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    分数阶微分方程积分边值问题正解的唯一性
    Uniqueness of Positive Solutions for the Fractional Differential Equation with Integral Boundary Value Problem

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作者:  

卢云雪,王 颖,吴英昭,李大伟:临沂大学数学与统计学院,山东 临沂

关键词:
分数阶微分方程正解积分边值问题不动点定理Fractional Differential Equation Positive Solutions Integral Boundary Value Problem The Fixed Point Theorem

摘要:

分数阶微分是可以处理任意阶数的微分、积分,是整数阶微积分的推广。本文主要研究分数阶微分方程积分边值问题的正解,应用Banach不动点定理,得到了方程正解的唯一性。

Fractional calculus is a kind of differential and integral which can deal with any order. It is a gen-eralization of integral calculus. In this paper, we mainly study the positive solutions for the frac-tional differential equation with integral boundary value problem. By using the Banach fixed point theorem, we obtain the uniqueness of the positive solutions of the equation.

1. 引言

分数阶微分方程在科学和工程领域的应用,特别是在流变学、电子网络、流体力学、粘弹性以及化学物理学等方面的应用,使得对分数阶微分方程的研究已经变得越来越重要,已成为人们研究的热点 [1] - [9],本文研究分数阶微分方程积分边值问题(BVP):

{ D 0 + α x ( t ) + λ f ( t , x ( t ) ) = 0 , 0 < t < 1 , x ( 0 ) = x ( 0 ) = = x ( n 2 ) = 0 , D 0 + α 1 x ( t ) = 0 1 h ( t ) x ( t ) d A ( t ) (1.1)

其中 n 1 < α n , n 2 中, D 0 + α 是Riemann-Liouville微分。 λ > 0 是参数, h : ( 0 , 1 ) [ 0 , + ) 是连续的并且 h L 1 ( 0 , 1 ) 0 1 h ( s ) x ( s ) d A ( s ) 表示具有广义测度的Riemann-Stieltjes积分, A : ( 0 , 1 ) ( , + ) 是有界变差函数, 0 1 h ( t ) t α 1 d A ( t ) < Γ ( α ) f : [ 0 , 1 ] × [ 0 , + ) [ 0 , + ) 是连续函数。

2. 预备知识

定义2.1: [10] [11] (Riemann-Liouville) α 阶积分定义为

I 0 + α x ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 x ( s ) d s ,

其中 n 1 < α n ,n为整数。

定义2.2: [10] [11] (Riemann-Liouville) α 阶导数定义为

D 0 + α x ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t ( t s ) n α 1 x ( s ) d s ,

其中 n 1 < α n ,n为整数。

引理2.1: [10] [11] 若 α > 0 x L ( 0 , 1 ) D 0 + α x L ( 0 , 1 ) ,则

I 0 + α D 0 + α x ( t ) = x ( t ) + c 1 t α 1 + c 2 t α 2 + + c n t α n ,

其中 c i ( , + ) i = 1 , 2 , , n n 1 < α n

引理2.2:假设 y C ( 0 , 1 ) L 1 ( 0 , 1 ) ,则分数阶微分方程

{ D 0 + α x ( t ) + y ( t ) = 0 , t ( 0 , 1 ) , n 1 < α n , n 2 , x ( 0 ) = x ( 0 ) = = x ( n 2 ) = 0 , D 0 + α 1 x ( 1 ) = 0 1 h ( t ) x ( t ) d A ( t ) (2.1)

有解

x ( t ) = 0 G ( t , s ) y ( s ) d s ,

其中

G ( t , s ) = G 0 ( t , s ) + G 1 ( t , s ) , (2.2)

G 0 ( t , s ) = 1 Γ ( α ) { t α 1 ( t s ) α 1 , 0 s t 1 , t α 1 , 0 t s 1 ,

G 1 ( t , s ) = t α 1 Γ ( α ) 0 1 h ( t ) t α 1 d A ( t ) 0 1 h ( t ) G 0 ( t , s ) d A ( t ) .

证明:由引理2.1,(2.1)中的方程可转化为等价于的积分方程

x ( t ) = I 0 + α y ( t ) + c 1 t α 1 + c 2 t α 2 + + c n t α n , c i ( , + ) , i = 1 , 2 , , n ,

x ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s + c 1 t α 1 + c 2 t α 2 + + c n t α n , c i ( , + ) , i = 1 , 2 , , n ,

由于 x ( 0 ) = x ( 0 ) = = x ( n 2 ) = 0 ,得 c 2 = c 3 = = c n = 0 ,因此有

x ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s + c 1 t α 1 ,

D 0 + α 1 x ( t ) = c 1 Γ ( α ) 0 t y ( s ) d s .

又有 D 0 + α 1 x ( 1 ) = 0 1 h ( t ) x ( t ) d A ( t )

D 0 + α 1 x ( 1 ) = c 1 Γ ( α ) 0 1 y ( s ) d s .

可得

c 1 = 1 Γ ( α ) ( 0 1 h ( t ) x ( t ) d A ( t ) + 0 1 y ( s ) d s ) ,

所以

x ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s + t α 1 1 Γ ( α ) ( 0 1 h ( t ) x ( t ) d A ( t ) + 0 1 y ( s ) d s ) = 0 1 G 0 ( t , s ) y ( s ) d s + t α 1 Γ ( α ) 0 1 h ( t ) x ( t ) d A ( t ) . (2.3)

对(2.3)式两端乘以 h ( t ) 并且求关于 A ( t ) 的积分,有

0 1 h ( t ) x ( t ) d A ( t ) = Γ ( α ) Γ ( α ) 0 1 h ( t ) t α 1 d A ( t ) 0 1 h ( t ) 0 1 G 0 ( t , s ) y ( s ) d s d A ( t ) ,

所以

x ( t ) = 0 1 G 0 ( t , s ) y ( s ) d s + t α 1 Γ ( α ) 0 1 h ( t ) x ( t ) d A ( t ) = 0 1 G 0 ( t , s ) y ( s ) d s + t α 1 Γ ( α ) Γ ( α ) Γ ( α ) 0 1 h ( t ) t α 1 d A ( t ) 0 1 h ( t ) 0 1 G 0 ( t , s ) y ( s ) d s d A ( t ) = 0 1 G 0 ( t , s ) y ( s ) d s + t α 1 Γ ( α ) 0 1 h ( t ) t α 1 d A ( t ) 0 1 h ( t ) 0 1 G 0 ( t , s ) y ( s ) d s d A ( t ) = 0 1 G 0 ( t , s ) y ( s ) d s + 0 1 G 1 ( t , s ) y ( s ) d s = 0 1 G ( t , s ) y ( s ) d s .

引理2.3:由(2.2)定义的 G ( t , s ) 有下列性质:

1) G ( t , s ) 0 ( t , s ) [ 0 , 1 ] × [ 0 , 1 ]

2) G ( t , s ) [ 0 , 1 ] × [ 0 , 1 ] 上连续。

3) G ( t , s ) ω ω = max { 1 Γ ( α ) , 0 1 h ( t ) d A ( t ) Γ ( α ) ( Γ ( α ) 0 1 h ( t ) t α 1 d A ( t ) ) }

证明:根据 G ( t , s ) 的定义,只需证明(3)成立。由于

G 0 ( t , s ) t α 1 Γ ( α ) 1 Γ ( α ) ,

G 1 ( t , s ) t α 1 Γ ( α ) ( Γ ( α ) 0 1 h ( t ) t α 1 d A ( t ) ) 0 1 h ( t ) d A ( t ) 1 Γ ( α ) ( Γ ( α ) 0 1 h ( t ) t α 1 d A ( t ) ) 0 1 h ( t ) d A ( t ) ,

所以 G ( t , s ) = G 0 ( t , s ) + G 1 ( t , s ) ω

3. 主要结果

X = C [ 0 , 1 ] ,定义范数 x = max 0 t 1 | x ( t ) | 。则X是Banach空间,记

K = { x X : x ( t ) 0 , t [ 0 , 1 ] } ,

因此K是X的一个锥。本文,我们假设下面的条件(H1)成立。

(H1) f : [ 0 , 1 ] × [ 0 , + ) [ 0 , + ) 是连续函数。

由(H1),定义积分算子 T : K X

( T x ) ( t ) = λ 0 1 G ( t , s ) f ( s , x ( s ) ) d s , t [ 0 , 1 ] (3.1)

显然BVP(1.1)有解x当且仅当 x K 是由(3.1)定义的算子T的不动点。

引理3.1:假设条件(H1)成立,则 T : K K 是全连续算子。

定理3.1:假设条件(H1)成立,并且存在函数 m ( t ) 0 满足下列条件。

(H2) | f ( t , x 2 ) f ( t , x 1 ) | m ( t ) | x 2 x 1 | , t [ 0 , 1 ] , x 1 , x 2 [ 0 , + )

0 1 m ( s ) d s < 1 λ ω ,

则BVP(1.1)有唯一正解。

证明:对任意的 x K ,由于 G ( t , s ) 0 f ( t , x ) 0 ,可得 T x ( t ) 0 ,因而 T ( K ) K

T x 2 T x 1 = max t [ 0 , 1 ] | T x 2 ( t ) T x 1 ( t ) | = max t [ 0 , 1 ] λ 0 1 G ( t , s ) | f ( s , x 2 ( s ) ) f ( s , x 1 ( s ) ) | d s λ 0 1 ω m ( s ) | x 2 ( s ) x 1 ( s ) | d s λ ω 0 1 m ( s ) d s x 2 x 1 < x 2 x 1

由引理3.1, T : K K 是全连续算子,根据Banach不动点理论,算子T在K中有唯一不动点,即为BVP(1.1)的唯一正解。

基金项目

本文受到临沂大学大学生创新创业训练计划项目(X201910452076)部分资助。

文章引用:
卢云雪, 王颖, 吴英昭, 李大伟. 分数阶微分方程积分边值问题正解的唯一性[J]. 理论数学, 2020, 10(1): 6-10. https://doi.org/10.12677/PM.2020.101002

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