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数学物理学报, 2021, 41(2): 402-414 doi:

论文

一类具有p-Laplacian算子的分数阶差分方程边值问题的正解

徐家发,1, 罗洪林,1, 刘立山,2

Positive Solutions for a Class of Fractional Difference Equations Boundary Value Problems with p-Laplacian Operator

Xu Jiafa,1, Luo Honglin,1, Liu Lishan,2

通讯作者: 徐家发, E-mail: xujiafa292@sina.com

收稿日期: 2020-03-8  

基金资助: 国家自然科学基金.  11871302
国家自然科学基金.  11601048
中国博士后项目.  2019M652348
重庆市自然科学基金.  cstc2020jcyj-msxmX0123
重庆市教委科技项目.  KJQN202000528
重庆市教委科技项目.  KJQN201900539
重庆师范大学青年拔尖人才资助项目.  02030307/0040
重庆师范大学数学科学学院重点实验室开放课题.  CSSXKFKTM202003

Received: 2020-03-8  

Fund supported: the NSFC.  11871302
the NSFC.  11601048
the Postdoctoral Program of China (2019M652348), the NSF of Chongqing.  2019M652348
the NSF of Chongqing.  cstc2020jcyj-msxmX0123
the Projects of Chongqing Commission of Science and Technology.  KJQN202000528
the Projects of Chongqing Commission of Science and Technology.  KJQN201900539
Young Talent of Chongqing Normal University.  02030307/0040
the Key Laboratory Open Issue of School of Mathematical Science, Chongqing Normal University.  CSSXKFKTM202003

作者简介 About authors

罗洪林,E-mail:luohonglin@cqnu.edu.cn , E-mail:luohonglin@cqnu.edu.cn

刘立山,E-mail:mathlls@163.com , E-mail:mathlls@163.com

Abstract

In this paper, we study a class of fractional difference equations boundary value problems with p-Laplacian operator. By virtue of the discrete Jensen inequalities, some relations between the considered problem and the corresponding problem without the p-Laplacian are established, and using the theory of fixed point index, the existence of positive solutions for the considered problem is obtained.

Keywords: Fractional difference equations ; Boundary value problem ; Positive solution ; Fixed point index

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本文引用格式

徐家发, 罗洪林, 刘立山. 一类具有p-Laplacian算子的分数阶差分方程边值问题的正解. 数学物理学报[J], 2021, 41(2): 402-414 doi:

Xu Jiafa, Luo Honglin, Liu Lishan. Positive Solutions for a Class of Fractional Difference Equations Boundary Value Problems with p-Laplacian Operator. Acta Mathematica Scientia[J], 2021, 41(2): 402-414 doi:

1 引言及预备知识

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

τ1Dr(t)+r(t)=τ1τ2Dν+1v(t)+τ1Dνv(t),

其中Dν,Dν+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

{Δνy(t)=λf(t+ν1,y(t+ν1)),t[0,T]Z,y(ν1)=y(ν+T)+Ni=1F(ti,y(ti)),

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

{Δνx(t)=f1(t+ν1,x(t+ν1),y(t+ν1)),t[0,T]Z,Δνy(t)=f2(t+ν1,x(t+ν1),y(t+ν1)),t[0,T]Z,x(ν1)=x(ν+T),y(ν1)=y(ν+T),

其中非线性项fi(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

{Δβ[ϕp(Δαy)](t)+f(t+α+β1,y(t+α+β1))=0,t[0,b+1]N0,Δαy(β1)=0,y(α+β3)=Δγy(α+β+b+1γ)=0,

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

{Δβ[ϕp(Δαy)](t)+f(α+β+t1,y(α+β+t1))=0,t[0,b]N0,Δαy(β2)=Δαy(β+b)=0,y(α+β4)=y(α+β+b)=0,

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

{Δνν1(ϕp(Δνν1y(t)))=f(t+ν1,y(t+ν1)),t[0,T]Z,y(ν1)=y(ν+T),Δνν1y(ν1)=Δνν1y(ν+T),
(1.1)

其中ν(0,1)是一实数, Δνν1是离散的分数阶差分, ϕp(s)=|s|p2sp-Laplacian算子, p>1,sR, 非线性项f满足条件

(H0) fC([0,T]Z×R+,R+), 其中R+=[0,+), [0,T]Z=[0,T]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)p1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tν_:=Γ(t+1)Γ(t+1ν), 对任意tν右边有定义. 如果t+1νGamma函数的极点, 而t+1不是Gamma函数的极点, 则tν_=0.

定义 1.2[5-6]  对ν>0, 函数fν阶和分定义为

Δνaf(t)=1Γ(ν)tνs=a(ts1)ν1_f(s),tNa+Nν,

其中Na+Nν={a+Nν,a+Nν+1,}.

ν>0, fν阶分数阶差分定义为

Δνaf(t)=ΔNΔνNaf(t),tNa+Nν,

其中NN: 0N1<νN.

ϕq=ϕ1p, 其中p,q是共轭数, 即1q+1p=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

y(t)=Ts=0G(t,s)Γ(ν)ϕq(Tr=0G(s+ν1,r)Γ(ν)f(r+ν1,y(r+ν1))),t[ν1,ν+T1]Zν1,
(1.2)

其中

G(t,s)={(ν+Ts1)ν1_tν1_Γ(ν)(ν+T)ν1_+(ts1)ν1_,0stνT,(ν+Ts1)ν1_tν1_Γ(ν)(ν+T)ν1_,tν<sT.
(1.3)

引理 1.1[7-8, 17]  令C=1+Γ(ν)(ν+T)ν1_(ν+T1)ν1_, 则对任意的(t,s)[ν1,ν+T1]Zν1×[0,T]Z, 我们有

0<(ν+T)ν1_Γ(ν)(ν+T)ν1_(ν+Ts1)ν1_G(t,s)CΓ(ν)Γ(ν)(ν+T)ν1_(ν+Ts1)ν1_.
(1.4)

引理 1.2[8]  令φ(t+ν1)=(ν+Tt1)ν1_,t[0,T]Z, φ(t)=(2ν+Tt2)ν1_,t[ν1,ν+T1]Zν1, 则对任意的[0,T]Z, 我们有

ν+T1t=ν1G(t,s)Γ(ν)φ(t)=Tt=0G(t+ν1,s)Γ(ν)φ(t+ν1)
(1.5)

κ1φ(s+ν1)ν+T1t=ν1G(t,s)Γ(ν)φ(t)κ2φ(s+ν1),
(1.6)

其中

κ1=Tt=0(ν+T)ν1_Γ(ν)(Γ(ν)(ν+T)ν1_)φ(t+ν1),[4mm]κ2=Tt=0CΓ(ν)(ν+T)ν1_φ(t+ν1).
(1.7)

E是从[ν1,ν+T1]Zν1R全体映射构成的集合, 并在其上赋予最大值范数, 即

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}}\},

P 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 .

  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 E 上的锥, \Omega 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, 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 E 上的锥, \Omega 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], 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

\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}
(1.8)

\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}
(1.9)

2 正解的存在性

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}}.

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.

  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 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

\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}
(2.1)

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

\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}
(2.2)

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},

\|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

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

另一方面证明

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

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

\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}
(2.5)

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

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

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 (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.

  首先证明

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

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

\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}
(2.8)

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

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

另一方面定义集合 W = \{y\in P: y = \lambda Ay, \lambda\in [0, 1]\} , 下证 W 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

\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}
(2.10)

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}},

\|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

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

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 (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.

  以下根据条件(H5)证明

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

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

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

并由引理1.5可知

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

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

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 (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 (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 .

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

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

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) .

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

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

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

\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,

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

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