一类分数阶三点边值问题解的存在性
Existence of Solutions for a Class of Fractional Three-Point Boundary Value Problems

作者: 李雯婧 :西北师范大学数学与统计学院,甘肃 兰州;

关键词: 分数阶微分方程解的存在性Leray-Schauder非线性抉择Green函数Fractional Differential Equation Existence of Positive Solutions Leray-Schauder Nonlinear Alternative Green Function

摘要:
本文运用Leray-Schauder二择一定理,研究了一类分数阶微分方程的三点边值问题 得到该问题解的存在性。进一步证明了限制非线性项的函数k(t),在被形如Btμ的函数控制后,该问题至少存在一个解。最后,通过实例验证了结论的有效性。

Abstract: In this paper, three-point boundary value problem for a class of fractional differential equations are studied by using Leray-Schauder nonlinear alternative. The existence of the solution is obtained. Furthermore, it is proved that the function k(t), which is limited to the nonlinear term, exists at least one solution of the problem when it is controlled by a function of the form Btμ. Finally, an example is given to verify the validity of the conclusion.

1. 引言

近年来,分数阶微分方程已经对数学模型的研究过程和其他学科领域的研究产生了深远的影响,例如电化学、热传导、地下水流动系统、多孔介质等。目前,分数阶边值问题的研究已取得了很多结果 [1] [2] [3] [4] [5]。

三点边值问题在微分方程理论中运用广泛 [6] [7]。庞杨等在文献 [8] 利用半序集上的不动点定理得到了Caputo分数阶三点边值问题

{ c D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = β u ( η )

解的存在性,其中 3 < α 4 0 < β 1 0 < η < 1 c D 0 + α 是标准Caputo分数阶导数。

Yan Sun在文献 [9] 等利用锥拉伸与压缩不动点定理得到了四阶三点边值问题

{ u ( 4 ) ( t ) + f ( t , u ( t ) ) = 0 , 0 t 1 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) α u ( η ) = λ

解的存在性,其中 η ( 0 , 1 ) α [ 0 , 1 η ) λ [ 0 , )

2015年,Batarfi和Jorge Losada在文献 [10] 运用Banach不动点定理和Krasnoselskii定理研究了分数阶三点边值问题

{ D α ( D + λ ) x ( t ) = f ( t , x ( t ) ) , t [ 0 , 1 ] , x ( 0 ) = 0 , x ( 0 ) = 0 , x ( 1 ) = β x ( η ) (1.1)

解的存在性,其中 D α α ( 1 , 2 ] 的一致分数阶导数, D 是一阶导数, f : [ 0 , 1 ] × R R 连续函数, λ , β R λ > 0 η ( 0 , 1 )

受上述文献的启发,本文运用Leray-Schauder非线性抉择研究问题(1.1)解的存在性,其中参数 λ > 1 β , η ( 0 , 1 )

2. 预备知识

定义1 [11] 设 α ( 0 , 1 ) 并且 f : [ 0 , ) R f α 阶分数导数可定义为

D α x ( t ) = lim ε 0 x ( t + ε t 1 α ) x ( t ) ε , t > 0

如果 f t ( 0 , a ) α 可微的, a > 0 lim t 0 + D α x ( t ) 存在,定义 D α x ( 0 ) = lim t 0 + D α x ( t )

定义 2 [11] 设 α ( n , n + 1 ] 并且 x t > 0 n 可微的,则 α 阶导数为

D α x ( t ) = lim ε 0 x ( [ α ] 1 ) ( t + ε t ( [ α ] α ) ) x ( [ α ] 1 ) ( t ) ε ,

其中 [ α ] 是大于或等于 α 的最小整数。如果 f t ( 0 , a ) α 可微的, a > 0 lim t 0 + D α x ( t ) 存在,定义 D α x ( 0 ) = lim t 0 + D α x ( t )

定义3 [11] 设 α ( 0 , 1 ) ,对于 t 0 α 阶积分记为 I α x ( t ) ,可定义为

D α x ( t ) I α x ( t ) = I 1 ( t α 1 x ) ( t ) = 0 t x ( s ) s 1 α d s

定义 4 [11] 设 α ( n 1 , n ) α 阶积分定义为

D α x ( t ) I α x ( t ) = I n ( t α [ α ] x ) ( t )

其中 I n 表示普通积分 I 1 进行 n 次积分。

引理1 [10] α ( 1 , 2 ] x 是定义在 I α 上的连续函数,当 t 0 时, D α I α x ( t ) = x ( t )

引理2 [12] 设 E 为Banach空间, Ω E 的有界开子集,且 0 Ω T : Ω ¯ E 是全连续算子,则必有下列结论之一成立。

1) 存在 u Ω θ > 1 ,使得 T u = θ u

2) 存在 T 上的不动点 u * Ω ¯

引理3 [2] 对于三点边值问题

{ D α ( D + λ ) x ( t ) = σ ( t ) , t [ 0 , 1 ] , x ( 0 ) = 0 , x ( 0 ) = 0 , x ( 1 ) = β x ( η ) ,

其中 α ( 1 , 2 ] σ C [ 0 , 1 ] 。当 β λ + e λ 1 λ η + e λ η 1 时,则该问题有唯一解

x ( t ) = 0 t e λ ( t s ) ( 0 s σ ( u ) u α 2 ( s u ) d u ) d s + A ( t ) [ β 0 η e λ ( η s ) ( 0 s σ ( u ) u α 2 ( s u ) d u ) d s 0 1 e λ ( 1 s ) ( 0 s σ ( u ) u α 2 ( s u ) d u ) d s ] ,

其中 A ( t ) = 1 Δ ( λ t + e λ t 1 ) Δ = λ + e λ 1 β ( λ η + e λ η 1 ) 0

E = C [ 0 , 1 ] ,对 x E E 在范数 x = max t [ 0 , 1 ] x Ω ¯ | x ( t ) | 下成为Banach空间。

定义算子 T : E E

( T x ) ( t ) = 0 t e λ ( t s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s + A ( t ) [ β 0 η e λ ( η s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s 0 1 e λ ( 1 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s ]

引理4 T E E 的全连续算子。

证明:设 Ω E 为有界开集,由 f 的连续性可知 f ( t , x ) [ 0 , 1 ] × Ω ¯ 上有界,记 R = max t [ 0 , 1 ] | f ( t , x ( t ) ) |

x Ω ¯ ,由算子 T 的定义可得

T x = max t [ 0 , 1 ] | 0 t e λ ( t s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s + A ( t ) [ β 0 η e λ ( η s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s 0 1 e λ ( 1 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s ] | 0 1 e λ ( 1 s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s + A ( 1 ) [ β 0 η e λ ( η s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s + 0 1 e λ ( 1 s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s ]

R [ ( 1 + A ( 1 ) ) 0 1 e λ ( 1 s ) ( 0 s u α 2 ( s u ) d u ) d s + A ( 1 ) β 0 η e λ ( η s ) ( 0 s u α 2 ( s u ) d u ) d s ] R α ( α 1 ) ( ( 1 + A ( 1 ) ) 0 1 e λ ( 1 s ) s α d s + A ( 1 ) β 0 η e λ ( η s ) s α d s ) R ( ( 1 + A ( 1 ) ) ( 1 e λ ) + A ( 1 ) β η α ( 1 e λ η ) ) α ( α 1 ) .

从而可知 T ( Ω ) 为一致有界的。

进一步,对任意的 x Ω 和任意的 t 1 t 2 [ 0 , 1 ] t 1 < t 2

| ( T x ) ( t 2 ) ( T x ) ( t 1 ) | = | 0 t 2 e λ ( t 2 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s + A ( t 2 ) [ β 0 η e λ ( η s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s 0 1 e λ ( 1 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s ] 0 t 1 e λ ( t 1 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s A ( t 1 ) [ β 0 η e λ ( η s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s 0 1 e λ ( 1 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s ] |

| 0 t 2 e λ ( t 2 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s 0 t 1 e λ ( t 2 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s | + | 0 t 1 e λ ( t 2 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s 0 t 1 e λ ( t 1 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s | + | A ( t 2 ) A ( t 1 ) | | [ β 0 η e λ ( η s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s 0 1 e λ ( 1 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s ] |

t 1 t 2 e λ ( t 2 s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s + 0 t 1 | e λ ( t 2 s ) e λ ( t 2 s ) | ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s + 2 λ ( t 2 t 1 ) Δ [ β 0 η e λ ( η s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s + 0 1 e λ ( 1 s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s ] R α ( α 1 ) [ t 1 t 2 e λ ( t 2 s ) s α d s + 0 t 1 | e λ ( t 2 s ) e λ ( t 2 s ) | s α d s + 2 λ ( t 2 t 1 ) Δ ( β 0 η e λ ( η s ) s α d s + 0 1 e λ ( 1 s ) s α d s ) ] = R α ( α 1 ) [ t 2 α t 1 α λ ( 1 e λ t 2 ) + t 1 α λ ( e λ ( t 2 s ) e λ ( t 2 s ) ) + 2 λ ( t 2 t 1 ) Δ β η α ( 1 e λ η ) + 1 e λ λ ] R α ( α 1 ) [ t 2 α t 1 α λ + ( t 2 t 1 ) + 2 ( β η α ( 1 e λ η ) + 1 e λ ) Δ ( t 2 t 1 ) ] .

再根据 t t α [ 0 , 1 ] 上的一致连续性,得 T ( Ω ) 是等度连续的。利用Arzela-Ascoli定理,可知上述引理成立。

3. 主要结果及证明

本文的主要结果是:

定理1设 f ( t , 0 ) 0 λ > 1 0 < β < 1 ,且存在非负函数 k h L 1 [ 0 , 1 ] ,使下列条件成立:

(C1) | f ( t , x ) | k ( t ) | x | + h ( t ) a . e . ( t , x ) [ 0 , 1 ] × R

(C2) 2 λ ( 1 β ) 0 1 ( 1 s ) s α 2 k ( s ) d s + β λ ( 1 β ) 0 η ( η s ) s α 2 k ( s ) d s < 1

则问题(1.1)至少有一个解 u * C [ 0 , 1 ]

证明:记

M = 2 λ ( 1 β ) 0 1 ( 1 s ) s α 2 k ( s ) d s + β λ ( 1 β ) 0 η ( η s ) s α 2 k ( s ) d s

N = 2 λ ( 1 β ) 0 1 ( 1 s ) s α 2 h ( s ) d s + β λ ( 1 β ) 0 η ( η s ) s α 2 h ( s ) d s

由条件(C2)可知, M < 1 。因为 f ( t , 0 ) 0 ,则存在一个区间 [ a , b ] [ 0 , 1 ] ,使得 min a t b | f ( t , 0 ) | > 0 。当

h ( t ) | f ( t , 0 ) | ,对 a . e . t [ 0 , 1 ] ,得到 N > 0

r = N ( 1 M ) 1 Ω = { x E : x < r } 。假设 x Ω θ > 1 ,使得 T x = θ x 。于是有

θ r = θ x = T x = max t [ 0 , 1 ] | 0 t e λ ( t s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s + A ( t ) [ β 0 η e λ ( η s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s 0 1 e λ ( 1 s ) ( 0 s f ( u , x ( u ) ) u α 2 ( s u ) d u ) d s ] | 0 1 e λ ( 1 s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s + A ( 1 ) [ β 0 η e λ ( η s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s + 0 1 e λ ( 1 s ) ( 0 s | f ( u , x ( u ) ) | u α 2 ( s u ) d u ) d s ]

( 1 + A ( 1 ) ) 0 1 e λ ( 1 s ) ( 0 s k ( u ) | x ( u ) | u α 2 ( s u ) d u ) d s + A ( 1 ) β 0 η e λ ( η s ) ( 0 s k ( u ) | x ( u ) | u α 2 ( s u ) d u ) d s + ( 1 + A ( 1 ) ) 0 1 e λ ( 1 s ) ( 0 s h ( u ) u α 2 ( s u ) d u ) d s + A ( 1 ) β 0 η e λ ( η s ) ( 0 s h ( u ) u α 2 ( s u ) d u ) d s 2 1 β 0 1 ( u 1 e λ ( 1 s ) ( s u ) d s ) k ( u ) | x ( u ) | u α 2 d u + β 1 β 0 η ( u η e λ ( η s ) ( s u ) d s ) k ( u ) | x ( u ) | u α 2 d u + 2 1 β 0 1 ( u 1 e λ ( 1 s ) ( s u ) d s ) h ( u ) u α 2 d u + β 1 β 0 η ( u η e λ ( η s ) ( s u ) d s ) h ( u ) u α 2 d u

2 λ ( 1 β ) 0 1 ( 1 s ) s α 2 k ( s ) | x ( s ) | d s + β 1 β 0 η ( 1 s ) s α 2 k ( s ) | x ( s ) | d s + 2 λ ( 1 β ) 0 1 ( 1 s ) s α 2 h ( s ) d s + β λ ( 1 β ) 0 η ( η s ) s α 2 h ( s ) d s = M x + N .

进一步,有

θ M + N r = M + N N ( 1 M ) 1 = M + ( 1 M ) = 1

这与 θ > 1 产生矛盾。由引理2可知, T 有不动点 u * Ω ¯ 。另一方面, f ( t , 0 ) 0 ,则问题(1.1)有非平凡解 u * C [ 0 , 1 ]

定理2设 f ( t , 0 ) 0 0 < β < 1 ,且存在非负函数 k h L 1 [ 0 , 1 ] ,使条件(C1),(C3)满足,

(C3)存在一个常数 μ > 0 ,使得

k ( s ) B s μ , a . e . s [ 0 , 1 ] ,

其中

B = λ ( 1 β ) ( α 1 + μ ) ( α + μ ) 2 + β η α + μ

则问题(1.1)至少有一个解 u * E

证明:由定理1的条件可知,要证明定理2,只需要证明 M < 1 即可。

M = 2 λ ( 1 β ) 0 1 ( 1 s ) s α 2 k ( s ) d s + β λ ( 1 β ) 0 η ( η s ) s α 2 k ( s ) d s B ( 2 λ ( 1 β ) 0 1 ( 1 s ) s α 2 + μ d s + β λ ( 1 β ) 0 η ( η s ) s α 2 + μ d s ) B ( 2 λ ( 1 β ) 1 ( α 1 + μ ) ( α + μ ) + β λ ( 1 β ) η α + μ ( α 1 + μ ) ( α + μ ) ) B 2 + β η α + μ λ ( 1 β ) ( α 1 + μ ) ( α + μ ) = 1.

M < 1 时,有 θ 1 。则结论得证。

4. 示例

考虑下面的分数阶微分方程边值问题:

{ D 4 3 ( D + 5 ) x ( t ) = t 5 | x | cos x 3 + 2 t + 1 , t [ 0 , 1 ] , x ( 0 ) = 0 , x ( 0 ) = 0 , x ( 1 ) = 1 2 x ( 1 3 )

解的存在性,这里 α = 4 3 λ = 5 β = 1 2 η = 1 3 。显然,

f ( t , x ) = t 5 | x | cos x 3 + 2 t + 1 ,

k ( t ) = t 2 , h ( t ) = 1

容易证得 k h L 1 [ 0 , 1 ] 为非负函数,且有

| f ( t , x ) | k ( t ) | x | + h ( t ) , a . e . ( t , x ) [ 0 , 1 ] × R ,

进一步,有

M = 2 5 × ( 1 1 2 ) 0 1 ( 1 s ) s 2 3 s 2 d s + 1 2 5 × ( 1 1 2 ) 0 1 3 ( 1 3 s ) s 2 3 s 2 d s 0 .13105 < 1.

故由定理1可知,该分数阶微分方程边值问题有解。

基金项目

国家自然科学基金(11561063)。

文章引用: 李雯婧 (2021) 一类分数阶三点边值问题解的存在性。 理论数学, 11, 1048-1054. doi: 10.12677/PM.2021.116118

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