应用数学进展  >> Vol. 10 No. 4 (April 2021)

具有左右分数阶导数和时滞的非瞬时脉冲微分方程非线性边值问题
Nonlinear Boundary Value Problems for Non-Instantaneous Pulse Differential Equations with Left-Right Fractional Derivatives and Delays

DOI: 10.12677/AAM.2021.104136, PDF, HTML, XML, 下载: 7  浏览: 26 

作者: 张雨馨:上海理工大学理学院,上海

关键词: 左右分数阶导数时滞非瞬时脉冲微分方程非线性边界条件上下解方法Left-Right Fractional Derivatives Time Delay Non-Instantaneous Pulse Nonlinear Boundary Conditions Upper and Lower Solution Method

摘要: 本文研究了一类特殊的具有左右分数阶导数和时滞的非瞬时脉冲微分方程,该方程具有交叉时滞,且带有非线性边界条件。并基于上下解方法得到多个正解存在性定理。
Abstract: In this paper, we study a class of special non-instantaneous impulsive differential equations with left and right fractional derivatives and delays. The equations have cross delays and nonlinear boundary conditions. Based on the upper and lower solution method, we obtain the existence theorems of multiple positive solutions.

文章引用: 张雨馨. 具有左右分数阶导数和时滞的非瞬时脉冲微分方程非线性边值问题[J]. 应用数学进展, 2021, 10(4): 1255-1269. https://doi.org/10.12677/AAM.2021.104136

1. 引言

分数阶微分方程非常适合刻画具有记忆和遗传性质的材料及过程,其对复杂系统的描述具有建模简单、描述准确、参数物理意义清楚等优势,因此也是复杂力学、物理过程数学建模的重要工具,如利用分数阶微积分在不同粘弹性流体的本构关系,在非牛顿流体中进行应用;分数阶SEIR传染病模型可以准确研究传染病、社交网络信息传播等方面的问题;在含未知参数的情况下,利用非线性分数阶系统状态估计分别含有分数阶有色过程噪声和有色测量噪声的连续时间问题。

在分数阶微分方程边值问题的研究 [1] [2] [3] [4] [5] 中,有时需要考虑左侧和右侧不同的定义,而同时带有左侧和右侧分数阶导数的微分方程相对于只含有分数阶右导数或左导数的分数阶微分方程,它的应用范围 [6] [7] [8] [9] 更加广泛,其在机械力学、生物工程、物理学、经济学等自然科学领域建立的数学模型中经常出现,并且具有很重要的作用,如用来分析空气中充满粒状材料时的室内外的温度数据等。

文献 [10] 研究了带有左右分数阶导数的微分方程边值问题:

{ D c b _ α D c a + α T ( t ) + λ T ( t ) = 0 , T ( a ) = T 0 , T ( b ) = T 1 ,

其中, T C [ 0 , 1 ] D c b _ α 为Caputo分数阶右导数, D c a + α 为Caputo分数阶左导数。作者利用分数阶微分方程的数值解针对实际问题进行分析。

文献 [3] 研究了带有左右阶导数的耦合微分方程边值问题:

{ ( D 0 + t α u ( t ) ) = f ( t , v ( t ) ) , t ( 0 , T ) , ( D t T β v ( t ) ) = g ( t , u ( t ) ) , t ( 0 , T ) , u ( 0 ) = 0 , D 0 + t α 1 u ( T ) = r 1 D t T β 1 v ( ξ ) , v ( T ) = 0 , D t T β 1 v ( 0 ) = r 2 D 0 + t α 1 u ( ξ ) ,

其中, D 0 + t α , D t T β 分别是 α 阶R-L分数阶左导数和 β 阶R-L分数阶右导数, 0 < α , β 2 ξ [ 0 , T ] 。作者运用上下解方法获得了边值问题解的存在性定理。

基于以上启发,本文研究含有左右分数阶导数和时滞的非瞬时脉冲微分方程非线性边值问题:

{ D t c ξ α u ( t ) = f 1 ( t , u ( t ) , u ( t + τ 1 ) ) , t [ 0 , ξ ] , D ξ + c t β u ( t ) = f 2 ( t , u ( t ) , u ( t τ 2 ) ) , t ( ξ , 1 ] , Δ u ( ξ ) = I ( ξ , u ( ξ ) ) , Δ u ( ξ ) = Q ( ξ , u ( ξ ) ) , h 0 ( u ( 0 ) , u ( 1 ) ) = 0 , h 1 ( D t c ξ α 1 u ( 0 ) , D ξ + c t β 1 u ( 1 ) ) = 0 (1)

解的存在性与多解性。其中, D t c ξ α 是右侧Caputo分数阶导数, D ξ + c t β 是左侧Caputo分数阶导数, 1 < α , β 2 , ξ ( 0 , 1 ) , u ( ξ + ) = lim ε 0 + u ( ξ + ε ) , u ( ξ ) = lim ε 0 u ( ξ + ε ) , τ 1 ( 0 , 1 ξ ) , τ 2 ( 0 , ξ ) , f 1 C ( [ 0 , ξ ] × + × + , + ) , f 2 C ( ( ξ , 1 ] × + × + , + ) , I , Q C ( , + ) , h 0 , h 1 C ( 2 , ) 为给定的非线性函数。

2. 线性边值问题

定义1 [11]:若 α > 0 a < b

I a + t α D a + c t α u ( t ) = u ( t ) + c 1 + c 2 ( t a ) + c 3 ( t a ) 2 + + c n ( t a ) n 1 ,

I t b α D t c b α u ( t ) = u ( t ) + d 1 + d 2 ( b t ) + d 3 ( b t ) 2 + + d n ( b t ) n 1 ,

其中 c i , d i , i = 1 , 2 , , n , n

引理1 [12]:令E为Banach空间,且 P E 是一个正规体锥。如果存在 α 1 β 1 α 2 β 2 P 使

α 1 β 1 α 2 β 2 ,

A : [ α 1 , β 2 ] E 是全连续算子,且为强增算子,使

α 1 A α 1 , A β 1 β 1 , α 2 A α 2 , A β 2 _ β 2 .

则算子A至少有三个不动点 x 1 , x 2 , x 3 使得

α 1 _ x 1 β 1 , α 2 x 2 _ β 2 , α 2 _ x 2 _ β 2 .

J = [ 0 , 1 ] J 0 = J \ ξ E = P C [ J , ] = { u : J : u J 0 , u ( ξ + ) u ( ξ ) u ( ξ ) = u ( ξ ) } 。显然E是Banach空间且定义其范数为

u = sup t [ 0 , 1 ] | u ( t ) | .

u [ 0 , ξ ] = sup t [ 0 , ξ ] | u ( t ) | , u ( ξ , 1 ] = sup t ( ξ , 1 ] | u ( t ) | ,则 u = max { u [ 0 , ξ ] , u ( ξ , 1 ] }

引理2:令 h C ( [ 0 , ξ ] , + ) , y C ( ( ξ , 1 ] , + ) ,对任意 m i , n i , i = 1 , 2 ,且 Δ 1 0 。则边值问题

{ D t c ξ α u ( t ) = h ( t ) , t ( 0 , ξ ) , D ξ + c t β u ( t ) = y ( t ) , t ( ξ , 1 ) , Δ u ( ξ ) = I , Δ u ( ξ ) = Q , m 1 u ( 0 ) + n 1 u ( 1 ) = γ 0 , m 2 D t c ξ α 1 u ( 0 ) + n 2 D ξ + c t β 1 u ( 1 ) = γ 1 (2)

E中存在唯一解

u ( t ) = { 0 ξ G 1 ( t , s ) h ( s ) d s + ξ 1 g 2 ( t , s ) y ( s ) d s + Δ 2 + t Δ 1 ( Q n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ) , t [ 0 , ξ ] , ξ 1 G 2 ( t , s ) y ( s ) d s + 0 ξ g 1 ( t , s ) h ( s ) d s + Δ 3 + t Δ 1 ( Q m 2 Γ ( 3 β ) ξ 2 α γ 1 ) , t ( ξ , 1 ] . (3)

其中

G 1 ( t , s ) = { g 1 ( t , s ) , 0 s t ξ , g 1 ( t , s ) + 1 Γ ( α ) ( s t ) α 1 , 0 t s ξ ; (4)

G 2 ( t , s ) = { g 2 ( t , s ) + 1 Γ ( β ) ( t s ) β 1 , ξ s t 1 , g 2 ( t , s ) , ξ t s 1 ; (5)

Δ 1 = m 2 ξ 2 α Γ ( 3 α ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) ; Δ 2 = 1 m 1 + n 1 ( n 1 I + n 1 ( 1 ξ + n 2 ( 1 ξ ) 2 β Δ 1 Γ ( 3 β ) ) Q n 1 Δ 1 γ 1 γ 0 ) ;

Δ 3 = 1 m 1 + n 1 ( m 1 I + ( m 1 ξ + n 1 + n 1 n 2 ( 1 ξ ) 2 β Δ 1 Γ ( 3 β ) ) Q n 1 Δ 1 γ 1 γ 0 ) ;

g 1 ( t , s ) = 1 m 1 + n 1 ( m 1 Γ ( α ) s α 1 + n 1 m 2 Δ 1 ) + t m 2 Δ 1 ; g 2 ( t , s ) = n 1 m 1 + n 1 ( 1 Γ ( β ) ( 1 s ) β 1 + n 2 Δ 1 ) t n 2 Δ 1 .

证明:设 u E 是边值问题(2)的解,则由定义1可知存在常数 c i , i = 0 , 1 , 2 , 3 使 D t C ξ α u ( t ) = h ( t ) 的解为:

u ( t ) = I t ξ α h ( t ) + c 0 + c 1 t = 1 Γ ( α ) t ξ ( s t ) α 1 h ( s ) d s + c 0 + c 1 t ,

u ( t ) = 1 Γ ( α 1 ) t ξ ( s t ) α 2 h ( s ) d s + c 1 ,

D t c ξ α 1 u ( t ) = t ξ h ( s ) d s c 1 Γ ( 3 α ) ( ξ t ) 2 α

D ξ + c t β u ( t ) = y ( t ) 的解为:

u ( t ) = I ξ + t β y ( t ) + c 2 + c 3 t = 1 Γ ( β ) ξ t ( t s ) β 1 y ( s ) d s + c 2 + c 3 t ,

u ( t ) = 1 Γ ( β 1 ) ξ t ( t s ) β 2 y ( s ) d s + c 3 ,

D ξ + c t β 1 u ( t ) = ξ t y ( s ) d s + c 3 Γ ( 3 β ) ( t ξ ) 2 β

由边值条件 Δ u ( ξ ) = I , Δ u ( ξ ) = Q

{ c 2 c 0 = I Q ξ , c 3 c 1 = Q .

再由边值条件 m 1 u ( 0 ) + n 1 u ( 1 ) = γ 0 m 2 D t c ξ α 1 u ( 0 ) + n 2 D ξ + c t β 1 u ( 1 ) = γ 1 ,可得

{ c 0 = 1 m 1 + n 1 ( 0 ξ ( m 1 Γ ( α ) s α 1 + n 1 m 2 Δ 1 ) h ( s ) d s + n 1 ξ 1 ( 1 Γ ( β ) ( 1 s ) β 1 n 2 Δ 1 ) y ( s ) d s + n 1 I + n 1 ( 1 ξ + n 2 ( 1 ξ ) 2 β Δ 1 Γ ( 3 β ) ) Q n 1 Δ 1 γ 1 γ 0 ) , c 1 = 1 Δ 1 ( m 2 0 ξ h ( s ) d s + n 2 ξ 1 y ( s ) d s + Q n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ) , c 2 = 1 m 1 + n 1 ( 0 ξ ( m 1 s α 1 Γ ( α ) + n 1 m 2 Δ 1 ) h ( s ) d s + n 1 ξ 1 ( ( 1 s ) β 1 Γ ( β ) n 2 Δ 1 ) y ( s ) d s m 1 I + ( m 1 ξ + n 1 + n 1 n 2 ( 1 ξ ) 2 β Δ 1 Γ ( 3 β ) ) Q n 1 Δ 1 γ 1 γ 0 ) , c 3 = 1 Δ 1 ( m 2 0 ξ h ( s ) d s + n 2 ξ 1 y ( s ) d s + Q m 2 Γ ( 3 β ) ξ 2 α γ 1 ) .

因此,当 t [ 0 , ξ ] 时,

u ( t ) = 1 Γ ( α ) t ξ ( s t ) α 1 h ( s ) d s 1 m 1 + n 1 ( 0 ξ ( m 1 Γ ( α ) s α 1 + n 1 m 2 Δ 1 ) h ( s ) d s + n 1 ξ 1 ( 1 Γ ( β ) ( 1 s ) β 1 n 2 Δ 1 ) y ( s ) d s + n 1 I + n 1 ( 1 ξ + n 2 ( 1 ξ ) 2 β Δ 1 Γ ( 3 β ) ) Q n 1 Δ 1 γ 1 γ 0 ) + t Δ 1 ( m 2 0 ξ h ( s ) d s + n 2 ξ 1 y ( s ) d s + Q n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ) = 0 ξ G 1 ( t , s ) h ( s ) d s + ξ 1 g 2 ( t , s ) y ( s ) d s + Δ 2 + t Δ 1 ( Q n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ) .

t ( ξ , 1 ] 时,

u ( t ) = 1 Γ ( β ) ξ t ( t s ) β 1 y ( s ) d s 1 m 1 + n 1 ( 0 ξ ( m 1 Γ ( α ) s α 1 + n 1 m 2 Δ 1 ) h ( s ) d s + n 1 ξ 1 ( 1 Γ ( β ) ( 1 s ) β 1 + n 2 Δ 1 ) y ( s ) d s m 1 I + ( m 1 ξ + n 1 + n 1 n 2 ( 1 ξ ) 2 β Δ 1 Γ ( 3 β ) ) Q n 1 Δ 1 γ 1 γ 0 ) + t Δ 1 ( m 2 0 ξ h ( s ) d s + n 2 ξ 1 y ( s ) d s + Q m 2 Γ ( 3 β ) ξ 2 α γ 1 ) = ξ 1 G 2 ( t , s ) y ( s ) d s + 0 ξ g 1 ( t , s ) h ( s ) d s + Δ 3 + t Δ 1 ( Q m 2 Γ ( 3 β ) ξ 2 α γ 1 ) .

易证(3)是方程(2)的解,反之亦然。

证毕。

为了以后证明,我们给出如下假设:

(H1) m i , n i ( i = 1 , 2 ) , n 1 > 0 , n 2 > 0 , n 1 > m 1 > n 1 ξ , γ 1 0 , γ 0 0

m 2 > max { Γ ( 3 α ) ξ α 1 n 2 Γ ( 3 β ) ( Γ ( 3 α ) Γ ( α ) ξ ) , Γ ( 3 α ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) ξ 2 α } .

引理3:假设(H1)成立,则由式(4)、(5)定义的函数 G i ( t , s ) , i = 1 , 2 满足以下性质:

1) 0 < G 1 ( 0 , s ) G 1 ( t , s ) G 1 ( ξ , s ) ,对任意 ( t , s ) [ 0 , ξ ] × [ 0 , ξ ]

2) 0 < G 2 ( ξ , s ) G 2 ( t , s ) G 2 ( 1 , s ) ,对任意 ( t , s ) [ ξ , 1 ] × [ ξ , 1 ]

证明:1) 显然 G i ( t , s ) , i = 1 , 2 为连续函数。 由(H1)知, Δ 1 > 0 m 2 > ξ α 1 Γ ( α ) Δ 1 ,则对于 t [ 0 , ξ ] ,当 0 s t ξ 时,

G 1 ( t , s ) = g 1 ( t , s ) = 1 m 1 + n 1 ( m 1 Γ ( α ) s α 1 + n 1 m 2 Δ 1 ) + t m 2 Δ 1 , g 1 ( t , s ) t = m 2 Δ 1 > 0 ;

0 s t ξ 时,由于 G 1 ( t , s ) = g 1 ( t , s ) + 1 Γ ( α ) ( s t ) α 1

G 1 ( t , s ) t = 1 Γ ( α 1 ) ( s t ) α 2 + m 2 Δ 1 , 2 G 1 ( t , s ) t 2 = α 2 Γ ( α 1 ) ( s t ) α 3 < 0 ,

G 1 ( t , s ) t G 1 ( s , s ) t = m 2 Δ 1 > 0 。因此, G 1 ( t , s ) 是关于t的单调递增函数,且

G 1 ( 0 , s ) G 1 ( t , s ) G 1 ( ξ , s ) .

G 1 ( 0 , s ) = g 1 ( 0 , s ) + 1 Γ ( α ) s α 1 = 1 m 1 + n 1 ( m 1 Γ ( α ) s α 1 + n 1 m 2 Δ 1 ) + 1 Γ ( α ) s α 1 = 1 m 1 + n 1 ( n 1 Γ ( α ) s α 1 + n 1 m 2 Δ 1 ) > n 1 m 1 + n 1 ( 1 Γ ( α ) ξ α 1 + m 2 Δ 1 ) > 0 ,

0 < G 1 ( 0 , s ) G 1 ( t , s ) G 1 ( ξ , s ) 成立。

2) 对于 t [ ξ , 1 ] ,当 ξ t s 1 时,

G 2 ( t , s ) = g 2 ( t , s ) = n 1 m 1 + n 1 ( 1 Γ ( β ) ( 1 s ) β 1 + n 2 Δ 1 ) + t n 2 Δ 1 , g 2 ( t , s ) t = n 2 Δ 1 > 0 ;

ξ s t 1 时,

G 2 ( t , s ) = g 2 ( t , s ) + 1 Γ ( β ) ( t s ) β 1 , G 2 ( t , s ) t = n 2 Δ 1 + 1 Γ ( β 1 ) ( t s ) β 2 > 0 ,

G 2 ( t , s ) 是关于t的单调递增函数,那么

G 2 ( ξ , s ) G 2 ( t , s ) G 2 ( 1 , s ) .

G 2 ( ξ , s ) = n 1 m 1 + n 1 ( 1 Γ ( β ) ( 1 s ) β 1 + n 2 Δ 1 ) + n 2 Δ 1 = 1 m 1 + n 1 ( n 1 Γ ( β ) ( 1 s ) β 1 + n 2 ( 1 ( m 1 + n 1 ) ξ ) Δ 1 ) > 0 ,

因此, 0 < G 2 ( ξ , s ) G 2 ( t , s ) G 2 ( 1 , s ) 成立。

证毕。

引理4:若(H1)成立, I , Q + Δ 1 0 。若u满足

{ D t c ξ α u ( t ) 0 , t ( 0 , ξ ) , D ξ + c t β u ( t ) 0 , t ( ξ , 1 ) , Δ u ( ξ ) = I , Δ u ( ξ ) = Q , m 1 u ( 0 ) + n 1 u ( 1 ) 0 , m 2 D t c ξ α 1 u ( 0 ) + n 2 D ξ + c t β 1 u ( 1 ) 0 , (4)

u ( t ) 0 t [ 0 , 1 ]

证明:对任意 h C ( [ 0 , ξ ] , + ) y C ( ( ξ , 1 ] , + ) ,由于 γ 1 0 γ 0 0 为常数。 考虑以下边值问题:

{ D t c ξ α u ( t ) = h ( t ) , t [ 0 , ξ ] , D ξ + c t β u ( t ) = y ( t ) , t ( ξ , 1 ] , Δ u ( ξ ) = I , Δ u ( ξ ) = Q , m 1 u ( 0 ) + n 1 u ( 1 ) = γ 0 , m 2 D t c ξ α 1 u ( 0 ) + n 2 D ξ + c t β 1 u ( 1 ) = γ 1 .

由引理2可得

u ( t ) = { 0 ξ G 1 ( t , s ) h ( s ) d s + ξ 1 g 2 ( t , s ) y ( s ) d s + Δ 2 + t Δ 1 ( Q n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ) , t [ 0 , ξ ] , ξ 1 G 2 ( t , s ) y ( s ) d s + 0 ξ g 1 ( t , s ) h ( s ) d s + Δ 2 + t Δ 1 ( Q m 2 Γ ( 3 β ) ξ 2 α γ 1 ) , t ( ξ , 1 ] .

由(H1)可得, Δ 1 , Δ 2 , Δ 3 > 0 Q n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 > 0 Q m 2 ξ 2 α Γ ( 3 β ) γ 1 > 0 t [ 0 , 1 ] 。再由引理3可得 u ( t ) 0 t [ 0 , 1 ] 显然成立。

证毕。

3. 分数阶微分方程的上下解方法

为方便叙述,我们假设下文满足以下假设:

(H2) 对任意 u 1 u 2 , v 1 v 2 f ( t , u 1 , v 1 ) f ( t , u 2 , v 2 ) , t [ 0 , ξ ]

g ( t , u 1 , v 1 ) g ( t , u 2 , v 2 ) , t ( ξ , 1 ] , I ( u 1 ) I ( u 2 ) , Q ( u 1 ) Q ( u 2 ) , t [ 0 , 1 ] .

对任意 u 1 < u 2 , v 1 < v 2 f ( t , u 1 , v 1 ) < f ( t , u 2 , v 2 ) , t [ 0 , ξ ]

g ( t , u 1 , v 1 ) < g ( t , u 2 , v 2 ) , t ( ξ , 1 ] , I ( u 1 ) < I ( u 2 ) , Q ( u 1 ) < Q ( u 2 ) , t [ 0 , 1 ] .

(H3) 若(H1)成立,且 x 1 , x 2 , y 1 , y 2 ,当 x 1 x 2 , y 1 y 2 时,

h 0 ( x 2 , y 2 ) h 0 ( x 1 , y 1 ) m 1 ( x 2 x 1 ) n 1 ( y 2 y 1 ) ,

h 1 ( x 2 , y 2 ) h 0 ( x 1 , y 1 ) m 2 ( x 2 x 1 ) n 2 ( y 2 y 1 ) .

P = { u E : u ( t ) 0 , t [ 0 , 1 ] } ,显然P为E中的正规体锥。且若 u ( t ) v ( t ) , t [ 0 , 1 ] ,则 u _ v P

对任意 u P ,考虑如下边值问题:

{ D t c ξ α u ( t ) = f 1 ( t , x ( t ) , x ( t + τ 1 ) ) , t ( 0, ξ ) , D ξ + c t β u ( t ) = f 2 ( t , x ( t ) , x ( t τ 2 ) ) , t ( ξ ,1 ) , Δ u ( ξ ) = I ( ξ , x ( ξ ) ) , Δ u ( ξ ) = Q ( ξ , x ( ξ ) ) , m 1 u ( 0 ) + n 1 u ( 1 ) = h 0 ( x ( 0 ) , x ( 1 ) ) + m 1 x ( 0 ) + n 1 x ( 1 ) : = γ 0 ( x ) , m 2 D t c ξ α 1 u ( 0 ) + n 2 D ξ + c t β 1 u ( 1 ) = h 1 ( D t c ξ α 1 x ( 0 ) , D ξ + c t β 1 x ( 1 ) ) + m 2 D t c ξ α 1 x ( 0 ) + n 2 D ξ + c t β 1 x ( 1 ) : = γ 1 ( x ) . (5)

由引理2知,边值问题(5)有唯一解

u ( t ) = { 0 ξ G 1 ( t , s ) f 1 ( s , x ( s ) , x ( s + τ 1 ) ) d s + ξ 1 g 2 ( t , s ) f 2 ( s , x ( s ) , x ( s τ 2 ) ) d s + Δ 2 x + t Δ 1 ( Q ( ξ , x ( ξ ) ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ( x ) ) , t [ 0 , ξ ] , ξ 1 G 2 ( t , s ) f 2 ( s , x ( s ) , x ( s τ 2 ) ) d s + 0 ξ g 1 ( t , s ) f 1 ( s , x ( s ) , x ( s + τ 1 ) ) d s + Δ 3 x + t Δ 1 ( Q ( ξ , x ( ξ ) ) m 2 Γ ( 3 β ) ξ 2 α γ 1 ( x ) ) , t ( ξ , 1 ] .

其中

Δ 2 x = 1 m 1 + n 1 ( n 1 I ( ξ , x ( ξ ) ) + n 1 ( 1 ξ + n 2 ( 1 ξ ) 2 β Δ 1 Γ ( 3 β ) ) Q ( ξ , x ( ξ ) ) n 1 Δ 1 γ 1 ( x ) γ 0 ( x ) ) ;

Δ 3 x = 1 m 1 + n 1 ( m 1 I ( ξ , x ( ξ ) ) + ( m 1 ξ + n 1 + n 1 n 2 Δ 1 Γ ( 3 β ) ) Q ( ξ , x ( ξ ) ) n 1 Δ 1 γ 1 ( x ) γ 0 ( x ) ) .

定义算子

T x ( t ) = { 0 ξ G 1 ( t , s ) f 1 ( s , x ( s ) , x ( s + τ 1 ) ) d s + ξ 1 g 2 ( t , s ) f 2 ( s , x ( s ) , x ( s τ 2 ) ) d s + Δ 2 x + t Δ 1 ( Q ( ξ , x ( ξ ) ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ( x ) ) , t [ 0 , ξ ] , ξ 1 G 2 ( t , s ) f 2 ( s , x ( s ) , x ( s τ 2 ) ) d s + 0 ξ g 1 ( t , s ) f 1 ( s , x ( s ) , x ( s + τ 1 ) ) d s + Δ 3 x + t Δ 1 ( Q ( ξ , x ( ξ ) ) m 2 ξ 2 α Γ ( 3 β ) γ 1 ( x ) ) , t ( ξ , 1 ] .

引理5:若(H1)成立,则T为全连续算子。

证明:由引理3,引理4知,对任意 x P ,当 t [ 0 , 1 ] 时, T x 0 显然成立。

因此, T : P P 是有意义的。

接下来,我们分两步证明:

第一步:T是连续算子。

设对任意 x n P , n = 1 , 2 , 存在 x P 使得当 n 时, x n x 0 。则存在 M ¯ 0 > 0 ,使得 x n M ¯ 0 , x M ¯ 0 。又由于 f 1 , f 2 连续, I , Q C ( , ) ,且 γ 0 ( x ) , γ 1 ( x ) ,则

lim n ( f 1 ( t , x n ( t ) , x n ( t + τ 1 ) ) f 1 ( t , x ( t ) , x ( t + τ 1 ) ) ) = 0 ,

lim n ( f 2 ( t , x n ( t ) , x n ( t τ 2 ) ) f 2 ( t , x ( t ) , x ( t τ 2 ) ) ) = 0 ,

lim n | I ( x n ( t ) ) I ( x ( t ) ) | = 0 , lim n | Q ( x n ( t ) ) Q ( x ( t ) ) | = 0 ,

lim n ( γ 0 ( x n ) γ 0 ( x ) ) = 0 , lim n ( γ 1 ( x n ) γ 1 ( x ) ) = 0 ,

且存在常数 M ¯ 1 > 0 ,使得 sup ( t , u , v ) A | f 1 ( t , u , v ) | M ¯ 1 , sup ( t , u , v ) B | f 2 ( t , u , v ) | M ¯ 1 ,其中

A = [ 0 , ξ ] × [ M ¯ 0 , M ¯ 0 ] × [ M ¯ 0 , M ¯ 0 ] , B = [ ξ , 1 ] × [ M ¯ 0 , M ¯ 0 ] × [ M ¯ 0 , M ¯ 0 ] .

再由引理3可得,当 t [ 0 , ξ ] 时,

| T ( x n ) T ( x ) | = | 0 ξ G 1 ( t , s ) ( f 1 ( s , x n ( s ) , x n ( s + τ 1 ) ) f 1 ( s , x ( t ) , x ( s + τ 1 ) ) ) d s + ξ 1 g 2 ( t , s ) ( f 2 ( s , x n ( s ) , x n ( s τ 2 ) ) f 2 ( s , x ( s ) , x ( s τ 2 ) ) ) d s + ( Δ 2 x n Δ 2 x ) + t Δ 1 ( ( Q ( ξ , x n ( ξ ) ) Q ( ξ , x ( ξ ) ) ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) ( γ 1 ( x n ) γ 1 ( x ) ) ) |

| T ( x n ) T ( x ) | = | 0 ξ G 1 ( t , s ) ( f 1 ( s , x n ( s ) , x n ( s + τ 1 ) ) f 1 ( s , x ( t ) , x ( s + τ 1 ) ) ) d s + ξ 1 g 2 ( t , s ) ( f 2 ( s , x n ( s ) , x n ( s τ 2 ) ) f 2 ( s , x ( s ) , x ( s τ 2 ) ) ) d s + ( Δ 2 x n Δ 2 x ) + t Δ 1 ( ( Q ( ξ , x n ( ξ ) ) Q ( ξ , x ( ξ ) ) ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) ( γ 1 ( x n ) γ 1 ( x ) ) ) |

则由Lebesgue控制收敛定理可知, lim n T u n T u [ 0 , ξ ] = 0 。同理可得, lim n T u n T u ( ξ , 1 ] = 0

因此,对任意 t [ 0 , 1 ] ,有 lim n T u n T u = 0 ,则算子T是连续算子。

第二步:T是紧的。

Ω P 为有界集,由 f 1 f 2 ,I,Q的连续性得,存在 M ¯ 2 > 0 ,使得对任意 t [ 0 , ξ ] , u , v Ω ,有 | f 1 ( t , u , v ) | M ¯ 2 ;对任意 t ( ξ , 1 ] u , v Ω | f 2 ( t , u , v ) | M ¯ 2 , | I | M ¯ 2 , | Q | M ¯ 2 , | γ 0 ( x ) | M ¯ 2 , | γ 1 ( x ) | M ¯ 2

| Δ 2 x | = 1 m 1 + n 1 ( n 1 ( 2 ξ + n 2 ( 1 ξ ) 2 β + Γ ( 3 β ) Δ 1 Γ ( 3 β ) ) 1 ) M ¯ 2 ;

| Δ 3 x | = 1 m 1 + n 1 ( m 1 ( ξ 1 ) + n 1 ( 1 + n 2 Γ ( 3 β ) Δ 1 Γ ( 3 β ) ) 1 ) M ¯ 2

sup t [ 0 , ξ ] | T u ( t ) | | 0 ξ G 1 ( ξ , s ) f 1 ( s , x ( s ) , x ( s + τ 1 ) ) d s + ξ 1 g 2 ( 1 , s ) f 2 ( s , x ( s ) , x ( s τ 2 ) ) d s + Δ 2 x + ξ Δ 1 ( Q ( ξ , x ( ξ ) ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ( x ) ) | ( 0 ξ G 1 ( ξ , s ) d s + ξ 1 g 2 ( 1 , s ) d s 1 m 1 + n 1 ( n 1 ( 2 ξ + n 2 ( 1 ξ ) 2 β Γ ( 3 β ) Δ 1 Γ ( 3 β ) ) 1 ) + ξ Δ 1 ( n 2 ( 1 ξ ) 2 β Γ ( 3 β ) 1 ) ) M ¯ 2 ,

sup t ( ξ , 1 ] | T u ( t ) | | ξ 1 G 2 ( 1 , s ) f 2 ( s , x ( s ) , x ( s τ 2 ) ) d s + 0 ξ g 1 ( ξ , s ) f 1 ( s , x ( s ) , x ( s + τ 1 ) ) d s + Δ 3 x + ξ Δ 1 ( Q ( ξ , x ( ξ ) ) m 2 Γ ( 3 β ) ξ 2 α γ 1 ( x ) ) | ( ξ 1 G 2 ( 1 , s ) d s + 0 ξ g 1 ( ξ , s ) d s 1 m 1 + n 1 ( m 1 ( ξ 1 ) + n 1 ( 1 + n 2 Γ ( 3 β ) Δ 1 Γ ( 3 β ) ) 1 ) + 1 Δ 1 ( m 2 ξ 2 α Γ ( 3 β ) 1 ) ) M ¯ 2 .

因此,算子 T ( Ω ) 一致有界。

由于 G 1 ( t , s ) , g 2 ( t , s ) [ 0 , ξ ] × [ 0 , ξ ] 上连续,所以 G 1 ( t , s ) , g 2 ( t , s ) [ 0 , ξ ] × [ 0 , ξ ] 上一致连续。因此对任意 ε > 0 ,存在 0 < δ 1 < ε Δ 1 Γ ( 3 β ) 2 | n 2 ( 1 ξ ) 2 β Γ ( 3 β ) ( 1 n 2 ) | M ¯ 2 ,当 | t 1 t 2 | < δ 1 时,有 | G 1 ( t 1 , s ) G 1 ( t 2 , s ) | < ε 4 M ¯ 2 , | g 2 ( t 1 , s ) g 2 ( t 2 , s ) | < ε 4 M ¯ 2 。因此,对任意的 t 1 , t 2 [ 0 , ξ ] | t 1 t 2 | < δ 1 u Ω

| T u ( t 2 ) T u ( t 1 ) | = | 0 ξ ( G 1 ( t 1 , s ) G 1 ( t 2 , s ) ) f 1 ( s , u ( s ) , u ( s + τ 1 ) ) d s + ξ 1 ( g 2 ( t 1 , s ) g 2 ( t 2 , s ) ) f 2 ( s , u ( s ) , u ( s τ 2 ) ) d s + t 1 t 2 Δ 1 ( Q ( ξ , x ( ξ ) ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ( x ) ) | M ¯ 2 ( 0 ξ | G 1 ( t 1 , s ) G 1 ( t 2 , s ) | d s + ξ 1 | g 2 ( t 1 , s ) g 2 ( t 2 , s ) | d s ) + | t 1 t 2 | | n 2 ( 1 ξ ) 2 β Γ ( 3 β ) ( 1 n 2 ) M ¯ 2 Δ 1 Γ ( 3 β ) < ε .

又由于 G 2 ( t , s ) g 1 ( t , s ) [ ξ , 1 ] × [ ξ , 1 ] 上连续,所以 G 2 ( t , s ) g 1 ( t , s ) [ ξ , 1 ] × [ ξ , 1 ] 上一致连续。

因此对上述 ε > 0 ,存在 0 < δ 2 < ε Δ 1 Γ ( 3 β ) 2 | m 2 ξ 2 α ( m 2 + 1 ) Γ ( 3 β ) | ,当 | t 3 t 4 | < δ 2 时,有 | G 2 ( t 3 , s ) G 2 ( t 4 , s ) | < ε 4 M ¯ 2 , | g 1 ( t 3 , s ) g 1 ( t 4 , s ) | < ε 4 M ¯ 2

因此,对任意的 t 3 , t 4 ( ξ , 1 ] | t 3 t 4 | < δ 2 u Ω ,有

| T u ( t 3 ) T u ( t 4 ) | = | ξ 1 ( G 2 ( t 3 , s ) G 1 ( t 4 , s ) ) f 2 ( s , u ( s ) , u ( s τ 2 ) ) d s + 0 ξ ( g 1 ( t 3 , s ) g 1 ( t 4 , s ) ) f 1 ( s , u ( s ) , u ( s + τ 1 ) ) d s + t 3 t 4 Δ 1 ( Q ( ξ , x ( ξ ) ) m 2 Γ ( 3 β ) ξ 2 α γ 1 ( x ) ) | M ¯ 2 ( ξ 1 | G 2 ( t 3 , s ) G 1 ( t 4 , s ) | d s + 0 ξ | g 1 ( t 3 , s ) g 1 ( t 4 , s ) | d s ) + | t 3 t 4 | | m 2 ξ 2 α ( m 2 + 1 ) Γ ( 3 β ) | Δ 1 Γ ( 3 β ) M ¯ 2 < ε .

因此, T ( Ω ) [ 0 , ξ ] , ( ξ , 1 ] 上等度连续,易知,当 t [ 0 , 1 ] 时,对任意 ε > 0 ,存在 δ 3 > 0 ,当 | t 5 t 6 | < δ 3 | T u ( t 5 ) T u ( t 6 ) | < ε ,因此, T ( Ω ) 是等度连续的。

由Arzela-Ascoli定理知 T ( Ω ) 相对列紧。又因为算子T是连续算子,所以算子T是全连续的。

证毕。

引理6:T为强增算子。

证明:对任意 x 1 , x 2 E , x 1 x 2 ,即 x 1 ( t ) x 2 ( t ) x 1 ( t ) x 2 ( t ) ,由(H2)可得,

f 1 ( t , x 2 ( t ) , x 2 ( t + τ 1 ) ) f 1 ( t , x 1 ( t ) , x 1 ( t + τ 1 ) ) 0 , t [ 0 , ξ ] ,

f 2 ( t , x 2 ( t ) , x 2 ( t τ 2 ) ) f 2 ( t , x 1 ( t ) , x 1 ( t τ 2 ) ) 0 , t ( ξ , 1 ] ,

( I ( ξ , x 2 ( ξ ) ) I ( ξ , x 1 ( ξ ) ) ) 0 , ( Q ( ξ , x 2 ( ξ ) ) Q ( ξ , x 1 ( ξ ) ) ) 0 , t [ 0 , 1 ] .

由于 x 1 ( t ) x 2 ( t ) ,则存在区间 [ a , b ] [ 0 , ξ ] [ a , b ] ( ξ , 1 ] 使得当 t [ a , b ] 时, x 1 ( t ) < x 2 ( t )

因此,当 [ a , b ] [ 0 , ξ ] 时,

f 1 ( t , x 2 ( t ) , x 2 ( t + τ 1 ) ) f 1 ( t , x 1 ( t ) , x 1 ( t + τ 1 ) ) > 0 ,

( I ( ξ , x 2 ( ξ ) ) I ( ξ , x 1 ( ξ ) ) ) > 0 , ( Q ( ξ , x 2 ( ξ ) ) Q ( ξ , x 1 ( ξ ) ) ) > 0 ,

且由(H3)可得

γ 0 ( x 2 ) γ 0 ( x 1 ) = h 0 ( x 2 ( 0 ) , x 2 ( 1 ) ) h 0 ( x 1 ( 0 ) , x 1 ( 1 ) ) + ( m 1 x 2 ( 0 ) + n 1 x 2 ( 1 ) ) ( m 1 x 1 ( 0 ) + n 1 x 1 ( 1 ) 0 , )

γ 1 ( x 2 ) γ 1 ( x 1 ) = h 1 ( D t c ξ α 1 x 2 ( 0 ) , D ξ + c t β 1 x 2 ( 1 ) ) h 1 ( D t c ξ α 1 x 2 ( 0 ) , D ξ + c t β 1 x 2 ( 1 ) ) + m 2 D t c ξ α 1 x 2 ( 0 ) + n 2 D ξ + c t β 1 x 2 ( 1 ) ( m 2 D t c ξ α 1 x 1 ( 0 ) + n 2 D ξ + c t β 1 x 1 ( 1 ) ) 0 ,

Δ 2 x 2 Δ 2 x 1 = 1 m 1 + n 1 ( n 1 ( I ( ξ , x 2 ( ξ ) ) I ( ξ , x 1 ( ξ ) ) ) ) + n 1 ( 1 ξ + n 2 ( 1 ξ ) 2 β Δ 1 Γ ( 3 β ) ) ( Q ( ξ , x 2 ( ξ ) ) Q ( ξ , x 1 ( ξ ) ) )

n 1 Δ 1 ( γ 1 ( x 2 ) γ 1 ( x 1 ) ) ( γ 0 ( x 2 ) γ 0 ( x 1 ) ) 0 ,

T x 2 ( t ) T x 1 ( t ) = 0 ξ G 1 ( t , s ) ( f 1 ( s , x 2 ( s ) , x 2 ( s + τ 1 ) ) f 1 ( s , x 1 ( s ) , x 1 ( s + τ 1 ) ) ) d s + Δ 2 x 2 Δ 2 x 1 + ξ 1 g 2 ( t , s ) ( f 2 ( t , x 2 ( t ) , x 2 ( t τ 2 ) ) f 2 ( t , x 1 ( t ) , x 1 ( t τ 2 ) ) ) d s + t Δ 1 ( ( Q ( ξ , x 2 ( ξ ) ) Q ( ξ , x 1 ( ξ ) ) ) n 2 ( 1 ξ ) 2 β Γ ( 3 β ) γ 1 ( x 2 ) γ 1 ( x 1 ) ) > 0 ξ G 1 ( t , s ) ( f 1 ( s , x 2 ( s ) , x 2 ( s + τ 1 ) ) f 1 ( s , x 1 ( s ) , x 1 ( s + τ 1 ) ) ) d s > 0.

同理当 [ a , b ] ( ξ , 1 ] 时,

( f 2 ( t , x 2 ( t ) , x 2 ( t τ 2 ) ) f 2 ( t , x 1 ( t ) , x 1 ( t τ 2 ) ) ) > 0 ,

Δ 3 x 2 Δ 3 x 1 = 1 m 1 + n 1 ( m 1 ( I ( ξ , x 2 ( ξ ) ) I ( ξ , x 1 ( ξ ) ) ) ) + ( m 1 ξ + n 1 + n 1 n 2 Δ 1 Γ ( 3 β ) ) ( Q ( ξ , x 2 ( ξ ) ) Q ( ξ , x 1 ( ξ ) ) ) n 1 Δ 1 ( γ 1 ( x 2 ) γ 1 ( x 1 ) ( γ 0 ( x 2 ) γ 0 ( x 1 ) ) ) 0 ,

T x 2 ( t ) T x 1 ( t ) = ξ 1 G 2 ( t , s ) ( f 2 ( s , x 2 ( s ) , x 2 ( s τ 2 ) ) f 2 ( s , x 1 ( s ) , x 1 ( s τ 2 ) ) ) d s + Δ 3 x 2 Δ 3 x 1 + 0 ξ g 1 ( t , s ) ( f 1 ( t , x 2 ( t ) , x 2 ( t + τ 1 ) ) f 1 ( t , x 1 ( t ) , x 1 ( t + τ 1 ) ) ) d s + t Δ 1 ( ( Q ( ξ , x 2 ( ξ ) ) Q ( ξ , x 1 ( ξ ) ) ) m 2 Γ ( 3 β ) ξ 2 α γ 1 ( x 2 ) γ 1 ( x 1 ) ) > ξ 1 G 2 ( t , s ) ( f 2 ( s , x 2 ( s ) , x 2 ( s τ 2 ) ) f 2 ( s , x 1 ( s ) , x 1 ( s τ 2 ) ) ) d s > 0.

综上所述,对任意 t [ 0 , 1 ] ( T x 2 ) ( t ) > ( T x 1 ) ( t ) ,则T为强增算子。

定义2:令 α , β E 。称 α 为边值问题(1)的一个下解,若 α 满足

{ D t c ξ α α ( t ) f 1 ( t , α ( t ) , α ( t + τ 1 ) ) , t [ 0 , ξ ] , D ξ + c t β α ( t ) f 2 ( t , α ( t ) , α ( t τ 2 ) ) , t ( ξ , 1 ] , Δ α ( ξ ) I ( ξ , u ( ξ ) ) , Δ α ( ξ ) Q ( ξ , α ( ξ ) ) , h 0 ( α ( 0 ) , α ( 1 ) ) 0 , h 1 ( D t c ξ α 1 α ( 0 ) , D ξ + c t β 1 α ( 1 ) ) 0.

β 为边值问题(1)的一个下解,若 β 满足

{ D t c ξ α β ( t ) f 1 ( t , β ( t ) , β ( t + τ 1 ) ) , t [ 0 , ξ ] , D ξ + c t β β ( t ) f 2 ( t , β ( t ) , β ( t τ 2 ) ) , t ( ξ , 1 ] , Δ β ( ξ ) I ( ξ , β ( ξ ) ) , Δ β ( ξ ) Q ( ξ , β ( ξ ) ) , h 0 ( β ( 0 ) , β ( 1 ) ) 0 , h 1 ( D t c ξ α 1 β ( 0 ) , D ξ + c t β 1 β ( 1 ) ) 0.

4. 主要结论

定理1:假设(H1)、(H2)、(H3)成立,且边值问题(1)存在两个下解 α 1 , α 2 和两个上解 β 1 , β 2 ,且 α 2 , β 1 不是边值问题(1)的解,

α 1 β 1 α 2 β 2 .

则边值问题(1)至少存在三个不同的解 u 1 u 2 u 3 满足:

α 1 _ u 1 β 1 , α 2 u 2 _ β 2 , α 2 _ u 3 _ β 1 .

证明:令算子T在 [ α 1 , β 1 ] 上, T | [ α 1 , β 1 ] 也记作T。由引理5和引理6可得, T : [ α 1 , β 1 ] P 是一个全连续强增算子。

通过定义算子T可得,

{ D t c ξ α ( T α 1 ) ( t ) = f 1 ( t , α 1 ( t ) , α 1 ( t + τ 1 ) ) , t ( 0 , ξ ) , D ξ + c t β ( T α 1 ) ( t ) = f 2 ( t , α 1 ( t ) , α 1 ( t τ 2 ) ) , t ( ξ , 1 ) , Δ ( T α 1 ) ( ξ ) = I ( ξ , α 1 ( ξ ) ) , Δ ( T α 1 ) ( ξ ) = Q ( ξ , α 1 ( ξ ) ) , m 1 ( T α 1 ) ( 0 ) + n 1 ( T α 1 ) ( 1 ) = h 0 ( α 1 ( 0 ) , α 1 ( 1 ) ) + m 1 α 1 ( 0 ) + n 1 α 1 ( 1 ) , m 2 D t c ξ α 1 ( T α 1 ) ( 0 ) + n 2 D ξ + c t β 1 ( T α 1 ) ( 1 ) = h 1 ( D t c ξ α 1 α 1 ( 0 ) , D ξ + c t β 1 α 1 ( 1 ) ) + m 2 D t c ξ α 1 α 1 ( 0 ) + n 2 D ξ + c t β 1 α 1 ( 1 ) .

α ( t ) = ( T α 1 ) ( t ) α 1 ( t ) 。由于 α 1 是边值问题(1)的一个下解,则

D t c ξ α α ( t ) = D t c ξ α ( T α 1 ) ( t ) D t c ξ α α 1 ( t ) = f 1 ( t , α 1 ( t ) , α 1 ( t + τ 1 ) ) D t c ξ α α 1 ( t ) 0 , t ( 0 , ξ ) ,

D ξ + c t β α ( t ) = D ξ + c t β ( T α 1 ) ( t ) D ξ + c t β α 1 ( t ) = f 2 ( t , α 1 ( t ) , α 1 ( t τ 2 ) ) D ξ + c t β α 1 ( t ) 0 , t ( ξ , 1 ) ,

m 1 α ( 0 ) + n 1 α ( 1 ) = m 1 ( ( T α 1 ) ( 0 ) α 1 ( 0 ) ) + n 1 ( ( T α 1 ) ( 1 ) α 1 ( 1 ) ) = ( m 1 ( T α 1 ) ( 0 ) + n 1 ( T α 1 ) ( 1 ) ) ( m 1 α 1 ( 0 ) + n 1 α 1 ( 1 ) ) = h 0 ( α 1 ( 0 ) , α 1 ( 1 ) ) + m 1 α 1 ( 0 ) + n 1 α 1 ( 1 ) ( m 1 α 1 ( 0 ) + n 1 α 1 ( 1 ) ) = h 0 ( α 1 ( 0 ) , α 1 ( 1 ) ) 0.

m 2 D t c ξ α 1 α ( 0 ) + n 2 D ξ + c t β 1 α ( 1 ) = m 2 D t c ξ α 1 ( ( T α 1 ) ( 0 ) α 1 ( 0 ) ) + n 2 D ξ + c t β 1 ( ( T α 1 ) ( 1 ) α 1 ( 1 ) ) = m 2 D t c ξ α 1 ( T α 1 ) ( 0 ) + n 2 D ξ + c t β 1 ( T α 1 ) ( 1 ) ( m 2 D t c ξ α 1 α 1 ( 0 ) + n 2 D ξ + c t β 1 α 1 ( 1 ) ) = h 1 ( D t c ξ α 1 α 1 ( 0 ) , D ξ + c t β 1 α 1 ( 1 ) ) + m 2 D t c ξ α 1 α 1 ( 0 ) + n 2 D ξ + c t β 1 α 1 ( 1 )

( m 2 D t c ξ α 1 α 1 ( 0 ) + n 2 D ξ + c t β 1 α 1 ( 1 ) ) = h 1 ( D t c ξ α 1 α 1 ( 0 ) , D ξ + c t β 1 α 1 ( 1 ) ) 0.

Δ α ( ξ ) = Δ ( T α 1 ) ( ξ ) Δ α 1 ( ξ ) = I ( ξ , α 1 ( ξ ) ) Δ α 1 ( ξ ) 0 ,

Δ α ( ξ ) = Δ ( T α 1 ) ( ξ ) Δ α 1 ( ξ ) = Q ( ξ , α 1 ( ξ ) ) Δ α 1 ( ξ ) 0.

由引理4可知, α ( t ) 0

因此, α 1 _ T α 1 。同理可得, α 2 _ T α 2

由于 α 2 是边值问题(1)的一个下解但不是边值问题(1)的解,则 ( T α 2 ) α 2 。因此,

α 2 T α 2 .

类似可得,

T β 1 β 1 , T β 2 _ β 2 .

由引理1可知,算子T至少有三个不动点 x 1 , x 2 , x 3 [ α 1 , β 2 ] 使得

α 1 _ x 1 β 1 , α 2 x 2 _ β 2 , α 2 _ x 2 _ β 2 .

因此,边值问题(1)至少有三个不同解。

证毕。

参考文献

参考文献

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