一类奇数阶泛函微分方程周期解的存在性 Existence of Periodic Solutions for a Class of Odd Order Functional Differential Equations

doi:10.12677/PM.2023.132038

Pure Mathematics
Vol. 13 No. 02 ( 2023 ), Article ID: 61992 , 9 pages
10.12677/PM.2023.132038

一类奇数阶泛函微分方程周期解的存在性

林梦媛，吴进，陈柏立^*

●How to Cite this Article

广东工业大学数学与统计学院，广东广州

收稿日期：2023年1月22日；录用日期：2023年2月21日；发布日期：2023年2月28日

摘要

本文运用Mawhin延拓定理，研究了一类奇数阶泛函微分方程周期解的存在性问题，得到了新的判定准则。

关键词

奇数阶，泛函微分方程，周期解，Mawhin延拓定理

Existence of Periodic Solutions for a Class of Odd Order Functional Differential Equations

Mengyuan Lin, Jin Wu, Baili Chen^*

School of Mathematics and Statistics, Guangdong University of Technology, Guangzhou Guangdong

Received: Jan. 22^nd, 2023; accepted: Feb. 21^st, 2023; published: Feb. 28^th, 2023

ABSTRACT

Using Mawhin’s continuation theorem we study the existence of periodic solutions for a class of odd order functional differential equations, and establish a new criterion.

Keywords:Odd Order, Functional Differential Equations, Periodic Solutions, Mawhin’s Continuation Theorem

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

由于泛函微分方程在许多数学模型中有着深刻的应用，因此吸引了国内外大量的学者进行研究 [1] [2] 。其中，周期解问题一直得到学者们的关注，但大部分都只是利用不动点理论、重合度理论或者临界点理论等等讨论了低阶的泛函微分方程 [3] - [15] ，涉及到高阶的却不多 [16] [17] [18] 。

本文将利用Mawhin延拓定理讨论一类奇数阶泛函微分方程

$x^{(2 n + 1)} (t) + \sum_{i = 0}^{2 n - 1} a_{i} (t) x^{(i)} (t) + g (x (t - τ (t))) = p (t)$ (1)

周期解的存在性，其中 $τ (t)$ 、 $a_{i} (t) (i = 0, 1, 2, \dots, n)$ 和 $p (t)$ 都是定义在R上具有正周期T的实连续函数。并且 $a_{0} (t) > 0$ ， $a_{2 k - 1} (t) > 0 (k = 1, 2, \dots, n)$ ， $g (x)$ 是定义在R上的实连续函数。

在第二节中，我们先给出一些预备知识，以及针对周期解存在性的条件做假设。在第三节中，我们将利用Mawhin延拓定理 [19] ，证明方程(1)的周期解存在性。

2. 预备知识

为了证明主要结果，我们需要介绍Mawhin延拓定理 [19] 。

令X和Y是两个Banach空间，且 $L : D o m L \subset X \to Y$ 是一个线性映射， $N : X \to Y$ 是一个连续映射。若映射L满足：1) $d i m K e r L = c o d i m I m L < + \infty$ ；2) ImL在Y上是闭的。则称映射是指标为零的Fredholm算子。若映射L是一个指标为零的Fredholm算子，则存在连续的投影算子 $P : X \to X$ 和 $Q : Y \to Y$ ，使得 $I m P = K e r L$ 和 $I m L = K e r Q = I m (I - Q)$ 成立。令 $K_{p}$ 表示 ${L |}_{D o m L \cap^{} K e r P} : (I - P) X \to I m L$ 的逆，若 $Q N (\bar{Ω})$ 是有界的且 $\bar{K_{p} (I - Q) N (\bar{Ω})}$ 是紧的，则称映射N是 $\bar{Ω}$ 上L-紧的，其中 $Ω$ 是X上的有界开子集。

引理2.1 ( 延拓定理) [19] 令L是一个具有零指标的Fredholm算子，N是一个在 $\bar{Ω}$ 上L-紧的非线性算子。如果

1) 对每个 $λ \in (0, 1)$ 及 $x \in \partial Ω, L x \neq λ N x$ ；

2) 对每个 $x \in \partial Ω \cap K e r L, Q N x \neq 0$ 及 $\deg (Q N, Ω \cap K e r L, 0) \neq 0$ ；

则方程 $L x = N x$ 在 $\bar{Ω} \cap D o m L$ 上至少有一个解。

在呈现我们主要结果之前，先作出如下假设：

(H₁)： $M_{2 k - 1} = \max_{t \in [0, T]} a_{2 k - 1} (t) \geq a_{2 k - 1} (t) \geq m_{2 k - 1} = \min_{t \in [0, T]} a_{2 k - 1} (t) > 0 (k = 1, 2, \dots, n)$ ；

(H₂)： $M_{2 k - 2} = \max_{t \in [0, T]} | a_{2 k - 2} (t) | (k = 2, \dots, n - 1), M_{0} = \max_{t \in [0, T]} a_{0} (t) \geq a_{0} (t) \geq m_{0} = \min_{t \in [0, T]} a_{0} (t) > 0$ ；

(H₃)： $M_{2 n - 1} < {(\frac{π}{T})}^{2}, \frac{M_{2 n - 2 i + 1}}{M_{2 n - 2 i - 1}} < {(\frac{π}{T})}^{2} (i = 2, 3, \dots, n)$ ；

(H₄)：存在正常数r使得 $\frac{m_{0} - r}{2} > \frac{T δ M_{0}^{2}}{T δ M_{0} + M_{1}}$ ，且 $A - D B > 0$ 和 $1 - A^{*} > 0$ 。其中 $δ = e^{\frac{M_{0}}{M_{1}} T} / e^{\frac{M_{0}}{M_{1}} T} - 1$ ， $A = 1 - A^{*}$ ，

$A^{*} = M_{2 n - 1} {(\frac{T}{2})}^{2} + M_{2 n - 2} {(\frac{T}{2})}^{3} + \dots + M_{1} {(\frac{T}{2})}^{2 n},$

$B = (M_{2 n - 1} - m_{2 n - 1}) (\frac{T}{2}) + M_{2 n - 2} {(\frac{T}{2})}^{2} + \dots + M_{2} {(\frac{T}{2})}^{2 n - 2} + (M_{1} - m_{1}) {(\frac{T}{2})}^{2 n - 1},$

$D = \frac{T}{m_{0} - r} (\frac{M_{0} M_{1}}{T δ M_{0} + M_{1}} + \frac{m_{0} + r}{2}) .$

3. 主要结果的证明

定理3.1 若假设(H₁)~(H₄)成立，且满足

$\lim_{| x | \to \infty} \sup | \frac{g (x)}{x} | \leq r,$ (2)

$\lim_{| x | \to \infty} sgn (x) g (x) = + \infty,$ (3)

则方程(1)至少存在一个T周期解。

为了证明定理3.1，我们要做如下准备工作。令

$X : = {x | x \in C^{2 n} (R, R), x (t + T) = x (t), \forall t \in R},$

和 $x^{(0)} (t) = x (t)$ ，并在空间X上定义如下范数：

$‖ x ‖ = \max_{0 \leq j \leq 2 n} \max_{t \in [0, T]} | x^{(2 n)} (t) | .$

类似地，令 $Y : = {y | y \in C (R, R), y (t + T) = y (t), \forall t \in R}$ ，在空间Y上定义如下范数：

${‖ y ‖}_{0} = \max_{t \in [0, T]} | y (t) | .$

显然， $(X, ‖ \cdot ‖)$ 和 $(Y, {‖ \cdot ‖}_{0})$ 都是Banach空间。

分别定义算子 $L : X \to Y$ 和 $N : X \to Y$ ，如下

$L x (t) = x^{(2 n + 1)} (t), t \in R; N x (t) = p (t) - \sum_{i = 0}^{2 n - 1} a_{i} (t) x^{(i)} (t) - g (x (t - τ (t))), t \in R .$ (4)

易知算子L是一个指标为零的Fredholm算子，再定义投影算子P、Q分别为

$P x (t) = x (0) = x (T), \forall t \in R, x \in X; Q y (t) = \frac{1}{T} \int_{0}^{T} y (s) d s, \forall t \in R, y \in Y .$ (5)

容易验证，算子N是一个在 $\bar{Ω}$ 上是L-紧的非线性算子。

考虑如下的辅助方程

$x^{(2 n + 1)} (t) + λ \sum_{i = 0}^{2 n - 1} a_{i} (t) x^{(i)} (t) + λ g (x (t - τ (t))) = λ p (t),$ (6)

其中 $0 < λ < 1$ 。

引理3.2 [20] 令 $x (t) \in C^{n} (R, R) \cap C_{T}$ ，则

$x^{(i)} \leq \frac{1}{2} \int_{0}^{T} | x^{(i + 1)} (s) | d s, i = 1, 2, \dots, n - 1.$

其中 $n \geq 2$ ，且 $C_{T} : = {x | x \in C (R, R), x (t + T) = x (t), \forall t \in R}$ 。

引理3.3 [21] 设M，λ是两个正数，且满足 $0 < M < {(\frac{π}{T})}^{2}$ 和 $0 < λ < 1$ ，则对任意的函数 $φ (t) (t \in [0, T])$ ，方程

${\begin{array}{l} x^{″} (t) + λ M x (t) = λ φ (t) \\ x (0) = x (T), x^{'} (0) = x^{'} ( T ) \end{array}$

有唯一解，其表达形式如下：

$x (t) = \int_{0}^{T} G (t, s) λ φ (s) d s,$

其中， $α = \sqrt{λ M}$ ，

$G (t, s) = {\begin{array}{l} ω (t - s), (k - 1) T \leq s \leq t \leq k T, \\ ω (T + t - s), (k - 1) T \leq t \leq s \leq k T (k \in N) . \end{array}$

和

$ω (t) = \frac{\cos α (t - \frac{T}{2})}{2 α \sin \frac{α T}{2}} .$

引理3.4 若定理3.1中的条件被满足，且 $x (t)$ 是方程(6)的一个T周期解，则存在独立于λ的常数 $D_{i} (i = 0, 1, \dots, 2 n)$ ，使得

$x^{(i)} \leq D_{i}, t \in [0, T], i = 0, 1, \dots, 2 n .$ (7)

证明：设 $x (t)$ 是方程(6)的一个T周期解。令

$ε = \frac{m_{0} - r}{2} - \frac{T δ M_{0}^{2}}{T δ M_{0} + M_{1}} .$

由(2)知，存在一个正常数 ${\bar{M}}_{1} > 0$ ，使得

$| g (x (t)) | \leq r | x (t) |, | x (t) | > {\bar{M}}_{1}, t \in R .$ (8)

令

$E_{1} = {t | t \in [0, T], | x (t) | 〉 {\bar{M}}_{1}}, E_{2} = [0, T] \ E_{1} .$ (9)

和

$ρ = \max_{E_{2}} g (x) .$ (10)

由(6)、(8)、(9)和(10)式和引理3.2，有

$\begin{matrix} {‖ x^{(2 n)} ‖}_{0} \leq \frac{1}{2} \int_{0}^{T} | x^{(2 n + 1)} (t) | d t \\ \leq \frac{λ}{2} \int_{0}^{T} [| \sum_{i = 0}^{2 n - 1} a_{i} (t) x^{(i)} (t) | + | g (x (t - τ (t))) | + | p (t) |] d t \\ \leq \frac{λ T}{2} [M_{2 n - 1} {‖ x^{(2 n - 1)} ‖}_{0} + M_{2 n - 2} {‖ x^{(2 n - 2)} ‖}_{0} + \dots + M_{0} {‖ x^{(0)} ‖}_{0}] + \frac{λ}{2} \int_{0}^{T} | g (x (t - τ (t))) | d t + \frac{λ T}{2} {‖ p ‖}_{0} \end{matrix}$

$\begin{matrix} \leq \frac{T}{2} [M_{2 n - 1} \frac{T}{2} + M_{2 n - 2} {(\frac{T}{2})}^{2} + \dots + M_{1} {(\frac{T}{2})}^{2 n - 1}] {‖ x^{(2 n)} ‖}_{0} + \frac{T}{2} M_{0} {‖ x ‖}_{0} \\ + \frac{1}{2} [\int_{E_{1}} | g (x (t - τ (t))) | d t + \int_{E_{2}} | g (x (t - τ (t))) | d t] + \frac{T}{2} {‖ p ‖}_{0} \\ \leq A^{*} {‖ x^{(2 n)} ‖}_{0} + \frac{T}{2} (M_{0} + r + ε) {‖ x ‖}_{0} + \frac{T}{2} (ρ + {‖ p ‖}_{0}) \\ = A^{*} {‖ x^{(2 n)} ‖}_{0} + \frac{m_{0} - r}{2} D {‖ x ‖}_{0} + \frac{T}{2} C . \end{matrix}$ (11)

其中， $C = ρ + {‖ p ‖}_{0}, D = \frac{T}{m_{0} - r} (\frac{M_{0} M_{1}}{T δ M_{0} + M_{1}} + \frac{m_{0} + r}{2})$ 。

由假设(H₄)和(11)式，有

${‖ x^{(2 n)} ‖}_{0} \leq {(1 - A^{*})}^{- 1} (\frac{m_{0} - r}{2} D {‖ x ‖}_{0} + \frac{T}{2} C) .$ (12)

由(6)式和引理3.3，得

$\begin{matrix} x^{(2 n - 1)} (t) = λ \int_{0}^{T} G_{1} (t, t_{1}) [(M_{2 n - 1} - a_{2 n - 1} (t_{1})) x^{(2 n - 1)} (t_{1}) + p (t_{1}) - g (x (t_{1} - τ (t_{1})))] d t_{1} \\ - λ \int_{0}^{T} G_{1} (t, t_{1}) \sum_{i = 0}^{2 n - 2} a_{i} (t_{1}) x^{(i)} (t_{1}) d t_{1}, \end{matrix}$ (13)

其中， $α_{1} = \sqrt{λ M_{2 n - 1}}$ ，

$G_{1} (t, t_{1}) = {\begin{array}{l} ω_{1} (t - t_{1}), (k - 1) T \leq t_{1} \leq t \leq k T, \\ ω_{1} (T + t - t_{1}), (k - 1) T \leq t \leq t_{1} \leq k T (k \in N) . \end{array}$

和

$ω_{1} (t) = \frac{\cos α_{1} (t - \frac{T}{2})}{2 α_{1} \sin \frac{α_{1} T}{2}}, \int_{0}^{T} G_{1} (t, t_{1}) d t_{1} = \frac{1}{λ M_{2 n - 1}} .$

由(13)式和引理3.3，得

$\begin{matrix} x^{(2 n - 3)} (t) = λ \int_{0}^{T} G_{2} (t, t_{1}) \int_{0}^{T} G_{1} (t_{1}, t_{2}) [p (t_{2}) - g (x (t_{2} - τ (t_{2})))] d t_{2} d t_{1} \\ + λ \int_{0}^{T} G_{2} (t, t_{1}) \int_{0}^{T} G_{1} (t_{1}, t_{2}) (M_{2 n - 1} - a_{2 n - 1} (t_{2})) x^{(2 n - 1)} (t_{2}) d t_{2} d t_{1} \\ + λ \int_{0}^{T} G_{2} (t, t_{1}) [\frac{M_{2 n - 3}}{M_{2 n - 1}} x^{(2 n - 3)} (t_{1}) - λ \int_{0}^{T} G_{1} (t_{1}, t_{2}) a_{2 n - 3} (t_{2}) x^{(2 n - 3)} (t_{2}) d t_{2}] d t_{1} \\ - λ \int_{0}^{T} G_{2} (t, t_{1}) \int_{0}^{T} G_{1} (t_{1}, t_{2}) [\sum_{i = 0}^{2 n - 4} a_{i} (t_{2}) x^{(i)} (t_{2}) + a_{2 n - 2} (t_{2}) x^{(2 n - 2)} (t_{2})] d t_{2} d t_{1}, \end{matrix}$ (14)

其中， $α_{2} = \sqrt{\frac{M_{2 n - 1}}{M_{2 n - 3}}}$ ，

$G_{2} (t, t_{2}) = {\begin{array}{l} ω_{2} (t - t_{2}), (k - 1) T \leq t_{2} \leq t \leq k T, \\ ω_{2} (T + t - t_{2}), (k - 1) T \leq t \leq t_{2} \leq k T (k \in N) . \end{array}$

和

$ω_{2} (t) = \frac{\cos α_{2} (t - \frac{T}{2})}{2 α_{2} \sin \frac{α_{2} T}{2}}, \int_{0}^{T} G_{2} (t, t_{2}) d t_{2} = \frac{M_{2 n - 1}}{M_{2 n - 3}} .$

利用数学归纳法，可以得到

$\begin{matrix} x^{'} (t) \\ = λ \int_{0}^{T} G_{n} (t, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [p (t_{n}) - g (x (t_{n} - τ (t_{n})))] d t_{n} \dots d t_{1} \\ + λ \int_{0}^{T} G_{n} (t, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) (M_{2 n - 1} - a_{2 n - 1} (t_{n})) x^{(2 n - 1)} (t_{n}) d t_{n} \dots d t_{1} \\ + \int_{0}^{T} G_{n} (t, t_{1}) \dots \int_{0}^{T} G_{2} (t_{n - 2}, t_{n - 1}) [\frac{M_{2 n - 3}}{M_{2 n - 1}} x^{(2 n - 3)} (t_{n - 1}) - λ \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) a_{2 n - 3} (t_{n}) x^{(2 n - 3)} (t_{n}) d t_{n}] d t_{n - 1} \dots d t_{1} \end{matrix}$

$\begin{matrix} + \int_{0}^{T} G_{n} (t, t_{1}) \dots \int_{0}^{T} G_{3} (t_{n - 3}, t_{n - 2}) [\frac{M_{2 n - 5}}{M_{2 n - 3}} x^{(2 n - 5)} (t_{n - 2}) - λ \int_{0}^{T} G_{2} (t_{n - 2}, t_{n - 1}) \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) a_{2 n - 5} (t_{n}) x^{(2 n - 5)} (t_{n}) d t_{n} d t_{n - 1}] d t_{n - 2} \dots d t_{1} \\ + \dots + \dots \\ + \int_{0}^{T} G_{n} (t, t_{1}) [\frac{M_{1}}{M_{3}} x^{(1)} (t_{1}) - λ \int_{0}^{T} G_{n - 1} (t_{1}, t_{2}) \int_{0}^{T} G_{n - 2} (t_{2}, t_{3}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) a_{1} (t_{n}) x^{(1)} (t_{n}) d t_{n} \dots d t_{2}] d t_{1} \\ - λ \int_{0}^{T} G_{n} (t, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [\sum_{k = 1}^{n - 1} a_{2 k} (t_{n}) x^{(2 k)} (t_{n})] d t_{n} \dots d t_{1} - λ \int_{0}^{T} G_{n} (t, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [a_{0} (t_{n}) x^{(0)} (t_{n})] d t_{n} \dots d t_{1} . \end{matrix}$ (15)

其中， $α_{i} = \sqrt{\frac{M_{2 n - 2 i + 1}}{M_{2 n - 2 i + 3}}}, 2 \leq i \leq n$ ，

$G_{i} (t, t_{1}) = {\begin{array}{l} ω_{i} (t - t_{i}), (k - 1) T \leq t_{i} \leq t \leq k T, \\ ω_{i} (T + t - t_{i}), (k - 1) T \leq t \leq t_{i} \leq k T (k \in N) . \end{array}$

和

$ω_{i} (t) = \frac{\cos α_{i} (t - \frac{T}{2})}{2 α_{i} \sin \frac{α_{i} T}{2}}, \int_{0}^{T} G_{i} (t, t_{i}) d t_{i} = \frac{M_{2 n - 2 i + 3}}{M_{2 n - 2 i + 1}}, 2 \leq i \leq n .$

再根据(15)式和常数变易法，

$\begin{matrix} x (t) = (e^{\frac{M_{0}}{M_{1}} T} - 1) \int_{0}^{t} Φ (t, s) {λ \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [p (t_{n}) - g (x (t_{n} - τ (t_{n})))] d t_{n} \dots d t_{1} \\ + λ \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) (M_{2 n - 1} - a_{2 n - 1} (t_{n})) x^{(2 n - 1)} (t_{n}) d t_{n} \dots d t_{1} \\ + \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{2} (t_{n - 2}, t_{n - 1}) [\frac{M_{2 n - 3}}{M_{2 n - 1}} x^{(2 n - 3)} (t_{n - 1}) - λ \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) a_{2 n - 3} (t_{n}) x^{(2 n - 3)} (t_{n}) d t_{n}] d t_{n - 1} \dots d t_{1} \\ + \dots + \dots \end{matrix}$

$\begin{matrix} + \int_{0}^{T} G_{n} (s, t_{1}) [\frac{M_{1}}{M_{3}} x^{(1)} (t_{1}) - λ \int_{0}^{T} G_{n - 1} (t_{1}, t_{2}) \int_{0}^{T} G_{n - 2} (t_{2}, t_{3}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) a_{1} (t_{n}) x^{(1)} (t_{n}) d t_{n} \dots d t_{2}] d t_{1} \\ - λ \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [\sum_{k = 1}^{n - 1} a_{2 k} (t_{n}) x^{(2 k)} (t_{n})] d t_{n} \dots d t_{1}} d s \\ + (e^{\frac{M_{0}}{M_{1}} T} - 1) \int_{o}^{t} Φ (t, s) {\frac{M_{0}}{M_{1}} x (s) - λ \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [a_{0} (t_{n}) x^{(0)} (t_{n})] d t_{n} \dots d t_{1}} d s \\ + \int_{0}^{T} Φ (t, s) {λ \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [p (t_{n}) - g (x (t_{n} - τ (t_{n})))] d t_{n} \dots d t_{1} \\ + λ \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) (M_{2 n - 1} - a_{2 n - 1} (t_{n})) x^{(2 n - 1)} (t_{n}) d t_{n} \dots d t_{1} \end{matrix}$

$\begin{array}{l} + \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{2} (t_{n - 2}, t_{n - 1}) [\frac{M_{2 n - 3}}{M_{2 n - 1}} x^{(2 n - 3)} (t_{n - 1}) - λ \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) a_{2 n - 3} (t_{n}) x^{(2 n - 3)} (t_{n}) d t_{n}] d t_{n - 1} \dots d t_{1} \\ + \dots + \dots \\ + \int_{0}^{T} G_{n} (s, t_{1}) [\frac{M_{3}}{M_{1}} x^{(1)} (t_{1}) - λ \int_{0}^{T} G_{n - 1} (t_{1}, t_{2}) \int_{0}^{T} G_{n - 2} (t_{2}, t_{3}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) a_{1} (t_{n}) x^{(1)} (t_{n}) d t_{n} \dots d t_{2}] d t_{1} \\ - λ \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [\sum_{k = 1}^{n - 1} a_{2 k} (t_{n}) x^{(2 k)} (t_{n})] d t_{n} \dots d t_{1}} d s \\ + \int_{0}^{T} Φ (t, s) {\frac{M_{0}}{M_{1}} x (s) - λ \int_{0}^{T} G_{n} (s, t_{1}) \dots \int_{0}^{T} G_{1} (t_{n - 1}, t_{n}) [a_{0} (t_{n}) x^{(0)} (t_{n})] d t_{n} \dots d t_{1}} d s . \end{array}$ (16)

其中

$Φ (t, s) = \frac{e^{\frac{M_{0}}{M_{1}} (s - t)}}{e^{\frac{M_{0}}{M_{1}} T} - 1}, Φ (t, s) \leq Φ (t, t + T) = \frac{e^{\frac{M_{0}}{M_{1}} T}}{e^{\frac{M_{0}}{M_{1}} T} - 1} = δ,$

且

$\int_{0}^{t} Φ (t, s) d s = \frac{M_{1} (1 - e^{- \frac{M_{0}}{M_{1}} T})}{M_{0} (e^{\frac{M_{0}}{M_{1}} T} - 1)} \leq \frac{M_{1}}{M_{0} (e^{\frac{M_{0}}{M_{1}} T} - 1)} .$

由(8)、(9)、(10)、(16)式和引理3.2，得

$\begin{matrix} {‖ x ‖}_{0} \leq (\frac{1}{M_{0}} + \frac{T δ}{M_{1}}) {{‖ p ‖}_{0} + ρ + [M_{0} - m_{0} + (r + ε)] {‖ x ‖}_{0} + [(M_{2 n - 1} - m_{2 n - 1}) (\frac{T}{2}) + M_{2 n - 2} {(\frac{T}{2})}^{2} \\ + (M_{2 n - 3} - m_{2 n - 3}) {(\frac{T}{2})}^{3} + M_{2 n - 4} {(\frac{T}{2})}^{4} + \dots + \dots (M_{3} - m_{3}) {(\frac{T}{2})}^{2 n - 3} + M_{2} {(\frac{T}{2})}^{2 n - 2} \\ + (M_{1} - m_{1}) {(\frac{T}{2})}^{2 n - 1}] {‖ x^{(2 n)} ‖}_{0}}, \end{matrix}$ (17)

经化简后，有

$\begin{array}{l} \frac{M_{0} M_{1}}{T δ M_{0} + M_{1}} {‖ x ‖}_{0} - (M_{0} - m_{0} + r + ε) {‖ x ‖}_{0} \leq {‖ p ‖}_{0} + ρ + [(M_{2 n - 1} - m_{2 n - 1}) (\frac{T}{2}) + M_{2 n - 2} {(\frac{T}{2})}^{2} \\ + (M_{2 n - 3} - m_{2 n - 3}) {(\frac{T}{2})}^{3} + \dots + M_{2} {(\frac{T}{2})}^{2 n - 2} + (M_{1} - m_{1}) {(\frac{T}{2})}^{2 n - 1}] {‖ x^{(2 n)} ‖}_{0}, \end{array}$ (18)

因此，根据假设(H₄)、(18)式和ε的取值，可得

${‖ x ‖}_{0} \leq \frac{2 (B {‖ x^{(2 n)} ‖}_{0} + C)}{m_{0} - r} .$ (19)

结合(12)式和(19)式，得

$(1 - A^{*}) {‖ x^{(2 n)} ‖}_{0} \leq D (B {‖ x^{(2 n)} ‖}_{0} + C) + \frac{T C}{2},$ (20)

因此有

$(A - D B) {‖ x^{(2 n)} ‖}_{0} \leq D C + \frac{T C}{2},$ (21)

其中， $C = ρ + {‖ p ‖}_{0}, D = \frac{T}{m_{0} - r} (\frac{M_{0} M_{1}}{T δ M_{0} + M_{1}} + \frac{m_{0} + r}{2}) .$

再由假设(H₄)和(21)式，得到

${‖ x^{(2 n)} ‖}_{0} \leq \frac{D C + T C / 2}{A - D B} = D_{2 n} .$ (22)

联立(19)和(22)式，得

${‖ x ‖}_{0} \leq \frac{2 (B D_{2 n} + C)}{m_{0} - r} = D_{0} .$ (23)

最后，由(22)、(23)式和引理3.2，得到

${‖ x^{(i)} ‖}_{0} \leq D_{i}, (0 \leq i \leq 2 n) .$

引理3.4得证。

定理3.1的证明：设 $x (t)$ 是方程(6)的一个T周期解。由引理3.4知，存在独立于λ的常数 $D_{i} (i = 0, 1, \dots, 2 n)$ ，使得(7)式成立。由(3)式知，存在正常数 $M_{2} > 0$ ，使得

$sgn (x) g (x) > 0, | x (t) | > M_{2}, t \in R .$ (24)

取一正常数 $\bar{D} > \max_{0 \leq i \leq 2 n} {D_{i}} + M_{2}$ ，令

$Ω : = {x \in X | ‖ x ‖ < \bar{D}} .$

此时是指标为零的Fredholm算子，N是在 $\bar{Ω}$ 上L-紧的非线性算子。对任意有界的周期解 $x (t)$ ，当 $x \in \partial Ω \cap D o m L$ ， $λ \in (0, 1)$ 时，有 $L x \neq λ N x$ 。

当 $x \in \partial Ω \cap K e r L$ 时，有 $x = \bar{D}$ 或 $x = - \bar{D}$ ，再结合(3)和(5)式可得

$\begin{array}{l} x^{(i)} (t) = 0, i = 1, 2, \dots, n - 1, \\ - \frac{1}{T} \int_{0}^{T} [\sum_{i = 0}^{2 n - 1} a_{0} (t) \bar{D} + g (\bar{D}) - p (t)] d t < 0, \\ \frac{1}{T} \int_{0}^{T} [\sum_{i = 0}^{2 n - 1} a_{0} (t) \bar{D} - g (- \bar{D}) + p (t)] d t > 0. \end{array}$ (25)

然后由上式可知，当 $x \in \partial Ω \cap K e r L$ 时，有

$Q N x = \frac{1}{T} \int_{0}^{T} [- \sum_{i = 0}^{2 n - 1} a_{i} (t) x^{(i)} (t) - g (x (t - τ (t))) + p (t)] d t \neq 0.$ (26)

故对任意的 $x \in \partial Ω \cap K e r L$ 和 $η \in (0, 1)$ ，有

$x H (x, η) = - η x^{2} + \frac{x}{T} (1 - η) \int_{0}^{T} [- \sum_{i = 0}^{2 n - 1} a_{i} (t) x^{(i)} (t) - g (x (t - τ (t))) + p (t)] d t \neq 0.$ (27)

因此， $H (x, η)$ 是一个同伦映射。进而有

$\begin{matrix} \deg (Q N, Ω \cap K e r L, 0) = \deg {\frac{1}{T} \int_{0}^{T} [- \sum_{i = 0}^{2 n - 1} a_{i} (t) x^{(i)} (t) - g (x (t - τ (t))) + p (t)] d t, Ω \cap K e r L, 0} \\ = \deg (- x, Ω \cap K e r L, 0) \neq 0. \end{matrix}$ (28)

根据引理2.1可知，方程 $L x = N x$ 在 $\bar{Ω} \cap D o m L$ 上至少存在一个解。因此，方程(1)至少存在一个周期解。定理3.1得证。

基金项目

广东省自然科学基金资助项目(2018A030313871)。

文章引用

林梦媛,吴进,陈柏立. 一类奇数阶泛函微分方程周期解的存在性
Existence of Periodic Solutions for a Class of Odd Order Functional Differential Equations[J]. 理论数学, 2023, 13(02): 345-353. https://doi.org/10.12677/PM.2023.132038

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NOTES

^*通讯作者。

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