一类带p-Laplacian算子的分数阶微分方程边值问题正解的存在性
Existence of Positive Solutions for Boundary Value Problem of Fractional Differential Equations with p-Laplacian Operator

作者: 段佳艳 * , 王文霞 , 郭晓珍 :太原师范学院数学系,山西 晋中;

关键词: p-Laplacian算子Leray-Schauder非线性抉择正解p-Laplacian Operator Leray-Schauder Nonlinear Alternative Theorem Positive Solution

摘要:
本文研究了一类带有p-Laplacian算子的分数阶微分方程边值问题的正解的存在性,利用Leray-Schauder非线性抉择,得出边值问题至少存在一个正解的充分条件,并给出了一个具体的例子。

Abstract: This paper is concerned with the existence of positive solution for a class of boundary value problems of fractional differential equations with p-Laplacian operator. By using  Leray-Schauder nonlinear choice, some sufficient conditions for the existence of at least one positive solution are obtained. In addition, an example is given to illustrate theoretical results.

1. 引言

分数阶微积分是在整数阶微积分的理论基础上发展而来,近年来分数阶微分方程在物理学、工程学、机械、医学、生物学等许多领域得到了广泛应用 [1] [2] [3] [4]。为了解决越来越多复杂的现象和问题,学者和专家开始研究带p-Laplacian算子的分数阶微分方程。文 [5] 研究如下带p-Laplacian算子的分数阶微分方程

{ D 0 + β ( φ p ( D 0 + α u ) ) ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 , D 0 + α u ( 0 ) = 0 ,

这里 D 0 + α , D 0 + β D 0 + γ 是标准的Riemann-Liouville型分数阶导数, 1 < α 2 0 < β 1 0 < γ 1 α γ 1 0 ,常数 σ 是一个正数, φ p ( s ) = | s | p 2 s , p > 1 。利用锥上的不动点定理,获得了正解的一些存在性和多重性结果。

文 [6] 研究如下带有p-Laplacian算子的Riemann-Liouville型分数阶微分方程边值问题

{ D α ( φ p ( D β u ( t ) ) ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = λ D γ u ( ξ ) , D β u ( 0 ) = 0 ,

应用凸锥上的不动点理获得了该问题正解的存在性结果。这里 α , β , γ 0 < α < 1 1 < β 2 γ = β 1 2 0 < ξ 1 2 λ [ 0 , + ) λ Γ ( β ) ξ ( β 1 ) / 2 < Γ ( β + 1 2 ) φ p ( s ) = | s | p 2 s , p > 1

基于上述文献中的研究,本文主要利用Leray-Schauder非线性抉择讨论如下带有p-Laplacian算子的Riemann-Liouville型分数阶微分方程边值问题

{ ( ϕ p ( D 0 + α u ( t ) ) ) + f ( t , u ( t ) , D 0 + α u ( t ) ) = 0 , t [ 0 , 1 ] , u ( 0 ) = u ( 1 ) = D 0 + α u ( 0 ) = 0 , (1)

解的存在性,其中 1 < α 2 ϕ p ( s ) = | s | p 2 s q > 1 ϕ p 1 = ϕ q 1 p + 1 q = 1

2. 预备知识

定义1 [7] 函数 y : ( 0 , + ) R α > 0 阶Riemann-Liouville分数阶积分为

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

等式的右端在 ( 0 , + ) 有定义,其中 Γ ( ) 为Gamma函数。

定义2 [7] 连续函数 y : ( 0 , + ) R α > 0 阶Riemann-Liouville分数阶导数为

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

等式的右端在 ( 0 , + ) 有定义,其中 n = min { m Z : m α } Γ ( ) 为Gamma函数。

引理1 [7] α > 0 u C ( 0 , 1 ) L ( 0 , 1 ) D 0 + α u C ( 0 , 1 ) L ( 0 , 1 ) ,则分数阶微分方程 D 0 + α u ( t ) = 0 有唯一解

u ( t ) = c 1 t α 1 + c 2 t α 2 + c n t α n ,

其中 n = min { m Z : m α } c i R , i = 1 , 2 , , n

引理2 [7] α > 0 u C ( 0 , 1 ) L ( 0 , 1 ) D 0 + α u C ( 0 , 1 ) L ( 0 , 1 ) ,则

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

其中 n = min { m Z : m α } c i R , i = 1 , 2 , , n

引理3 若 y ( t ) C [ 0 , 1 ] ,且 1 < α 2 ,则分数阶微分方程

{ ( ϕ p ( D 0 + α u ( t ) ) ) + y ( t ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = D 0 + α u ( 0 ) = 0 ,

有唯一解

u ( t ) = 0 1 G ( t , s ) ϕ q ( 0 s y ( τ ) d τ ) d s ,

其中

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

证明 由 ( ϕ p ( D 0 + α u ( t ) ) ) + y ( t ) = 0 可得

ϕ p ( D 0 + α u ( t ) ) = 0 t y ( s ) d s + c 0 ,

于是

D 0 + α u ( t ) = ϕ q ( 0 t y ( s ) d s + c 0 ) .

根据引理2可得

u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 ϕ q ( 0 s y ( τ ) d τ + c 0 ) d s + c 1 t α 1 + c 2 t α 2 .

D 0 + α u ( 0 ) = 0 c 0 = 0 ,由 u ( 0 ) = 0 ,得 c 2 = 0 ,由 u ( 1 ) = 0 ,得

c 1 = 1 Γ ( α ) 0 1 ( 1 s ) α 1 ϕ q ( 0 s y ( τ ) d τ ) d s .

所以

u ( t ) = 1 Γ ( α ) 0 1 t α 1 ( 1 s ) α 1 ϕ q ( 0 s y ( τ ) d τ ) d s 1 Γ ( α ) 0 t ( t s ) α 1 ϕ q ( 0 s y ( τ ) d τ ) d s = 1 Γ ( α ) 0 t [ t α 1 ( 1 s ) α 1 ( t s ) α 1 ] ϕ q ( 0 s y ( τ ) d τ ) d s + 1 Γ ( α ) t 1 t α 1 ( 1 s ) α 1 ϕ q ( 0 s y ( τ ) d τ ) d s = 0 1 G ( t , s ) ϕ q ( 0 s y ( τ ) d τ ) d s .

证毕。

注 由引理3的证明容易看到 D 0 + α u ( t ) = ϕ q ( 0 t y ( s ) d s ) , t [ 0 , 1 ]

引理4 [8] G ( t , s ) 有下面的性质

1) G ( t , s ) 0 ,对 s , t ( 0 , 1 )

2)存在正函数 r C [ 0 , 1 ] 使得

min 1 / 4 t 3 / 4 G ( t , s ) r ( s ) max t [ 0 , 1 ] G ( t , s ) = r ( s ) G ( s , s ) , 0 < s < 1 ;

3) max t [ 0 , 1 ] 0 1 G ( t , s ) d s = 1 2 2 ( α 1 ) Γ ( α )

引理5 [8] (Leray-Schauder非线性抉择)假设 Ω 是线性赋范空间X中包含原点的开集, F : Ω ¯ X 全连续,并且满足边界条件,即当 x Ω 0 < λ < 1 F λ x ,则F在 Ω ¯ 上至少有一个不动点。

3. 主要结果

下节将用到如下假设

(H1) f : [ 0 , 1 ] × [ 0 , + ) × R [ 0 , + ) 为连续函数,假设存在非负连续函数 j ( t ) , l ( t ) , w ( t ) 使得

| f ( t , u , v ) | j ( t ) ϕ p ( | u | ) + l ( t ) ϕ p ( | v | ) + w ( t ) , t [ 0 , 1 ] .

X = { u ( t ) C [ 0 , 1 ] | D 0 + α u ( t ) C [ 0 , 1 ] } ,定义范数 u = max t [ 0 , 1 ] | u ( t ) | + max t [ 0 , 1 ] | D 0 + α u ( t ) | ,容易证明X是Banach空间。定义算子

T u ( t ) = 0 1 G ( t , s ) ϕ q ( 0 s f ( τ , u ( τ ) , D 0 + α u ( τ ) ) d τ ) d s , t [ 0 , 1 ] .

定理1假设(H1)成立。若(H1)中的函数 j , l , w 满足存在常数 ρ > 0 使得

2 2 ( α 1 ) Γ ( α ) ρ [ ρ p 1 ( 0 1 j ( τ ) d τ + 0 1 l ( τ ) d τ ) + 0 1 w ( τ ) d τ ] q 1 1 , (2.1)

则边值问题(1)存在解 u = u ( t ) 且满足 0 < u < ρ

证明 首先证明 T : X X 为全连续算子。由函数f的连续性容易证明T是连续算子。

以下证明T为紧的。设 Ω 是X中的有界子集,于是存在正数 M > 0 使得 u Ω u M 。从而对于任意的 u Ω

| T u ( t ) | = | 0 1 G ( t , s ) ϕ q ( 0 s f ( τ , u ( τ ) , D 0 + α u ( τ ) ) d τ ) d s | 0 1 G ( t , s ) ϕ q ( | 0 s f ( τ , u ( τ ) , D 0 + α u ( τ ) ) d τ | ) d s 0 1 G ( t , s ) ϕ q ( 0 s | f ( τ , u ( τ ) , D 0 + α u ( τ ) ) | d τ ) d s 0 1 G ( t , s ) ϕ q ( 0 1 ( j ( τ ) ϕ p ( | u ( τ ) | ) + l ( τ ) ϕ p ( | D 0 + α u ( τ ) | ) + w ( τ ) ) d τ ) d s 0 1 G ( t , s ) ϕ q ( 0 1 ( j ( τ ) ϕ p ( u ) + l ( τ ) ϕ p ( u ) + w ( τ ) ) d τ ) d s 1 2 2 ( α 1 ) Γ ( α ) [ M p 1 ( 0 1 j ( τ ) d τ + 0 1 l ( τ ) d τ ) + 0 1 w ( τ ) d τ ] q 1 , t [ 0 , 1 ] ,

| D 0 + α T u ( t ) | = | ϕ q ( 0 t f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) | ϕ q ( 0 t | f ( s , u ( s ) , D 0 + α u ( s ) ) | d s ) ϕ q ( 0 1 ( j ( s ) ϕ p ( | u ( s ) | ) + l ( s ) ϕ p ( | D 0 + α u ( s ) | ) + w ( s ) ) d s ) [ ϕ q ( ϕ p ( u ) 0 1 j ( s ) d s + ϕ p ( u ) 0 1 l ( s ) d s + 0 1 w ( s ) d s ) ] [ M p 1 ( 0 1 j ( s ) d s + 0 1 l ( s ) d s ) + 0 1 w ( s ) d s ] q 1 , t [ 0 , 1 ] .

因此 T ( Ω ) { D 0 + α T u | u Ω } 皆为一致有界的子集合。

再证 T ( Ω ) { D 0 + α T u | u Ω } 皆是等度连续的。对任意的 0 t 1 < t 2 1 u Ω

| T u ( t 2 ) T u ( t 1 ) | = | 0 1 G ( t 2 , s ) ϕ q ( 0 s f ( τ , u ( τ ) , D 0 + α u ( τ ) ) d τ ) d s 0 1 G ( t 1 , s ) ϕ q ( 0 s f ( τ , u ( τ ) , D 0 + α u ( τ ) ) d τ ) d s | = | 0 1 [ G ( t 2 , s ) G ( t 1 , s ) ] ϕ q ( 0 s f ( τ , u ( τ ) , D 0 + α u ( τ ) ) d τ ) d s | [ M p 1 ( 0 1 j ( τ ) d τ + 0 1 l ( τ ) d τ ) + 0 1 w ( τ ) d τ ] q 1 0 1 | G ( t 2 , s ) G ( t 1 , s ) | d s .

注意到 G ( t , s ) [ 0 , 1 ] × [ 0 , 1 ] 上的一致连续性可知, T ( Ω ) 是等度连续的。此外

| D 0 + α T u ( t 2 ) D 0 + α T u ( t 1 ) | = | [ ϕ q ( 0 t 2 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) ϕ q ( 0 t 1 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) ] | = | ( 0 t 2 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) q 1 ( 0 t 1 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) q 1 | ,

0 < q 1 < 1 时,根据不等式 b m + c m ( b + c ) m , b 0 , c 0 , 0 < m < 1 可得

0 ( 0 t 2 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) q 1 ( 0 t 1 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) q 1 ( t 1 t 2 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) q 1 ( t 1 t 2 ( j ( s ) ϕ p ( | u ( s ) | ) + l ( s ) ϕ p ( | D 0 + α u ( s ) | ) + w ( s ) ) d s ) q 1 ( ϕ p ( u ) t 1 t 2 j ( s ) d s + ϕ p ( u ) t 1 t 2 l ( s ) d s + t 1 t 2 w ( s ) d s ) q 1 ( M p 1 ( t 1 t 2 j ( s ) d s + t 1 t 2 l ( s ) d s ) + t 1 t 2 w ( s ) d s ) q 1 ,

再根据积分第一中值定理,因 j ( s ) , l ( s ) , w ( s ) [ t 1 , t 2 ] 上连续,则至少存在三点 μ , η , κ [ t 1 , t 2 ] ,使得

t 1 t 2 j ( s ) d s = j ( μ ) ( t 2 t 1 ) , t 1 t 2 l ( s ) d s = l ( η ) ( t 2 t 1 ) , t 1 t 2 w ( s ) d s = w ( κ ) ( t 2 t 1 ) ,

于是

| D 0 + α T u ( t 2 ) D 0 + α T u ( t 1 ) | [ M p 1 ( j ( μ ) + l ( η ) ) + w ( κ ) ] q 1 ( t 2 t 1 ) q 1 ,

q 1 1 时,由拉格朗日中值定理有

0 ( 0 t 2 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) q 1 ( 0 t 1 f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) q 1 = ( q 1 ) ( 0 ξ f ( s , u ( s ) , D 0 + α u ( s ) ) d s ) q 2 f ( ξ , u ( ξ ) , D 0 + α u ( ξ ) ) ( t 2 t 1 ) ( q 1 ) [ M p 1 ( 0 ξ j ( s ) d s + 0 ξ l ( s ) d s ) + 0 ξ w ( s ) d s ] q 2 ( M p 1 ( j ( ξ ) + l ( ξ ) ) + w ( ξ ) ) ( t 2 t 1 ) ,

其中 ξ ( t 1 , t 2 ) 之间某个确定的值。故

| D 0 + α T u ( t 2 ) D 0 + α T u ( t 1 ) | ( q 1 ) [ M p 1 ( 0 ξ j ( s ) d s + 0 ξ l ( s ) d s ) + 0 ξ w ( s ) d s ] q 2 ( M p 1 ( j ( ξ ) + l ( ξ ) ) + w ( ξ ) ) ( t 2 t 1 ) .

由此可得 { D 0 + α T u | u Ω } 是等度连续的。

既然集合 T ( Ω ) { D 0 + α T u | u Ω } 都是一致有界且等度连续的,根据Arzela-Ascoli定理可知 T ( Ω ) { D 0 + α T u | u Ω } 皆为 C [ 0 , 1 ] 中的相对紧集,进而可知T是全连续算子。

U = { u X | u ρ } ,有 U X ,由上述证明可知 T : U ¯ X 是全连续的。我们断言当 u U λ ( 0,1 ) u λ T u 。如若不然存在 u 0 U λ 0 ( 0,1 ) 使 u 0 = λ 0 T u 0 。于是有

ρ = u 0 = λ 0 T u 0 1 2 2 ( α 1 ) Γ ( α ) [ ρ p 1 ( 0 1 j ( τ ) d τ + 0 1 l ( τ ) d τ ) + 0 1 w ( τ ) d τ ] q 1 ,

进而有

2 2 ( α 1 ) Γ ( α ) ρ [ ρ p 1 ( 0 1 j ( τ ) d τ + 0 1 l ( τ ) d τ ) + 0 1 w ( τ ) d τ ] q 1 1 ,

此与(2.1)式矛盾。由引理5可知边值问题(1)存在解 u ( t ) 使得 0 u ρ 。证毕。

4. 举例

考虑下列具有p-laplacian算子分数阶微分方程边值问题

{ ( ϕ 3 ( D 0 + 3 / 2 u ( t ) ) ) + ( t 2 + u 2 ( t ) 1 + t 2 + e t | D 0 + 3 / 2 u ( t ) | 2 ) = 0 , 0 t 1 , u ( 0 ) = u ( 1 ) = D 0 + 3 / 2 u ( 0 ) = 0 ,

其中 p = 3 , α = 3 / 2

f ( t , u , v ) = ( t 2 + u 2 1 + t 2 + e t | v | 2 ) , 0 t 1 ,

选取 j ( t ) = e t , l ( t ) = e t , w ( t ) = t ,有

| f ( t , u , v ) | e t ϕ 3 ( | u | ) + e t ϕ 3 ( | v | ) + t , 0 t 1 ,

ρ = 1 ,有

2 2 ( α 1 ) Γ ( α ) ρ [ ρ p 1 ( 0 1 j ( τ ) d τ + 0 1 l ( τ ) d τ ) + 0 1 w ( τ ) d τ ] q 1 > 1 ,

定理1的条件皆满足,所以该边值问题至少存在一个解。

基金项目

国家自然科学基金(11361047)。

NOTES

*通讯作者。

文章引用: 段佳艳 , 王文霞 , 郭晓珍 (2020) 一类带p-Laplacian算子的分数阶微分方程边值问题正解的存在性。 应用数学进展, 9, 2301-2307. doi: 10.12677/AAM.2020.912269

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