分数阶微积分在数学和工程方面已经成为人们特别熟知的概念,其是整数阶微积分的推广。分数阶微积分有好多种形式,譬如,Riemann-Liouville、Caputo分数阶微积分,带有一个函数的分数阶微积分是Riemann-Liouville分数阶微积分的推广形式。在本文中,基于带有一个函数的分数阶微积分的基本性质和Picard迭代方法,我们将讨论一类以带有一个函数的分数阶导数表示的微分方程初值问题解的存在唯一性。同时通过本文的研究,我们不仅将Picard迭代法应用于一类以带有一个函数的分数阶导数表示的微分方程初值问题解的存在唯一性的论证中,还提供了求解此类分数阶微分方程初值问题近似解的一种思路。 Fractional calculus has become a particularly well-known concept in mathematics and engineering, and is a generalization of integer-order calculus. There are many forms of fractional calculus, for example, Riemann-Liouville, Caputo fractional calculus, and fractional calculus with a function is a generalized form of Riemann-Liouville fractional calculus. In this article, based on the fundamental properties of fractional calculus with a function and the Picard iterative method, we will discuss the uniqueness of a class of solutions to the initial value problem of differential equations represented by fractional derivatives with a function. At the same time, through the research of this paper, we not only apply the Picard iterative method to the demonstration of the uniqueness of the solution of the initial value problem of differential equations represented by a fractional derivative with a function, but also provide an idea for solving the approximate solution of the initial value problem of such fractional order differential equations.
分数阶微积分在数学和工程方面已经成为人们特别熟知的概念,其是整数阶微积分的推广。分数阶微积分有好多种形式,譬如,Riemann-Liouville、Caputo分数阶微积分,带有一个函数的分数阶微积分是Riemann-Liouville分数阶微积分的推广形式。在本文中,基于带有一个函数的分数阶微积分的基本性质和Picard迭代方法,我们将讨论一类以带有一个函数的分数阶导数表示的微分方程初值问题解的存在唯一性。同时通过本文的研究,我们不仅将Picard迭代法应用于一类以带有一个函数的分数阶导数表示的微分方程初值问题解的存在唯一性的论证中,还提供了求解此类分数阶微分方程初值问题近似解的一种思路。
分数阶微分方程,初值问题,Picard迭代法,存在性,唯一性
Yuling Yang, Junwei Liang, Jian Li
School of Science, China University of Mining & Technology, Beijing
Received: Feb. 15th, 2023; accepted: Mar. 14th, 2023; published: Mar. 22nd, 2023
Fractional calculus has become a particularly well-known concept in mathematics and engineering, and is a generalization of integer-order calculus. There are many forms of fractional calculus, for example, Riemann-Liouville, Caputo fractional calculus, and fractional calculus with a function is a generalized form of Riemann-Liouville fractional calculus. In this article, based on the fundamental properties of fractional calculus with a function and the Picard iterative method, we will discuss the uniqueness of a class of solutions to the initial value problem of differential equations represented by fractional derivatives with a function. At the same time, through the research of this paper, we not only apply the Picard iterative method to the demonstration of the uniqueness of the solution of the initial value problem of differential equations represented by a fractional derivative with a function, but also provide an idea for solving the approximate solution of the initial value problem of such fractional order differential equations.
Keywords:Fractional Differential Equations, Initial Value Problems, Picard Iterative Method, Existence, Uniqueness
Copyright © 2023 by author(s) and Hans Publishers Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).
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自二十世纪九十年代以来,很多领域都涉及到分数阶微分方程,例如电子工程 [
在文献 [
{ D t 0 + α x ( t ) = f ( t , x ( t ) ) , t > t 0 lim t → t 0 ( t − t 0 ) 1 − α x ( t ) = x 0 .
其中 D t 0 + α x ( t ) 是Rimann-Liouville分数阶导数, α ∈ ( 0 , 1 ) , x 0 ∈ R 是一个初值,函数f在 t = t 0 处可能是奇
异的,将著名的Picard迭代技术推广到分数阶微分方程,得到了上述问题解的存在唯一性。
受到文献 [
{ D a + , h α u ( x ) = f ( x , u ( x ) ) I a + , h 1 − α u ( x ) | x = a = u 0 (1)
其中 D a + , h α u ( x ) 被称为带有一个函数的 α 阶Rimann-Liouville分数阶导数, I a + , h 1 − α u ( x ) 被称为带有一个函数的 1 − α 阶Rimann-Liouville分数阶积分, u 0 ∈ R , α ∈ ( 0 , 1 ) , h ( x ) > 0 , x ∈ ( a , b ] , h ′ ∈ C ( a , b ) , h ′ ( x ) > 0 , x ∈ ( a , b ) 。
本文结构安排如下:在第二部分,我们首先给出一些必要的定义和引理;在第三部分,我们将该微分方程的初值问题转换成与其等价的积分方程,通过Picard迭代法,提供了一致收敛到所讨论问题的解的可计算序列,得到了分数阶微分方程初值问题(1)解的存在唯一性,并且建立了一致逼近解的迭代格式。
微分方程解的存在唯一性是研究微分方程解的适定性的一个关键问题。由于缺乏求解分数阶和随机微积分控制的非线性动力系统的通用技术,在采用离散化方法获得近似解之前,研究解的存在唯一性是前提与基础。常用研究方法有Picard逐步逼近法和运用Banach不动点定理,其中Picard逐步逼近法是较为重要的近似计算方法 [
但由于Picard迭代法是最早在数学上完善处理这样的逐次逼近的函数序列的方法,它的缺点也非常明显:证明过程繁琐且满足条件的函数序列不易得出,也无法借助计算机验证。且在数值计算时,Picard迭代法的收敛性和速度都得不到保证,求解常微分方程的迭代方法仍有较大发展空间。
在这一部分里,我们给出带有一个函数的分数阶微积分的定义和基本性质,具体的请参见文献 [
定义1 [
I a + , h α f ( x ) = 1 Γ ( α ) ∫ a x [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t ) d t .
I a + , h α f ( x ) 被称为带有一个函数的 α 阶Riemann-Liouville分数阶积分。
定义2 [
D a + , h α f ( x ) = 1 Γ ( 1 − α ) ( 1 h ′ ( x ) ) d d x ∫ a x [ h ( x ) − h ( t ) ] − α h ′ ( t ) f ( t ) d t .
引理1 [
1) 若函数 u ( x ) 在 [ a , b ] 上连续,则对任意的 x ∈ [ a , b ] ,有
D a + , h α I a + , h α u ( x ) = u ( x ) .
2) 若函数 u ( x ) 在 [ a , b ] 上连续,且存在可积的 D a + , h α u ( x ) ,则对任意的 x ∈ ( a , b ] ,有
I a + , h α D a + , h α u ( x ) = u ( x ) − c [ h ( x ) − h ( a ) ] α − 1
c = I a + , h 1 − α u ( x ) | x = a .
在本部分,我们将基于带有一个函数的分数阶微积分的基本性质,利用Picard迭代方法,讨论分数阶微分方程初值问题(1)解的存在唯一性。
令:
a 1 > 0 , b > 0 , J = ( a , a + a 1 ] , B = { u | u ∈ C [ a , a + a 1 ] , | u − u 0 Γ ( α ) | ≤ b } ,
B t = { u | | [ h ( x ) − h ( a ) ] 1 − α u ∈ C [ a , a + a 1 ] , [ h ( x ) − h ( a ) ] 1 − α u − u 0 Γ ( α ) | ≤ b , x ∈ [ a , a + a 1 ] } ,
h ( x ) > 0 , x ∈ J , h ′ ∈ C [ a , a + a 1 ] , h ′ ( x ) > 0 , x ∈ J .
下面是本论文的主要结果。
定理 假设如下条件成立:
1) 对于任意的 x ∈ J , x → f ( x , [ h ( x ) − h ( a ) ] α − 1 u ) 在 B t 上是连续的,对于任意的 u ∈ B t , u → f ( x , [ h ( x ) − h ( a ) ] α − 1 u ) 在J上是可测的。
2) 存在 K > − 1 , M ≥ 0 使得对于任意的 x ∈ J , u ∈ B ,有
| f ( x , [ h ( x ) − h ( a ) ] α − 1 u ) | ≤ M [ h ( x ) − h ( a ) ] k ,
3) 存在 K > − 1 , L ≥ 0 使得对于任意的 x ∈ J , u 1 , u 2 ∈ B 有
| f ( x , [ h ( x ) − h ( a ) ] α − 1 u 1 ) − f ( x , [ h ( x ) − h ( a ) ] α − 1 u 2 | ≤ L [ h ( x ) − h ( a ) ] k | u 1 − u 2 | .
则分数阶微分方程初值问题(1)在J上存在唯一的连续解 ϕ ( x ) ,其中 l = min { a 1 , 1 M h ( b M Γ ( α ) B ( α , k + 1 ) ) 1 k + 1 } , M h = max a ≤ x ≤ a + a 1 | h ′ ( x ) | 。
证明我们将用Picard迭代法来证明该结论,为了完成该定理的证明,我们需要下面的5个结论。
结论1在连续函数空间 C ( a , a + l ] 上,微分方程初值问题(1)可转换成如下等价的积分方程:
u ( x ) = 1 Γ ( α ) ∫ a x [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , u ( t ) ) d t + u 0 Γ ( α ) [ h ( x ) − h ( a ) ] α − 1 , x ∈ ( a , a + l ] . (2)
证明若 u ∈ C ( a , a + h ] 满足微分方程初值问题(1),即
D a + , h α u ( x ) = f ( x , u ( x ) ) . (3)
对(3)两边同时进行 α 阶积分得到
I a + , h α D a + , h α u ( x ) = I a + , h α f ( x , u ( x ) ) .
则由引理1可得
u ( x ) − c [ h ( x ) − h ( a ) ] α − 1 = 1 Γ ( α ) ∫ a x [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , u ( t ) ) d t , x ∈ ( a , a + l ] .
再将 I a + , h 1 − α u ( x ) | x = a = u 0 代入得到
u ( x ) = 1 Γ ( α ) ∫ a x [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , u ( t ) ) d t + u 0 Γ ( α ) [ h ( x ) − h ( a ) ] α − 1 , x ∈ ( a , a + l ] .
反之,若 u ∈ C ( a , a + l ] 满足积分方程,即
u ( x ) = 1 Γ ( α ) ∫ a x [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , u ( t ) ) d t + u 0 Γ ( α ) [ h ( x ) − h ( a ) ] α − 1 , x ∈ ( a , a + l ] .
则对此方程两边同时求 α 阶导数得到
D a + , h α u ( x ) = D a + , h α [ 1 Γ ( α ) ∫ a x ( h ( x ) − h ( t ) ) α − 1 h ′ ( t ) f ( t , u ( t ) ) d t + u 0 Γ ( α ) ( h ( x ) − h ( a ) ) α − 1 ] , x ∈ ( a , a + l ] .
即
D a + , h α u ( x ) = D a + , h α I a + , h α f ( x , u ( x ) ) + u 0 Γ ( α ) D a + , h α ( h ( x ) − h ( a ) ) α − 1 , a < x ≤ a + l .
由引理1和文献 [
D a + , h α u ( x ) = f ( x , u ( x ) ) .
并且通过计算可得: I a + , h 1 − α u ( x ) | x = a = u 0 。命题1得证。
我们选取
ϕ 0 ( x ) = u 0 Γ ( α ) [ h ( x ) − h ( a ) ] α − 1 , x ∈ ( a , a + l ] ,
ϕ n ( x ) = u 0 Γ ( α ) [ h ( x ) − h ( a ) ] α − 1 + ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , ϕ n − 1 ( t ) ) d t , x ∈ ( a , a + l ] , n = 1 , 2 , ⋯ .
作为Picard迭代序列。
结论2对于所有的n,函数 ϕ n ( x ) 在 a < x ≤ a + l 上有定义、连续且满足不等式:
| [ h ( x ) − h ( a ) ] 1 − α ϕ n ( x ) − u 0 | ≤ b , x ∈ [ a , a + l ] .
证明由函数h的性质和条件(1)、(2)知 ϕ 0 ( x ) 与 ϕ 1 ( x ) 在 a < x ≤ a + l 上连续,且
| [ h ( x ) − h ( a ) ] 1 − α ϕ 1 ( x ) − u 0 Γ ( α ) | = | ∫ a x [ h ( x ) − h ( a ) ] 1 − α Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , ϕ 0 ( t ) ) d t | ≤ [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) | f ( t , ϕ 0 ( t ) ) | d t = [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) | f ( t , [ h ( t ) − h ( a ) ] α − 1 [ h ( t ) − h ( a ) ] 1 − α ϕ 0 ( t ) ) | d t ≤ M [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) [ h ( t ) − h ( a ) ] k d t = M [ h ( x ) − h ( a ) ] k + 1 B ( α , k + 1 ) Γ ( α ) ≤ M ( M h l ) k + 1 B ( α , k + 1 ) Γ ( α ) ≤ b .
假设 ϕ n ∈ C ( a , a + l ] 且对 ∀ x ∈ [ x 0 , x 0 + h ] 满足不等式:
| [ h ( x ) − h ( a ) ] 1 − α ϕ n ( x ) − u 0 Γ ( α ) | ≤ b .
由函数h的性质和条件(1)、(2)知 ϕ n + 1 ∈ C ( a , a + l ] 且有:
| [ h ( x ) − h ( a ) ] 1 − α ϕ n + 1 ( x ) − u 0 Γ ( α ) | ≤ | ∫ a x [ h ( x ) − h ( a ) ] 1 − α Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , ϕ n ( t ) ) d t | ≤ [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) | f ( t , ϕ n ( t ) ) | d t ≤ M [ h ( x ) − h ( a ) ] k + 1 B ( α , k + 1 ) Γ ( α ) ≤ M ( M h l ) k + 1 B ( α , k + 1 ) Γ ( α ) ≤ b ,
所以 ϕ n + 1 ( x ) 也满足上述不等式,由数学归纳法知,结论2对所有n均成立,结论2证毕。
结论3函数序列 { [ h ( x ) − h ( a ) ] 1 − α ϕ n ( x ) } 在 x 0 ≤ x ≤ x 0 + h 上是一致收敛的。
证明考虑
[ h ( x ) − h ( a ) ] 1 − α ϕ 0 ( x ) + [ h ( x ) − h ( a ) ] 1 − α [ ϕ 1 ( x ) − ϕ 0 ( x ) ] + ⋯ + [ h ( x ) − h ( a ) ] 1 − α [ ϕ n ( x ) − ϕ n − 1 ( x ) ] + ⋯ , x ∈ [ a , a + l ] .
由条件(1)可得:
[ h ( x ) − h ( a ) ] 1 − α | ϕ 1 ( x ) − ϕ 0 ( x ) | ≤ M [ h ( x ) − h ( a ) ] k + 1 B ( α , k + 1 ) Γ ( α ) , x ∈ [ a , a + l ] .
由利普希茨条件可得:
[ h ( x ) − h ( a ) ] 1 − α | ϕ 2 ( x ) − ϕ 1 ( x ) | ≤ L [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) [ h ( t ) − h ( a ) ] k + 1 − α | ϕ 1 ( t ) − ϕ 0 ( t ) | d t ≤ L M B ( α , k + 1 ) Γ ( α ) [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) [ h ( t ) − h ( a ) ] 1 + 2 k d t = L M B ( α , k + 1 ) Γ ( α ) B ( α , 2 k + 2 ) Γ ( α ) [ h ( x ) − h ( a ) ] 2 + 2 k , x ∈ [ a , a + l ] .
下面证明对任意正整数n,有
[ h ( x ) − h ( a ) ] 1 − α | ϕ n + 1 ( x ) − ϕ n ( x ) | ≤ L n M [ h ( x ) − h ( a ) ] ( n + 1 ) ( k + 1 ) ∏ i = 1 n B ( α , ( k + 1 ) ( i + 1 ) ) Γ ( α ) , x ∈ [ a , a + l ] (4)
已知该式子对于 n = 1 成立,假设其对于 n = m 也成立,由利普希茨条件,当 a ≤ x ≤ a + l 时,有:
[ h ( x ) − h ( a ) ] 1 − α | ϕ m + 2 ( x ) − ϕ m + 1 ( x ) | ≤ [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) | f ( t , ϕ m + 1 ( t ) ) − f ( t , ϕ m ( t ) ) | d t ≤ L [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) [ h ( t ) − h ( a ) ] k + 1 − α | ϕ m + 1 ( x ) − ϕ m ( x ) | d t ≤ [ h ( x ) − h ( a ) ] 1 − α ∫ a x [ h ( x ) − h ( t ) ] α − 1 Γ ( α ) h ′ ( t ) L [ h ( t ) − h ( a ) ] k L m M [ h ( x ) − h ( a ) ] ( m + 2 ) ( k + 1 ) × ∏ i = 1 m B ( α , ( k + 1 ) ( i + 1 ) ) Γ ( α ) d t = L m + 1 M [ h ( x ) − h ( a ) ] ( m + 2 ) ( k + 1 ) ∏ i = 1 m + 1 B ( α , ( k + 1 ) ( i + 1 ) ) Γ ( α ) , x ∈ [ a , a + l ] .
故其对 m + 1 也是成立的,由数学归纳法可知对于任意正整数n,式(4)都成立。
考虑
∑ n = 1 ∞ C n = ∑ n = 1 ∞ M L n + 1 ( M h l ) ( n + 2 ) ( k + 1 ) ∏ i = 1 n + 1 Γ ( ( k + 1 ) ( i + 1 ) ) Γ ( α + ( i + 1 ) ( k + 1 ) ) .
则有
C n + 1 C n = L ( M h l ) k + 1 Γ ( ( n + 2 ) ( k + 1 ) ) Γ ( α + ( n + 2 ) ( k + 1 ) ) → 0 ( n → ∞ ) .
由比式判别法可知, ∑ n = 1 ∞ C n 在 a ≤ x ≤ a + l 上一致收敛,因此序列 { [ h ( x ) − h ( a ) ] 1 − α ϕ n ( x ) } 也在 a ≤ x ≤ a + l 上一致收敛,结论3证毕。
结论4 ϕ ( x ) 是积分方程(2)的定义于 a < x ≤ a + l 上的连续解。
证明由结论2知
| f ( x , φ n ( x ) ) − f ( x , φ ( x ) ) | ≤ L [ h ( x ) − h ( a ) ] k [ h ( x ) − h ( a ) ] 1 − α | ϕ ( x ) − ϕ ( x ) | , x ∈ ( a , a + l ] .
则
[ h ( x ) − h ( a ) ] − k | f ( x , ϕ n ( x ) ) − f ( x , ϕ ( x ) ) | ≤ L [ h ( x ) − h ( a ) ] 1 − α | ϕ n ( x ) − ϕ ( x ) | , x ∈ [ a , a + l ] .
由结论3知函数序列 { L [ h ( x ) − h ( a ) ] 1 − α | ϕ n ( x ) − ϕ ( x ) | } 在 a ≤ x ≤ a + l 上一致收敛于0,因此,对函数
序列两边取极限,得到
[ h ( x ) − h ( a ) ] 1 − α ϕ ( x ) = lim n → ∞ [ h ( x ) − h ( a ) ] 1 − α ϕ n ( x ) = lim n → ∞ [ u 0 Γ ( α ) + [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , φ n − 1 ( t ) ) d t ] = u 0 Γ ( α ) + [ h ( x ) − h ( a ) ] 1 − α lim n → ∞ ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , ϕ n − 1 ( t ) ) d t
= u 0 Γ ( α ) + [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) [ h ( x ) − h ( a ) ] k × lim n → ∞ [ h ( x ) − h ( a ) ] − k f ( t , ϕ n − 1 ( t ) ) d t = u 0 Γ ( α ) + [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , φ ( t ) ) d t , x ∈ [ a , a + l ] .
即
φ ( x ) = u 0 Γ ( α ) [ h ( x ) − h ( a ) ] α − 1 + ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , φ ( t ) ) d t , x ∈ ( a , a + l ] .
这就是说, φ ( x ) 是积分方程(2)的定义于 ( a , a + l ] 上的连续解,命题4证毕。
结论5设 ψ ( x ) 是积分方程(2)的定义于 a < x ≤ a + l 上的另一个连续解,则
φ ( x ) = ψ ( x ) , x ∈ ( a , a + l ] .
证明 我们证明 ψ ( x ) 也是序列 { φ n ( x ) } 的一致收敛极限函数。
因为 ψ ( x ) 也是积分方程(2)的解,则有
ψ ( x ) = u 0 Γ ( α ) [ h ( x ) − h ( a ) ] α − 1 + ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , ψ ( t ) ) d t , x ∈ ( a , a + l ] .
由条件(2)知
| f ( x , ψ ( x ) ) | = | f ( x , [ h ( x ) − h ( a ) ] α − 1 [ h ( x ) − h ( a ) ] 1 − α ψ ( x ) ) | ≤ M [ h ( x ) − h ( a ) ] k , x ∈ ( a , a + l ] .
故
[ h ( x ) − h ( a ) ] 1 − α | φ 0 ( x ) − ψ ( x ) | = [ h ( x ) − h ( a ) ] 1 − α | ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) f ( t , ψ ( t ) ) d t | ≤ [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) | f ( t , ψ ( t ) ) | d t ≤ [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) M [ h ( t ) − h ( a ) ] k d t = M [ h ( x ) − h ( a ) ] k + 1 B ( α , k + 1 ) Γ ( α ) , x ∈ [ a , a + l ] .
且
[ h ( x ) − h ( a ) ] 1 − α | ϕ 1 ( x ) − ψ ( x ) | = [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) | f ( t , ϕ 0 ( t ) ) − f ( t , ψ ( t ) ) | d t ≤ [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) L [ h ( t ) − h ( a ) ] k [ h ( x ) − h ( a ) ] 1 − α | ϕ 0 ( t ) − ψ ( t ) | d t ≤ [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) L [ h ( t ) − h ( a ) ] k M [ h ( t ) − h ( a ) ] k + 1 B ( α , k + 1 ) Γ ( α ) d t = M L [ h ( x ) − h ( a ) ] 2 k + 2 B ( α , k + 1 ) Γ ( α ) B ( α , 2 k + 2 ) Γ ( α ) , x ∈ ( a , a + l ] .
现设
[ h ( x ) − h ( a ) ] 1 − α | ϕ n ( x ) − ψ ( x ) | ≤ L n M [ h ( x ) − h ( a ) ] ( n + 1 ) ( k + 1 ) ∏ i = 1 n B ( α , ( k + 1 ) ( i + 1 ) ) Γ ( α ) , x ∈ [ a , a + l ] .
又
[ h ( x ) − h ( a ) ] 1 − α | ϕ n + 1 ( x ) − ψ ( x ) | = [ h ( x ) − h ( a ) ] 1 − α ∫ a x 1 Γ ( α ) [ h ( x ) − h ( t ) ] α − 1 h ′ ( t ) L [ h ( t ) − h ( a ) ] k [ h ( x ) − h ( a ) ] 1 − α | φ n ( t ) − ψ ( t ) | d t ≤ M L n + 1 [ h ( x ) − h ( a ) ] ( n + 2 ) ( k + 1 ) ∏ i = 1 n + 1 B ( α , ( k + 1 ) ( i + 1 ) ) Γ ( α ) , x ∈ [ a , a + l ] .
故
[ h ( x ) − h ( a ) ] 1 − α | ϕ n + 1 ( x ) − ψ ( x ) | ≤ M L n + 1 ( M h l ) ( n + 2 ) ( k + 1 ) ∏ i = 1 n + 1 B ( α , ( k + 1 ) ( i + 1 ) ) Γ ( α ) ≤ M L n + 1 ( M h l ) ( n + 2 ) ( k + 1 ) ∏ i = 1 n + 1 Γ ( ( k + 1 ) ( i + 1 ) ) Γ ( α + ( k + 1 ) ( i + 1 ) ) → 0 ( n → ∞ ) .
则
lim n → ∞ [ h ( x ) − h ( a ) ] 1 − α ϕ n ( x ) = [ h ( x ) − h ( a ) ] 1 − α ψ ( x ) , x ∈ [ a , a + l ] .
即
ψ ( x ) ≡ ϕ ( x ) , x ∈ ( a , a + l ] .
这就证明了 φ ( x ) 的唯一性,结论5证毕。
故微分方程初值问题(1)在 ( a , a + l ] 上存在唯一的连续解 ϕ ( x ) ,其中 l = min { a 1 , 1 M h ( b M Γ ( α ) B ( α , k + 1 ) ) 1 k + 1 } 。
注:本文所得结论可适用于分数阶微分方程初值问题(1)一些具体实例,我们可以参照文献 [
感谢同学们在论文编写过程中给予的大力支持和帮忙,给我带来极大的动力。也要感谢参考文献中的作者们,他们的研究文章使我们受益匪浅。
本论文得到了中国矿业大学(北京)大学生创新训练项目(202207002)的资助。
杨钰翎,梁俊玮,李 健. 一类分数阶微分方程初值问题解的存在唯一性The Existence of Uniqueness in the Solution of a Class of Initial Value Problem for Fractional Differential Equations[J]. 理论数学, 2023, 13(03): 476-485. https://doi.org/10.12677/PM.2023.133052