ S $的有界性. 从而取$ R_1>\max\{M_1, r_1\} $ ($ r_1 $见条件(H2)), 则当$ y\in \partial B_{R_1}\cap P, \lambda\ge 0 $时, 我们可得$ y\not = Ay+\lambda y_0 $, 从而根据不动点指数的缺方向性(引理1.4)知

(2.3) $ \begin{equation} i\big(A, B_{R_1} \cap P, P\big) = 0. \end{equation} $

另一方面证明

(2.4) $ \begin{equation} y\not = \lambda Ay, y\in \partial B_{r_1}\cap P, \lambda\in [0, 1], \end{equation} $

其中$ r_1 $见条件(H2). 事实上, 若(2.4)式不成立, 则存在$ y\in \partial B_{r_1}\cap P, \lambda\in [0, 1] $使得$ y = \lambda Ay $, 这表明

(2.5) $ \begin{eqnarray} y^{p^*}(t)& = &[\lambda (Ay)(t)]^{p^*}\le [(Ay)(t)]^{p^*}{}\\ & = & \left[ \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\kappa_2\Gamma(\nu)} \kappa_2\bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\kappa_2\Gamma(\nu)}\kappa_2 f(r+\nu-1, y(r+\nu-1))\bigg)^{\frac{1}{p-1}}\right]^{p^*}{}\\ & \le& \kappa_2^{\frac{pp^*}{p-1}} (T+1)^{p^*-1} \sum\limits_{s = 0}^{T} \bigg( \frac{G(t, s)}{\kappa_2\Gamma(\nu)} \bigg)^{p^*} {}\\ &&\times \bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\kappa_2\Gamma(\nu)} f(r+\nu-1, y(r+\nu-1))\bigg)^{\frac{p^*}{p-1}}{}\\ & \le& \kappa_2^{\frac{pp^*}{p-1}} (T+1)^{\frac{pp^*}{p-1}-2} {}\\ &&\times\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\kappa_2\Gamma(\nu)} \sum\limits_{r = 0}^{T} \bigg(\frac{G(s+\nu-1, r)}{\kappa_2\Gamma(\nu)}\bigg)^{\frac{p^*}{p-1}} f^{\frac{p^*}{p-1}}(r+\nu-1, y(r+\nu-1)){}\\ & \le &\kappa_2^{\frac{pp^*}{p-1}-2} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f^{\frac{p^*}{p-1}}(r+\nu-1, y(r+\nu-1)).{}\\ \end{eqnarray} $

将(H2)代入(2.5)式, 我们可得到

$ y^{p^*}(t)\le a^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}-2} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} y^{p^*}(r+\nu-1). $

在上式两端乘以$ \varphi^* $, 并在区间$ [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $上求和可得

$ \begin{eqnarray*} && \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p^*}(t) \varphi^*(t)\\ & \le &a^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}-2} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{t = \nu-1}^{\nu+T-1} \varphi^*(t) \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} y^{p^*}(r+\nu-1)\\ & \le& a^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{s = 0}^{T} \varphi(s+\nu-1) \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} y^{p^*}(r+\nu-1)\\ & \le & a^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{r = 0}^{T} y^{p^*}(r+\nu-1)\varphi(r+\nu-1). \end{eqnarray*} $

注意到条件(H2)中$ a_2 $的取值范围可知$ a^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}} (T+1)^{\frac{pp^*}{p-1}-2}<1 $, 从而解上述不等式得

$ \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p^*}(t) \varphi^*(t) = 0, $

又注意到$ \varphi^* $在区间$ [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $上不恒为0, 从而$ y^{p^*}(t)\equiv 0, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $, 这与$ y\in \partial B_{r_1}\cap P $矛盾. 矛盾表明(2.4)式成立, 从而根据不动点指数的同伦不变性(引理1.5)可得

(2.6) $ \begin{equation} i\big(A, B_{r_1} \cap P, P\big) = 1. \end{equation} $

根据(2.3)和(2.6)式, 我们可推出

$ i(A, (B_{R_1} \backslash \overline{B}_{r_1}) \cap P, P) = i(A, B_{R_1} \cap P, P)-i(A, B_{r_1} \cap P, P) = -1. $

从而算子$ A $在$ (B_{R_1} \backslash \overline{B}_{r_1}) \cap P $内至少有一个不动点, 这也表明(1.1)在$ (B_{R_1} \backslash \overline{B}_{r_1}) \cap P $内至少有一个解, 并且该解也是正解.

定理 2.2  若条件(H0), (H3)–(H4)成立, 则方程(1.1)至少有一个正解.

证  首先证明

(2.7) $ \begin{equation} y\not = Ay+\lambda \overline{y}_0, y\in \partial B_{r_2}\cap P, \lambda \ge 0, \end{equation} $

其中$ \overline{y}_0 $是锥$ P $中某一固定元素, $ r_2 $见条件(H3). 事实上, 若(2.7)式不成立, 则存在$ y\in \partial B_{r_2}\cap P, \lambda \ge 0 $使得$ y = Ay+\lambda \overline{y}_0 $, 结合条件(H3)和(2.1)式我们有

(2.8) $ \begin{eqnarray} y^{p_*}(t) & \ge& \kappa_2^{\frac{pp_*}{p-1}-2}(T+1)^{\frac{pp_*}{p-1}-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)}\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f^{\frac{p_*}{p-1}}(r+\nu-1, y(r+\nu-1)){}\\ & \ge &a^{\frac{p_*}{p-1}}_3 \kappa_2^{\frac{pp_*}{p-1}-2}(T+1)^{\frac{pp_*}{p-1}-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)}\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} y^{p_*}(r+\nu-1). \end{eqnarray} $

在上式两端乘以$ \varphi^* $, 并在区间$ [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $上求和可得

$ \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p_*}(t) \varphi^*(t) \ge a^{\frac{p_*}{p-1}}_3 \kappa_2^{\frac{pp_*}{p-1}-2} \kappa_1^2 (T+1)^{\frac{pp_*}{p-1}-2} \sum\limits_{r = 0}^{T} y^{p_*}(r+\nu-1)\varphi(r+\nu-1). $

注意到条件(H3)中$ a_3 $的取值范围, 我们很容易推出

$ \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p_*}(t) \varphi^*(t) = 0, $

又注意到$ \varphi^* $在区间$ [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $上不恒为0, 从而$ y^{p_*}(t)\equiv 0, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $, 这与$ y\in \partial B_{r_2}\cap P $矛盾. 矛盾表明(2.7)式成立, 从而根据不动点指数的同伦不变性(引理1.5)可得

(2.9) $ \begin{equation} i\big(A, B_{r_2} \cap P, P\big) = 0. \end{equation} $

另一方面定义集合$ W = \{y\in P: y = \lambda Ay, \lambda\in [0, 1]\} $, 下证$ W $在$ P $中有界. 若$ y\in W $, 则$ y(t)\le \lambda (Ay)(t)\le (Ay)(t), t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $. 结合条件(H4)和(2.5)式我们有

(2.10) $ \begin{eqnarray} y^{p^*}(t)& \le &\kappa_2^{\frac{pp^*}{p-1}-2} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f^{\frac{p^*}{p-1}}(r+\nu-1, y(r+\nu-1)){}\\ & \le& \kappa_2^{\frac{pp^*}{p-1}-2} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)}(a_4y^{p-1}(r+\nu-1)+c_2)^{\frac{p^*}{p-1}} {}\\ & \le& \kappa_2^{\frac{pp^*}{p-1}-2} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)}{}\\ &&\times\left[a^{\frac{p^*}{p-1}}_4y^{p^*}(r+\nu-1)+c^{\frac{p^*}{p-1}}_2\right] . \end{eqnarray} $

在上式两端乘以$ \varphi^* $, 并在区间$ [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $上求和可得

$ \begin{eqnarray*} \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p^*}(t) \varphi^*(t)& \le &\kappa_2^{\frac{pp^*}{p-1}} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{r = 0}^{T} \varphi(r+\nu-1)\left[a^{\frac{p^*}{p-1}}_4y^{p^*}(r+\nu-1)+c^{\frac{p^*}{p-1}}_2\right]\\ & \le &a^{\frac{p^*}{p-1}}_4 \kappa_2^{\frac{pp^*}{p-1}} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2} \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p^*}(t) \varphi^*(t)\\ &&+c^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}+1} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2}\frac{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}}{C^{*}}. \end{eqnarray*} $

解上述不等式得

$ \sum\limits_{t = \nu-1}^{\nu+T-1} y^{p^*}(t) \varphi^*(t) \le \frac{c^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}+1} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2}\frac{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}}{C^{*}}}{1-a^{\frac{p^*}{p-1}}_4 \kappa_2^{\frac{pp^*}{p-1}} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2}}. $

注意到$ y\in P $, 从而由引理1.3知$ y\in P_0 $. 则有

$ \bigg(\frac{(\nu+T)^{\underline{\nu-1}}}{C^{*} \Gamma(\nu)}\bigg)^{p^*}\|y\|^{p^*} \sum\limits_{t = \nu-1}^{\nu+T-1} \varphi^*(t)\le \frac{c^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}+1} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2}\frac{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}}{C^{*}}}{1-a^{\frac{p^*}{p-1}}_4 \kappa_2^{\frac{pp^*}{p-1}} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2}}, $

即

$ \|y\|\le \sqrt[p^*]{\frac{c^{\frac{p^*}{p-1}}_2 \kappa_2^{\frac{pp^*}{p-1}} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2}}{1-a^{\frac{p^*}{p-1}}_4 \kappa_2^{\frac{pp^*}{p-1}} 2^{\frac{p^*}{p-1}-1} (T+1)^{\frac{pp^*}{p-1}-2}}}\cdot \frac{C^{*} \Gamma(\nu)}{(\nu+T)^{\underline{\nu-1}}}: = M_2. $

这就证明了集合$ W $的有界性. 此时我们若取$ R_2>\max\{M_2, r_2\} $, ($ r_2 $见条件(H3)), 则有

$ y \not = \lambda Ay, y\in \partial B_{R_2}\cap P, \lambda\in [0, 1], $

从而根据不动点指数的同伦不变性(引理1.5)可得

(2.11) $ \begin{equation} i\big(A, B_{R_2} \cap P, P\big) = 1. \end{equation} $

根据(2.9)和(2.11)式, 我们可推出

$ i(A, (B_{R_2} \backslash \overline{B}_{r_2}) \cap P, P) = i(A, B_{R_2} \cap P, P)-i(A, B_{r_2} \cap P, P) = 1. $

从而算子$ A $在$ (B_{R_2} \backslash \overline{B}_{r_2}) \cap P $内至少有一个不动点, 这也表明方程(1.1)在$ (B_{R_2} \backslash \overline{B}_{r_2}) \cap P $内至少有一个正解.

定理 2.3  若条件(H0), (H1), (H3), (H5)成立, 则方程(1.1)至少有两个正解.

证  以下根据条件(H5)证明

(2.12) $ \begin{equation} \|Ay\|<\|y\|, y\in \partial B_{\rho_1}\cap P. \end{equation} $

以下分两种情况讨论. 第一种情况: 当$ p\ge 2 $, 即$ p-1\ge 1 $时, 我们有

$ \begin{eqnarray*} [(Ay)(t)]^{p-1} & = &\left[\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\kappa_2\Gamma(\nu)} \kappa_2 \bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(r+\nu-1, y(r+\nu-1))\bigg)^{\frac{1}{p-1}}\right]^{p-1}\\ & \le& \kappa_2^{p-1} (T+1)^{p-2} \sum\limits_{s = 0}^{T} \bigg( \frac{G(t, s)}{\kappa_2\Gamma(\nu)} \bigg)^{p-1} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} \\ &&\times f(r+\nu-1, y(r+\nu-1)) \\ & \le& \kappa_2^{p-2} (T+1)^{p-2} \sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} f(r+\nu-1, y(r+\nu-1))\\ & \le& \kappa_2^{p-2} (T+1)^{p-2} \sum\limits_{s = 0}^{T} \frac{C^{*}\varphi(s+\nu-1) }{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} \delta_1 \rho_1^{p-1} \\ & \le& \kappa_2^{p} (T+1)^{p-2} \delta_1 \rho_1^{p-1}\\ & <& \rho_1^{p-1}. \end{eqnarray*} $

第二种情况: 当$ p\in (1, 2) $, 即$ \frac{1}{p-1}>1 $时, 我们有

$ \begin{eqnarray*} (Ay)(t)& = &\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \bigg(\sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\kappa_2\Gamma(\nu)} \kappa_2 f(r+\nu-1, y(r+\nu-1))\bigg)^{\frac{1}{p-1}}\\ & \le &\kappa_2^{\frac{1}{p-1}} (T+1)^{\frac{1}{p-1}-1}\sum\limits_{s = 0}^{T} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{r = 0}^{T} \bigg( \frac{G(s+\nu-1, r)}{\kappa_2\Gamma(\nu)} \bigg)^{\frac{1}{p-1}} \\ &&\times f^{\frac{1}{p-1}}(r+\nu-1, y(r+\nu-1))\\ & \le& \kappa_2^{\frac{1}{p-1}-1} (T+1)^{\frac{1}{p-1}-1}\sum\limits_{s = 0}^{T} \frac{C^{*}\varphi(s+\nu-1) }{\Gamma(\nu)-(\nu+T)^{\underline{\nu-1}}} \sum\limits_{r = 0}^{T} \frac{G(s+\nu-1, r)}{\Gamma(\nu)} \big(\delta_1 \rho_1^{p-1} \big)^{\frac{1}{p-1}}\\ & \le& \kappa_2^{\frac{1}{p-1}+1} (T+1)^{\frac{1}{p-1}-1} \delta^{\frac{1}{p-1}}_1 \rho_1\\ & <& \rho_1. \end{eqnarray*} $

注意到以上两种情况的最后一个不等式右端均为常数, 故而可得(2.12)式成立. 这亦表明

(2.13) $ \begin{equation} y \not = \lambda Ay, y\in \partial B_{\rho_1}\cap P, \lambda\in [0, 1], \end{equation} $

并由引理1.5可知

(2.14) $ \begin{equation} i\big(A, B_{\rho_1} \cap P, P\big) = 1. \end{equation} $

我们仅需证明(2.13)式, 若该式不成立, 则存在$ y\in \partial B_{\rho_1}\cap P, \lambda\in [0, 1] $使得$ y = \lambda Ay $, 即

$ y(t) = \lambda (Ay)(t)\le (Ay)(t) \le \|Ay\|, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}, $

这表明

$ \|y\|\le \|Ay\|, y\in \partial B_{\rho_1}\cap P. $

这与(2.12)式是矛盾的, 从而(2.13)式成立.

注意到(H1)中我们可以取$ R_1>\max\{M_1, \rho_1, r_2\} $ ($ r_2 $见条件(H3)), 从而由(2.3), (2.9)和(2.14)式可得

$ i(A, (B_{R_1} \backslash \overline{B}_{\rho_1}) \cap P, P) = i(A, B_{R_1} \cap P, P)-i(A, B_{\rho_1} \cap P, P) = -1 $

和

$ i(A, (B_{\rho_1} \backslash \overline{B}_{r_2}) \cap P, P) = i(A, B_{\rho_1} \cap P, P)-i(A, B_{r_2} \cap P, P) = 1. $

从而算子$ A $分别在$ (B_{R_1} \backslash \overline{B}_{\rho_1}) \cap P $和$ (B_{\rho_1} \backslash \overline{B}_{r_2}) \cap P $内至少有一个不动点, 故方程(1.1)分别在$ (B_{R_1} \backslash \overline{B}_{\rho_1}) \cap P $和$ (B_{\rho_1} \backslash \overline{B}_{r_2}) \cap P $内至少有一个正解, 即至少有两个正解.

例 2.1  令$ f(t, y) = y^\alpha, y\in {{\Bbb R}} ^+, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $, 其中$ \alpha>p-1 $. 则

$ \liminf\limits_{y\to +\infty}\frac{f(t, y)}{y^{p-1}} = \liminf\limits_{y\to +\infty}\frac{ y^\alpha}{y^{p-1}} = +\infty $

和

$ \limsup\limits_{y\to 0^+}\frac{f(t, y)}{y^{p-1}} = \limsup\limits_{y\to 0^+}\frac{ y^\alpha}{y^{p-1}} = 0, $

对$ t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $一致成立. 从而条件(H1)–(H2)满足.

例 2.2  令$ f(t, y) = y^\beta, y\in {{\Bbb R}} ^+, t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $, 其中$ \beta\in (0, p-1) $. 则

$ \liminf\limits_{y\to 0^+}\frac{f(t, y)}{y^{p-1}} = \liminf\limits_{y\to 0^+}\frac{ y^\beta}{y^{p-1}} = +\infty $

和

$ \limsup\limits_{y\to +\infty}\frac{f(t, y)}{y^{p-1}} = \limsup\limits_{y\to +\infty}\frac{ y^\beta}{y^{p-1}} = 0, $

对$ t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $一致成立. 从而条件(H3)–(H4)满足.

例 2.3  令$ \rho_1 = 1 $, 有

$ f(t, y) = \left\{\begin{array}{ll}\delta_1y^\beta , &y\in [0, 1], t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}, \beta\in (0, p-1) , \\ \delta_1 y^\alpha, &y\in [1, +\infty), t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}}, \alpha>p-1 .\end{array}\right. $

则当$ y\in [0, 1], t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $时, $ f(t, y)\le \delta_1 $. 从而条件(H5)满足. 另外

$ \liminf\limits_{y\to +\infty}\frac{f(t, y)}{y^{p-1}} = \liminf\limits_{y\to +\infty}\frac{\delta_1 y^\alpha}{y^{p-1}} = +\infty $

和

$ \liminf\limits_{y\to 0^+}\frac{f(t, y)}{y^{p-1}} = \liminf\limits_{y\to 0^+}\frac{\delta_1 y^\beta}{y^{p-1}} = +\infty, $

对$ t\in [\nu-1, \nu+T-1]_{{\Bbb Z}_{\nu-1}} $一致成立. 从而条件(H1), (H3)满足.