带有瞬时和非瞬时脉冲的分数阶微分方程边值问题的变分结构
Variational Structure to Boundary Problem of Fractional Differential Equations with Instantaneous and Non-Instantaneous Impulses
摘要: 本文首次将瞬时脉冲,非瞬时脉冲和Sturm-Liouville边界条件同时放在分数阶微分方程问题中研究,使用变分法建立了问题的变分结构。此外,由于在同一数学模型中同时考虑瞬时脉冲、非瞬时脉冲和Sturm-Liouville边界条件,我们克服了问题中弱解是经典解的困难。
Abstract: The instantaneous impulsive, non-instantaneous impulsive and Sturm-Liouville boundary condition are studied in the fractional differential equation problem for the first time. The variational structure of problem is established by using the variational method. In addition, considering in-stantaneous impulses condition, non-instantaneous impulses condition and Sturm-Liouville bound-ary condition in the same mathematical model, we overcome the difficulty that the weak solution is the classical solution of problem.
文章引用:蔡颖杰, 田玉. 带有瞬时和非瞬时脉冲的分数阶微分方程边值问题的变分结构[J]. 理论数学, 2021, 11(11): 1841-1849. https://doi.org/10.12677/PM.2021.1111207

1. 引言

本文运用变分法研究以下带有瞬时和非瞬时脉冲的分数阶微分方程边值问题的新的变分结构

{ d d t ( 1 2 D 0 t β u ( t ) + 1 2 D t t 1 β u ( t ) ) = f 1 ( t , u ( t ) ) , t ( 0 , t 1 ] , d d t ( 1 2 D s 1 t β u ( t ) + 1 2 D t T β u ( t ) ) = f 2 ( t , u ( t ) ) , t ( s 1 , T ] , 1 2 ( D t s 1 β u ) | t = t 1 + 1 2 ( D 0 t β u ) | t = t 1 = I ( u ( t 1 ) ) , D t 1 t β u ( t ) + D t s 1 β u ( t ) = ( D t s 1 β u ) | t = t 1 + , t ( t 1 , s 1 ] ( D t 1 t β u ) | t = s 1 = ( D t T β u ) | t = s 1 + , a u ( 0 ) b 2 ( D t t 1 β u ) | t = 0 = 0 , c u ( T ) + d 2 ( D s 1 t β u ) | t = T = 0 , (1.1)

其中 D 0 t β , D t 1 t β D s 1 t β 是左Riemann-Liouville型分数阶积分, D t t 1 β , D t s 1 β D t T β 是右Riemann-Liouville型分数阶积分, 0 β < 1 a , c > 0 b , d 0 I C ( R , R ) f 1 C ( ( 0 , t 1 ] × R , R ) f 2 C ( ( s 1 , T ] × R , R ) 。如果 β = 0 b = d = 0 ,问题(1.1)退化为以下标准的带有瞬时脉冲、非瞬时脉冲和Dirichlet边界条件的二阶微分方程边值问题

{ u ( t ) = f 1 ( t , u ( t ) ) , t ( 0 , t 1 ] , u ( t ) = f 2 ( t , u ( t ) ) , t ( s 1 , T ] , u ( t 1 + ) u ( t 1 ) = I ( u ( t 1 ) ) , u ( t ) = u ( t 1 + ) , t ( t 1 , s 1 ] , u ( s 1 ) = u ( s 1 + ) , u ( 0 ) = u ( T ) = 0. (1.2)

在科学研究中,许多物理、化学和生物现象都可以用微分方程来描述,微分方程的研究备受关注,特别是带有脉冲效应的微分方程。脉冲微分系统最突出的特点是它可以充分考虑突变对状态的影响,可以更深刻地反映事物的变化规律。在现实生活中,受外界不确定性影响的许多现象会突然发生变化,根据变化的持续时间相对于整个过程的持续时间是短还是长,脉冲可分为瞬时脉冲和非瞬时脉冲 [1] - [10]。V’Milman和A’Myshkis [6] 在1960年首次提出了瞬时脉冲,然而,在许多实际应用中,瞬时脉冲无法描述所有现象。在2013年,Hernandez和O’Regan [2] 第一次介绍了非瞬时脉冲,从那以后,非瞬时脉冲引起了研究者的关注。

另外,近年来分数阶微分方程在粘弹性、神经元和电化学等领域得到了广泛的应用,建议读者阅读参考文献 [11] - [16]。

据我们所知,具有Dirichlet边界条件的整数阶脉冲微分方程边值问题的变分结构是近些年才得到的 [5] [8]。然而,研究分数阶微分方程边值问题变分结构的文献很少,为了弥补这一空白,本文研究了同时带有瞬时脉冲,非瞬时脉冲和Sturm-Liouville边界条件的分数阶微分方程边值问题,利用变分法首次建立了问题(1.1)的变分结构,并且克服了问题(1.1)中弱解是经典解的困难。

本文共分为四节:在第二节中,我们给出了分数阶微积分的一些基本定义和性质;在第三节中,我们构造了问题(1.1)在 b , d > 0 情况下的新变分结构。此外,我们还证明了问题(1.1)中的弱解是经典解的结论;在第四节中,我们讨论了 b d = 0 的三种情况,并得到了其变分结构。

2. 预备知识

在本节中,我们将回顾分数微积分的一些基本定义和性质。有关分数阶微积分的更多内容请读者参阅文献 [11] - [16]。

定义2.1 (左、右Riemann-Liouville型分数阶积分 [12] [13] )。设f为 [ a , b ] 上定义的函数和 0 < γ < 1 D a t γ 表示左Riemann-Liouville型分数阶积分,定义如下:

D a t γ f ( t ) = 1 Γ ( γ ) a t ( t s ) γ 1 f ( s ) d s , t [ a , b ] .

右Riemann-Liouville型分数阶积分用 D t b γ 表示,定义如下:

D t b γ f ( t ) = 1 Γ ( γ ) t b ( s t ) γ 1 f ( s ) d s , t [ a , b ] .

其中 Γ > 0 是经典伽马函数。

定义2.2 (左、右Riemann-Liouville型分数阶微分 [12] [13] )。设f为 [ a , b ] 上定义的函数和 0 < γ < 1 D a t γ 表示左Riemann-Liouville型分数阶微分,定义如下:

D a t γ f ( t ) = d d t D a t γ 1 f ( t ) = 1 Γ ( 1 γ ) d d t ( a t ( t s ) γ f ( s ) d s ) , t [ a , b ] .

同样地, D t b γ 表示右Riemann-Liouville分数阶微分,定义如下:

D t b γ f ( t ) = d d t D t b γ 1 f ( t ) = 1 Γ ( 1 γ ) d d t ( t b ( s t ) γ f ( s ) d s ) , t [ a , b ] .

定义2.3 (左、右 Caputo型分数微分 [12] )。设 0 < γ < 1 f A C ( [ a , b ] ) ,其中 A C ( [ a , b ] ) 表示区间 [ a , b ] 上绝对连续函数的空间。左Caputo型分数阶微分的定义如下:

D a c t γ f ( t ) = D a t γ 1 f ( t ) = 1 Γ ( 1 γ ) ( a t ( t s ) γ f ( s ) d s ) , t [ a , b ] .

右Caputo型分数阶微分的定义如下:

D t c b γ f ( t ) = D t b γ 1 f ( t ) = 1 Γ ( 1 γ ) ( t b ( s t ) γ f ( s ) d s ) , t [ a , b ] .

性质2.1 ( [12] )左、右Riemann-Liouville型分数阶积分算子具有半群的性质,即, f C ( [ a , b ] , R ) 以及 [ a , b ] 上几乎处处满足 f L 1 ( [ a , b ] , R ) 的函数,对于所有 t [ a , b ] γ 1 , γ 2 > 0 ,有

D a t γ 1 ( D a t γ 2 f ( t ) ) = D a t γ 1 γ 2 f ( t ) D t b γ 1 ( D t b γ 2 f ( t ) ) = D t b γ 1 γ 2 f ( t ) .

性质2.2 ( [12] [13] )我们有以下分数阶积分的性质

a b [ D a t γ f ( t ) ] g ( t ) d t = a b [ D t b γ g ( t ) ] f ( t ) d t , γ > 0 ,

其中 f L P ( [ a , b ] , R ) g L q ( [ a , b ] , R ) 以及 p 1 q 1 1 / p + 1 / q 1 + γ 或者 p 1 q 1 1 / p + 1 / q = 1 + γ

3. 对于 b , d > 0 的变分结构

在本节中,我们构造了问题(1.1)的能量泛函,并证明了问题(1.1)中弱解是经典解的结论。

定义3.1令 α ( 1 2 , 1 ] p [ 1 , + ) ,分数阶微分空间

X = E α , 2 = { u : [ 0 , T ] R : u D 0 c t α u L P ( [ 0 , T ] , R ) }

C ( [ 0 , T ] , R ) 的闭包定义。

α = 1 β 2 b , d > 0 ,问题(1.1)新的能量泛函 φ : X R 定义如下:

φ ( u ) = 1 2 0 t 1 ( D 0 c t α u , D t c t 1 α u ) d t 1 2 t 1 s 1 ( D t 1 c t α u , D t c s 1 α u ) d t 1 2 s 1 T ( D s 1 c t α u , D t c T α u ) d t + a 2 b ( u ( 0 ) ) 2 + c 2 d ( u ( T ) ) 2 0 t 1 F 1 ( t , u ( t ) ) d t s 1 T F 2 ( t , u ( t ) ) d t + 0 u ( t 1 ) I ( s ) d s , (3.1)

其中 F i ( t , u ) = 0 u f i ( t , s ) d s ( i = 1 , 2 ) 。由于 f 1 , f 2 和I是连续的函数,我们可以得到 φ C 1 ( X , R ) 以及

φ ( u ) , v = 1 2 0 t 1 ( D 0 t β u + D t t 1 β u , v ) d t + 1 2 t 1 s 1 ( D t 1 t β u + D t s 1 β u , v ) d t + 1 2 s 1 T ( D s 1 t β u + D t T β u , v ) d t + a b u ( 0 ) v ( 0 ) + c d u ( T ) v ( T ) 0 t 1 f 1 ( t , u ( t ) ) v ( t ) d t s 1 T f 2 ( t , u ( t ) ) v ( t ) d t + I ( u ( t 1 ) ) v ( t 1 ) . (3.2)

事实上,对于所有 u , v X ,有

φ ( u ) , v = 1 2 0 t 1 ( D 0 c t α u , D t c t 1 α v ) + ( D t c t 1 α u , D 0 c t α v ) d t 1 2 t 1 s 1 ( D t 1 c t α u , D t c s 1 α v ) + ( D t c s 1 α u , D t 1 c t α v ) d t 1 2 s 1 T ( D s 1 c t α u , D t c T α v ) + ( D t c T α u , D s 1 c t α v ) d t + a b u ( 0 ) v ( 0 ) + c d u ( T ) v ( T ) 0 t 1 f 1 ( t , u ( t ) ) v ( t ) d t s 1 T f 2 ( t , u ( t ) ) v ( t ) d t + I ( u ( t 1 ) ) v ( t 1 ) . (3.3)

结合定义2.3、性质2.1和性质2.2,可以得到

1 2 0 t 1 ( D 0 c t α u , D t c t 1 α v ) + ( D t c t 1 α u , D 0 c t α v , ) d t = 1 2 0 t 1 ( D 0 c t α u , D t t 1 α 1 v ) + ( D t c t 1 α u , D 0 t α 1 v ) d t = 1 2 0 t 1 ( D 0 t α 1 ( D 0 c t α u ) , v ) + ( D t t 1 α 1 ( D t c t 1 α u ) , v ) d t = 1 2 0 t 1 ( D 0 t α 1 ( D 0 t α 1 u ) , v ) + ( D t t 1 α 1 ( D t t 1 α 1 u ) , v ) d t = 1 2 0 t 1 ( D 0 t β u , v ' ) + ( D t t 1 β u , v ) d t = 1 2 0 t 1 ( D 0 t β u + D t t 1 β u , v ) d t . (3.4)

同理,还可以得到

1 2 t 1 s 1 ( D t 1 c t α u , D t c s 1 α v ) + ( D t c s 1 α u , D t 1 c t α v ) d t = 1 2 t 1 s 1 ( D t 1 t β u + D t s 1 β u , v ) d t (3.5)

以及

1 2 s 1 T ( D s 1 c t α u , D t c T α v ) + ( D t c T α u , D s 1 c t α v ) d t = 1 2 s 1 T ( D s 1 t β u + D t T β u , v ) d t . (3.6)

根据(3.3)~(3.6)式,得知(3.2)式成立。

定义3.2若对于所有的 v X ,u满足 φ ( u ) , v = 0 ,则称函数 u X 为问题(1.1)的弱解。

定义3.3若函数u使得 D 0 t β u D t t 1 β u C 1 ( 0 , t 1 ] D s 1 t β u D t T β u C 1 ( s 1 , T ] 且满足问题(1.1)中的瞬时脉冲,非瞬时脉冲和Sturm-Liouville边界条件,则称函数 u X 为问题(1.1)的经典解。

定理3.1如果 u X 是问题(1)的弱解,那么 u X 是问题(1.1)的经典解。

证明:如果u是问题(1.1)的弱解,则对于所有的 v X 都有 φ ( u ) , v = 0 ,分三步完成证明。

步骤1. 证明u满足问题(1.1)中的分数阶微分方程。

不失一般性,假设 v C 0 ( 0 , t 1 ] v C 0 ( 0 , t 1 ] ,且对于 t { 0 } ( t 1 , T ] v 0 。将 v ( t ) 代入(3.2)式中得

1 2 0 t 1 ( D 0 t β u + D t t 1 β u , v ) d t = 0 t 1 f 1 ( t , u ( t ) ) v ( t ) d t ,

即,

0 = 0 t 1 ( 1 2 ( D 0 t β u + D t t 1 β u ) v f 1 ( t , u ( t ) ) v ( t ) ) d t = 0 t 1 ( 1 2 ( D 0 t β u + D t t 1 β u ) v v ( t ) d d t 0 t f 1 ( t , u ( s ) ) d s ) d t = 0 t 1 1 2 ( D 0 t β u + D t t 1 β u ) v d t v ( t ) 0 t f 1 ( t , u ( s ) ) d s | 0 t 1 + 0 t 1 v ( t ) 0 t f 1 ( t , u ( s ) ) d s d t = 0 t 1 ( 1 2 ( D 0 t β u + D t t 1 β u ) + 0 t f 1 ( t , u ( s ) ) d s ) v ( t ) d t

利用Dubois-Reymond引理,对于所有 v ( t ) C 0 ( 0 , t 1 ] ,可得

1 2 ( D 0 t β u + D t t 1 β u ) + 0 t f 1 ( t , u ( s ) ) d s = .

因为 f 1 C ( ( 0 , t 1 ] × R , R ) ,有

d d t ( 1 2 D 0 t β u + 1 2 D t t 1 β u ) + f 1 ( t , u ( t ) ) = 0 ,

这意味着

d d t ( 1 2 D 0 t β u ( t ) + 1 2 D t t 1 β u ( t ) ) = f 1 ( t , u ( t ) ) , t ( 0 , t 1 ] , (3.7)

因此问题(1.1)中的第一个分数阶微分方程成立。同理,取 v C 0 ( s 1 , T ] v C 0 ( s 1 , T ] ,且对于 t [ 0 , s 1 ] v 0 ,那么第二个分数阶微分方程成立,即,

d d t ( 1 2 D s 1 t β u ( t ) + 1 2 D t T β u ( t ) ) = f 2 ( t , u ( t ) ) , t ( s 1 , T ] . (3.8)

根据(3.7)、(3.8)、 f 1 C ( ( 0 , t 1 ] × R , R ) 以及 f 2 C ( ( s 1 , T ] × R , R ) 可知

1 2 D 0 t β u + 1 2 D t t 1 β u C 1 ( 0 , t 1 ] 1 2 D s 1 t β u + 1 2 D t T β u C 1 ( s 1 , T ] .

第二步,证明u满足瞬时脉冲条件和非瞬时脉冲条件。

由(3.7)和(3.8),可得

0 t 1 f 1 ( t , u ( t ) ) v ( t ) d t s 1 T f 2 ( t , u ( t ) ) v ( t ) d t = 0 t 1 d d t ( 1 2 D 0 t β u ( t ) + 1 2 D t t 1 β u ( t ) ) v ( t ) d t + s 1 T d d t ( 1 2 D s 1 t β u ( t ) + 1 2 D t T β u ( t ) ) v ( t ) d t = 1 2 0 t 1 v ( t ) d [ D 0 t β u ( t ) + D t t 1 β u ( t ) ] + 1 2 s 1 T v ( t ) d [ D s 1 t β u ( t ) + D t T β u ( t ) ] = 1 2 v ( t ) [ D 0 t β u ( t ) + D t t 1 β u ( t ) ] | 0 t 1 1 2 0 t 1 ( D 0 t β u ( t ) + D t t 1 β u ( t ) , v ( t ) ) d t + 1 2 v ( t ) [ D s 1 t β u ( t ) + D t T β u ( t ) ] | s 1 + T 1 2 s 1 T ( D s 1 t β u ( t ) + D t T β u ( t ) , v ( t ) ) d t

= 1 2 v ( t 1 ) ( D 0 t β u ) | t = t 1 1 2 v ( 0 ) ( D t t 1 β u ) | t = 0 1 2 0 t 1 ( D 0 t β u ( t ) + D t t 1 β u ( t ) , v ( t ) ) d t + 1 2 v ( T ) ( D s 1 t β u ) | t = T 1 2 v ( s 1 ) ( D t T β u ) | t = s 1 + 1 2 s 1 T ( D s 1 t β u ( t ) + D t T β u ( t ) , v ( t ) ) d t , (3.9)

将(3.9)代到(3.2)中,有

0 = 1 2 t 1 s 1 ( D t 1 t β u + D t s 1 β u , v ) d t + 1 2 v ( t 1 ) ( D 0 t β u ) | t = t 1 1 2 v ( 0 ) ( D t t 1 β u ) | t = 0 + 1 2 v ( T ) ( D s 1 t β u ) | t = T 1 2 v ( s 1 ) ( D t T β u ) | t = s 1 + + a b u ( 0 ) v ( 0 ) + c d u ( T ) v ( T ) + I ( u ( t 1 ) ) v ( t 1 ) . (3.10)

不失一般性,假设 v C 0 ( t 1 , s 1 ] v C 0 ( t 1 , s 1 ] ,且对于 t [ 0 , t 1 ] ( s 1 , T ] v 0 。将 v ( t ) 带入(3.10),可得

1 2 t 1 s 1 ( D t 1 t β u + D t s 1 β u , v ) d t = 0.

由Dubois-Reymond引理,可知对于 t ( t 1 , s 1 ] g ( t ) : = D t 1 t β u ( t ) + D t s 1 β u ( t ) = ,这意味着 g : = g ( t ) = g ( t 1 + ) = g ( s 1 ) ,即,

D t 1 t β u ( t ) + D t s 1 β u ( t ) = ( D t 1 t β u ) | t = s 1 = ( D t s 1 β u ) | t = t 1 + = g , t ( t 1 , s 1 ] . (3.11)

这表明问题(1.1)中的第四个等式成立。

将(3.11)代入到(3.10),则

0 = 1 2 t 1 s 1 g v d t + 1 2 v ( t 1 ) ( D 0 t β u ) | t = t 1 1 2 v ( 0 ) ( D t t 1 β u ) | t = 0 + 1 2 v ( T ) ( D s 1 t β u ) | t = T 1 2 v ( s 1 ) ( D t T β u ) | t = s 1 + + a b u ( 0 ) v ( 0 ) + c d u ( T ) v ( T ) + I ( u ( t 1 ) ) v ( t 1 ) = [ 1 2 g + 1 2 ( D 0 t β u ) | t = t 1 + I ( u ( t 1 ) ) ] v ( t 1 ) + 1 2 [ g ( D t T β u ) | t = s 1 + ] v ( s 1 ) + [ 1 2 ( D t t 1 β u ) | t = 0 + a b u ( 0 ) ] v ( 0 ) + [ 1 2 ( D s 1 t β u ) | t = T + c d u ( T ) ] v ( T ) . (3.12)

不失一般性,假设 v ( s 1 ) = v ( 0 ) = v ( T ) = 0 ,这意味着瞬时脉冲条件

1 2 ( D t s 1 β u ) | t = t 1 + 1 2 ( D 0 t β u ) | t = t 1 = I ( u ( t 1 ) )

成立。另外,取 v ( t 1 ) = v ( 0 ) = v ( T ) = 0 ,有

( D t 1 t β u ) | t = s 1 = ( D t T β u ) | t = s 1 + .

因此非瞬时脉冲条件成立。

步骤3. 证明u满足问题(1.1)的Sturm-Liouville边界条件。

利用(3.12),假设 v ( t 1 ) = v ( s 1 ) = v ( 0 ) = 0 v ( t 1 ) = v ( s 1 ) = v ( T ) = 0 ,可得

{ a u ( 0 ) b 2 ( D t t 1 β u ) | t = 0 = 0 , c u ( T ) + d 2 ( D s 1 t β u ) | t = T = 0.

所以u满足问题(1.1)的Sturm-Lionville边界条件,证毕。

4. 对于 b d = 0 的变分结构

情况1. 如果 b = 0 , d > 0 ,则问题(1.1)中的边界条件退化为

u ( 0 ) = 0 , c u ( T ) d 2 ( D s 1 t β u ) | t = T = 0. (4.1)

定义分数阶微分空间为 X 1 = { u X : u ( 0 ) = 0 } 以及泛函 φ 1 : X 1 R

φ 1 ( u ) = 1 2 0 t 1 ( D 0 c t α u , D t c t 1 α u ) d t 1 2 t 1 s 1 ( D t 1 c t α u , D t c s 1 α u ) d t 1 2 s 1 T ( D s 1 c t α u , D t c T α u ) d t + c 2 d ( u ( T ) ) 2 0 t 1 F 1 ( t , u ( t ) ) d t s 1 T F 2 ( t , u ( t ) ) d t + 0 u ( t 1 ) I ( s ) d s .

情况2. 如果 b > 0 , d = 0 ,则问题(1.1)中的边界条件退化为

a u ( 0 ) b 2 ( D t t 1 β u ) | t = 0 = 0 , u ( T ) = 0. (4.2)

定义分数阶微分空间为 X 2 = { u X : u ( T ) = 0 } 以及泛函 φ 2 : X 2 R

φ 2 ( u ) = 1 2 0 t 1 ( D 0 c t α u , D t c t 1 α u ) d t 1 2 t 1 s 1 ( D t 1 c t α u , D t c s 1 α u ) d t 1 2 s 1 T ( D s 1 c t α u , D t c T α u ) d t + a 2 b ( u ( 0 ) ) 2 0 t 1 F 1 ( t , u ( t ) ) d t s 1 T F 2 ( t , u ( t ) ) d t + 0 u ( t 1 ) I ( s ) d s .

情况3. 如果 b = d = 0 ,则问题(1.1)中的边界条件退化为

u ( 0 ) = u ( T ) = 0. (4.3)

定义分数阶微分空间为 X 3 = { u X : u ( 0 ) = u ( T ) = 0 } 以及泛函 φ 3 : X 3 R

φ 3 ( u ) = 1 2 0 t 1 ( D 0 c t α u , D t c t 1 α u ) d t 1 2 t 1 s 1 ( D t 1 c t α u , D t c s 1 α u ) d t 1 2 s 1 T ( D s 1 c t α u , D t c T α u ) d t 0 t 1 F 1 ( t , u ( t ) ) d t s 1 T F 2 ( t , u ( t ) ) d t + 0 u ( t 1 ) I ( s ) d s .

推论:当问题(1.1)的边界条件退化为(4.1),(4.2)或者(4.3)时,显然,带有边界条件(4.1),(4.2)或者(4.3)的问题(1.1)中弱解是经典解的结论依然成立。

基金项目

2021研究生专业课程建设项目,2021北京邮电大学“高新课程”建设项目。

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