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含p-Laplacian算子高阶分数阶微分方程的唯一迭代正解①

2017-11-22仲秋艳张兴秋

关键词:边值仲秋边值问题

仲秋艳 张兴秋

(1.济宁医学院 信息技术中心,山东 济宁 272067;2.济宁医学院 医学信息工程学院,山东 日照 276826)

含p-Laplacian算子高阶分数阶微分方程的唯一迭代正解①

仲秋艳1张兴秋2

(1.济宁医学院 信息技术中心,山东 济宁 272067;2.济宁医学院 医学信息工程学院,山东 日照 276826)

利用单调迭代技巧,得到了一类含p-Laplacian算子的高阶分数阶微分方程非局部问题迭代正解的唯一性结果,同时给出了解的迭代程序和误差估计.

分数阶方程, 单调迭代,p-Laplacian算子,正解,唯一性

0 引言

近年来,由于在复杂介质的电动力学、电磁学、聚合物流变学、分数控制系统与分数控制器、神经的分数模型以及分数回归模型等诸多方面的应用[1-3],分数阶微分方程的研究得到了广泛的关注和迅速的发展.文献[4-16],在不同的边值条件下,获得了分数阶非局部边值问题正解以及多个正解的存在性.

如何求出分数阶微分方程的正解在应用上具有重要的意义.本文主要目的是给出下列具有P-Laplacian算子和积分边值条件的分数阶微分方程(下称PFDE)

(1)

与文献[4-7]相比,本文具有以下特征:首先边值条件更加广泛;其次,微分方程含有P-Laplacian,这使得考虑的问题更加一般;最后,本文不但给出了解的存在性结果,而且给出了解的迭代程序及误差估计.显然,稍加修改,本文结果可用于含Riemann-Stieltjes积分的边值问题.

1 引理

有关分数阶微积分的基本定义和性质可在文献[1-4]找到,从略.

(2)

引理1[17]对y∈C[0,1],y≥0,(2)的唯一解为

其中

G(t,s)=G1(t,s)+G2(t,s),

(3)

(4)

(5)

显然,G(t,s)在[0,1]×[0,1]上连续.

(a1)G(t,s)≥m1tα-1s(1-s)α-1-i,∀t,s∈[0,1];

(a2)G(t,s)≤M1tα-2s(1-s)α-1-i,∀t,s∈[0,1];

(a3)G(t,s)>M1tα-1(1-s)α-1-i,∀t,s∈[0,1];

(a4)G(t,s)>0,∀t,s∈(0,1);

2 主要结果

记e(t)=tα-1,本文使用以下条件.

(H)f∈C((0,1)×R+,R+),对于固定的t∈(0,1)以及c∈(0,1),f(t,u)关于u非减且存在常数0

f(t,cu)≥φp[c(1+η(c))]f(t,u),

(6)

这里η(c)=m(c-r-1).

注1 如果c≥1,那么(6)式等价于

(7)

设E=C[0,1],令P={x∈C[0,1]∶x(t)≥0,t∈[0,1]}.则P是Banach空间E中正规常数为1的正规锥.定义E的子集D如下

D={u∈P∶存在正数lu<1

(8)

注意到e(t)∈P,易知D非空.定义算A如

(9)

定理1 设条件(H)满足.如果

(10)

那么PFDE(1)有唯一正解w*∈D,且存在常数

使得ke(t)≤w*(t)≤Ke(t),t∈[0,1],

证明首先证明算子A映D到D.

事实上,对任意u∈D,存在正数Lu>1>lu满足

lue(t)≤u(t)≤Lue(t),t∈[0,1].

(11)

由条件(H),引理2,(7),(10),(11)诸式得

(12)

(13)

这里,

于是

(14)

则0<δ<1.由(14)式得

[1+η(δ)]-1w0≤Aw0≤[1+η(δ)]w0.

(15)

un=Aun-1,vn=Avn-1,n=1,2,…,

(16)

注意到f(t,u)关于u非减,由(6)和(7)两式得

(17)

(18)

根据(15)-(18)诸式,我们有

u1=Au0≥δ[1+η(δ)]Aw0≥δw0=u0,

(19)

(20)

由(19)、(20)式,利用数学归纳法,我们有

u0≤u1≤…≤un≤…≤vn≤…≤v1≤v0.

(21)

此外,由(17)式得

(22)

注意到u0=δ2v0,我们有

(23)

由于P为正规常数为1的正规锥,故对于任意正整数P,由un+p-un≤vn-un,我们有

(24)

即{un}是一个Cauchy列.于是un收敛于某w*∈D.显然

‖vn-w*‖≤‖vn-un‖+‖un-w*‖,

结合(24)式,我们有vn→w*及un≤w*≤vn,故,un+1=Aun≤Aw*≤Avn=vn+1,n=1,2,…,令n→+∞取极限得w*=Aw*,即,w*∈D是A的一个不动点.

取初值w0=e(t), 则le≤Ae≤Le,这里

u1=Au0≥k[1+η(k)]Ae≥k[1+η(k)]le≥ke=u0,

(25)

v1=Av0≤K[1+η(K-1)]-1Ae≤K[1+η(K-1)]-1Le≤Ke=v0.

(26)

根据(25)及(26)两式可知唯一解w*满足ke(t)≤w*(t)≤Ke(t),t∈[0,1].

(27)

下面在比条件(H)更强的条件下研究PFDE(1).列出以下条件:

(H*)f∈C((0,1)×R+,R+),对于任意固定的t∈(0,1),c∈(0,1),f(t,u)关于u非减,且存在常数0<λ*<1使得对于任意(t,u)∈(0,1)×R+,f(t,cu)≥φp(cλ*)f(t,u).

(H0)f∈C((0,1)×R+,R+),对于任意固定的t∈(0,1),c∈(0,1),f(t,u)关于u非减,且存在常数0<λ<1使得对于任意(t,u)∈(0,1)×R+,f(t,cu)≥cλf(t,u).

(H1)f∈C(0,1)×R+,R+),存在常数0<λ1≤λ2<1使得对于任意(t,u)∈(0,1)×R+,cλ2f(t,u)≤f(t,cu)≤cλ1f(t,u),0

证由条件(H*),对任意c∈(0,1),(t,u)∈(0,1)×R+,我们有

f(t,cu)≥φp(cλ*)f(t,u)=φp(cc-1(1-λ*))f(t,u)≥φp[c(1+η(c))]f(t,u),

这里η(c)=m(c-r-1),m=1,r=1-λ*,于是,由定理1可证定理2.

注2 从定理2的证明过程可知,条件(H)弱于(H*).另外,由文献[12]注 3.4可知(H0)弱于(H1).文献[12],[14]利用条件(H0)研究微分方程解的存在唯一性,而文献[18-20]则利用条件(H1)研究某些整数阶微分方程正解存在的充要条件.易知当p>2时,(H*)弱于(H0).

3 应用举例

考察下列分数阶微分方程

(28)

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[14] 仲秋艳,张兴秋.一类具有p-Laplacian算子的奇异分数阶微分方程无穷点边值问题的正解[J]. 聊城大学学报:自然科学版,2016, 29(2):25-32.

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UniqueIterativePositiveSolutionforHigher-orderFractionalDifferentialEquationswithp-Laplacian

ZHONG Qiu-yan ZHANG Xing-qiu

(1.Department of Information Technology, Jining Medical College, Jining 272067, China; 2.School of Medical Information Engineering, Jining Medical College, Rizhao 276826, China)

By means of monotone iterative technique, uniqueness results of positive solutions for a class of higher nonlocal fractional differential equations withp-Laplacian are obtained. The iterative sequences of solution and error estimation are also considered.

fractional differential equations,monotone iterative technique,p-Laplacian, positive solution, uniqueness

2017-05-12

国家自然科学基金项目(11571296,11571197);山东省高校科技发展计划项目(J15LI16);山东省自然科学基金项目(ZR2015AL002)资助

张兴秋,E-mail:zhxq197508@163.com.

O175.8

A

1672-6634(2017)03-0028-07

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