对任意的$ e, d\in \bf{{\Bbb R}} $且$ e<d $,令$ {\Bbb N}_{e} = \{e, e+1, e+2, \cdots \}, [e, d]_{ {\Bbb N}_{e}} = [e, d] \cap {\Bbb N}_{e} $.本文研究以下具有半正非线性项的分数阶差分方程组边值问题正解的存在性

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

其中$ 1<\nu\le 2 $, $ b\in {\Bbb N}_1 $, $ \Delta^\nu $为Riemann-Liouville分数阶差分算子,非线性项$ f_{i} \in C([\nu-1, $$ \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^{+} \times {{\Bbb R}}^+, {{\Bbb R}} )\ (i = 1, 2) $满足以下半正型条件

(C1)存在正常数$ M>0 $使得

$ f_{i}(t, x, y)+M \geq 0, (t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^{+} \times {{\Bbb R}} ^{+}, i = 1, 2. $

分数阶方程是近年来研究的热点问题.数学家们研究发现运用分数阶模型能更精确地模拟现实问题,其应用领域涵盖流体力学、流变学、粘弹性力学、电分析化学、生物系统的电传导等方面,虽然存在许多争论,但其应用的广泛性与有效性则是不争的事实^[1].然而我们注意到,绝大多数学者研究的问题都是微分方程,对于分数差分方程却鲜有问津.郑祖庥教授^{[1, 2]}指出:"对于分数微分方程来说,离散化或者问题提出时便是离散的分数差分方程是不可避免的.迄今只作为近似解计算的出发点,没有对分数差分方程的专门研究,因此,无论从理论还是应用的角度看,这都是极大的缺憾".然而令人喜悦的是,近期已有很多工作致力于研究分数差分方程,参见文献[2-15]及其所附参考文献,其中专著[2-3]研究总结了近期这方面一些杰出的工作.

在文献[4]中,作者运用范数型拉伸与压缩不动点定理^[16]研究了具有半正非线性项的分数差分方程边值问题

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

其中$ \nu\in (0, 1) $, $ \lambda $为一正参数.

在文献[5]中,作者运用拓扑度理论16],研究了问题(1.2)非平凡解的存在性($ \lambda = 1, F\equiv 0 $),并在非线性项非负条件下,运用单调有界原理获得了唯一正解的存在性,给出了迭代序列.

对于分数差分方程组的研究,可参见文献[6-8]及其所附参考文献.在文献[6]中,作者运用范数型拉伸与压缩不动点定理研究了分数阶差分方程组边值问题正解的存在性

$ \left\{\begin{array}{l}{-\Delta^{\nu_{1}} y_{1}(t) = \lambda_{1} f_{1}\left(t+\nu_{1}-1, y_{1}\left(t+\nu_{1}-1\right), y_{2}\left(t+\nu_{2}-1\right)\right), t \in[1, b+1]_{{\Bbb N}}}, \\ {-\Delta^{\nu_{2}} y_{2}(t) = \lambda_{2} f_{2}\left(t+\nu_{2}-1, y_{1}\left(t+\nu_{1}-1\right), y_{2}\left(t+\nu_{2}-1\right)\right), t \in[1, b+1]_{{\Bbb N}}}, \\ {y_{1}\left(\nu_{1}-2\right) = y_{1}\left(\nu_{1}+b+1\right) = 0} , \\ {y_{2}\left(\nu_{2}-2\right) = y_{2}\left(\nu_{2}+b+1\right) = 0, }\end{array}\right. $

其中$ \nu_1, \nu_2\in (1, 2] $, $ \lambda_1, \lambda_2 $是两个正参数.他们使用的半正非线性项$ f_i(i = 1, 2) $的超线性增长条件是

(1.3) $ \begin{equation} \lim _{y_{1}+y_{2} \rightarrow+\infty} \frac{f_{i}(t, x, y)}{y_{1}+y_{2}} = +\infty, {\qquad} \lim _{y_{1}+y_{2} \rightarrow 0^{+}} \frac{f_{i}(t, x, y)}{y_{1}+y_{2}} = 0, \end{equation} $

对$ t \in\left[\nu_{i}, \nu_{i}+b\right]_{{\Bbb N}_{\nu_{i}-2}} $一致成立.

受上述文献的启发,本文运用不动点指数理论研究分数差分方程组边值问题(1.1)正解的存在性.创新点在于:(l)非线性项可以做到下方有界; (2)运用凹凸函数刻画非线性项之间的耦合行为; (3)所使用的条件优于条件(1.3),详见第2节将给出的(C2).

以下我们给出本文所使用到的分数阶差分计算相关的定义及引理.

定义1.1^{[7, 9, 14]}  对任意有意义的$ t, \nu\in {{\Bbb R}} $,定义$ t^{\underline{\nu}} = \frac{\Gamma(t+1)}{\Gamma(t+1-\nu)} $.如果$ t+1-\nu $是伽马函数的一个奇点且$ t + 1 $不是奇点,则$ t^{\underline{\nu}} = 0 $.

定义1.2^[9]  当$ \nu>0 $时,函数$ f $的$ \nu $阶和分定义为

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

由此定义$ f $的$ \nu $阶分数差分为

$ \Delta^{\nu} f(t) = \Delta^{N} \Delta^{\nu-N} f(t), $

其中$ t \in {\Bbb N}_{a+\nu}, N \in {\Bbb N} $满足: $ 0 \leq N-1<\nu \leq N $.

引理1.3^[9]  若对于实数$ t, \nu $使得$ t^{\underline{\nu}}, t^{\underline{\nu-1}} $均是良定义的,则$ \Delta t^{\underline{\nu}} = \nu t^{\underline{\nu-1}} $.

引理1.4^[9]  若$ N $满足：$ 0 \leq N-1<\nu \leq N $,则对于$ c_i\in {{\Bbb R}} , i = 1, 2, \cdots, N, $我们有

$ \Delta^{-\nu} \Delta^{\nu} y(t) = y(t)+c_{1} t^{\underline{\nu-1}}+c_{2} t^{\underline{\nu-2}}+\cdots+c_{N} t^{\underline{\nu-N}} . $

为了获得方程组(1.1)正解的存在性,我们首先考虑如下的辅助问题

(1.4) $ \begin{equation} \left\{\begin{array}{l}{\Delta^{\nu} x(t) = f(t+\nu-1, x(t+\nu-1)), t \in[0, b]_{{\Bbb N}_{0}}}, \\ {x(\nu-2) = 0, \Delta x(\nu-2) = \Delta x(\nu+b-1), }\end{array}\right. \end{equation} $

其中$ 1<\nu\le 2 $, $ b\in {\Bbb N}_1 $, $ f \in C([\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+, {{\Bbb R}} ) $满足以下半正型条件

(C1) $ ' $存在正常数$ M>0 $使得

$ f(t, x)+M \geq 0, (t, x) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^{+} . $

引理1.5^{[15,引理3.1]}  分数阶差分方程边值问题(1.4)的解可表示为

(1.5) $ \begin{equation} x(t) = \frac{1}{\Gamma(\nu)} \sum\limits_{s = 0}^{b} G(t, s) f(s+\nu-1, x(s+\nu-1)), (t, s) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times[0, b]_{{\Bbb N}_{0}}, \end{equation} $

其中格林函数$ G $定义为

(1.6) $ \begin{equation} G(t, s) = \left\{ \begin{array}{ll} { } \frac{(b+\nu-s-2)^{\underline{\nu-2}} t^{\underline{\nu-1}} } {\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}+(t-s-1)^{\underline{\nu-1}}, & 0 \leq s \leq t-\nu \leq b, \\ { } \frac{(b+\nu-s-2)^{\underline{\nu-2}} t^{\underline{\nu-1}}} {\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}, & 0 \leq t-\nu<s \leq b. \end{array} \right. \end{equation} $

引理1.6^{[15,引理3.2]}  格林函数$ G $满足以下不等式

(1) $ 0\le G(t, s)\le \left[1+\frac{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}{(\nu+b-2)^{\underline{\nu-2}}}\right] \frac{(\nu+b)^{\underline{\nu-1}} (\nu+b-s-2)^{\underline{\nu-2}}}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}} $, $ (t, s) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times[0, b]_{{\Bbb N}_{0}} $,

(2) $ \min \limits_{t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu}-1}} G(t, s)\ge \frac{\Gamma(\nu)(\nu+b-s-2)^{\underline{\nu-2}}}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}>0, s\in [0, b]_{{\Bbb N}_0} $.

引理1.7  令$ \phi(t+\nu-1) = (\nu+b-t-2)^{\underline{\nu-2}}, $$ t \in[0, b]_{{\Bbb N}_{0}} $, $ \phi^{*}(t) = (2 \nu+b-t-3)^{\underline{\nu-2}}, $$ t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $,从而对任意的$ s\in [0, b]_{{\Bbb N}_0} $,我们有

(1.7) $ \begin{equation} \kappa_{1} \phi^{*}(s) \leq \sum\limits_{t = \nu-1}^{b+\nu-1} G(t, s) \phi^{*}(t) \leq \kappa_{2} \phi^{*}(s), \end{equation} $

其中

$ \kappa_{1} = \sum\limits_{t = 0}^{b} \frac{\Gamma(\nu) \phi(t+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}, $

$ \kappa_{2} = \sum\limits_{t = 0}^{b}\left[1+\frac{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}} {(\nu+b-2)^{\underline{\nu-2}}}\right] \frac{(\nu+b)^{\underline{\nu-1}} \phi(t+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}. $

证  根据$ \phi, \phi^* $的定义,我们有

$ \sum\limits_{t = \nu-1}^{b+\nu-1} G(t, s) \phi^{*}(t) = \sum\limits_{t = 0}^{b} G(t+\nu-1, s) \phi(t+\nu-1), s \in[0, b]_{\mathrm{N}_{0}}. $

从而根据引理1.6,对任意的$ s\in [0, b]_{{\Bbb N}_0} $,我们可以得到如下的不等式

(1.8) $ \begin{eqnarray} \sum\limits_{t = 0}^{b} G(t+\nu-1, s) \phi(t+\nu-1) &\geq &\sum\limits_{t = 0}^{b} \frac{\Gamma(\nu) \phi(t+\nu-1)} {\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}} \cdot \phi(s+\nu-1){}\\ & = &\kappa_{1} \phi^{*}(s) \end{eqnarray} $

和

(1.9) $ \begin{eqnarray} & &\sum\limits_{t = 0}^{b} G(t+\nu-1, s) \phi(t+\nu-1){}\\ & \le& \sum\limits_{t = 0}^{b}\left[1+\frac{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}{(\nu+b-2)^{\underline{\nu-2}}}\right] \frac{(\nu+b)^{\underline{\nu-1}} \phi(t+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}} \cdot \phi(s+\nu-1){}\\ & = & \kappa_{2} \phi^{*}(s). \end{eqnarray} $

证毕.

令$ E $是从$ [\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $到实数集$ {{\Bbb R}} $上的全体映射构成的集合,并在其上赋予最大模范数$ \|x\| = \max \left\{|x(t)| : t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}\right\} $.则$ (E, \|\cdot\|) $是一Banach空间.定义集合

$ P = \{x\in E: x(t)\ge 0, t\in [\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}\}, $

$ P_{0} = \left\{x \in E : x(t) \geq \frac{\kappa_{1}}{\kappa_{2}}\|x\|, t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}\right\}, $

则$ P, P_0 $均是$ E $上的锥.

若$ x\in P $,定义如下算子

$ (L x)(t) = \frac{1}{\Gamma(\nu)} \sum\limits_{s = 0}^{b} G(t, s) x(s+\nu-1), (t, s) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times[0, b]_{{\Bbb N}_{0}}, $

则可得以下的引理.

引理1.8  对任意的$ t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $, $ L(P)\subset P_0 $.

证  根据引理1.6,对任意的$ t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $,我们有

$ (L x)(t) \leq \frac{1}{\Gamma(\nu)} \sum\limits_{s = 0}^{b}\left[1+\frac{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}{(\nu+b-2)^{\underline{\nu-2}}}\right] \frac{(\nu+b)^{\underline{\nu-1}} \phi(s+\nu-1) x(s+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}. $

另一方面,对任意的$ t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $,我们可得到

$ \begin{eqnarray*} (L x)(t)& \geq& \frac{1}{\Gamma(\nu)} \sum\limits_{s = 0}^{b} \frac{\Gamma(\nu) \phi(s+\nu-1) x(s+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}\\ & = & \frac{\left[1+\frac{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}{(\nu+b-2) ^{\underline{\nu-2}}}\right]^{-1}\Gamma(\nu)}{\Gamma(\nu)(\nu+b) ^{\underline{\nu-1}}} \sum\limits_{s = 0}^b \left[1+\frac{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}{(\nu+b-2) ^{\underline{\nu-2}}}\right]\\ & & \times \frac{(\nu+b)^{\underline{\nu-1}} \phi(s+\nu-1) x(s+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}\\ & \ge & \frac{\left[1+\frac{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}{(\nu+b-2) ^{\underline{\nu-2}}}\right]^{-1}\Gamma(\nu)}{(\nu+b) ^{\underline{\nu-1}}}\|Lx\|\\ & \ge & \frac{\kappa_1}{\kappa_2}\|Lx\|. \end{eqnarray*} $

证毕.

定义$ w $是以下分数阶差分方程边值问题的解

(1.10) $ \begin{equation} \left\{\begin{array}{l}{\Delta^{\nu} w(t) = M, t \in[0, b]_{{\Bbb N}_{0}}}, \\ {w(\nu-2) = 0, \Delta w(\nu-2) = \Delta w(\nu+b-1), }\end{array}\right. \end{equation} $

其中$ 1<\nu\le 2 $, $ b\in {\Bbb N}_1 $, $ M $由条件(C1)确定.由引理1.5知

(1.11) $ \begin{equation} w(t) = \frac{M}{\Gamma(\nu)} \sum\limits_{s = 0}^{b} G(t, s), (t, s) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times[0, b]_{{\Bbb N}_{0}}. \end{equation} $

根据引理1.6(1),我们可得到

$ w(t) \leq \frac{M}{\Gamma(\nu)} \sum\limits_{s = 0}^{b}\left[1+\frac{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}}{(\nu+b-2)^{\underline{\nu-2}}}\right] \frac{(\nu+b)^{\underline{\nu-1}} \phi(s+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\underline{\nu-2}}} = \frac{M \kappa_{2}}{\Gamma(\nu)}. $

因此,若$ x\in P_0 $,则对任意的$ t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $,当$ \|x\|\ge \frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)} $时,我们有

(1.12) $ \begin{equation} x(t)-w(t) \geq \frac{\kappa_{1}}{\kappa_{2}}\|x\|-\frac{M \kappa_{2}}{\Gamma(\nu)} \geq 0. \end{equation} $

我们将问题(1.4)做如下变形

(1.13) $ \begin{equation} \left\{\begin{array}{l}{\Delta^{\nu} x(t) = f(t+\nu-1, \max \{x(t+\nu-1)-w(t+\nu-1), 0\})+M, t \in[0, b]_{{\Bbb N}_{0}}}, \\ {x(\nu-2) = 0, \Delta x(\nu-2) = \Delta x(\nu+b-1), }\end{array}\right. \end{equation} $

其中$ w $由(1.11)式定义.关于问题(1.4)和(1.13),我们有如下引理.

引理1.9  (1)若$ w $和$ x $分别是问题(1.10)和(1.13)的解,并且对任意的$ t\in [0, b]_{{\Bbb N}_{0}} $,有$ x(t+\nu-1)-w(t+\nu-1)\ge 0 $,则$ x-w $是问题(1.4)的正解.

(2)若$ x $是问题(1.4)的正解,则$ x+w $是问题(1.13)的正解.

由此可知我们仅需要找寻问题(1.13)超过$ w $的解,并且注意到(1.12)式的计算,又可知该解的范数超过$ \frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)} $即可.

定义算子$ A: E\to E $如下

$ (A x)(t) = \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}[f(s+\nu-1, \max \{x(s+\nu-1)-w(s+\nu-1), 0\})+M], $

$ t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}. $

由引理1.5知,问题(1.13)解的存在性等价于算子$ A $不动点的存在性.并且根据引理1.8的证明,引理1.6和条件(C1)$ ' $知, $ A(P)\subset P, A(P)\subset P_0 $.

引理1.10^[16]  设$ E $为实Banach空间, $ P $为$ E $中的锥, $ \Omega $是$ E $中的有界开集, $ A:\overline{\Omega}\cap P \to P $是一全连续算子.若存在$ x_0\in P\backslash \{0\} $使得

$ x \neq A x+\lambda x_{0}, \forall x \in \partial \Omega \cap P, \lambda \geq 0, $

则$ i(A, \Omega\cap P, P) = 0 $,其中$ i $为锥$ P $上的不动点指数.

引理1.11^[16]  设$ E $为实Banach空间, $ P $为$ E $中的锥, $ \Omega $是$ E $中的有界开集, $ 0\in \Omega $, $ A:\overline{\Omega}\cap P \to P $是一全连续算子.若

$ x \neq \lambda A x, \forall x \in \partial \Omega \cap P, \lambda \in[0, 1], $

则$ i(A, \Omega\cap P, P) = 1 $.

根据第1节的讨论知,我们需要将问题(1.1)做如下变形

(2.1) $ \begin{equation} \left\{ \begin{array}{ll} \Delta^{\nu} x(t)& = f_{1}(t+\nu-1, \max \{x(t+\nu-1)-w(t+\nu-1), 0\}, \\ & \max \{y(t+\nu-1)-w(t+\nu-1), 0\} )+M, t \in[0, b]_{{\Bbb N}_{0}}, \\ \Delta^{\nu} y(t)& = f_{2}(t+\nu-1, \max \{x(t+\nu-1)-w(t+\nu-1), 0\}, \\ &\max\{y(t+\nu-1)-w(t+\nu-1), 0\} )+M, t \in[0, b]_{{\Bbb N}_{0}}, \\ & x(\nu-2) = 0, \Delta x(\nu-2) = \Delta x(\nu+b-1), \\ & y(\nu-2) = 0, \Delta y(\nu-2) = \Delta y(\nu+b-1). \end{array} \right. \end{equation} $

为方便计,以下简记

$ {f_{i}^{*}(t+\nu-1, x(t+\nu-1), y(t+\nu-1)) = f_{i}(t+\nu-1, \max \{x(t+\nu-1)-w(t+\nu-1), 0\}, } $

$ {\max \{y(t+\nu-1)-w(t+\nu-1), 0\} )+M, t \in[0, b]_{{\Bbb N}_{0}}, i = 1, 2.} $

从而

$ f_{i}^{*}(t, x(t), y(t)) = f_{i}(t, \max \{x(t)-w(t), 0\}, $

$ \max \{y(t)-w(t), 0\} )+M, t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}, i = 1, 2. $

根据引理1.5,我们可以将(2.1)式转化为如下的和式

(2.2) $ \begin{equation} \left\{\begin{array}{l} { }{x(t) = \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} f_{1}^{*}(s+\nu-1, x(s+\nu-1), y(s+\nu-1)), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}}, \\ { } {y(t) = \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} f_{2}^{*}(s+\nu-1, x(s+\nu-1), y(s+\nu-1)), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}.}\end{array}\right. \end{equation} $

定义算子$ T_i: P\times P\to P $, $ T:P\times P\to P\times P $如下

$ T_i(x, y)(t) = \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} f_{i}^{*}(s+\nu-1, x(s+\nu-1), y(s+\nu-1)), $

$ t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}, i = 1, 2 $

和

$ T(x, y)(t) = \left(T_{1}, T_{2}\right)(x, y)(t), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}. $

根据第1节的讨论知,算子$ T $不动点的存在性等价于方程组(2.1)解的存在性.再根据引理1.9,若$ (x, y) $是$ T $的不动点且$ x(t)\ge w(t), y(t)\ge w(t), t\in [\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $,则$ (x-w, y-w) $是方程组(1.1)的正解.从而接下来的任务是在合适的条件下找寻算子$ T $范数超过$ \frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)} $的不动点.

为简便记,我们使用$ d_1, d_2, \cdots $表示不同的正实数(在后面的证明过程中不具体计算出来),接下来给出非线性项满足的增长性条件

(C2)存在函数$ p, q\in C({{\Bbb R}} ^+, {{\Bbb R}} ^+) $和常数$ \gamma_{1}>\frac{\Gamma^{2}(\nu)}{\kappa_{1}^{2}}, d_{1}>0 $使得

(ⅰ) $ p $是$ {{\Bbb R}} ^+ $上的严格增的凹函数,且$ { }\lim_{y\to +\infty}p(y) = +\infty $,

(ⅱ) $ f_{1}^{*}(t, x, y) \geq p(y)-d_{1}, f_{2}^{*}(t, x, y) \geq q(x)-d_{1}, \forall(t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+ \times {{\Bbb R}} ^+, $

$ p\left(\frac{\kappa_{2}}{\Gamma(\nu)} q(x)\right) \geq \frac{\kappa_{2}}{\Gamma(\nu)} \gamma_{1} x-d_{1}, \forall x \in {{\Bbb R}} ^+. $

(C3) $ f_{i}^{*}(t, x, y)<\frac{M \kappa_{2}}{\kappa_{1}}, \forall(t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times\left[0, \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right] \times\left[0, \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right], i = 1, 2. $

(C4)存在函数$ \xi, \eta\in C({{\Bbb R}} ^+, {{\Bbb R}} ^+) $和常数$ \gamma_{2} \in\left(0, \frac{\Gamma^{2}(\nu)}{\kappa_{2}}\right), d_{2}>0 $使得

(ⅰ) $ \xi $是$ {{\Bbb R}} ^+ $上的严格增的凸函数,且$ { }\lim_{y\to +\infty}\xi(y) = +\infty $;

(ⅱ) $ f_{1}^{*}(t, x, y) \leq \xi(y), f_{2}^{*}(t, x, y) \leq \eta(x), \forall(t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+ \times {{\Bbb R}} ^+ $,

$ \xi\left(\frac{\kappa_{2}}{\Gamma(\nu)} \eta(x)\right) \leq \frac{\kappa_{2}}{\Gamma(\nu)} \gamma_{2} x+d_{2}, x \in {{\Bbb R}} ^+. $

(C5) $ f_{i}^{*}(t, x, y)>\frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}}, \forall(t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times\left[0, \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right] \times\left[0, \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right], i = 1, 2. $

例2.1  令$ p(y) = y^{\frac{4}{5}}, q(x) = x^{2}, x, y \in {{\Bbb R}} ^+ $.则

$ \lim\limits _{x \rightarrow+\infty} \frac{p\left(\frac{\kappa_{2}}{\Gamma(\nu)} q(x)\right)}{x} = \lim _{x \rightarrow+\infty} \frac{\left(\frac{\kappa_{2}}{\Gamma(\nu)}\right)^{\frac{4}{5}} x^{\frac{8}{5}}}{x} = +\infty, $

从而$ p, q $满足条件(C2).另一方面,选取$ f_1^*, f_2^* $如下

$ f_{1}^{*}(t, x, y) = \frac{1}{\beta_{1}+e^{|\sin t x|}} \frac{\Gamma(\nu)}{\kappa_{2}} y, \quad f_{2}^{*}(t, x, y) = \frac{1}{\beta_{2}+e^{|\cos t y|}} \frac{\Gamma(\nu)} {\kappa_{2}}\left(\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right)^{1-\beta_{3}} x^{\beta_{3}}, $

其中$ \beta_1>0, \beta_2>0, \beta_3>2, (t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+ \times {{\Bbb R}} ^+ $.下证$ f_1^*, f_2^* $满足条件(C2)(ⅱ)和(C3).若$ (t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times\left[0, \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right] \times\left[0, \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right] $,则

$ f_{1}^{*}(t, x, y)<\frac{\Gamma(\nu)}{\kappa_{2}} \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)} = \frac{M \kappa_{2}}{\kappa_{1}}, f_{2}^{*}(t, x, y)< \frac{\Gamma(\nu)}{\kappa_{2}}\left(\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right)^{1-\beta_{3}}\left(\frac{M \kappa_{2}^{2}} {\kappa_{1} \Gamma(\nu)}\right)^{\beta_{3}} = \frac{M \kappa_{2}}{\kappa_{1}}. $

则条件(C3)成立.另一方面

$ \begin{eqnarray*} \liminf _{y \rightarrow+\infty} \frac{f_{1}^{*}(t, x, y)}{p(y)}& = &\liminf _{y \rightarrow+\infty} \frac{\frac{1}{\beta_{1}+e^{|\sin t x|} }\frac{\Gamma(\nu)}{\kappa_{2}} y}{y^{\frac{4}{5}}}\\ & = &+\infty, {\qquad} (t, x) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+, \\ \liminf _{x \rightarrow+\infty} \frac{f_{2}^{*}(t, x, y)}{q(x)} & = &\liminf _{x \rightarrow+\infty} \frac{\frac{1}{\beta_{2}+e^{|\cos t y|}} \frac{\Gamma(\nu)}{\kappa_{2}}\left(\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right)^{1-\beta_{3}} x^{\beta_{3}}}{x^{2}}\\ & = &+\infty, {\qquad} (t, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+. \end{eqnarray*} $

从而条件(C2)(ⅱ)成立.

例2.2  令$ \xi(y) = y^{2}, \eta(x) = \ln (x+1), x, y \in {{\Bbb R}} ^+ $.则

$ \lim _{x \rightarrow+\infty} \frac{\xi\left(\frac{\kappa_{2}}{\Gamma(\nu)} \eta(x)\right)}{x} = \lim _{x \rightarrow+\infty} \frac{\left(\frac{\kappa_{2}}{\Gamma(\nu)}\right)^{2} \ln ^{2}(x+1)}{x} = 0, $

从而$ \xi, \eta $满足条件(C4).另一方面,选取$ f_1^*, f_2^* $如下

$ f_{1}^{*}(t, x, y) = \left(\frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}} e^{\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}}+\beta_{4}+|\sin t x|\right) e^{-y}, $

$ f_{2}^{*}(t, x, y) = \left(\frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}} e^{\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}}+\beta_{5}+|\cos t y|\right) e^{-x}, $

其中$ \beta_4, \beta_5>0, (t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+ \times {{\Bbb R}} ^+ $.下证$ f_1^*, f_2^* $满足条件(C4)(ii)和(C5)若$ (t, x, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times\left[0, \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right] \times\left[0, \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}\right] $,则

$ f_{1}^{*}(t, x, y)>\frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}} e^{\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}} e^{-y} \geq \frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}} e^{\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)} }e^{-\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}} = \frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}}, $

$ f_{2}^{*}(t, x, y)>\frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}} e^{\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}} e^{-x} \geq \frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}} e^{\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}} e^{-\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}} = \frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}}. $

则条件(C5)成立.另一方面

$ \begin{eqnarray*} \limsup _{y \rightarrow+\infty} \frac{f_{1}^{*}(t, x, y)}{\xi(y)}& = &\limsup _{y \rightarrow+\infty} \frac{\left(\frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}} e^{\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}}+\beta_{4}+|\sin t x|\right) e^{-y}}{y^{2}}\\ & = &0, {\quad} (t, x) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+, \\ \limsup _{x \rightarrow+\infty} \frac{f_{2}^{*}(t, x, y)}{\eta(x)}& = &\limsup _{x \rightarrow+\infty} \frac{\left(\frac{M \kappa_{2}^{2}}{\kappa_{1}^{2}} e^{\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)}}+\beta_{5}+|\cos t y|\right) e^{-x}}{\ln (x+1)}\\ & = &0, {\quad}(t, y) \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \times {{\Bbb R}} ^+. \end{eqnarray*} $

从而条件(C4)(ⅱ)成立.

定义$ E $中的开球$ B_\rho = \{x\in E: \|x\|<\rho\}, \rho>0 $.

定理2.3  若条件(C1)–(C3)成立,则差分方程组(1.1)至少有一个正解.

证  第一步    证明存在$ R>\frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)} $使得

(2.3) $ \begin{equation} (x, y) \neq T(x, y)+\lambda\left(x_{0}, x_{0}\right), \forall(x, y) \in \partial B_{R} \cap(P \times P), \lambda \geq 0, \end{equation} $

其中$ x_0\in P_0 $是一给定的函数.反证法,若上式不成立,则存在$ (x, y) \in \partial B_{R} \cap(P \times P), \lambda \geq 0 $使得$ (x, y) = T(x, y)+\lambda\left(x_{0}, x_{0}\right) $,从而$ x(t) \geq T_{1}(x, y)(t), y(t) \geq T_{2}(x, y)(t), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu}-1} $.根据条件(C2)(ⅱ),我们有

(2.4) $ \begin{eqnarray} x(t) \geq T_{1}(x, y)(t) & \geq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}\left[p(y(s+\nu-1)-w(s+\nu-1))-d_{1}\right] {} \\ &\geq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} p(y(s+\nu-1)-w(s+\nu-1))-d_{3}{} \\ & \geq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}[p(y(s+\nu-1))-p(w(s+\nu-1))]-d_{3} {}\\ & \geq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} p(y(s+\nu-1))-d_{4}, t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \end{eqnarray} $

和

(2.5) $ \begin{eqnarray} y(t) &\geq &T_{2}(x, y)(t) \geq \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}\left[q(x(s+\nu-1)-w(s+\nu-1))-d_{1}\right] {} \\ & \geq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} q(x(s+\nu-1)-w(s+\nu-1))-d_{3}, t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}.} \end{eqnarray} $

注意到$ \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}\le \frac{\kappa_2}{\Gamma(\nu)}, \forall t\in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}. $从而根据(2.5)式和条件(C2)(ⅱ),对任意的$ t\in [0, b]_{{\Bbb N}_{0}} $,我们可以推出

$ \begin{eqnarray*} p(y(t+\nu-1))+p\left(d_{3}\right) & \geq & p\left(y(t+\nu-1)+d_{3}\right) \\ & \geq & p\left[\sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} q(x(s+\nu-1)-w(s+\nu-1))\right] \\ & = & p\left[\sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} \frac{\kappa_{2}}{\Gamma_{2}} \frac{\kappa_{2}}{\Gamma(\nu)} q(x(s+\nu-1)-w(s+\nu-1))\right] \\ & \geq & \sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} \frac{\Gamma(\nu)}{\kappa_{2}} p\left(\frac{\kappa_{2}}{\Gamma(\nu)} q(x(s+\nu-1)-w(s+\nu-1))\right) \\ & \geq & \sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} \frac{\Gamma(\nu)}{\Gamma(\nu)} \gamma_{1}(x(s+\nu-1)-w(s+\nu-1))-d_{5} \\ & \geq & \gamma_{1} \sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)}(x(s+\nu-1)-w(s+\nu-1))-d_{5} \\ & \geq & \gamma_{1} \sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)} {\Gamma(\nu)} x(s+\nu-1)-d_{6} . \end{eqnarray*} $

从而可得

$ p(y(t+\nu-1)) \geq \gamma_{1} \sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} x(s+\nu-1)-d_{7}, \forall t \in[0, b]_{{\Bbb N}_{0}}. $

因此,将上式带入(2.4)式可得

(2.6) $ \begin{eqnarray} x(t) & \geq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} p(y(s+\nu-1))-d_{4} {}\\ & \geq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}\left[\gamma_{1} \sum\limits_{\tau = 0}^{b} \frac{G(s+\nu-1, \tau)}{\Gamma(\nu)} x(\tau+\nu-1)-d_{7}\right]-d_{4}{} \\ & \geq & \gamma_{1} \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{\tau = 0}^{b} \frac{G(s+\nu-1, \tau)}{\Gamma(\nu)} x(\tau+\nu-1)-d_{8}, t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}.} {\quad} \end{eqnarray} $

将(2.6)式两边分别乘以$ \phi^*(t) $并从$ \nu-1 $到$ \nu+b-1 $求和,借助不等式(1.7)和(1.8),可得

$ \begin{eqnarray*} \sum\limits_{t = \nu-1}^{\nu+b-1} x(t) \phi^{*}(t) & \geq & \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t)\left[\gamma_{1} \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{\tau = 0}^{b} \frac{G(s+\nu-1, \tau)}{\Gamma(\nu)} x(\tau+\nu-1)-d_{8}\right] \\ & \geq & \frac{\gamma_{1}}{\Gamma^{2}(\nu)} \kappa_{1}^{2} \sum\limits_{t = 0}^{b} x(t+\nu-1) \phi(t+\nu-1)-d_{9} \\ & = & \frac{\gamma_{1} \kappa_{1}^{2}}{\Gamma^{2}(\nu)} \sum\limits_{t = \nu-1}^{\nu+b-1} x(t) \phi^{*}(t)-d_{9}. \end{eqnarray*} $

由于$ T_1(P, P)\subset P_0, x_0\in P_0 $,则$ x\in P_0 $.从而

$ \frac{\kappa_{1}}{\kappa_{2}}\|x\| \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t) \leq \sum\limits_{t = \nu-1}^{\nu+b-1} x(t) \phi^{*}(t) \leq \frac{d_{9} \Gamma^{2}(\nu)}{\gamma_{1} \kappa_{1}^{2}-\Gamma^{2}(\nu)}, $

由此解得

$ \|x\| \leq \frac{d_{9} \Gamma^{2}(\nu)}{\gamma_{1} \kappa_{1}^{2}-\Gamma^{2}(\nu)}\left[\frac{\kappa_{1}}{\kappa_{2}} \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t)\right]^{-1} : = M_{1}. $

另一方面,注意到$ T_2(P, P)\subset P_0, x_0\in P_0 $,则$ y\in P_0 $.并且不妨设$ y(t)\not \equiv 0, t\in [\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $(否则$ \|y\| = 0 $,已然有界).从而$ \|y\|>0 $且$ p(\|y\|)>0 $.再根据函数$ p $的凹性,我们有

$ \begin{eqnarray*} \frac{\kappa_{1}}{\kappa_{2}}\|y\| \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t) \leq \sum\limits_{t = \nu-1}^{\nu+b-1} y(t) \phi^{*}(t) & = & \frac{\|y\|}{p(\|y\|)} \sum\limits_{t = \nu-1}^{\nu+b-1} \frac{y(t)}{\|y\|} p(\|y\|) \phi^{*}(t) \\ & \leq & \frac{\|y\|}{p(\|y\|)} \sum\limits_{t = \nu-1}^{\nu+b-1} p(y(t)) \phi^{*}(t). \end{eqnarray*} $

在(2.6)式两边乘以$ \phi^*(t) $并从$ \nu-1 $到$ \nu+b-1 $求和,借助不等式(1.7)可得

$ \begin{eqnarray*} \sum\limits_{t = \nu-1}^{\nu+b-1}\left(x(t)+d_{4}\right) \phi^{*}(t) & \geq & \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t) \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} p(y(s+\nu-1)) \\ & \geq & \frac{\kappa_{1}}{\Gamma(\nu)} \sum\limits_{t = 0}^{b} p(y(t+\nu-1)) \phi(t+\nu-1) \\ & = & \frac{\kappa_{1}}{\Gamma(\nu)} \sum\limits_{t = \nu-1}^{\nu+b-1} p(y(t)) \phi^{*}(t). \end{eqnarray*} $

结合上述两个不等式,我们可得

$ \begin{eqnarray*} p(\|y\|) & \leq & \left[\frac{\kappa_{1}}{\kappa_{2}} \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t)\right]^{-1} \sum\limits_{t = \nu-1}^{\nu+b-1} p(y(t)) \phi^{*}(t) \\ & \leq & \left[\frac{\kappa_{1}}{\kappa_{2}} \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t)\right]^{-1} \frac{\Gamma(\nu)}{\kappa_{1}} \sum\limits_{t = \nu-1}^{\nu+b-1}\left(x(t)+d_{4}\right) \phi^{*}(t) \\ & \leq & \left[\frac{\kappa_{1}}{\kappa_{2}} \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t)\right]^{-1} \frac{\Gamma(\nu)}{\kappa_{1}}\left(\frac{d_{9} \Gamma^{2}(\nu)}{\gamma_{1} \kappa_{1}^{2}-\Gamma^{2}(\nu)}\left[\frac{\kappa_{1}}{\kappa_{2}} \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t)\right]^{-1}+d_{4}\right) \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t). \end{eqnarray*} $

根据条件(C2)(ⅰ), $ { }\lim_{z\to +\infty}p(z) = +\infty $,因此存在$ M_2>0 $使得$ \|y\|\le M_2 $.

综上所述$ \|x\|\le M_1, $$ \|y\|\le M_2 $.取$ R>\max\{\frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)}, M_1, M_2\} $使得(2.3)式成立.从而根据引理1.10可知

(2.7) $ \begin{equation} i\left(T, B_{R} \cap(P \times P), P \times P\right) = 0. \end{equation} $

第二步{\quad}证明

(2.8) $ \begin{equation} (x, y) \neq \lambda T(x, y), \forall(x, y) \in \partial B_{\Theta} \cap(P \times P), \lambda \in[0, 1], \end{equation} $

其中$ \Theta = \frac{M \kappa^2_{2}}{\kappa_1 \Gamma(\nu)} $.反证法,若上式不成立,则存在$ (x, y) \in \partial B_{\Theta} \cap(P \times P), \lambda_0 \in[0, 1] $使得$ (x, y) = \lambda_0 T(x, y) $.这表明

$ x(t) \leq T_{1}(x, y)(t), y(t) \leq T_{2}(x, y)(t), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}. $

从而

(2.9) $ \begin{equation} \|x\| \leq\left\|T_{1}(x, y)\right\|, \|y\| \leq\left\|T_{2}(x, y)\right\|. \end{equation} $

然而根据条件(C3),我们有

$ \begin{eqnarray*} T_{1}(x, y)(t) & = & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}\left[f_{1}^{*}(s+\nu-1, x(s+\nu-1), y(s+\nu-1))+M\right] \\ &< & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} \frac{\Gamma(\nu)}{\kappa_{2}} \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)} \leq \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)} = \|x\|, t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}, } \end{eqnarray*} $

这表明$ \|T_1(x, y)\|<\|x\| $.同理可证$ \|T_2(x, y)\|<\|y\| $.这与(2.9)式矛盾,故(2.8)式成立.由引理1.11知

(2.10) $ \begin{equation} i\left(T, B_{\Theta} \cap(P \times P), P \times P\right) = 1. \end{equation} $

通过(2.7)和(2.10)式计算可得

$ i\left(T, \left(B_{R} \backslash \overline{B}_{\Theta}\right) \cap(P \times P), P \times P\right) = 0-1 = -1. $

因此算子$ T $在$ \left(B_{R} \backslash \overline{B}_{\Theta}\right) \cap(P \times P) $中至少有一个不动点$ (x, y) $,又由于$ \|x\|, \|y\|\ge \frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)} $,则$ (x-w, y-w) $是问题(1.1)的一个正解.证毕. \hfill\rule{0.8mm}{3.5mm}

定理2.4  若条件(C1), (C4)–(C5)成立,则差分方程组(1.1)至少有一个正解.

证  第一步    证明存在$ R>\frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)} $使得

(2.11) $ \begin{equation} (x, y) \neq \lambda T(x, y), \forall(x, y) \in \partial B_{R} \cap(P \times P), \lambda \in[0, 1]. \end{equation} $

反证法.若上式不成立,则存在$ (x, y) \in \partial B_{R} \cap(P \times P), \lambda_0 \in[0, 1] $使得$ (x, y) = \lambda_0 T(x, y) $.从而,我们有

$ x(t) \leq T_{1}(x, y)(t), y(t) \leq T_{2}(x, y)(t), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}. $

根据条件(C4)(ⅱ)我们可得

(2.12) $ \begin{eqnarray} x(t) & \leq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} \xi(y(s+\nu-1)-w(s+\nu-1)){} \\ & \leq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}[\xi(y(s+\nu-1))-\xi(w(s+\nu-1))] {}\\ & \leq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} \xi(y(s+\nu-1)), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} \end{eqnarray} $

和

(2.13) $ \begin{equation} y(t) \leq \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} \eta(x(s+\nu-1)-w(s+\nu-1)), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}. \end{equation} $

从而,对任意的$ t\in [0, b]_{{\Bbb N}_0} $,我们有

(2.14) $ \begin{eqnarray} \xi(y(t+\nu-1)) & \leq & \xi\left[\sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} \eta(x(s+\nu-1)-w(s+\nu-1))\right] {}\\ & = & \xi\left[\sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} \frac{\Gamma(\nu)}{\kappa_{2}} \frac{\kappa_{2}}{\Gamma(\nu)} \eta(x(s+\nu-1)-w(s+\nu-1))\right] {}\\ & \leq & \sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} \frac{\Gamma(\nu)}{\kappa_{2}} \xi\left(\frac{\kappa_{2}}{\Gamma(\nu)} \eta(x(s+\nu-1)-w(s+\nu-1))\right){} \\ & \leq & \sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} \frac{\Gamma(\nu)}{\kappa_{2}}\left[\frac{\kappa_{2}}{\Gamma(\nu)} \gamma_{2}(x(s+\nu-1)-w(s+\nu-1))+d_{2}\right] {}\\ & \leq & \gamma_{2} \sum\limits_{s = 0}^{b} \frac{G(t+\nu-1, s)}{\Gamma(\nu)} x(s+\nu-1)+d_{10}. \end{eqnarray} $

结合(2.12)式,我们可推出

$ \begin{eqnarray*} x(t) & \leq & \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}\left[\gamma_{2} \sum\limits_{\tau = 0}^{b} \frac{G(s+\nu-1, \tau)}{\Gamma(\nu)} x(\tau+\nu-1)+d_{10}\right] \\ & \leq & \gamma_{2} \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{\tau = 0}^{b} \frac{G(s+\nu-1, \tau)}{\Gamma(\nu)} x(\tau+\nu-1)+d_{11}, t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}.} \end{eqnarray*} $

在上式两边乘以$ \phi^*(t) $并从$ \nu-1 $到$ \nu+b-1 $求和,借助不等式(1.7)和(1.9)可得

$ \begin{eqnarray*} \sum\limits_{t = \nu-1}^{\nu+b-1} x(t) \phi^{*}(t) & \leq & \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t)\left[\gamma_{2} \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} \sum\limits_{\tau = 0}^{b} \frac{G(s+\nu-1, \tau)}{\Gamma(\nu)} x(\tau+\nu-1)+d_{11}\right] \\ & \leq & \gamma_{2} \frac{\kappa_{2}^{2}}{\Gamma^{2}(\nu)} \sum\limits_{t = \nu-1}^{\nu+b-1} x(t) \phi^{*}(t)+d_{11} \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t). \end{eqnarray*} $

注意到$ T_1(P, P)\subset P_0 $,则$ x\in P_0 $.从而由上式可解得

$ \left(1-\gamma_{2} \frac{\kappa_{2}^{2}}{\Gamma^{2}(\nu)}\right) \frac{\kappa_{1}}{\kappa_{2}}\|x\| \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t) \leq\left(1-\gamma_{2} \frac{\kappa_{2}^{2}}{\Gamma^{2}(\nu)}\right) \sum\limits_{t = \nu-1}^{\nu+b-1} x(t) \phi^{*}(t) \leq d_{11} \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t). $

解该不等式得

$ \|x\| \leq d_{11}\left[\left(1-\gamma_{2} \frac{\kappa_{2}^{2}}{\Gamma^{2}(\nu)}\right) \frac{\kappa_{1}}{\kappa_{2}}\right]^{-1} = M_{3}. $

在(2.14)式两边乘以$ \phi^*(t) $并从$ \nu-1 $到$ \nu+b-1 $求和,借助不等式(1.7)和(1.9)可得

$ \begin{eqnarray*} \sum\limits_{t = \nu-1}^{\nu+b-1} \xi(y(t)) \phi^{*}(t) & \leq & \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t)\left[\gamma_{2} \sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)} x(s+\nu-1)+d_{10}\right] \\ & \leq & \gamma_{2} \frac{\kappa_{2}}{\Gamma(\nu)} \sum\limits_{t = \nu-1}^{\nu+b-1} x(t) \phi^{*}(t)+d_{10} \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t) \\ & \leq & \left[\gamma_{2} d_{11} \frac{\kappa_{2}}{\Gamma(\nu)}\left[\left(1-\gamma_{2} \frac{\kappa_{2}^{2}}{\Gamma^{2}(\nu)}\right) \frac{\kappa_{1}}{\kappa_{2}}\right] ^{-1}+d_{10}\right] \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t) . \end{eqnarray*} $

注意到$ T_2(P, P)\subset P_0 $,则$ y\in P_0 $.从而根据函数$ \xi $的严格增长性,我们有

$ \begin{eqnarray*} \sum\limits_{t = \nu-1}^{\nu+b-1} \xi\left(\frac{\kappa_{1}}{\kappa_{2}}\|y\|\right) \phi^{*}(t) &\leq &\sum\limits_{t = \nu-1}^{\nu+b-1} \xi(y(t)) \phi^{*}(t)\\ & \leq&\left[\gamma_{2} d_{11} \frac{\kappa_{2}}{\Gamma(\nu)}\left[\left(1-\gamma_{2} \frac{\kappa_{2}^{2}}{\Gamma^{2}(\nu)}\right) \frac{\kappa_{1}} {\kappa_{2}}\right]^{-1}+d_{10}\right] \sum\limits_{t = \nu-1}^{\nu+b-1} \phi^{*}(t), \end{eqnarray*} $

从而可得

$ \xi\left(\frac{\kappa_{1}}{\kappa_{2}}\|y\|\right) \leq \gamma_{2} d_{11} \frac{\kappa_{2}}{\Gamma(\nu)}\left[\left(1-\gamma_{2} \frac{\kappa_{2}^{2}}{\Gamma^{2}(\nu)}\right) \frac{\kappa_{1}}{\kappa_{2}}\right]^{-1}+d_{10}. $

再根据条件(C4)(ⅰ),存在$ M_4>0 $使得$ \|y\|\le M_4 $.

综上所述$ \|x\|\le M_3, $$ \|y\|\le M_4 $.取$ R>\max\{\frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)}, M_3, M_4\} $使得(2.11)式成立.由此,结合引理1.11可得

(2.15) $ \begin{equation} i\left(T, B_{R} \cap(P \times P), P \times P\right) = 1. \end{equation} $

第二步    证明

(2.16) $ \begin{equation} (x, y) \neq T(x, y)+\lambda\left(x_{0}, x_{0}\right), \forall(x, y) \in \partial B_{\Theta} \cap(P \times P), \lambda \geq 0, \end{equation} $

其中$ x_0\in P $是一给定的函数, $ \Theta = \frac{M \kappa^2_{2}}{\kappa_1 \Gamma(\nu)} $.

反证法.若上式不成立,则存在$ (x, y) \in \partial B_{\Theta} \cap(P \times P), \lambda_0\ge 0 $使得$ (x, y) = T(x, y)+\lambda_0 (x_0, y_0) $.这表明

$ x(t) \geq T_{1}(x, y)(t), y(t) \geq T_{2}(x, y)(t), t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}}. $

因此

(2.17) $ \begin{equation} \|x\| \geq\left\|T_{1}(x, y)\right\|, \|y\| \geq\left\|T_{2}(x, y)\right\|. \end{equation} $

然而根据条件(C5),对任意的$ t \in[\nu-1, \nu+b-1]_{{\Bbb N}_{\nu-1}} $,我们有

$ \begin{eqnarray*} T_{1}(x, y)(t) & = &\sum\limits_{s = 0}^{b} \frac{G(t, s)}{\Gamma(\nu)}\left[f_{1}^{*}(s+\nu-1, x(s+\nu-1), y(s+\nu-1))+M\right] \\ &> & \sum\limits_{s = 0}^{b} \frac{\phi(s+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\nu-2}}\left[\sum\limits_{s = 0}^{b} \frac{\phi(s+\nu-1)}{\Gamma(\nu-1)-(b+\nu-1)^{\nu-2}}\right]^{-1} \frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)} \\ & = &\frac{M \kappa_{2}^{2}}{\kappa_{1} \Gamma(\nu)} \\ & = & \|x\|. \end{eqnarray*} $

这表明$ \|T_1(x, y)\|>\|x\| $.同理可证$ \|T_2(x, y)\|>\|y\| $.这与(2.17)式矛盾.矛盾表明(2.16)式成立.由引理1.10可得

(2.18) $ \begin{equation} i\left(T, B_{\Theta} \cap(P \times P), P \times P\right) = 0. \end{equation} $

根据(2.15)和(2.18)式可知

$ i\left(T, \left(B_{R} \backslash \overline{B}_{\Theta}\right) \cap(P \times P), P \times P\right) = 1-0 = 1. $

因此算子$ T $在$ \left(B_{R} \backslash \overline{B}_{\Theta}\right) \cap(P \times P) $中至少有一个不动点$ (x, y) $,又由于$ \|x\|, \|y\|\ge \frac{M \kappa^2_{2}}{\kappa_1\Gamma(\nu)} $,则$ (x-w, y-w) $是问题(1.1)的一个正解.证毕.