具有左右分数阶导数和时滞的非瞬时脉冲微分方程非线性边值问题 Nonlinear Boundary Value Problems for Non-Instantaneous Pulse Differential Equations with Left-Right Fractional Derivatives and Delays

doi:10.12677/AAM.2021.104136

Advances in Applied Mathematics
Vol. 10 No. 04 ( 2021 ), Article ID: 42069 , 15 pages
10.12677/AAM.2021.104136

具有左右分数阶导数和时滞的非瞬时脉冲微分方程非线性边值问题

张雨馨

●How to Cite this Article

上海理工大学理学院，上海

收稿日期：2021年3月27日；录用日期：2021年4月15日；发布日期：2021年4月29日

摘要

本文研究了一类特殊的具有左右分数阶导数和时滞的非瞬时脉冲微分方程，该方程具有交叉时滞，且带有非线性边界条件。并基于上下解方法得到多个正解存在性定理。

关键词

左右分数阶导数，时滞，非瞬时脉冲微分方程，非线性边界条件，上下解方法

Nonlinear Boundary Value Problems for Non-Instantaneous Pulse Differential Equations with Left-Right Fractional Derivatives and Delays

Yuxin Zhang

College of Science, University of Shanghai for Science and Technology, Shanghai

Received: Mar. 27^th, 2021; accepted: Apr. 15^th, 2021; published: Apr. 29^th, 2021

ABSTRACT

In this paper, we study a class of special non-instantaneous impulsive differential equations with left and right fractional derivatives and delays. The equations have cross delays and nonlinear boundary conditions. Based on the upper and lower solution method, we obtain the existence theorems of multiple positive solutions.

Keywords:Left-Right Fractional Derivatives, Time Delay, Non-Instantaneous Pulse, Nonlinear Boundary Conditions, Upper and Lower Solution Method

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

1. 引言

分数阶微分方程非常适合刻画具有记忆和遗传性质的材料及过程，其对复杂系统的描述具有建模简单、描述准确、参数物理意义清楚等优势，因此也是复杂力学、物理过程数学建模的重要工具，如利用分数阶微积分在不同粘弹性流体的本构关系，在非牛顿流体中进行应用；分数阶SEIR传染病模型可以准确研究传染病、社交网络信息传播等方面的问题；在含未知参数的情况下，利用非线性分数阶系统状态估计分别含有分数阶有色过程噪声和有色测量噪声的连续时间问题。

在分数阶微分方程边值问题的研究 [1] [2] [3] [4] [5] 中，有时需要考虑左侧和右侧不同的定义，而同时带有左侧和右侧分数阶导数的微分方程相对于只含有分数阶右导数或左导数的分数阶微分方程，它的应用范围 [6] [7] [8] [9] 更加广泛，其在机械力学、生物工程、物理学、经济学等自然科学领域建立的数学模型中经常出现，并且具有很重要的作用，如用来分析空气中充满粒状材料时的室内外的温度数据等。

文献 [10] 研究了带有左右分数阶导数的微分方程边值问题：

${\begin{cases} {}^{c}D_{b^{_}}^{α} {}^{c}D_{a^{+}}^{α} T (t) + λ T (t) = 0, \\ T (a) = T_{0}, T (b) = T_{1}, \end{cases}$

其中， $T \in C [0, 1]$ ， ${}^{c}D_{b^{_}}^{α}$ 为Caputo分数阶右导数， ${}^{c}D_{a^{+}}^{α}$ 为Caputo分数阶左导数。作者利用分数阶微分方程的数值解针对实际问题进行分析。

文献 [3] 研究了带有左右阶导数的耦合微分方程边值问题：

${\begin{cases} - ({}_{0^{+}}D_{t}^{α} u (t)) = f (t, v (t)), t \in (0, T), \\ - ({}_{t}D_{T^{-}}^{β} v (t)) = g (t, u (t)), t \in (0, T), \\ u (0) = 0, {}_{0^{+}}D_{t}^{α - 1} u (T) = r_{1} {}_{t}D_{T^{-}}^{β - 1} v (ξ), \\ v (T) = 0, {}_{t}D_{T^{-}}^{β - 1} v (0) = r_{2} {}_{0^{+}}D_{t}^{α - 1} u (ξ), \end{cases}$

其中， ${}_{0^{+}}D_{t}^{α}, {}_{t}D_{T^{-}}^{β}$ 分别是 $α$ 阶R-L分数阶左导数和 $β$ 阶R-L分数阶右导数， $0 < α, β \leq 2$ ， $ξ \in [0, T]$ 。作者运用上下解方法获得了边值问题解的存在性定理。

基于以上启发，本文研究含有左右分数阶导数和时滞的非瞬时脉冲微分方程非线性边值问题：

${\begin{cases} {}_{t}^{c}D_{ξ^{-}}^{α} u (t) = f_{1} (t, u (t), u (t + τ_{1})), t \in [0, ξ], \\ {}_{ξ^{+}}^{c}D_{t}^{β} u (t) = f_{2} (t, u (t), u (t - τ_{2})), t \in (ξ, 1], \\ Δ u (ξ) = I (ξ, u (ξ)), Δ u^{'} (ξ) = Q (ξ, u (ξ)), \\ h_{0} (u (0), u (1)) = 0, h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} u (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} u (1)) = 0 \end{cases}$ (1)

解的存在性与多解性。其中， ${}_{t}^{c}D_{ξ^{-}}^{α}$ 是右侧Caputo分数阶导数， ${}_{ξ^{+}}^{c}D_{t}^{β}$ 是左侧Caputo分数阶导数， $1 < α, β \leq 2$ , $ξ \in (0, 1)$ , $u (ξ^{+}) = \lim_{ε \to 0^{+}} u (ξ + ε)$ , $u (ξ^{-}) = \lim_{ε \to 0^{-}} u (ξ + ε)$ , $τ_{1} \in (0, 1 - ξ)$ , $τ_{2} \in (0, ξ)$ , $f_{1} \in C ([0, ξ] \times ℝ^{+} \times ℝ^{+}, ℝ^{+})$ , $f_{2} \in C ((ξ, 1] \times ℝ^{+} \times ℝ^{+}, ℝ^{+})$ , $I, Q \in C (ℝ, ℝ^{+})$ , $h_{0}, h_{1} \in C (ℝ^{2}, ℝ)$ 为给定的非线性函数。

2. 线性边值问题

定义1 [11]：若 $α > 0$ ， $a < b \in ℝ$ 则

${}_{a^{+}}I_{t}^{α} {}_{{}^{a^{+}}}^{c}D_{t}^{α} u (t) = u (t) + c_{1} + c_{2} (t - a) + c_{3} {(t - a)}^{2} + \dots + c_{n} {(t - a)}^{n - 1},$

${}_{t}I_{b^{-}}^{α} {}_{t}^{c}D_{b^{-}}^{α} u (t) = u (t) + d_{1} + d_{2} (b - t) + d_{3} {(b - t)}^{2} + \dots + d_{n} {(b - t)}^{n - 1},$

其中 $c_{i}, d_{i} \in ℝ, i = 1, 2, \dots, n, n \in ℕ$ 。

引理1 [12]：令E为Banach空间，且 $P \subset E$ 是一个正规体锥。如果存在 $α_{1}$ ， $β_{1}$ ， $α_{2}$ ， $β_{2} \in P$ 使

$α_{1} ≺ β_{1} ≺ α_{2} ≺ β_{2}$ ,

且 $A : [α_{1}, β_{2}] \to E$ 是全连续算子，且为强增算子，使

$α_{1} ≺ A α_{1}$ , $A β_{1} ≺ β_{1}$ , $α_{2} ≺ A α_{2}$ , $A β_{2} \underline{≺} β_{2}$ .

则算子A至少有三个不动点 $x_{1}, x_{2}, x_{3}$ 使得

$α_{1} \underline{≺} x_{1} ≺ ≺ β_{1}$ , $α_{2} ≺ ≺ x_{2} \underline{≺} β_{2}$ , $α_{2} \underline{≺} x_{2} \underline{≺} β_{2}$ .

令 $J = [0, 1]$ ， $J_{0} = J \ ξ$ ， $E = P C [J, ℝ] = {u : J \to ℝ : u 在 J_{0} 上是连续的, u (ξ^{+}) 与 u (ξ^{-}) 存在且 u (ξ^{-}) = u (ξ)}$ 。显然E是Banach空间且定义其范数为

$‖ u ‖ = \sup_{t \in [0, 1]} | u (t) |$ .

令 ${‖ u ‖}_{[0, ξ]} = \sup_{t \in [0, ξ]} | u (t) |$ , ${‖ u ‖}_{(ξ, 1]} = \sup_{t \in (ξ, 1]} | u (t) |$ ，则 $‖ u ‖ = \max {{‖ u ‖}_{[0, ξ]}, {‖ u ‖}_{(ξ, 1]}}$ 。

引理2：令 $h \in C ([0, ξ], ℝ^{+}), y \in C ((ξ, 1], ℝ^{+})$ ，对任意 $m_{i}, n_{i} \in ℝ, i = 1, 2$ ，且 $Δ_{1} \neq 0$ 。则边值问题

${\begin{cases} {}_{t}^{c}D_{ξ^{-}}^{α} u (t) = h (t), t \in (0, ξ), \\ {}_{ξ^{+}}^{c}D_{t}^{β} u (t) = y (t), t \in (ξ, 1), \\ Δ u (ξ) = I, Δ u^{'} (ξ) = Q, \\ m_{1} u (0) + n_{1} u (1) = γ_{0}, m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} u (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} u (1) = γ_{1} \end{cases}$ (2)

在E中存在唯一解

$u (t) = {\begin{cases} \int_{0}^{ξ} G_{1} (t, s) h (s) d s + \int_{ξ}^{1} g_{2} (t, s) y (s) d s + Δ_{2} + \frac{t}{Δ_{1}} (- \frac{Q n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1}), t \in [0, ξ], \\ \int_{ξ}^{1} G_{2} (t, s) y (s) d s + \int_{0}^{ξ} g_{1} (t, s) h (s) d s + Δ_{3} + \frac{t}{Δ_{1}} (\frac{Q m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1}), t \in (ξ, 1] . \end{cases}$ (3)

其中

$G_{1} (t, s) = {\begin{cases} g_{1} (t, s), 0 \leq s \leq t \leq ξ, \\ g_{1} (t, s) + \frac{1}{Γ (α)} {(s - t)}^{α - 1}, 0 \leq t \leq s \leq ξ; \end{cases}$ (4)

$G_{2} (t, s) = {\begin{cases} g_{2} (t, s) + \frac{1}{Γ (β)} {(t - s)}^{β - 1}, ξ \leq s \leq t \leq 1, \\ g_{2} (t, s), ξ \leq t \leq s \leq 1; \end{cases}$ (5)

记

$Δ_{1} = \frac{m_{2} ξ^{2 - α}}{Γ (3 - α)} - \frac{n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)}; Δ_{2} = - \frac{1}{m_{1} + n_{1}} (n_{1} I + n_{1} (1 - ξ + \frac{n_{2} {(1 - ξ)}^{2 - β}}{Δ_{1} Γ (3 - β)}) Q - \frac{n_{1}}{Δ_{1}} γ_{1} - γ_{0});$

$Δ_{3} = - \frac{1}{m_{1} + n_{1}} (- m_{1} I + (m_{1} ξ + n_{1} + \frac{n_{1} n_{2} {(1 - ξ)}^{2 - β}}{Δ_{1} Γ (3 - β)}) Q - \frac{n_{1}}{Δ_{1}} γ_{1} - γ_{0});$

$g_{1} (t, s) = - \frac{1}{m_{1} + n_{1}} (\frac{m_{1}}{Γ (α)} s^{α - 1} + \frac{n_{1} m_{2}}{Δ_{1}}) + \frac{t m_{2}}{Δ_{1}}; g_{2} (t, s) = - \frac{n_{1}}{m_{1} + n_{1}} (\frac{1}{Γ (β)} {(1 - s)}^{β - 1} + \frac{n_{2}}{Δ_{1}}) - \frac{t n_{2}}{Δ_{1}} .$

证明：设 $u \in E$ 是边值问题(2)的解，则由定义1可知存在常数 $c_{i} \in ℝ$ , $i = 0, 1, 2, 3$ 使 ${}_{t}^{C}D_{ξ^{-}}^{α} u (t) = h (t)$ 的解为：

$u (t) = {}_{t}I_{ξ -}^{α} h (t) + c_{0} + c_{1} t = \frac{1}{Γ (α)} \int_{t}^{ξ} {(s - t)}^{α - 1} h (s) d s + c_{0} + c_{1} t,$

$u^{'} (t) = - \frac{1}{Γ (α - 1)} \int_{t}^{ξ} {(s - t)}^{α - 2} h (s) d s + c_{1},$

${}_{t}^{c}D_{ξ^{-}}^{α - 1} u (t) = \int_{t}^{ξ} h (s) d s - \frac{c_{1}}{Γ (3 - α)} {(ξ - t)}^{2 - α}$

${}_{ξ^{+}}^{c}D_{t}^{β} u (t) = y (t)$ 的解为：

$u (t) = {}_{ξ^{+}}I_{t}^{β} y (t) + c_{2} + c_{3} t = \frac{1}{Γ (β)} \int_{ξ}^{t} {(t - s)}^{β - 1} y (s) d s + c_{2} + c_{3} t,$

$u^{'} (t) = \frac{1}{Γ (β - 1)} \int_{ξ}^{t} {(t - s)}^{β - 2} y (s) d s + c_{3},$

${}_{ξ^{+}}^{c}D_{t}^{β - 1} u (t) = \int_{ξ}^{t} y (s) d s + \frac{c_{3}}{Γ (3 - β)} {(t - ξ)}^{2 - β}$

由边值条件 $Δ u (ξ) = I, Δ u^{'} (ξ) = Q$ 得

${\begin{cases} c_{2} - c_{0} = I - Q ξ, \\ c_{3} - c_{1} = Q . \end{cases}$

再由边值条件 $m_{1} u (0) + n_{1} u (1) = γ_{0}$ ， $m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} u (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} u (1) = γ_{1}$ ，可得

${\begin{cases} c_{0} = - \frac{1}{m_{1} + n_{1}} (\int_{0}^{ξ} (\frac{m_{1}}{Γ (α)} s^{α - 1} + \frac{n_{1} m_{2}}{Δ_{1}}) h (s) d s + n_{1} \int_{ξ}^{1} (\frac{1}{Γ (β)} {(1 - s)}^{β - 1} - \frac{n_{2}}{Δ_{1}}) y (s) d s \\ + n_{1} I + n_{1} (1 - ξ + \frac{n_{2} {(1 - ξ)}^{2 - β}}{Δ_{1} Γ (3 - β)}) Q - \frac{n_{1}}{Δ_{1}} γ_{1} - γ_{0}), \\ c_{1} = \frac{1}{Δ_{1}} (m_{2} \int_{0}^{ξ} h (s) d s + n_{2} \int_{ξ}^{1} y (s) d s + \frac{Q n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1}), \\ c_{2} = - \frac{1}{m_{1} + n_{1}} (\int_{0}^{ξ} (\frac{m_{1} s^{α - 1}}{Γ (α)} + \frac{n_{1} m_{2}}{Δ_{1}}) h (s) d s + n_{1} \int_{ξ}^{1} (\frac{{(1 - s)}^{β - 1}}{Γ (β)} - \frac{n_{2}}{Δ_{1}}) y (s) d s \\ - m_{1} I + (m_{1} ξ + n_{1} + \frac{n_{1} n_{2} {(1 - ξ)}^{2 - β}}{Δ_{1} Γ (3 - β)}) Q - \frac{n_{1}}{Δ_{1}} γ_{1} - γ_{0}), \\ c_{3} = \frac{1}{Δ_{1}} (m_{2} \int_{0}^{ξ} h (s) d s + n_{2} \int_{ξ}^{1} y (s) d s + \frac{Q m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1}) . \end{cases}$

因此，当 $t \in [0, ξ]$ 时，

$\begin{matrix} u (t) = \frac{1}{Γ (α)} \int_{t}^{ξ} {(s - t)}^{α - 1} h (s) d s - \frac{1}{m_{1} + n_{1}} (\int_{0}^{ξ} (\frac{m_{1}}{Γ (α)} s^{α - 1} + \frac{n_{1} m_{2}}{Δ_{1}}) h (s) d s \\ + n_{1} \int_{ξ}^{1} (\frac{1}{Γ (β)} {(1 - s)}^{β - 1} - \frac{n_{2}}{Δ_{1}}) y (s) d s + n_{1} I + n_{1} (1 - ξ + \frac{n_{2} {(1 - ξ)}^{2 - β}}{Δ_{1} Γ (3 - β)}) Q - \frac{n_{1}}{Δ_{1}} γ_{1} - γ_{0}) \\ + \frac{t}{Δ_{1}} (m_{2} \int_{0}^{ξ} h (s) d s + n_{2} \int_{ξ}^{1} y (s) d s + \frac{Q n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1}) \\ = \int_{0}^{ξ} G_{1} (t, s) h (s) d s + \int_{ξ}^{1} g_{2} (t, s) y (s) d s + Δ_{2} + \frac{t}{Δ_{1}} (\frac{Q n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1}) . \end{matrix}$

当 $t \in (ξ, 1]$ 时，

$\begin{matrix} u (t) = \frac{1}{Γ (β)} \int_{ξ}^{t} {(t - s)}^{β - 1} y (s) d s - \frac{1}{m_{1} + n_{1}} (\int_{0}^{ξ} (\frac{m_{1}}{Γ (α)} s^{α - 1} + \frac{n_{1} m_{2}}{Δ_{1}}) h (s) d s \\ + n_{1} \int_{ξ}^{1} (\frac{1}{Γ (β)} {(1 - s)}^{β - 1} + \frac{n_{2}}{Δ_{1}}) y (s) d s - m_{1} I + (m_{1} ξ + n_{1} + \frac{n_{1} n_{2} {(1 - ξ)}^{2 - β}}{Δ_{1} Γ (3 - β)}) Q - \frac{n_{1}}{Δ_{1}} γ_{1} - γ_{0}) \\ + \frac{t}{Δ_{1}} (m_{2} \int_{0}^{ξ} h (s) d s + n_{2} \int_{ξ}^{1} y (s) d s + \frac{Q m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1}) \\ = \int_{ξ}^{1} G_{2} (t, s) y (s) d s + \int_{0}^{ξ} g_{1} (t, s) h (s) d s + Δ_{3} + \frac{t}{Δ_{1}} (\frac{Q m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1}) . \end{matrix}$

易证(3)是方程(2)的解，反之亦然。

证毕。

为了以后证明，我们给出如下假设：

(H1) $m_{i}, n_{i} \in ℝ (i = 1, 2)$ , $n_{1} > 0, n_{2} > 0, - n_{1} > m_{1} > - \frac{n_{1}}{ξ}, γ_{1} \leq 0, γ_{0} \leq 0$ ，

$m_{2} > \max {\frac{Γ (3 - α) ξ^{α - 1} n_{2}}{Γ (3 - β) (Γ (3 - α) Γ (α) - ξ)}, - \frac{Γ (3 - α) n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β) ξ^{2 - α}}} .$

引理3：假设(H1)成立，则由式(4)、(5)定义的函数 $G_{i} (t, s), i = 1, 2$ 满足以下性质：

1) $0 < G_{1} (0, s) \leq G_{1} (t, s) \leq G_{1} (ξ, s)$ ，对任意 $(t, s) \in [0, ξ] \times [0, ξ]$ ；

2) $0 < G_{2} (ξ, s) \leq G_{2} (t, s) \leq G_{2} (1, s)$ ，对任意 $(t, s) \in [ξ, 1] \times [ξ, 1]$ 。

证明：1) 显然 $G_{i} (t, s), i = 1, 2$ 为连续函数。由(H1)知， $Δ_{1} > 0$ ， $m_{2} > \frac{ξ^{α - 1}}{Γ (α)} Δ_{1}$ ，则对于 $t \in [0, ξ]$ ，当 $0 \leq s \leq t \leq ξ$ 时，

$G_{1} (t, s) = g_{1} (t, s) = - \frac{1}{m_{1} + n_{1}} (\frac{m_{1}}{Γ (α)} s^{α - 1} + \frac{n_{1} m_{2}}{Δ_{1}}) + \frac{t m_{2}}{Δ_{1}}, \frac{\partial g_{1} (t, s)}{\partial t} = \frac{m_{2}}{Δ_{1}} > 0;$

当 $0 \leq s \leq t \leq ξ$ 时，由于 $G_{1} (t, s) = g_{1} (t, s) + \frac{1}{Γ (α)} {(s - t)}^{α - 1}$ ，

$\frac{\partial G_{1} (t, s)}{\partial t} = - \frac{1}{Γ (α - 1)} {(s - t)}^{α - 2} + \frac{m_{2}}{Δ_{1}}, \frac{\partial^{2} G_{1} (t, s)}{\partial t^{2}} = \frac{α - 2}{Γ (α - 1)} {(s - t)}^{α - 3} < 0,$

则 $\frac{\partial G_{1} (t, s)}{\partial t} \geq \frac{\partial G_{1} (s, s)}{\partial t} = \frac{m_{2}}{Δ_{1}} > 0$ 。因此， $G_{1} (t, s)$ 是关于t的单调递增函数，且

$G_{1} (0, s) \leq G_{1} (t, s) \leq G_{1} (ξ, s) .$

又

$\begin{array}{l} G_{1} (0, s) = g_{1} (0, s) + \frac{1}{Γ (α)} s^{α - 1} = - \frac{1}{m_{1} + n_{1}} (\frac{m_{1}}{Γ (α)} s^{α - 1} + \frac{n_{1} m_{2}}{Δ_{1}}) + \frac{1}{Γ (α)} s^{α - 1} \\ = - \frac{1}{m_{1} + n_{1}} (\frac{- n_{1}}{Γ (α)} s^{α - 1} + \frac{n_{1} m_{2}}{Δ_{1}}) > - \frac{n_{1}}{m_{1} + n_{1}} (\frac{- 1}{Γ (α)} ξ^{α - 1} + \frac{m_{2}}{Δ_{1}}) > 0, \end{array}$

则 $0 < G_{1} (0, s) \leq G_{1} (t, s) \leq G_{1} (ξ, s)$ 成立。

2) 对于 $t \in [ξ, 1]$ ，当 $ξ \leq t \leq s \leq 1$ 时，

$G_{2} (t, s) = g_{2} (t, s) = - \frac{n_{1}}{m_{1} + n_{1}} (\frac{1}{Γ (β)} {(1 - s)}^{β - 1} + \frac{n_{2}}{Δ_{1}}) + \frac{t n_{2}}{Δ_{1}}, \frac{\partial g_{2} (t, s)}{\partial t} = \frac{n_{2}}{Δ_{1}} > 0;$

当 $ξ \leq s \leq t \leq 1$ 时，

$G_{2} (t, s) = g_{2} (t, s) + \frac{1}{Γ (β)} {(t - s)}^{β - 1}, \frac{\partial G_{2} (t, s)}{\partial t} = \frac{n_{2}}{Δ_{1}} + \frac{1}{Γ (β - 1)} {(t - s)}^{β - 2} > 0,$

则 $G_{2} (t, s)$ 是关于t的单调递增函数，那么

$G_{2} (ξ, s) \leq G_{2} (t, s) \leq G_{2} (1, s) .$

又

$G_{2} (ξ, s) = - \frac{n_{1}}{m_{1} + n_{1}} (\frac{1}{Γ (β)} {(1 - s)}^{β - 1} + \frac{n_{2}}{Δ_{1}}) + \frac{n_{2}}{Δ_{1}} = \frac{1}{m_{1} + n_{1}} (\frac{n_{1}}{Γ (β)} {(1 - s)}^{β - 1} + \frac{n_{2} (1 - (m_{1} + n_{1}) ξ)}{Δ_{1}}) > 0,$

因此， $0 < G_{2} (ξ, s) \leq G_{2} (t, s) \leq G_{2} (1, s)$ 成立。

证毕。

引理4：若(H1)成立， $I, Q \in ℝ^{+}$ ， $Δ_{1} \neq 0$ 。若u满足

${\begin{cases} {}_{t}^{c}D_{ξ^{-}}^{α} u (t) \geq 0, t \in (0, ξ), \\ {}_{ξ^{+}}^{c}D_{t}^{β} u (t) \geq 0, t \in (ξ, 1), \\ Δ u (ξ) = I, Δ u^{'} (ξ) = Q, \\ m_{1} u (0) + n_{1} u (1) \leq 0, m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} u (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} u (1) \geq 0, \end{cases}$ (4)

则 $u (t) \geq 0$ ， $t \in [0, 1]$ 。

证明：对任意 $h \in C ([0, ξ], ℝ^{+})$ ， $y \in C ((ξ, 1], ℝ^{+})$ ，由于 $γ_{1} \leq 0$ ， $γ_{0} \leq 0$ 为常数。考虑以下边值问题:

${\begin{cases} {}_{t}^{c}D_{ξ^{-}}^{α} u (t) = h (t), t \in [0, ξ], \\ {}_{ξ^{+}}^{c}D_{t}^{β} u (t) = y (t), t \in (ξ, 1], \\ Δ u (ξ) = I, Δ u^{'} (ξ) = Q, \\ m_{1} u (0) + n_{1} u (1) = γ_{0}, m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} u (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} u (1) = γ_{1} . \end{cases}$

由引理2可得

$u (t) = {\begin{cases} \int_{0}^{ξ} G_{1} (t, s) h (s) d s + \int_{ξ}^{1} g_{2} (t, s) y (s) d s + Δ_{2} + \frac{t}{Δ_{1}} (\frac{Q n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1}), t \in [0, ξ], \\ \int_{ξ}^{1} G_{2} (t, s) y (s) d s + \int_{0}^{ξ} g_{1} (t, s) h (s) d s + Δ_{2} + \frac{t}{Δ_{1}} (\frac{Q m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1}), t \in (ξ, 1] . \end{cases}$

由(H1)可得， $Δ_{1}, Δ_{2}, Δ_{3} > 0$ ， $\frac{Q n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1} > 0$ ， $\frac{Q m_{2} ξ^{2 - α}}{Γ (3 - β)} - γ_{1} > 0$ ， $t \in [0, 1]$ 。再由引理3可得 $u (t) \geq 0$ ， $t \in [0, 1]$ 显然成立。

证毕。

3. 分数阶微分方程的上下解方法

为方便叙述，我们假设下文满足以下假设：

(H2) 对任意 $u_{1} \leq u_{2}, v_{1} \leq v_{2}$ ， $f (t, u_{1}, v_{1}) \leq f (t, u_{2}, v_{2}), t \in [0, ξ]$ ，

$g (t, u_{1}, v_{1}) \leq g (t, u_{2}, v_{2}), t \in (ξ, 1]$ , $I (u_{1}) \leq I (u_{2}), Q (u_{1}) \leq Q (u_{2}), t \in [0, 1]$ .

对任意 $u_{1} < u_{2}, v_{1} < v_{2}$ ， $f (t, u_{1}, v_{1}) < f (t, u_{2}, v_{2}), t \in [0, ξ]$ ，

$g (t, u_{1}, v_{1}) < g (t, u_{2}, v_{2}), t \in (ξ, 1]$ , $I (u_{1}) < I (u_{2}), Q (u_{1}) < Q (u_{2}), t \in [0, 1]$ .

(H3) 若(H1)成立，且 $x_{1}, x_{2}, y_{1}, y_{2} \in ℝ$ ，当 $x_{1} \leq x_{2}, y_{1} \leq y_{2}$ 时，

$h_{0} (x_{2}, y_{2}) - h_{0} (x_{1}, y_{1}) \leq - m_{1} (x_{2} - x_{1}) - n_{1} (y_{2} - y_{1})$ ,

$h_{1} (x_{2}, y_{2}) - h_{0} (x_{1}, y_{1}) \geq - m_{2} (x_{2} - x_{1}) - n_{2} (y_{2} - y_{1})$ .

记 $P = {u \in E : u (t) \geq 0, t \in [0, 1]}$ ，显然P为E中的正规体锥。且若 $u (t) \leq v (t)$ , $t \in [0, 1]$ ，则 $u \underline{≺} v \in P$ 。

对任意 $u \in P$ ，考虑如下边值问题：

${\begin{cases} {}_{t}^{c}D_{ξ^{-}}^{α} u (t) = f_{1} (t, x (t), x (t + τ_{1})), t \in (0, ξ), \\ {}_{ξ^{+}}^{c}D_{t}^{β} u (t) = f_{2} (t, x (t), x (t - τ_{2})), t \in (ξ,1), \\ Δ u (ξ) = I (ξ, x (ξ)), Δ u^{'} (ξ) = Q (ξ, x (ξ)), \\ m_{1} u (0) + n_{1} u (1) = h_{0} (x (0), x (1)) + m_{1} x (0) + n_{1} x (1) : = γ_{0} (x), \\ m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} u (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} u (1) \\ = h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} x (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} x (1)) + m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} x (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} x (1) : = γ_{1} (x) . \end{cases}$ (5)

由引理2知，边值问题(5)有唯一解

$u (t) = {\begin{cases} \int_{0}^{ξ} G_{1} (t, s) f_{1} (s, x (s), x (s + τ_{1})) d s + \int_{ξ}^{1} g_{2} (t, s) f_{2} (s, x (s), x (s - τ_{2})) d s \\ + Δ_{2}^{x} + \frac{t}{Δ_{1}} (\frac{Q (ξ, x (ξ)) n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1} (x)), t \in [0, ξ], \\ \int_{ξ}^{1} G_{2} (t, s) f_{2} (s, x (s), x (s - τ_{2})) d s + \int_{0}^{ξ} g_{1} (t, s) f_{1} (s, x (s), x (s + τ_{1})) d s \\ + Δ_{3}^{x} + \frac{t}{Δ_{1}} (\frac{Q (ξ, x (ξ)) m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1} (x)), t \in (ξ, 1] . \end{cases}$

其中

$Δ_{2}^{x} = - \frac{1}{m_{1} + n_{1}} (n_{1} I (ξ, x (ξ)) + n_{1} (1 - ξ + \frac{n_{2} {(1 - ξ)}^{2 - β}}{Δ_{1} Γ (3 - β)}) Q (ξ, x (ξ)) - \frac{n_{1}}{Δ_{1}} γ_{1} (x) - γ_{0} (x));$

$Δ_{3}^{x} = - \frac{1}{m_{1} + n_{1}} (- m_{1} I (ξ, x (ξ)) + (m_{1} ξ + n_{1} + \frac{n_{1} n_{2}}{Δ_{1} Γ (3 - β)}) Q (ξ, x (ξ)) - \frac{n_{1}}{Δ_{1}} γ_{1} (x) - γ_{0} (x)) .$

定义算子

$T x (t) = {\begin{cases} \int_{0}^{ξ} G_{1} (t, s) f_{1} (s, x (s), x (s + τ_{1})) d s + \int_{ξ}^{1} g_{2} (t, s) f_{2} (s, x (s), x (s - τ_{2})) d s \\ + Δ_{2}^{x} + \frac{t}{Δ_{1}} (\frac{Q (ξ, x (ξ)) n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1} (x)), t \in [0, ξ], \\ \int_{ξ}^{1} G_{2} (t, s) f_{2} (s, x (s), x (s - τ_{2})) d s + \int_{0}^{ξ} g_{1} (t, s) f_{1} (s, x (s), x (s + τ_{1})) d s \\ + Δ_{3}^{x} + \frac{t}{Δ_{1}} (\frac{Q (ξ, x (ξ)) m_{2} ξ^{2 - α}}{Γ (3 - β)} - γ_{1} (x)), t \in (ξ, 1] . \end{cases}$

引理5：若(H1)成立，则T为全连续算子。

证明：由引理3，引理4知，对任意 $x \in P$ ，当 $t \in [0, 1]$ 时， $T x \geq 0$ 显然成立。

因此， $T : P \to P$ 是有意义的。

接下来，我们分两步证明：

第一步：T是连续算子。

设对任意 $x_{n} \in P$ , $n = 1, 2, \dots$ 存在 $x \in P$ 使得当 $n \to \infty$ 时， $‖ x_{n} - x ‖ \to 0$ 。则存在 ${\bar{M}}_{0} > 0$ ，使得 $‖ x_{n} ‖ \leq {\bar{M}}_{0}$ , $‖ x ‖ \leq {\bar{M}}_{0}$ 。又由于 $f_{1}, f_{2}$ 连续， $I, Q \in C (ℝ, ℝ)$ ，且 $γ_{0} (x), γ_{1} (x) \in ℝ$ ，则

$\lim_{n \to \infty} (f_{1} (t, x_{n} (t), x_{n} (t + τ_{1})) - f_{1} (t, x (t), x (t + τ_{1}))) = 0,$

$\lim_{n \to \infty} (f_{2} (t, x_{n} (t), x_{n} (t - τ_{2})) - f_{2} (t, x (t), x (t - τ_{2}))) = 0,$

$\lim_{n \to \infty} | I (x_{n} (t)) - I (x (t)) | = 0, \lim_{n \to \infty} | Q (x_{n} (t)) - Q (x (t)) | = 0,$

$\lim_{n \to \infty} (γ_{0} (x_{n}) - γ_{0} (x)) = 0, \lim_{n \to \infty} (γ_{1} (x_{n}) - γ_{1} (x)) = 0,$

且存在常数 ${\bar{M}}_{1} > 0$ ，使得 $\sup_{(t, u, v) \in A} | f_{1} (t, u, v) | \leq {\bar{M}}_{1}$ , $\sup_{(t, u, v) \in B} | f_{2} (t, u, v) | \leq {\bar{M}}_{1}$ ，其中

$A = [0, ξ] \times [- {\bar{M}}_{0}, {\bar{M}}_{0}] \times [- {\bar{M}}_{0}, {\bar{M}}_{0}], B = [ξ, 1] \times [- {\bar{M}}_{0}, {\bar{M}}_{0}] \times [- {\bar{M}}_{0}, {\bar{M}}_{0}] .$

再由引理3可得，当 $t \in [0, ξ]$ 时，

$\begin{array}{l} | T (x_{n}) - T (x) | \\ = | \int_{0}^{ξ} G_{1} (t, s) (f_{1} (s, x_{n} (s), x_{n} (s + τ_{1})) - f_{1} (s, x (t), x (s + τ_{1}))) d s \begin{matrix} \overset{}{} \end{matrix} \\ + \int_{ξ}^{1} g_{2} (t, s) (f_{2} (s, x_{n} (s), x_{n} (s - τ_{2})) - f_{2} (s, x (s), x (s - τ_{2}))) d s \\ + (Δ_{2}^{x_{n}} - Δ_{2}^{x}) + \frac{t}{Δ_{1}} (\frac{(Q (ξ, x_{n} (ξ)) - Q (ξ, x (ξ))) n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - (γ_{1} (x_{n}) - γ_{1} (x))) | \end{array}$

则由Lebesgue控制收敛定理可知， $\lim_{n \to \infty} {‖ T u_{n} - T u ‖}_{[0, ξ]} = 0$ 。同理可得， $\lim_{n \to \infty} {‖ T u_{n} - T u ‖}_{(ξ, 1]} = 0$ 。

因此，对任意 $t \in [0, 1]$ ，有 $\lim_{n \to \infty} ‖ T u_{n} - T u ‖ = 0$ ，则算子T是连续算子。

第二步：T是紧的。

令 $Ω \subset P$ 为有界集，由 $f_{1}$ ， $f_{2}$ ，I，Q的连续性得，存在 ${\bar{M}}_{2} > 0$ ，使得对任意 $t \in [0, ξ]$ , $u, v \in Ω$ ，有 $| f_{1} (t, u, v) | \leq {\bar{M}}_{2}$ ；对任意 $t \in (ξ, 1]$ ， $u, v \in Ω$ 有 $| f_{2} (t, u, v) | \leq {\bar{M}}_{2}$ , $| I | \leq {\bar{M}}_{2}$ , $| Q | \leq {\bar{M}}_{2}$ , $| γ_{0} (x) | \leq {\bar{M}}_{2}$ , $| γ_{1} (x) | \leq {\bar{M}}_{2}$ 。

则

$| Δ_{2}^{x} | = - \frac{1}{m_{1} + n_{1}} (n_{1} (2 - ξ + \frac{n_{2} {(1 - ξ)}^{2 - β} + Γ (3 - β)}{Δ_{1} Γ (3 - β)}) - 1) {\bar{M}}_{2};$

$| Δ_{3}^{x} | = - \frac{1}{m_{1} + n_{1}} (m_{1} (ξ - 1) + n_{1} (1 + \frac{n_{2} - Γ (3 - β)}{Δ_{1} Γ (3 - β)}) - 1) {\bar{M}}_{2}$

$\begin{matrix} \sup_{t \in [0, ξ]} | T u (t) | \leq | \int_{0}^{ξ} G_{1} (ξ, s) f_{1} (s, x (s), x (s + τ_{1})) d s + \int_{ξ}^{1} g_{2} (1, s) f_{2} (s, x (s), x (s - τ_{2})) d s \begin{matrix} \overset{}{} \end{matrix} \\ + Δ_{2}^{x} + \frac{ξ}{Δ_{1}} (\frac{Q (ξ, x (ξ)) n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1} (x)) | \\ \leq (\int_{0}^{ξ} G_{1} (ξ, s) d s + \int_{ξ}^{1} g_{2} (1, s) d s - \frac{1}{m_{1} + n_{1}} (n_{1} (2 - ξ + \frac{n_{2} {(1 - ξ)}^{2 - β} - Γ (3 - β)}{Δ_{1} Γ (3 - β)}) - 1) \\ + \frac{ξ}{Δ_{1}} (\frac{n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - 1)) {\bar{M}}_{2}, \end{matrix}$

$\begin{matrix} \sup_{t \in (ξ, 1]} | T u (t) | \leq | \int_{ξ}^{1} G_{2} (1, s) f_{2} (s, x (s), x (s - τ_{2})) d s + \int_{0}^{ξ} g_{1} (ξ, s) f_{1} (s, x (s), x (s + τ_{1})) d s \begin{matrix} \overset{}{} \end{matrix} \\ + Δ_{3}^{x} + \frac{ξ}{Δ_{1}} (\frac{Q (ξ, x (ξ)) m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1} (x)) | \\ \leq (\int_{ξ}^{1} G_{2} (1, s) d s + \int_{0}^{ξ} g_{1} (ξ, s) d s - \frac{1}{m_{1} + n_{1}} (m_{1} (ξ - 1) + n_{1} (1 + \frac{n_{2} - Γ (3 - β)}{Δ_{1} Γ (3 - β)}) - 1) \\ + \frac{1}{Δ_{1}} (\frac{m_{2} ξ^{2 - α}}{Γ (3 - β)} - 1)) {\bar{M}}_{2} . \end{matrix}$

因此，算子 $T (Ω)$ 一致有界。

由于 $G_{1} (t, s)$ , $g_{2} (t, s)$ 在 $[0, ξ] \times [0, ξ]$ 上连续，所以 $G_{1} (t, s)$ , $g_{2} (t, s)$ 在 $[0, ξ] \times [0, ξ]$ 上一致连续。因此对任意 $ε > 0$ ，存在 $0 < δ_{1} < \frac{ε Δ_{1} Γ (3 - β)}{2 | n_{2} {(1 - ξ)}^{2 - β} - Γ (3 - β) (1 - n_{2}) | {\bar{M}}_{2}}$ ，当 $| t_{1} - t_{2} | < δ_{1}$ 时，有 $| G_{1} (t_{1}, s) - G_{1} (t_{2}, s) | < \frac{ε}{4 {\bar{M}}_{2}}$ , $| g_{2} (t_{1}, s) - g_{2} (t_{2}, s) | < \frac{ε}{4 {\bar{M}}_{2}}$ 。因此，对任意的 $t_{1}, t_{2} \in [0, ξ]$ ， $| t_{1} - t_{2} | < δ_{1}$ ， $u \in Ω$ 有

$\begin{matrix} | T u (t_{2}) - T u (t_{1}) | = | \int_{0}^{ξ} (G_{1} (t_{1}, s) - G_{1} (t_{2}, s)) f_{1} (s, u (s), u (s + τ_{1})) d s \begin{matrix} \overset{}{} \end{matrix} \\ + \int_{ξ}^{1} (g_{2} (t_{1}, s) - g_{2} (t_{2}, s)) f_{2} (s, u (s), u (s - τ_{2})) d s \\ + \frac{t_{1} - t_{2}}{Δ_{1}} (\frac{Q (ξ, x (ξ)) n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1} (x)) | \\ \leq {\bar{M}}_{2} (\int_{0}^{ξ} | G_{1} (t_{1}, s) - G_{1} (t_{2}, s) | d s + \int_{ξ}^{1} | g_{2} (t_{1}, s) - g_{2} (t_{2}, s) | d s) \\ + | t_{1} - t_{2} | \frac{| n_{2} {(1 - ξ)}^{2 - β} - Γ (3 - β) (1 - n_{2}) {\bar{M}}_{2}}{Δ_{1} Γ (3 - β)} \\ < ε . \end{matrix}$

又由于 $G_{2} (t, s)$ ， $g_{1} (t, s)$ 在 $[ξ, 1] \times [ξ, 1]$ 上连续，所以 $G_{2} (t, s)$ ， $g_{1} (t, s)$ 在 $[ξ, 1] \times [ξ, 1]$ 上一致连续。

因此对上述 $ε > 0$ ，存在 $0 < δ_{2} < \frac{ε Δ_{1} Γ (3 - β)}{2 | m_{2} ξ^{2 - α} - (m_{2} + 1) Γ (3 - β) |}$ ，当 $| t_{3} - t_{4} | < δ_{2}$ 时，有 $| G_{2} (t_{3}, s) - G_{2} (t_{4}, s) | < \frac{ε}{4 {\bar{M}}_{2}}$ , $| g_{1} (t_{3}, s) - g_{1} (t_{4}, s) | < \frac{ε}{4 {\bar{M}}_{2}}$ 。

因此，对任意的 $t_{3}, t_{4} \in (ξ, 1]$ ， $| t_{3} - t_{4} | < δ_{2}$ ， $u \in Ω$ ，有

$\begin{matrix} | T u (t_{3}) - T u (t_{4}) | = | \int_{ξ}^{1} (G_{2} (t_{3}, s) - G_{1} (t_{4}, s)) f_{2} (s, u (s), u (s - τ_{2})) d s \begin{matrix} \overset{}{} \end{matrix} \\ + \int_{0}^{ξ} (g_{1} (t_{3}, s) - g_{1} (t_{4}, s)) f_{1} (s, u (s), u (s + τ_{1})) d s \\ + \frac{t_{3} - t_{4}}{Δ_{1}} (\frac{Q (ξ, x (ξ)) m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1} (x)) | \\ \leq {\bar{M}}_{2} (\int_{ξ}^{1} | G_{2} (t_{3}, s) - G_{1} (t_{4}, s) | d s + \int_{0}^{ξ} | g_{1} (t_{3}, s) - g_{1} (t_{4}, s) | d s) \\ + | t_{3} - t_{4} | \frac{| m_{2} ξ^{2 - α} - (m_{2} + 1) Γ (3 - β) |}{Δ_{1} Γ (3 - β)} {\bar{M}}_{2} \\ < ε . \end{matrix}$

因此， $T (Ω)$ 在 $[0, ξ], (ξ, 1]$ 上等度连续，易知，当 $t \in [0, 1]$ 时，对任意 $ε > 0$ ，存在 $δ_{3} > 0$ ，当 $| t_{5} - t_{6} | < δ_{3}$ 时 $| T u (t_{5}) - T u (t_{6}) | < ε$ ，因此， $T (Ω)$ 是等度连续的。

由Arzela-Ascoli定理知 $T (Ω)$ 相对列紧。又因为算子T是连续算子，所以算子T是全连续的。

证毕。

引理6：T为强增算子。

证明：对任意 $x_{1}, x_{2} \in E, x_{1} ≺ x_{2}$ ，即 $x_{1} (t) \leq x_{2} (t)$ 且 $x_{1} (t) \equiv x_{2} (t)$ ，由(H2)可得，

$f_{1} (t, x_{2} (t), x_{2} (t + τ_{1})) - f_{1} (t, x_{1} (t), x_{1} (t + τ_{1})) \geq 0, t \in [0, ξ],$

$f_{2} (t, x_{2} (t), x_{2} (t - τ_{2})) - f_{2} (t, x_{1} (t), x_{1} (t - τ_{2})) \geq 0, t \in (ξ, 1],$

$(I (ξ, x_{2} (ξ)) - I (ξ, x_{1} (ξ))) \geq 0, (Q (ξ, x_{2} (ξ)) - Q (ξ, x_{1} (ξ))) \geq 0, t \in [0, 1] .$

由于 $x_{1} (t) \equiv x_{2} (t)$ ，则存在区间 $[a, b] \subset [0, ξ]$ 或 $[a, b] \subset (ξ, 1]$ 使得当 $t \in [a, b]$ 时， $x_{1} (t) < x_{2} (t)$ 。

因此，当 $[a, b] \subset [0, ξ]$ 时，

$f_{1} (t, x_{2} (t), x_{2} (t + τ_{1})) - f_{1} (t, x_{1} (t), x_{1} (t + τ_{1})) > 0,$

$(I (ξ, x_{2} (ξ)) - I (ξ, x_{1} (ξ))) > 0, (Q (ξ, x_{2} (ξ)) - Q (ξ, x_{1} (ξ))) > 0,$

且由(H3)可得

$\begin{matrix} γ_{0} (x_{2}) - γ_{0} (x_{1}) = h_{0} (x_{2} (0), x_{2} (1)) - h_{0} (x_{1} (0), x_{1} (1)) \\ + (m_{1} x_{2} (0) + n_{1} x_{2} (1)) - (\begin{matrix} m_{1} x_{1} (0) + n_{1} x_{1} (1) \\ \leq 0, \end{matrix}) \end{matrix}$

$\begin{matrix} γ_{1} (x_{2}) - γ_{1} (x_{1}) = h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} x_{2} (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} x_{2} (1)) - h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} x_{2} (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} x_{2} (1)) \\ + m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} x_{2} (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} x_{2} (1) - (m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} x_{1} (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} x_{1} (1)) \\ \leq 0, \end{matrix}$

$\begin{matrix} Δ_{2}^{x_{2}} - Δ_{2}^{x_{1}} = - \frac{1}{m_{1} + n_{1}} (n_{1} (I (ξ, x_{2} (ξ)) - I (ξ, x_{1} (ξ)))) \\ + n_{1} (1 - ξ + \frac{n_{2} {(1 - ξ)}^{2 - β}}{Δ_{1} Γ (3 - β)}) (Q (ξ, x_{2} (ξ)) - Q (ξ, x_{1} (ξ))) \end{matrix}$

$- \frac{n_{1}}{Δ_{1}} (γ_{1} (x_{2}) - γ_{1} (x_{1})) - (γ_{0} (x_{2}) - γ_{0} (x_{1})) \geq 0,$

$\begin{array}{l} T x_{2} (t) - T x_{1} (t) \\ = \int_{0}^{ξ} G_{1} (t, s) (f_{1} (s, x_{2} (s), x_{2} (s + τ_{1})) - f_{1} (s, x_{1} (s), x_{1} (s + τ_{1}))) d s + Δ_{2}^{x_{2}} - Δ_{2}^{x_{1}} \\ + \int_{ξ}^{1} g_{2} (t, s) (f_{2} (t, x_{2} (t), x_{2} (t - τ_{2})) - f_{2} (t, x_{1} (t), x_{1} (t - τ_{2}))) d s \\ + \frac{t}{Δ_{1}} (\frac{(Q (ξ, x_{2} (ξ)) - Q (ξ, x_{1} (ξ))) n_{2} {(1 - ξ)}^{2 - β}}{Γ (3 - β)} - γ_{1} (x_{2}) - γ_{1} (x_{1})) \\ > \int_{0}^{ξ} G_{1} (t, s) (f_{1} (s, x_{2} (s), x_{2} (s + τ_{1})) - f_{1} (s, x_{1} (s), x_{1} (s + τ_{1}))) d s \\ > 0. \end{array}$

同理当 $[a, b] \subset (ξ, 1]$ 时，

$(f_{2} (t, x_{2} (t), x_{2} (t - τ_{2})) - f_{2} (t, x_{1} (t), x_{1} (t - τ_{2}))) > 0,$

$\begin{matrix} Δ_{3}^{x_{2}} - Δ_{3}^{x_{1}} = - \frac{1}{m_{1} + n_{1}} (- m_{1} (I (ξ, x_{2} (ξ)) - I (ξ, x_{1} (ξ)))) \\ + (m_{1} ξ + n_{1} + \frac{n_{1} n_{2}}{Δ_{1} Γ (3 - β)}) (Q (ξ, x_{2} (ξ)) - Q (ξ, x_{1} (ξ))) \\ - \frac{n_{1}}{Δ_{1}} (γ_{1} (x_{2}) - γ_{1} (x_{1}) - (γ_{0} (x_{2}) - γ_{0} (x_{1}))) \\ \geq 0, \end{matrix}$

$\begin{array}{l} T x_{2} (t) - T x_{1} (t) \\ = \int_{ξ}^{1} G_{2} (t, s) (f_{2} (s, x_{2} (s), x_{2} (s - τ_{2})) - f_{2} (s, x_{1} (s), x_{1} (s - τ_{2}))) d s + Δ_{3}^{x_{2}} - Δ_{3}^{x_{1}} \\ + \int_{0}^{ξ} g_{1} (t, s) (f_{1} (t, x_{2} (t), x_{2} (t + τ_{1})) - f_{1} (t, x_{1} (t), x_{1} (t + τ_{1}))) d s \\ + \frac{t}{Δ_{1}} (\frac{(Q (ξ, x_{2} (ξ)) - Q (ξ, x_{1} (ξ))) m_{2}}{Γ (3 - β)} ξ^{2 - α} - γ_{1} (x_{2}) - γ_{1} (x_{1})) \\ > \int_{ξ}^{1} G_{2} (t, s) (f_{2} (s, x_{2} (s), x_{2} (s - τ_{2})) - f_{2} (s, x_{1} (s), x_{1} (s - τ_{2}))) d s \\ > 0. \end{array}$

综上所述，对任意 $t \in [0, 1]$ 有 $(T x_{2}) (t) > (T x_{1}) (t)$ ，则T为强增算子。

定义2：令 $α, β \in E$ 。称 $α$ 为边值问题(1)的一个下解，若 $α$ 满足

${\begin{cases} {}_{t}^{c}D_{ξ^{-}}^{α} α (t) \leq f_{1} (t, α (t), α (t + τ_{1})), t \in [0, ξ], \\ {}_{ξ^{+}}^{c}D_{t}^{β} α (t) \leq f_{2} (t, α (t), α (t - τ_{2})), t \in (ξ, 1], \\ Δ α (ξ) \leq I (ξ, u (ξ)), Δ α^{'} (ξ) \leq Q (ξ, α (ξ)), \\ h_{0} (α (0), α (1)) \leq 0, h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} α (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} α (1)) \geq 0. \end{cases}$

称 $β$ 为边值问题(1)的一个下解，若 $β$ 满足

${\begin{cases} {}_{t}^{c}D_{ξ^{-}}^{α} β (t) \geq f_{1} (t, β (t), β (t + τ_{1})), t \in [0, ξ], \\ {}_{ξ^{+}}^{c}D_{t}^{β} β (t) \geq f_{2} (t, β (t), β (t - τ_{2})), t \in (ξ, 1], \\ Δ β (ξ) \geq I (ξ, β (ξ)), Δ β^{'} (ξ) \geq Q (ξ, β (ξ)), \\ h_{0} (β (0), β (1)) \geq 0, h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} β (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} β (1)) \leq 0. \end{cases}$

4. 主要结论

定理1：假设(H1)、(H2)、(H3)成立，且边值问题(1)存在两个下解 $α_{1}, α_{2}$ 和两个上解 $β_{1}, β_{2}$ ，且 $α_{2}, β_{1}$ 不是边值问题(1)的解，

$α_{1} ≺ β_{1} ≺ α_{2} ≺ β_{2} .$

则边值问题(1)至少存在三个不同的解 $u_{1}$ ， $u_{2}$ ， $u_{3}$ 满足：

$α_{1} \underline{≺} u_{1} ≺ ≺ β_{1}, α_{2} ≺ ≺ u_{2} \underline{≺} β_{2}, α_{2} \underline{≺} u_{3} \underline{≺} β_{1} .$

证明：令算子T在 $[α_{1}, β_{1}]$ 上， $T |_{[α_{1}, β_{1}]}$ 也记作T。由引理5和引理6可得， $T : [α_{1}, β_{1}] \to P$ 是一个全连续强增算子。

通过定义算子T可得，

${\begin{cases} {}_{t}^{c}D_{ξ^{-}}^{α} (T α_{1}) (t) = f_{1} (t, α_{1} (t), α_{1} (t + τ_{1})), t \in (0, ξ), \\ {}_{ξ^{+}}^{c}D_{t}^{β} (T α_{1}) (t) = f_{2} (t, α_{1} (t), α_{1} (t - τ_{2})), t \in (ξ, 1), \\ Δ (T α_{1}) (ξ) = I (ξ, α_{1} (ξ)), Δ {(T α_{1})}^{'} (ξ) = Q (ξ, α_{1} (ξ)), \\ m_{1} (T α_{1}) (0) + n_{1} (T α_{1}) (1) = h_{0} (α_{1} (0), α_{1} (1)) + m_{1} α_{1} (0) + n_{1} α_{1} (1), \\ m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} (T α_{1}) (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} (T α_{1}) (1) \\ = h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} α_{1} (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} α_{1} (1)) + m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} α_{1} (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} α_{1} (1) . \end{cases}$

令 $α (t) = (T α_{1}) (t) - α_{1} (t)$ 。由于 $α_{1}$ 是边值问题(1)的一个下解，则

${}_{t}^{c}D_{ξ^{-}}^{α} α (t) = {}_{t}^{c}D_{ξ^{-}}^{α} (T α_{1}) (t) - {}_{t}^{c}D_{ξ^{-}}^{α} α_{1} (t) = f_{1} (t, α_{1} (t), α_{1} (t + τ_{1})) - {}_{t}^{c}D_{ξ^{-}}^{α} α_{1} (t) \geq 0, t \in (0, ξ),$

${}_{ξ^{+}}^{c}D_{t}^{β} α (t) = {}_{ξ^{+}}^{c}D_{t}^{β} (T α_{1}) (t) - {}_{ξ^{+}}^{c}D_{t}^{β} α_{1} (t) = f_{2} (t, α_{1} (t), α_{1} (t - τ_{2})) - {}_{ξ^{+}}^{c}D_{t}^{β} α_{1} (t) \geq 0, t \in (ξ, 1),$

且

$\begin{matrix} m_{1} α (0) + n_{1} α (1) = m_{1} ((T α_{1}) (0) - α_{1} (0)) + n_{1} ((T α_{1}) (1) - α_{1} (1)) \\ = (m_{1} (T α_{1}) (0) + n_{1} (T α_{1}) (1)) - (m_{1} α_{1} (0) + n_{1} α_{1} (1)) \\ = h_{0} (α_{1} (0), α_{1} (1)) + m_{1} α_{1} (0) + n_{1} α_{1} (1) - (m_{1} α_{1} (0) + n_{1} α_{1} (1)) \\ = h_{0} (α_{1} (0), α_{1} (1)) \\ \leq 0. \end{matrix}$

$\begin{array}{l} m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} α (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} α (1) \\ = m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} ((T α_{1}) (0) - α_{1} (0)) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} ((T α_{1}) (1) - α_{1} (1)) \\ = m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} (T α_{1}) (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} (T α_{1}) (1) - (m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} α_{1} (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} α_{1} (1)) \\ = h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} α_{1} (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} α_{1} (1)) + m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} α_{1} (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} α_{1} ( 1 ) \end{array}$

$- (m_{2} {}_{t}^{c}D_{ξ^{-}}^{α - 1} α_{1} (0) + n_{2} {}_{ξ^{+}}^{c}D_{t}^{β - 1} α_{1} (1)) = h_{1} ({}_{t}^{c}D_{ξ^{-}}^{α - 1} α_{1} (0), {}_{ξ^{+}}^{c}D_{t}^{β - 1} α_{1} (1)) \geq 0.$

$Δ α (ξ) = Δ (T α_{1}) (ξ) - Δ α_{1} (ξ) = I (ξ, α_{1} (ξ)) - Δ α_{1} (ξ) \geq 0,$

$Δ α^{'} (ξ) = Δ {(T α_{1})}^{'} (ξ) - Δ {α^{'}}_{1} (ξ) = Q (ξ, α_{1} (ξ)) - Δ {α^{'}}_{1} (ξ) \geq 0.$

由引理4可知， $α (t) \geq 0$ 。

因此， $α_{1} \underline{≺} T α_{1}$ 。同理可得， $α_{2} \underline{≺} T α_{2}$ 。

由于 $α_{2}$ 是边值问题(1)的一个下解但不是边值问题(1)的解，则 $(T α_{2}) \neq α_{2}$ 。因此，

$α_{2} ≺ T α_{2} .$

类似可得，

$T β_{1} ≺ β_{1}, T β_{2} \underline{≺} β_{2} .$

由引理1可知，算子T至少有三个不动点 $x_{1}, x_{2}, x_{3} \in [α_{1}, β_{2}]$ 使得

$α_{1} \underline{≺} x_{1} ≺ ≺ β_{1}, α_{2} ≺ ≺ x_{2} \underline{≺} β_{2}, α_{2} \underline{≺} x_{2} \underline{≺} β_{2} .$

因此，边值问题(1)至少有三个不同解。

证毕。

文章引用

张雨馨. 具有左右分数阶导数和时滞的非瞬时脉冲微分方程非线性边值问题
Nonlinear Boundary Value Problems for Non-Instantaneous Pulse Differential Equations with Left-Right Fractional Derivatives and Delays[J]. 应用数学进展, 2021, 10(04): 1255-1269. https://doi.org/10.12677/AAM.2021.104136

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