具p-Laplacian算子的分数阶脉冲微分方程边值问题解的存在性 Existence of Solutions for Fractional Impulsive Differential Equations with p-Laplacian Operator

doi:10.12677/PM.2017.76057

Pure Mathematics
Vol.07 No.06(2017), Article ID:22650,10 pages
10.12677/PM.2017.76057

Existence of Solutions for Fractional Impulsive Differential Equations with p-Laplacian Operator

Yuanbin Liu, Xiujuan Wang, Weimin Hu

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School of Mathematics and Statistics, Yili Normal University, Yining Xinjiang

Received: Oct. 22^nd, 2017; accepted: Nov. 5^th, 2017; published: Nov. 13^th, 2017

ABSTRACT

In this paper, we discuss the existence of solutions of boundary value problems for a class of fractional impulsive differential equations. Some fixed point theorems are used to obtain sufficient conditions for the existence of solutions of Impulsive Differential Equations.

Keywords:Fractional Difference Equation, Impulsive, Fixed-Point, Boundary Problem

具p-Laplacian算子的分数阶脉冲微分方程边值问题解的存在性

刘元彬，汪秀娟，胡卫敏

伊犁师范学院，新疆伊宁

收稿日期：2017年10月22日；录用日期：2017年11月5日；发布日期：2017年11月13日

摘要

本文讨论了一类分数阶脉冲微分方程的边值问题解的存在性，应用一些不动点定理给出了脉冲微分方程解的存在性的充分条件。

关键词 :分数阶，脉冲：不动点定理，边值问题

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

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

1. 引言

分数阶微分方程在科技、经济和工程等领域都得到了广泛的应用，引起了广大数学学者的关注和研究，并在分数阶微分系统方面得到了很多的研究成果 [1] - [9] ，含有p-Laplacian算子的微分方程也逐渐热门 [1] [2] [3] ，而对非线性分数阶脉冲微分方程脉冲边值问题的研究相对而言比较少，在文献 [4] - [9] 中也研究了含有脉冲项的分数阶微分方程的边值问题，因此本文主要研究含有p-Laplacian算子的分数阶脉冲微分方程边值问题解的存在性。

在文献 [6] 中，作者运用Banach压缩映射原理和Leray-Schauder不动点定理研究了分数阶差分方程边值问题

${\begin{cases} ^{C} D_{0 +}^{q} u (t) = f (t, u (t)), 1 < q \leq 2, t \in J^{'}, \\ Δ (u (t_{k})) = I_{k} (u (t_{k})), Δ (u^{'} (t_{k})) = Q_{k} (u (t_{k})), \\ a u (0) - b u^{'} (0) = x_{0}, c u (1) + d u^{'} (1) = x_{1}, \end{cases}$ (1)

解的存在性，这里是标准的Caputo分数阶导数，其中 $f \in (C \times R, R)$ ， $I_{k}, Q_{k} \in C (R, R)$ ，是连续函数。

文献 [2] 中，作者运用不动点定理研究了如下边值问题

${\begin{cases} ^{C} D_{0 +}^{β} φ_{p} (^{C} D_{0 +}^{β} x (t)) = f (t, x (t)), 1 < α \leq 2, t \in J^{'}, \\ x (0) + x (1) = 0, {}^{C}D_{0 +}^{α} x (0) + {}^{C}D_{0 +}^{α} x (1) = 0, \end{cases}$ (2)

解的存在性，这里 $^{C} D_{0 +}^{β}$ 是标准的Caputo分数阶导数。

受上述文献启发，本文应用不动点定理研究含p-Laplacian算子的分数阶脉冲微分方程边值问题

${\begin{cases} φ_{p} (^{C} D_{0 +}^{α} u (t)) = f (t, u (t)), t \in J^{'} \\ Δ (u (t_{k})) = I_{k} (u (t_{k})), Δ (u^{'} (t_{k})) = Q_{k} (u (t_{k})), \\ a u (0) - b u^{'} (0) = x_{0}, c u (0) + d {}^{C}D_{0 +}^{β} u (1) = x_{1} \end{cases}$ (3)

解的存在性，其中 $^{C} D_{0 +}^{α}$ 是标准的Caputo分数阶导数，

$1 < α \leq 2$ ， $α - 1 < β \leq 1$ ， $a \geq 0$ ， $b > 0$ ， $c \geq 0$ ， $d > 0$ ，

$(δ = b c + a (c + \frac{d}{Γ (2 - β)}))$ ， $x_{0}, x_{1} \in R$ ， $f \in C (J \times R, R)$ ， $I_{k}, J_{k} \in R$ ，

$J = [0, 1]$ ， $0 = t_{0} < t_{1} < \dots < t_{k} < \dots < t_{m} < t_{m + 1} = 1$ ， $J^{'} = J \ {t_{1}, t_{2}, \dots, t_{m}}$

其脉冲项为

$Δ u (t_{k}) = u (t_{k}^{+}) - u (t_{k}^{-})$ ， $Δ u^{'} (t_{k}) = u^{'} (t_{k}^{+}) - u^{'} (t_{k}^{-})$

这里 $u (t_{k}^{+})$ 与 $u (t_{k}^{-})$ 分别为 $u (t_{k})$ 在 $t = t_{k}$ 处的右左极限， $k = 1, 2, \dots, m$ ，其中 $φ_{p} (s) = {| s |}^{p - 2} s$ 为p-Laplacian算子， $p > 1$ ，并且 $φ_{p} (s)$ 的逆算子 $φ_{q} (s)$ ， $\frac{1}{p} + \frac{1}{q} = 1$ 。

2. 相关定义及引理

在这一节，为了后面的讨论，我们给出一些定义及相关引理。

令 $J = [0, 1]$ ， $0 = t_{0} < t_{1} < \dots < t_{k} < \dots < t_{m} < t_{m + 1} = 1$ ， $J_{0} = [0, t_{1}]$ ， $J_{1} = (t_{1}, t_{2}]$ ， $\dots$ ， $J_{m} = (t_{m}, 1]$ ， $P C (J, R) = {u : J \to R | u \in C (J_{k}), k = 0, 1, \dots, m, 并且 u (t_{k}^{+}) 存在}$ ，其空间上范数为 $‖ u ‖ = \sup_{t \in J} | u (t) |$ ，有

$P C (J, R) = {u : J \to R | u \in C (J_{k}), k = 0, 1, \dots, m, 并且 u (t_{k}^{+}), u^{'} (t_{k}^{+}) 存在}$ ，与其范数 ${‖ u ‖}_{P C^{'}} = \max_{t \in J} {‖ u ‖, ‖ u^{'} ‖}$ ，显然 $P C$ 与 $P C^{'}$ 为Banach空间。

定理1：设 $E$ 为Banach空间， $Ω$ 为 $E$ 上的非空有界闭凸子集，算子 $T : \bar{Ω} \to E$ 为全连续算子，使得 $‖ T u ‖ \leq ‖ u ‖$ ， $\forall u \in \bar{Ω}$ ，则 $T$ 在 $\bar{Ω}$ 存在不动点。

定理2：设 $E$ 为Banach空间， $T : E \to E$ 为全连续算子， $V = {u \in E | u = μ T u, 0 < μ < 1}$ ，为有界集，则 $T$ 在 $\bar{Ω}$ 存在不动点。

引理1： [1] 令 $α > 0$ ，则分数阶微分方程 $^{C} D_{0 +}^{α} u (t) = 0$ 有唯一解

$u (t) = c_{0} + c_{1} t + \dots + c_{n - 1} t^{n - 1}, c_{i} \in R, n = [α] + 1$

引理2： [1] 令 $α > 0$ ，则有

$I_{0 +}^{α} {}^{C}D_{0 +}^{α} u (t) = u (t) + c_{0} + c_{1} t + \dots + c_{n - 1} t^{n - 1}, c_{i} \in R, i = 0, 1, \dots, n - 1, n = [α] + 1$

对于边值问题(3)，可通过 $φ_{p} (s)$ 的逆算子 $φ_{q} (s)$ 将其转化为等价的边值问题

${\begin{cases} ^{C} D_{0 +}^{α} u (t) = φ_{q} (f (t, u (t))), t \in J^{'} \\ Δ (u (t_{k})) = I_{k} (u (t_{k})), Δ (u^{'} (t_{k})) = Q_{k} (u (t_{k})), \\ a u (0) - b u^{'} (0) = x_{0}, c u (0) + d^{C} D_{0 +}^{β} u (1) = x_{1} \end{cases}$ (4)

引理3：对于给定函数 $y \in C [0, 1]$ ， $u (t)$ 为如下边值问题的解

${\begin{cases} ^{C} D_{0 +}^{α} u (t) = φ_{q} (y (t)), t \in J^{'} \\ Δ (u (t_{k})) = I_{k} (u (t_{k})), Δ (u^{'} (t_{k})) = Q_{k} (u (t_{k})), \\ a u (0) - b u^{'} (0) = x_{0}, c u (0) + d^{C} D_{0 +}^{β} u (1) = x_{1} \end{cases}$ (5)

当且仅当

$u (t) = {\begin{cases} \frac{1}{Γ (α)} \int_{0}^{t} {(t - s)}^{α - 1} φ_{q} (y (s)) d t - c_{1} - c_{2} t, t \in J_{0} \\ \frac{1}{Γ (α)} \int_{t_{k}}^{t} {(t - s)}^{α - 1} φ_{q} (y (s)) d t + \sum_{i = 1}^{k} \frac{t - t_{k}}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (y (s)) d t \\ + \sum_{i = 1}^{k - 1} \frac{t_{k} - t_{i}}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (y (s)) d t + \sum_{i = 1}^{k} I_{i} (u (t_{i})) + \sum_{i = 1}^{k} (t - t_{k}) Q_{i} (u (t_{i})) \\ + \sum_{i = 1}^{k - 1} (t_{k} - t_{i}) Q_{i} (u (t_{i})) - c_{1} - c_{2} t, t \in J_{k} \end{cases}$

其中

$\begin{array}{l} c_{1} = \frac{1}{δ} (\frac{b c}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} φ_{q} (y (s)) d s + \frac{b c (1 - t_{m})}{Γ (α - 1)} \sum_{i = 1}^{m} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s \\ + \sum_{i = 1}^{m - 1} \frac{b c (t_{m} - t_{i})}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s + \sum_{i = 1}^{m} b c I_{i} (u (t_{i})) + \sum_{i = 1}^{m} b c (1 - t_{m}) Q_{i} (u (t_{i})) \\ + \sum_{i = 1}^{m - 1} b c (t_{m} - t_{i}) Q_{i} (u (t_{i})) + \sum_{i = 1}^{m + 1} \frac{b d}{Γ (α - 1) Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s \\ + \frac{b d}{Γ (2 - β)} \sum_{i = 1}^{m} Q_{i} (u (t_{i})) - b x_{1} - c x_{0} - \frac{d x_{0}}{Γ (2 - β)}) \end{array}$

$\begin{array}{l} c_{2} = \frac{1}{δ} (\frac{a c}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} φ_{q} (y (s)) d s + \frac{a c (1 - t_{m})}{Γ (α - 1)} \sum_{i = 1}^{m} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s \\ + \sum_{i = 1}^{m - 1} \frac{a c (t_{m} - t_{i})}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s + \sum_{i = 1}^{m} a c I_{i} (u (t_{i})) + \sum_{i = 1}^{m} a c (1 - t_{m}) Q_{i} (u (t_{i})) \\ + \sum_{i = 1}^{m - 1} a c (t_{m} - t_{i}) Q_{i} (u (t_{i})) + \sum_{i = 1}^{m + 1} \frac{a d}{Γ (α - 1) Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s \\ + \frac{a d}{Γ (2 - β)} \sum_{i = 1}^{m} Q_{i} (u (t_{i})) - a x_{1} - c x_{0}) \end{array}$

证明：设 $u (t)$ 为(5)的解，由(2)可得

$u (t) = I_{0 +}^{α} y (t) - c_{1} - c_{2} t = \frac{1}{Γ (α)} \int_{0}^{t} {(t - s)}^{α - 1} φ_{q} (y (s)) d t - c_{1} - c_{2} t, t \in J_{0}, c_{1}, c_{2} \in R$

则有

$u^{'} (t) = \frac{1}{Γ (α - 1)} \int_{0}^{t} {(t - s)}^{α - 1} φ_{q} (y (s)) d t - c_{2}$ (6)

若 $t \in J_{1}$ ，有

$u (t) = \frac{1}{Γ (α)} \int_{t_{1}}^{t} {(t - s)}^{α - 1} φ_{q} (y (s)) d s - d_{1} - d_{2} (t - t_{1})$

$u^{'} (t) = \frac{1}{Γ (α - 1)} \int_{t_{1}}^{t} {(t - s)}^{α - 2} φ_{q} (y (s)) d s - d_{2}$

其中 $d_{1}, d_{2} \in R$ ，可得

$u (t^{-}) = \frac{1}{Γ (α)} \int_{t_{1}}^{t} {(t - s)}^{α - 1} φ_{q} (y (s)) d s - c_{1} - c_{2} t_{1}, u (t^{+}) = - d_{1}$

$u^{'} (t^{-}) = \frac{1}{Γ (α - 1)} \int_{t_{1}}^{t} {(t - s)}^{α - 2} φ_{q} (y (s)) d s - c_{2}, u^{'} (t^{+}) = - d_{2}$

由上述定义 $Δ u (t_{k}) = u (t_{k}^{+}) - u (t_{k}^{-})$ ， $Δ u^{'} (t_{k}) = u^{'} (t_{k}^{+}) - u^{'} (t_{k}^{-})$ ，则

$- d_{1} = \frac{1}{Γ (α)} \int_{0}^{t_{1}} {(t_{1} - s)}^{α - 1} φ_{q} (y (s)) d s - c_{1} - c_{2} t_{1} + I_{1} (u (t_{1}))$

$- d_{2} = \frac{1}{Γ (α - 1)} \int_{0}^{t_{1}} {(t_{1} - s)}^{α - 2} φ_{q} (y (s)) d s - c_{2} + Q_{1} (u (t_{1}))$

因此

$\begin{array}{l} u (t) = \frac{1}{Γ (α)} \int_{t_{k}}^{t} {(t - s)}^{α - 1} φ_{q} (y (s)) d t + \sum_{i = 1}^{k} \frac{t - t_{k}}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (y (s)) d t \\ + \sum_{i = 1}^{k - 1} \frac{t_{k} - t_{i}}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (y (s)) d t + \sum_{i = 1}^{k} I_{i} (u (t_{i})) + \sum_{i = 1}^{k} (t - t_{k}) Q_{i} (u (t_{i})) \\ + \sum_{i = 1}^{k - 1} (t_{k} - t_{i}) Q_{i} (u (t_{i})) - c_{1} - c_{2} t \end{array}$

由边值条件 $a u (0) - b u^{'} (0) = x_{0}$ ， $c u (0) + d {}^{C}D_{0 +}^{β} u (1) = x_{1}$ ，

$\begin{array}{l} - a c_{1} + b c_{2} = x_{0} \\ c c_{1} + (c + \frac{1}{Γ (2 - β)}) \\ = \frac{c}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} φ_{q} (y (s)) d s + \sum_{i = 1}^{k} \frac{c (1 - t_{i})}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s \\ + \sum_{i = 1}^{m} c I_{i} (u (t_{i})) + \sum_{i = 1}^{m} c (1 - t_{i}) Q_{i} (u (t_{i})) + \sum_{i = 1}^{m - 1} b c (t_{m} - t_{i}) Q_{i} (u (t_{i})) \\ + \sum_{i = 1}^{m} \frac{d}{Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s + \frac{d}{Γ (2 - β)} \sum_{i = 1}^{m} Q_{i} (u (t_{i})) - x_{1} \end{array}$

可得

命题得证。

3. 主要结果

定义算子 $T : P C (J, R) \to P C (J, R)$ 如下

$\begin{array}{l} T u (t) = \frac{1}{Γ (α)} \int_{t_{k}}^{t} {(t - s)}^{α - 1} φ_{q} (f (s, u)) d t + \sum_{i = 1}^{k} \frac{1}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (f (s, u)) d t \\ + \sum_{i = 1}^{k - 1} \frac{t - t_{k}}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (f (s, u)) d t + \sum_{i = 1}^{k - 1} \frac{t_{k} - t_{i}}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (f (s, u)) d t \\ + \sum_{i = 1}^{k} I_{i} (u (t_{i})) + \sum_{i = 1}^{k} (t - t_{k}) Q_{i} (u (t_{i})) + \sum_{i = 1}^{k - 1} (t_{k} - t_{i}) Q_{i} (u (t_{i})) - c_{1} - c_{2} t \end{array}$

其中

$\begin{array}{l} c_{1} = \frac{1}{δ} (\frac{b c}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} φ_{q} (f (s, u)) d s + \frac{b c (1 - t_{m})}{Γ (α - 1)} \sum_{i = 1}^{m} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (f (s, u)) d s \\ + \sum_{i = 1}^{m - 1} \frac{b c (t_{m} - t_{i})}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (f (s, u)) d s + \sum_{i = 1}^{m} b c I_{i} (u (t_{i})) + \sum_{i = 1}^{m} b c (1 - t_{m}) Q_{i} (u (t_{i})) \\ + \sum_{i = 1}^{m - 1} b c (t_{m} - t_{i}) Q_{i} (u (t_{i})) + \sum_{i = 1}^{m + 1} \frac{b d}{Γ (α - 1) Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (y (s)) d s \\ + \frac{b d}{Γ (2 - β)} \sum_{i = 1}^{m} Q_{i} (u (t_{i})) - b x_{1} - c x_{0} - \frac{d x_{0}}{Γ (2 - β)}) \end{array}$

$\begin{array}{l} c_{2} = \frac{1}{δ} (\frac{a c}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} φ_{q} (f (s, u)) d s + \frac{a c (1 - t_{m})}{Γ (α - 1)} \sum_{i = 1}^{m} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (f (s, u)) d s \\ + \sum_{i = 1}^{m - 1} \frac{a c (t_{m} - t_{i})}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (f (s, u)) d s + \sum_{i = 1}^{m} a c I_{i} (u (t_{i})) + \sum_{i = 1}^{m} a c (1 - t_{m}) Q_{i} (u (t_{i})) \\ + \sum_{i = 1}^{m - 1} a c (t_{m} - t_{i}) Q_{i} (u (t_{i})) + \sum_{i = 1}^{m + 1} \frac{a d}{Γ (α - 1) Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} φ_{q} (f (s, u)) d s \\ + \frac{a d}{Γ (2 - β)} \sum_{i = 1}^{m} Q_{i} (u (t_{i})) - a x_{1} - c x_{0}) \end{array}$

由引理3，令 $y (t) = f (t, u (t))$ 则边值问题(3)有解等价存在不动点，即 $u = T u$ ，于是(3)有解当且仅当 $u = T u$ 。

定理3：令 $\lim_{u \to 0} \frac{φ_{q} (f (t, u))}{u} = 0$ ， $\lim_{u \to 0} \frac{I_{k} (u)}{u} = 0$ ， $\lim_{u \to 0} \frac{Q_{k} (u)}{u} = 0$ ，并且有

$\begin{array}{l} Λ = \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} (m + 1) (c (a + b) + δ) + \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} ((2 m - 1) (c (a + b) + δ) + \frac{(m + 1) (a + b) d}{Γ (2 - β)}) \\ + \frac{m (c (a + b) + δ)}{δ} δ_{2} + \frac{1}{δ} ((2 m - 1) (c (a + b) + δ) + \frac{m d (a + b)}{Γ (2 - β)}) δ_{3} \\ + \frac{1}{δ} ((a + b) | x_{1} | + \frac{2 c + d}{Γ (2 - β)} | x_{0} |) \leq 1 \end{array}$

证明：首先，证明 $T : P C (J, R) \to P C (J, R)$ 是全连续算子， $T$ 作用于 $f, I_{k}, Q_{k}$ 均连续，

设 $Ω$ 为 $T : P C (J, R) \to P C (J, R)$ 的有界子集，则存在常数 $L_{i} > 0 (i = 1, 2, 3, 4)$ 使得 $| f (t, u) | \leq L_{1}$ ， $| I_{k} (u) | \leq L_{2}$ ， $| Q_{k} (u) | \leq L_{3}$ ， $| φ_{q} (f (t, u)) | \leq L_{4}$ ，又

$\begin{array}{l} | c_{1} | \leq \frac{1}{δ} (\frac{b c}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} | φ_{q} (f (s, u)) | d s + \frac{b c (1 - t_{m})}{Γ (α - 1)} \sum_{i = 1}^{m} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} | φ_{q} (f (s, u)) | d s \\ + \sum_{i = 1}^{m - 1} \frac{b c (t_{m} - t_{i})}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} | φ_{q} (f (s, u)) | d s + \sum_{i = 1}^{m} b c | I_{i} (u (t_{i})) | + \sum_{i = 1}^{m} b c (1 - t_{m}) | Q_{i} (u (t_{i})) | \\ + \sum_{i = 1}^{m - 1} b c (t_{m} - t_{i}) | Q_{i} (u (t_{i})) | + \sum_{i = 1}^{m + 1} \frac{b d}{Γ (α - 1) Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} | φ_{q} (f (s, u)) | d s \\ + \frac{b d}{Γ (2 - β)} \sum_{i = 1}^{m} | Q_{i} (u (t_{i})) | + b | x_{1} | + c | x_{0} | + \frac{d | x_{0} |}{Γ (2 - β)}) \end{array}$

$\begin{array}{l} \leq \frac{1}{δ} (\frac{b c L_{4}}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} d s + \frac{b c L_{4}}{Γ (α - 1)} \sum_{i = 1}^{m} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} d s + \sum_{i = 1}^{m - 1} \frac{b c L_{4}}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} d s \\ + \sum_{i = 1}^{m} b c L_{2} + \sum_{i = 1}^{m} b c (1 - t_{m}) L_{3} + \sum_{i = 1}^{m - 1} b c (t_{m} - t_{i}) L_{3} + \sum_{i = 1}^{m + 1} \frac{b d L_{4}}{Γ (α - 1) Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} d s \\ + \frac{b d}{Γ (2 - β)} \sum_{i = 1}^{m} L_{3} + b | x_{1} | + c | x_{0} | + \frac{d | x_{0} |}{Γ (2 - β)}) \\ \leq \frac{1}{δ} \frac{b c (m + 1) L_{4}}{Γ (α + 1)} + \frac{1}{δ} \frac{b c m L_{4}}{Γ (α)} + \frac{1}{δ} \sum_{i = 1}^{m - 1} \frac{b c m L_{4}}{Γ (α)} + \frac{b c m L_{2}}{δ} + \frac{b c m L_{3}}{δ} + \frac{b c (m - 1) L_{3}}{δ} \\ + \frac{1}{δ} \frac{b d (m + 1) L_{4}}{Γ (α) Γ (2 - β)} + \frac{1}{δ} \frac{b d m L_{3}}{Γ (2 - β)} + \frac{1}{δ} b | x_{1} | + \frac{1}{δ} c | x_{0} | + \frac{1}{δ} \frac{d | x_{0} |}{Γ (2 - β)} \end{array}$

$\begin{array}{l} | c_{2} | \leq \frac{1}{δ} (\frac{a c}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} | φ_{q} (f (s, u)) | d s + \frac{a c (1 - t_{m})}{Γ (α - 1)} \sum_{i = 1}^{m} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} | φ_{q} (f (s, u)) | d s \\ + \sum_{i = 1}^{m - 1} \frac{a c (t_{m} - t_{i})}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} | φ_{q} (f (s, u)) | d s + \sum_{i = 1}^{m} a c | I_{i} (u (t_{i})) | + \sum_{i = 1}^{m} a c (1 - t_{m}) | Q_{i} (u (t_{i})) | \\ + \sum_{i = 1}^{m - 1} a c (t_{m} - t_{i}) | Q_{i} (u (t_{i})) | + \sum_{i = 1}^{m + 1} \frac{a d}{Γ (α - 1) Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} | φ_{q} (f (s, u)) | d s \\ + \frac{a d}{Γ (2 - β)} \sum_{i = 1}^{m} | Q_{i} (u (t_{i})) | - a | x_{1} | - c | x_{0} |) \end{array}$

$\begin{array}{l} \leq \frac{1}{δ} (\frac{a c L_{4}}{Γ (α)} \sum_{i = 1}^{m + 1} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 1} d s + \frac{a c L_{4}}{Γ (α - 1)} \sum_{i = 1}^{m} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} d s + \sum_{i = 1}^{m - 1} \frac{a c L_{4}}{Γ (α - 1)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} d s \\ + \sum_{i = 1}^{m} a c L_{2} + \sum_{i = 1}^{m} a c (1 - t_{m}) L_{3} + \sum_{i = 1}^{m - 1} a c (t_{m} - t_{i}) L_{3} + \sum_{i = 1}^{m + 1} \frac{a d L_{4}}{Γ (α - 1) Γ (2 - β)} \int_{t_{i - 1}}^{t_{i}} {(t_{i} - s)}^{α - 2} d s \\ + \frac{a d}{Γ (2 - β)} \sum_{i = 1}^{m} L_{3} - a | x_{1} | - c | x_{0} |) \\ \leq \frac{1}{δ} \frac{a c (m + 1) L_{4}}{Γ (α + 1)} + \frac{1}{δ} \frac{a c m L_{4}}{Γ (α)} + \frac{1}{δ} \frac{a c (m - 1) L_{4}}{Γ (α)} + \frac{a c m L_{2}}{δ} + \frac{a c m L_{3}}{δ} \\ + \frac{a c (m - 1) L_{3}}{δ} + \frac{1}{δ} \frac{a d (m + 1) L_{4}}{Γ (α) Γ (2 - β)} + \frac{1}{δ} \frac{a d m L_{3}}{Γ (2 - β)} + \frac{1}{δ} a | x_{1} | + \frac{1}{δ} c | x_{0} | \end{array}$

$\begin{array}{l} | T u (t) | \leq \frac{1}{δ} \frac{L_{4}}{Γ (α + 1)} (m + 1) (c (a + b) + δ) + \frac{1}{δ} \frac{L_{4}}{Γ (α + 1)} ((2 m - 1) (c (a + b) + δ) + \frac{(m + 1) (a + b) d}{Γ (2 - β)}) \\ + \frac{m (c (a + b) + δ)}{δ} L_{2} + \frac{m}{δ} ((2 p - 1) (δ + c (a + b)) + \frac{m d (a + b)}{Γ (2 - β)}) L_{3} + \frac{1}{δ} ((a + b) | x_{1} | + \frac{2 c + d}{Γ (2 - β)} | x_{0} |) : = L \end{array}$

因此 $‖ T u ‖ \leq L$ 。

另一方面

$| T u^{'} (t) | \leq \frac{1}{δ} \frac{a c (m + 1) L_{4}}{Γ (α + 1)} + \frac{a c m L_{2}}{δ} + \frac{L_{3}}{δ} ((m + 1) (δ + \frac{a d}{Γ (2 - β)}) + a c (2 m - 1)) + \frac{(a | x_{1} | + c | x_{0} |)}{δ} : = \bar{L}$

因此对于 $t_{1}, t_{2} \in J_{k}, t_{1} \leq t_{2}, 0 \leq k \leq m$ 有

$| T u (t_{2}) - T u (t_{1}) | \leq \int_{t_{1}}^{t_{2}} {(T u)}^{'} (s) d s \leq \bar{L} (t_{2} - t_{1})$

则 $t$ 在每一个子空间等度连续，由Arzela-Ascoli定理， $T$ 为全连续算子，由条件 $\lim_{u \to 0} \frac{φ_{q} (f (t, u))}{u} = 0$ ， $\lim_{u \to 0} \frac{I_{k} (u)}{u} = 0$ ， $\lim_{u \to 0} \frac{Q_{k} (u)}{u} = 0$ ，存在常数 $r > 0, δ_{i} > 0 (i = 1, 2, 3)$ 有 $| φ_{q} (f (t, u)) | \leq δ_{1} | u |$ ， $| I_{i} (u) | \leq δ_{2} | u |$ ， $| Q_{i} (u) | \leq δ_{3} | u |$ ，由

$\begin{array}{l} Λ = \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} (m + 1) (c (a + b) + δ) + \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} ((2 m - 1) (c (a + b) + δ) + \frac{(m + 1) (a + b) d}{Γ (2 - β)}) \\ + \frac{m (c (a + b) + δ)}{δ} δ_{2} + \frac{1}{δ} ((2 m - 1) (c (a + b) + δ) + \frac{m d (a + b)}{Γ (2 - β)}) δ_{3} + \frac{1}{δ} ((a + b) | x_{1} | + \frac{2 c + d}{Γ (2 - β)} | x_{0} |) \leq 1 \end{array}$

对于 $u \in P C (J, R)$ ，定义 $Ω = {u \in P C (J, R) | ‖ u ‖ < r}$ ， $u \in \partial Ω$ 时， $‖ u ‖ = r$ ，有

$\begin{array}{l} | T u (t) | \leq {\frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} (m + 1) (c (a + b) + δ) + \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} ((2 m - 1) (c (a + b) + δ) + \frac{(m + 1) (a + b) d}{Γ (2 - β)}) \\ + \frac{m (c (a + b) + δ)}{δ} δ_{2} + \frac{1}{δ} ((2 m - 1) (c (a + b) + δ) + \frac{m d (a + b)}{Γ (2 - β)}) δ_{3} + \frac{1}{δ} ((a + b) | x_{1} | + \frac{2 c + d}{Γ (2 - β)} | x_{0} |)} {‖ u ‖}_{P C^{'}} \end{array}$

有 ${‖ T u ‖}_{P C^{'}} \leq {‖ u ‖}_{P C^{'}}$ ，由定理1算子 $T$ 至少存在一个不动点，则(3)至少存在一个解。

定理4：假设存在正常数 $L_{i} \geq 0 (i = 1, 2, 3, 4)$ 使得 $| f (t, u) | \leq L_{1}$ ， $| I_{k} (u) | \leq L_{2}$ ， $| Q_{k} (u) | \leq L_{3}$ ， $| φ_{q} (f (t, u)) | \leq L_{4}$ ，则(3)至少存在一个解。

证明：由定理3，算子 $T$ 为全连续算子，此时假设 $V = {u \in E | u = μ T u, 0 < μ < u}$ 为有界集，令 $u \in V$ ，我们有

$\begin{array}{l} u (t) = \frac{μ}{Γ (α)} \int_{t_{k}}^{t} {(t - s)}^{α - 1} φ_{q} (f (s, u)) d t + \frac{μ}{Γ (α)} \sum_{i = 1}^{k} \int_{t_{k}}^{t} {(t - s)}^{α - 1} φ_{q} (f (s, u)) d t \\ + \sum_{i = 1}^{k} \frac{μ (t - t_{k})}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (f (s, u)) d t + \sum_{i = 1}^{k - 1} \frac{μ (t_{k} - t_{i})}{Γ (α - 1)} \int_{t_{- 1 k}}^{t_{k}} {(t - s)}^{α - 2} φ_{q} (f (s, u)) d t \\ + \sum_{i = 1}^{k} μ I_{i} (u (t_{i})) + \sum_{i = 1}^{k} μ (t - t_{k}) Q_{i} (u (t_{i})) + \sum_{i = 1}^{k - 1} μ (t_{k} - t_{i}) Q_{i} (u (t_{i})) - μ c_{1} - μ c_{2} t \end{array}$

由条件，有

$\begin{array}{l} | u (t) | = μ | T u (t) | \\ \leq \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} (m + 1) (c (a + b) + δ) + \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} ((2 m - 1) (c (a + b) + δ) + \frac{(m + 1) (a + b) d}{Γ (2 - β)}) \\ + \frac{m (c (a + b) + δ)}{δ} δ_{2} + \frac{1}{δ} ((2 m - 1) (c (a + b) + δ) + \frac{m d (a + b)}{Γ (2 - β)}) δ_{3} + \frac{1}{δ} ((a + b) | x_{1} | + \frac{2 c + d}{Γ (2 - β)} | x_{0} |) \end{array}$

则对 $t \in J$ ，

$\begin{array}{l} {‖ u (t) ‖}_{P C^{'}} \leq \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} (m + 1) (c (a + b) + δ) + \frac{1}{δ} \frac{δ_{1}}{Γ (α + 1)} ((2 m - 1) (c (a + b) + δ) + \frac{(m + 1) (a + b) d}{Γ (2 - β)}) \\ + \frac{m (c (a + b) + δ)}{δ} δ_{2} + \frac{1}{δ} ((2 m - 1) (c (a + b) + δ) + \frac{m d (a + b)}{Γ (2 - β)}) δ_{3} + \frac{1}{δ} ((a + b) | x_{1} | + \frac{2 c + d}{Γ (2 - β)} | x_{0} |) \end{array}$

又由 $V$ 为有界集，故由定理2算子 $T$ 至少存在一个不动点，则(3)至少存在一个解。

4. 例子

对于如下具p-Laplacian算子的非线性分数阶脉冲微分方程边值问题

${\begin{cases} φ_{5 / 3} (^{C} D_{0 +}^{\frac{5}{4}} u (t)) = \frac{\cos t}{{(t + 5 \sqrt{2})}^{2}} \frac{u (t)}{1 + u (t)}, \\ Δ (u (\frac{1}{2})) = \frac{| u (\frac{1}{2}) |}{20 + | u (\frac{1}{2}) |}, Δ (u^{'} (\frac{1}{2})) = Q_{k} (\frac{| u (\frac{1}{2}) |}{20 + | u (\frac{1}{2}) |}), \\ u (0) - u^{'} (0) = 0, u (0) + {}^{C}D_{0 +}^{\frac{1}{2}} u (1) = 0 \end{cases}$

这里 $p = \frac{5}{3}, α = \frac{5}{4}, a = b = c = d = m = 1, β = \frac{1}{2}, x_{0} = x_{1} = 0$ ，取 $L_{1} = \frac{1}{50}, L_{2} = \frac{1}{20}, L_{3} = \frac{1}{40}$ 有 $Λ \leq 1$ 满足条件，由定理3和定理4，边值问题(3)至少存在一个解。

基金项目

新疆维吾尔自治区研究生科研创新项目(XJGRI2016136)，新疆高校科研计划重点项目(XJEDU2014I040)。

文章引用

刘元彬,汪秀娟,胡卫敏. 具p-Laplacian算子的分数阶脉冲微分方程边值问题解的存在性
Existence of Solutions for Fractional Impulsive Differential Equations with p-Laplacian Operator[J]. 理论数学, 2017, 07(06): 437-446. http://dx.doi.org/10.12677/PM.2017.76057

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