具有脉冲的非线性耦合积分–微分系统的周期性
Periodic and Asymptotically Periodic Solutions on Nonlinear Coupled Integro-Differential Systems with Impulses
DOI: 10.12677/AAM.2019.81015, PDF, HTML, XML, 下载: 1,076  浏览: 3,330  国家自然科学基金支持
作者: 陈秋凤, 李建利:湖南师范大学,数学与统计学院,湖南 长沙
关键词: 脉冲微分方程Schauder不动点定理周期解渐近周期解Impulsive Differential Equation Schauder’s Fixed Point Theorem Periodic Solutions Asymptotic Periodic Solutions
摘要: 该文研究了具有脉冲的非线性耦合积分—微分系统的周期性。利用Schauder不动点定理,证明了具有脉冲的非线性耦合积分—微分系统至少存在一个周期解和一个渐近周期解,我们的结果推广和改进了相关文献的结果。
Abstract: In this paper, we study the existence of periodic and asymptotically periodic solutions for a coupled nonlinear Volterra integro-differential equation with impulses. By using Schauder’s fixed point theorem, we obtain that the system has at least one periodic solution and an asymptotically periodic solution.
文章引用:陈秋凤, 李建利. 具有脉冲的非线性耦合积分–微分系统的周期性[J]. 应用数学进展, 2019, 8(1): 135-144. https://doi.org/10.12677/AAM.2019.81015

1. 引言

脉冲效应指在一定时间内状态突然发生改变,脉冲微分方程广泛应用于种群生物学,医学,工程,化学等领域 [1] [2] [3] [4] [5] 。耦合积分-微分方程也广泛应用于生物和环境科学的许多领域。该文中,我们研究一类具有脉冲的非线性耦合积分-微分方程周期解和渐近周期解的存在性。

在文献 [6] [7] [8] [9] 中,作者研究了Volterra积分–微分方程线性系统渐近周期解的存在性。在文献 [10] 中,作者考虑了无脉冲时周期解和渐近周期解的存在性,该文改进了文献 [10] 中的一些条件,进而得到周期解和渐近周期解的存在性结果。

该文考虑下面的耦合Volterra脉冲微分方程

{ x ( t ) = h 1 ( t ) x ( t ) + h 2 ( t ) y ( t ) + t a ( t , s ) f ( x ( s ) , y ( s ) ) , t t k y ( t ) = p 1 ( t ) y ( t ) + p 2 ( t ) x ( t ) + t b ( t , s ) g ( x ( s ) , y ( s ) ) , t t k x ( t k + ) = x ( t k ) + I k ( x ( t k ) ) , t = t k y ( t k + ) = y ( t k ) + J k ( y ( t k ) ) , t = t k (1.1)

其中a,b,f,g, h i , p i , i = 1 , 2 I k , J k , k N + 是连续函数。

假设存在一个正常数 T > 0 ,使得对 t , s R

a ( t + T , s + T ) = a ( t , s ) , b ( t + T , s + T ) = b ( t , s ) , p i ( t + T ) = p i ( t ) , h i ( t + T ) = h i ( t ) , i = 1 , 2 (1.2)

且存在常数 q > 0 ,使得

t k + q = t k + T , I k + q ( ) = I k ( ) , J k + q ( ) = J k ( ) , k N + (1.3)

此外,假设

0 T h 1 ( s ) d s 0 , 0 T p 1 ( s ) d s 0 (1.4)

定义 P T = { ( φ , ψ ) ( t + T ) = ( φ , ψ ) ( t ) } ,其中 φ , ψ 是实值连续函数。定义范数

( x , y ) = max { max t [ 0 , T ] | x ( t ) | , max t [ 0 , T ] | y ( t ) | }

那么, P T 是Banach空间。

引理1.1 假设(1.2)~(1.4)成立,如果 ( x , y ) P T ,那么 x ( t ) y ( t ) 是(1.1)的一个周期解当且仅当

x ( t ) = { x ( t ) , t [ 0 , T ] x ( t n T ) , t [ n T , ( n + 1 ) T ]

y ( t ) = { y ( t ) , t [ 0 , T ] y ( t n T ) , t [ n T , ( n + 1 ) T ]

其中

x * ( t ) = 0 T G 1 ( t , s ) ( h 2 ( s ) y ( s ) + s a ( s , u ) f ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < T G 1 ( t , t k ) I k ( x ( t k ) )

G 1 ( t , s ) = 1 1 e 0 T h 1 ( s ) d s { e s t h 1 ( u ) d u , 0 s < t T e s t + T h 1 ( u ) d u , 0 t s T

y * ( t ) = 0 T G 2 ( t , s ) ( p 2 ( s ) x ( s ) + s b ( s , u ) g ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < T G 2 ( t , t k ) J k ( y ( t k ) )

G 2 ( t , s ) = 1 1 e 0 T p 1 ( s ) d s { e s t p 1 ( u ) d u , 0 s < t T e s t + T p 1 ( u ) d u , 0 t s T

证明:设 ( x , y ) P T 是(1.1)的解,对(1.1)的第一个方程从 [ 0 , t ] 积分得

x ( t ) e 0 t h 1 ( s ) d s x ( 0 ) = 0 t e 0 s h 1 ( u ) d u ( h 2 ( s ) y ( s ) + s a ( s , u ) f ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < t e 0 t k h 1 ( s ) d s I k ( x ( t k ) )

x ( 0 ) = x ( T )

x * ( t ) = x ( 0 ) e 0 t h 1 ( u ) d u + 0 t e 0 t h 1 ( u ) d u ( h 2 ( s ) y ( s ) + s a ( s , u ) f ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < T e t k t h 1 ( s ) d s I k ( x ( t k ) ) = 1 1 e 0 T h 1 ( s ) d s 0 t e s t h 1 ( u ) d u ( h 2 ( s ) y ( s ) + s a ( s , u ) f ( x ( u ) , y ( u ) ) d u ) d s + e 0 T h 1 ( s ) d s 1 e 0 T h 1 ( s ) d s t T e s t h 1 ( u ) d u ( h 2 ( s ) y ( s ) + s a ( s , u ) f ( x ( u ) , y ( u ) ) d u ) d s + 1 1 e 0 T h 1 ( s ) d s 0 < t k < t e t k t h 1 ( s ) d s I k ( x ( t k ) ) + e 0 T h 1 ( s ) d s 1 e 0 T h 1 ( s ) d s t < t k < T e t k t h 1 ( s ) d s I k ( x ( t k ) ) = 0 T G 1 ( t , s ) ( h 2 ( s ) y ( s ) + s a ( s , u ) f ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < T G 1 ( t , t k ) I k ( x ( t k ) )

同理可证,

y * ( t ) = 0 T G 2 ( t , s ) ( p 2 ( s ) x ( s ) + s b ( s , u ) g ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < T G 2 ( t , t k ) J k ( y ( t k ) )

反之亦然,因此,引理1.1得证。

2. 周期解

定理2.1 (Schuder不动点定理)设X是Banach空间,K是X中有界闭凸子集。如果 T : K K 是全连续的,则T在K中有一个不动点。

本文给出以下假设。

假设存在正常数 L 1 , L 2 , L 1 , L 2 , M 1 , M 2 , N 1 , N 2 , b k , c k , ( k N + ) ,使得

| f ( x , y ) | M 1 , | g ( x , y ) | M 2 (2.1)

0 T | G 1 ( t , s ) h 2 ( s ) | d s L 1 , 0 T | G 2 ( t , s ) p 2 ( s ) | d s L 2 (2.2)

0 T | G 1 ( t , s ) | s | a ( s , u ) | d u d s N 1 , 0 T | G 2 ( t , s ) | s | b ( s , u ) | d u d s N 2 (2.3)

{ | I k ( x ) | b k | x | , | k = 1 p G 1 ( t , t k ) b k | L 1 , t [ 0 , T ] | J k ( x ) | c k | y | , | k = 1 p G 2 ( t , t k ) c k | L 2 , t [ 0 , T ] (2.4)

Ω M = { ( x , y ) : ( x , y ) P T , ( x , y ) M } P T 的子集,则 Ω M P T 中的有界闭子集。

定义算子 F : Ω M P T

F ( x , y ) = ( F 1 ( x , y ) ( t ) , F 2 ( x , y ) (t))

其中

F 1 ( x , y ) ( t ) = 0 T G 1 ( t , s ) ( h 2 ( s ) y ( s ) + s a ( s , u ) f ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < T G 1 ( t , t k ) I k ( x ( t k ) )

F 2 ( x , y ) ( t ) = 0 T G 2 ( t , s ) ( p 2 ( s ) x ( s ) + s b ( s , u ) g ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < T G 2 ( t , t k ) J k ( y ( t k ) )

定理2.2 假设(1.2)~(1.4),(2.1)~(2.4)成立,且

L 1 + L 1 < 1 , L 2 + L 2 < 1 (2.5)

则(1.1)至少有一个T-周期解。

证明:若 x ( t ) , y ( t ) 是(1.1)的周期解,由引理1.1可知, F ( x , y ) ( t + T ) = F ( x , y ) ( t ) 。设

M = max { N 1 M 1 1 L 1 L 1 , N 2 M 2 1 L 2 L 2 }

对于 ( x , y ) Ω M , t [ 0 , T ]

| F 1 ( x , y ) ( t ) | 0 T | G 1 ( t , s ) h 2 ( s ) | | y ( s ) | d s + 0 < t k < T | G 1 ( t , t k ) | | I k ( x ( t k ) ) | + 0 T | G 1 ( t , s ) | s | a ( s , u ) | | f ( x ( u ) , y ( u ) ) | d u d s M L 1 + N 1 M 1 + M L 1 M

| F 2 ( x , y ) ( t ) | 0 T | G 2 ( t , s ) p 2 ( s ) | | x ( s ) | d s + 0 < t k < T | G 2 ( t , t k ) | | J k ( y ( t k ) ) | + 0 T | G 2 ( t , s ) | s | b ( s , u ) | | g ( x ( u ) , y ( u ) ) | d u d s M L 2 + N 2 M 2 + M L 2 M

所以, F ( Ω M ) Ω M

下证F连续。取 Ω M 中的一个序列 { ( x n , y n ) } ,且

lim n ( x n , y n ) ( x , y ) = 0

因为 Ω M 是闭的,我们有 ( x , y ) Ω M 。由F的定义,有

F ( x n , y n ) F ( x , y ) = max { max t [ 0 , T ] | F 1 ( x n , y n ) ( t ) F 1 ( x , y ) ( t ) | , max t [ 0 , T ] | F 2 ( x n , y n ) ( t ) F 2 ( x , y ) ( t ) | }

| F 1 ( x n , y n ) ( t ) F 1 ( x , y ) ( t ) | = | 0 T G 1 ( t , s ) ( h 2 ( s ) y n ( s ) + s a ( s , u ) f ( x n ( u ) , y n ( u ) ) d u ) d s 0 T G 1 ( t , s ) ( h 2 ( s ) y ( s ) + s a ( s , u ) f ( x ( u ) , y ( u ) ) d u ) d s + 0 < t k < T G 1 ( t , t k ) ( I k ( x n ( t k ) ) I k ( x ( t k ) ) ) | 0 T | G 1 ( t , s ) | ( s | a ( s , u ) | | f ( x n ( u ) , y n ( u ) ) f ( x ( u ) , y ( u ) ) | d u ) d s + 0 T | G 1 ( t , s ) h 2 ( s ) | | y n ( s ) y ( s ) | d s + 0 < t k < T | G 1 ( t , t k ) | | I k ( x n ( t k ) ) I k ( x ( t k ) ) |

f , I k 的连续性和Lebesgue控制收敛定理,有

lim n max t [ 0 , T ] | F 1 ( x n , y n ) ( t ) F 1 ( x , y ) ( t ) | = 0

同理可证,

lim n max t [ 0 , T ] | F 2 ( x n , y n ) ( t ) F 2 ( x , y ) ( t ) | = 0

所以,F是连续映射。因为 F ( Ω M ) Ω M ,所以 F ( Ω M ) 是一致有界的。易证对任意的 ( x , y ) Ω M , t t k 存在一个常数 L > 0 ,使得 | d d t F 1 ( x , y ) ( t ) | L , | d d t F 2 ( x , y ) ( t ) | L ,即 | d d t F ( x , y ) ( t ) | L 。所以, F ( Ω M ) Ω M 是拟等度连续的。由Arzela-Ascoli’s定理, F ( Ω M ) 是列紧的。综上,F是全连续的。

由定理2.1可知,F在 Ω M 中有一个不动点,即存在 ( x , y ) Ω M ,使得 ( x , y ) = F ( x , y ) 。由引理1.1可知,(1.1)有一个T-周期解。

定理2.3 假设(1.2)~(1.4),(2.3)~(2.5)成立,且存在连续函数 G , W ,及正常数 Q 1 , Q 2 使得

| f ( x , y ) | Q 1 W ( | y | ) , W ( u ) u 1 L 1 L 1 N 1 Q 1 , u > 0

| g ( x , y ) | Q 2 G ( | x | ) , G ( u ) u 1 L 2 L 2 N 2 Q 2 , u > 0

那么(1.1)至少有一个T-周期解。

证明:令

M = max { W ( M ) N 1 Q 1 1 L 1 L 1 , G ( M ) N 2 Q 2 1 L 2 L 2 }

( x , y ) Ω M , t [ 0 , T ]

| F 1 ( x , y ) ( t ) | 0 T | G 1 ( t , s ) h 2 ( s ) | | y ( s ) | d s + 0 < t k < T | G 1 ( t , t k ) | | I k ( x ( t k ) ) | + 0 T | G 1 ( t , s ) | s | a ( s , u ) | | f ( x ( u ) , y ( u ) ) | d u d s 0 T | G 1 ( t , s ) h 2 ( s ) | | y ( s ) | d s + 0 < t k < T | G 1 ( t , t k ) | | I k ( x ( t k ) ) | + Q 1 0 T | G 1 ( t , s ) | s | a ( s , u ) | W ( | y ( u ) | ) d u d s M L 1 + M L 1 + 1 L 1 L 1 N 1 Q 1 N 1 Q 1 | y ( u ) | M L 1 + M L 1 + ( 1 L 1 L 1 ) M = M

类似地,

| F 2 ( x , y ) ( t ) | 0 T | G 2 ( t , s ) p 2 ( s ) | | x ( s ) | d s + 0 < t k < T | G 2 ( t , t k ) | | J k ( y ( t k ) ) | + 0 T | G 2 ( t , s ) | s | b ( s , u ) | | g ( x ( u ) , y ( u ) ) | d u d s 0 T | G 2 ( t , s ) p 2 ( s ) | | x ( s ) | d s + 0 < t k < T | G 2 ( t , t k ) | | J k ( y ( t k ) ) | + Q 2 0 T | G 2 ( t , s ) | s | b ( s , u ) | G ( | x ( u ) | ) d u d s M L 2 + M L 2 + 1 L 2 L 2 N 2 Q 2 N 2 Q 2 | x ( u ) | M L 2 + M L 2 + ( 1 L 2 L 2 ) M = M

其余的证明类似定理2.2的证明。

3. 渐近周期解

定义3.1 一个函数 x ( t ) 称为渐近T-周期的,如果存在两个函数 x 1 ( t ) x 2 ( t ) ,使得 x ( t ) = x 1 ( t ) + x 2 ( t ) ,其中 x 1 ( t ) 是T-周期的, lim t x 2 ( t ) = 0

假设 h 1 ( t ) , p 1 ( t ) 是T-周期的,且

0 T h 1 ( s ) d s = 0 , 0 T p 1 ( s ) d s = 0 (3.1)

从而存在常数 m k , M k , k = 1 , 2 ,使得

m 1 e 0 t h 1 ( s ) d s M 1 , m 2 e 0 t p 1 ( s ) d s M 2 (3.2)

假设存在正数A,B使得

0 s | a ( s , u ) | d u d s A , 0 s | b ( s , u ) | d u d s B (3.3)

另外,假设

t | h 2 ( s ) | d s M 3 , t | p 2 ( s ) | d s M 4 (3.4)

b k 0 , c k 0 , k = 1 b k < , k = 1 c k < (3.5)

定理3.1 假设(2.4),(3.1)~(3.5)成立,存在正常数 Q ¯ 1 , Q ¯ 2 ,使得

| f ( x , y ) | Q ¯ 1 | y | , | g ( x , y ) | Q ¯ 2 | x |

1 M 3 M 1 m 1 1 M 1 m 1 1 A Q ¯ 1 M 1 M 5 > 0 , 1 M 4 M 2 m 2 1 M 2 m 2 1 B Q ¯ 2 M 2 M 6 > 0

则系统(1.1)有渐近T-周期解 ( x , y ) ,满足

x ( t ) = x 1 ( t ) + x 2 ( t ) , y ( t ) = y 1 ( t ) + y 2 (t)

其中,

x 1 ( t ) = c 1 e 0 t h 1 ( s ) d s , y 1 ( t ) = c 2 e 0 t p 1 ( s ) d s , t R

c 1 , c 2 为固定的非零常数,

lim t x 2 ( t ) = lim t y 2 ( t ) = 0 .

证明:定义 H T = { ( φ , ψ ) : φ = φ 1 + φ 2 , ψ = ψ 1 + ψ 2 , ( φ 1 , ψ 1 ) ( t + T ) = ( φ 1 , ψ 1 ) ( t ) , ( φ 2 , ψ 2 ) ( t ) ( 0 , 0 ) , t } 。定义范数

( x , y ) = max { max t [ 0 , T ] | x ( t ) | , max t [ 0 , T ] | y ( t ) | }

H T 是一个Banach空间。记 Ω M * = { ( x , y ) : ( x , y ) H T , ( x , y ) M * } ,则 Ω M * H T 中的有界闭凸子集。

( x , y ) Ω M * ,定义算子 E : Ω M * H T

E ( x , y ) ( t ) = ( E 1 ( y ) ( t ) , E 2 ( x ) (t))

其中

E 1 ( y ) ( t ) = c 1 e 0 t h 1 ( s ) d s t e 0 t h 1 ( l ) d l e 0 s h 1 ( l ) d l h 2 ( s ) y ( s ) d s t s e 0 t h 1 ( l ) d l e 0 s h 1 ( l ) d l a ( s , u ) f ( x ( u ) , y ( u ) ) d u d s t < t k < e t k t h 1 ( s ) d s I k ( x ( t k ) )

E 2 ( x ) ( t ) = c 2 e 0 t p 1 ( s ) d s t e 0 t p 1 ( l ) d l e 0 s p 1 ( l ) d l p 2 ( s ) x ( s ) d s t s e 0 t p 1 ( l ) d l e 0 s p 1 ( l ) d l b ( s , u ) g ( x ( u ) , y ( u ) ) d u d s t < t k < e t k t p 1 ( s ) d s J k ( y ( t k ) )

下证映射E在 Ω M * 中有一个不动点。由(3.5)可知,存在正常数 M 5 , M 6 使得

| k = 1 b k | M 5 , | k = 1 c k | M 6

M * = { c 1 M 1 1 M 3 M 1 m 1 1 M 1 m 1 1 A Q ¯ 1 M 1 M 5 , c 2 M 2 1 M 4 M 2 m 2 1 M 2 m 2 1 B Q ¯ 2 M 2 M 6 }

| E 1 ( y ) ( t ) c 1 e 0 t h 1 ( s ) d s | M * M 3 M 1 m 1 1 + M 1 m 1 1 Q ¯ 1 t s | a ( s , u ) | | x | d u d s + t < t k < e t k t h 1 ( s ) d s | I k ( x ( t k ) ) | M * M 3 M 1 m 1 1 + M 1 m 1 1 Q ¯ 1 x 0 s | a ( s , u ) | d u d s + | t < t k < b k | M 1 M * M * M 3 M 1 m 1 1 + M 1 m 1 1 A Q ¯ 1 M * + M 1 M * M 5

同理,

| E 2 ( x ) ( t ) c 2 e 0 t p 1 ( s ) d s | M * M 4 M 2 m 2 1 + M 2 m 2 1 B Q ¯ 2 M * + M 2 M * M 6

因此,

| E 1 ( y ) ( t ) | M * M 3 M 1 m 1 1 + M 1 m 1 1 A Q ¯ 1 M * + M 1 M * M 5 + c 1 M 1 M *

| E 2 ( x ) ( t ) | M * M 4 M 2 m 2 1 + M 2 m 2 1 B Q ¯ 2 M * + M 2 M * M 6 + c 2 M 2 M *

x 1 ( t ) = c 1 e 0 t h 1 ( s ) d s , y 1 ( t ) = c 2 e 0 t p 1 ( s ) d s , t R

x 2 ( t ) = t e 0 t h 1 ( l ) d l e 0 s h 1 ( l ) d l h 2 ( s ) y ( s ) d s t s e 0 t h 1 ( l ) d l e 0 s h 1 ( l ) d l a ( s , u ) f ( x ( u ) , y ( u ) ) d u d s t < t k < e t k t h 1 ( s ) d s I k ( x ( t k ) )

y 2 ( t ) = t e 0 t p 1 ( l ) d l e 0 s p 1 ( l ) d l p 2 ( s ) x ( s ) d s t s e 0 t p 1 ( l ) d l e 0 s p 1 ( l ) d l b ( s , u ) g ( x ( u ) , y ( u ) ) d u d s t < t k < e t k t p 1 ( s ) d s J k ( y ( t k ) )

计算可知,

x 1 ( t + T ) = c 1 e 0 t + T h 1 ( s ) d s = c 1 e 0 t h 1 ( s ) d s e t t + T h 1 ( s ) d s = c 1 e 0 t h 1 ( s ) d s

y 1 ( t + T ) = c 2 e 0 t + T p 1 ( s ) d s = c 2 e 0 t p 1 ( s ) d s e t t + T p 1 ( s ) d s = c 2 e 0 t p 1 ( s ) d s

由(3.4)~(3.5)可知,

lim t | x 2 ( t ) | M * M 1 m 1 1 lim t t | h 2 ( s ) | d s + M 1 m 1 1 Q ¯ 1 M * lim t t s | a ( s , u ) | d u d s + M * M 1 lim t | t < t k < b k | = 0

lim t | y 2 ( t ) | M * M 2 m 2 1 lim t t | p 2 ( s ) | d s + M 2 m 2 1 Q ¯ 2 M * lim t t s | b ( s , u ) | d u d s + M * M 2 lim t | t < t k < c k | = 0

因此, lim t x 2 ( t ) = 0 , lim t y 2 ( t ) = 0 。故 E ( Ω M * ) Ω M *

类似定理2.2可证E是全连续的。因此,由Schauder不动点定理,存在一个不动点 ( x , y ) Ω M * ,使得

E ( x , y ) ( t ) = ( E 1 ( y ) ( t ) , E 2 ( x ) ( t ) ) = ( x ( t ) , y (t))

易知该不动点是(1.1)的解。所以 ( x , y ) 是(1.1)的解,定理3.2得证。

例3.1 考虑下面的系统

{ x ( t ) = cos ( t ) x ( t ) + 2 t ( t 2 + 10 ) 2 y ( t ) + t e 6 t + 4 s sin ( x ( t ) + y ( t ) ) d s , t t k y ( t ) = cos ( t ) y ( t ) + 2 t ( t 2 + 10 ) 2 x ( t ) + t e 6 t + 4 s cos ( x ( t ) + y ( t ) ) d s , t t k x ( t k + ) = x ( t k ) + ( 1 12 ) k ( x ( t k ) ) , t = t k y ( t k + ) = y ( t k ) + ( 1 13 ) k ( y ( t k ) ) , t = t k

经过简单的计算知,定理3.1中的条件均满足,因此,该系统有一个渐近2π-周期解。

基金项目

国家自然科学基金(No. 11571088, No. 11471109);

浙江省自然科学项目(No. LY14A010024);

湖南省教育厅项目(No. 14A098)。

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