无穷区间上分数阶积分边值问题的研究_薛婷婷.pdf

高校应用数学学报2023,38(1):73-84无穷区间上分数阶积分边值问题的研究薛婷婷,樊小琳(新疆工程学院 数理学院,新疆乌鲁木齐 830000)摘要:利用Krasnoselskii不动点定理,研究一类无穷区间上分数阶微分方程积分边值问题,得到几个关于该问题正解存在的结果,并给出一个例子用于验证主要定理.关键词:分数阶微分方程;积分边值问题;无穷区间;正解中图分类号:O175.8文献标识码:A文章编号:1000-4424(2023)01-0073-121引言近年来,随着分数阶微积分理论的发展,分数阶微分方程在许多领域被广泛的应用,如:物理力学领域,反常扩散研究领域,自动控制领域,生物医学领域等.所以对分数阶微分方程边值问题正解存在的研究引起人们的关注,得到了许多重要的结果(参见1-6).其中,对于无穷区间上的分数阶微分方程边值问题,已有一些重要的研究结果(参见7-20).例如13利用Sch auder不动点定理研究问题D0+u(t)=f(t,u(t),D10+u(t),t 0,),u(0)=0,limtD10+u(t)=u,u R正解的存在性,其中1 2,f C(0,)R2,R).17通过Sch auder不动点定理和Banach压缩映射原理,研究问题D0+u(t)=f(t,u(t),D10+u(t),t 0,),u(0)=0,D10+u()=m2i=1iu(i)的可解性,其中1 2.对此,很自然的想法是:能否把导数的阶数提高,即由1 2提高到2 3?能否研究边界条件更一般的问题(如积分边值问题)?收稿日期:2021-09-22修回日期:2022-08-12基金项目:新疆自治区科技厅青年科学基金(2021D01B35;2021D01A65);新疆自治区高校科研计划自然科学项目(XJEDU2021Y048);新疆工程学院博士启动基金(2020 xgy012302)DOI:10.13299/ki.amjcu.00224674高 校 应 用 数 学 学 报第38卷第1期基于此,本文研究如下一类无穷区间上分数阶微分方程积分边值问题(简记ISLBVP)D0+u(t)+f(t,u(t),D10+u(t)=0,t (0,+),u(0)=0,D20+u(0)D10+u(0)=+0g1(s)u(s)ds+,D10+u(+)=+0g2(s)u(s)ds+(1)正解的存在性,其中D0+是Riemann-Liouville型分数阶导数,2 0,存在函数hr L1(R+,R+)使得|f(t,(1+t1)u,v)|hr(t),a.e.t R+,max|u|,|v|r.则称f满足L1-Carath eodory条件.本文利用Krasnoselskii不动点定理,得到几个关于该问题正解存在的结果.与上述文献所研究问题相比,本文所研究的积分边值问题更具一般性并且分数阶导数的阶数由1 2提高到2 3,在一定程度上推广了已有文献的工作.2 预备知识引引引理理理2.115令n 1 1,=i,i=1,2,+1,则D0+t=(+1)(+1)t,(ii)D0+ti=0,i=1,2,+1.定义空间X=u:u C(R+,R),D10+u C(R+,R),suptR+|u(t)|1+t1+,suptR+D10+u(t)0且满足ISLBVP(1),则称u是ISLBVP(1)的一个正解.薛婷婷等:无穷区间上分数阶积分边值问题的研究75引引引理理理2.3假设y L1(R+,R+)且y(t)0,t R+.令u P,则问题D0+u(t)+y(t)=0,t (0,+),2 3,u(0)=0,D20+u(0)D10+u(0)=+0g1(s)u(s)ds+,D10+u(+)=+0g2(s)u(s)ds+(2)有唯一正解u(t)=+0G(t,s)y(s)ds+0K(t,s)u(s)ds+t1()+(+)t2(1),(3)其中G(t,s)=1()t1+(1)t2(t s)1,0 s t +,t1+(1)t2,0 t s 0;(2)对任意的(t,s)R+R+,0 G(t,s)1+t11+(1)();(3)对任意的(t,s)R+R+,0 K(t,s)1+t11()(1+(1)g2(s)+(1)g1(s).引引引理理理2.5(见16)(Krasnoselskii不动点定理)设是Banach空间X中的一个非空有界闭凸集,假设映像A:X和T:X具有下述性质(i)-(iii).(i)对x,y ,Ax+Ty ;(ii)A是压缩映射;(iii)T是全连续的,则A+T在中必有不动点.引引引理理理2.6(见13)设Z X是一个有界集,如果满足条件(a)对u Z,有u(t)1+t1和D10+u(t)在紧区间R+上是等度连续的;(b)对任意给定的 0,都存在常数k=k()0,使得对t1,t2 k和u Z,有u(t1)1+t11u(t2)1+t21,D10+u(t1)D10+u(t2)0,使得|un|r,|u|r.即suptR+un(t)1+t1 r,suptR+u(t)1+t1 r,suptR+D10+un(t)r,suptR+D10+u(t)r.故对a.e.s R+有f(s,un(s),D10+un(s)f(s,u(s),D10+u(s)0,n 和f(s,un(s),D10+un(s)f(s,u(s),D10+u(s)=f(s,(1+s1)un(s)1+s1,D10+un(s)f(s,(1+s1)u(s)1+s1,D10+u(s)2hr(s).由Lebesgue控制收敛定理可知+0f(s,un(s),D10+un(s)f(s,u(s),D10+u(s)ds 0,n .因此由引理2.4知|Tun(t)Tu(t)|1+t1=+0G(t,s)1+t1(f(s,un(s),D10+un(s)f(s,u(s),D10+u(s)ds)+0G(t,s)1+t1f(s,un(s),D10+un(s)f(s,u(s),D10+u(s)ds1+(1)()+0f(s,un(s),D10+un(s)f(s,u(s),D10+u(s)ds 0,n +.由(8)可知D10+Tun(t)D10+Tu(t)=+t(f(s,un(s),D10+un(s)f(s,u(s),D10+u(s)ds)+0f(s,un(s),D10+un(s)f(s,u(s),D10+u(s)ds 0,n +.故|Tun Tu|0(n ),这意味着算子T是连续的.接下来,分三步验证算子T是紧的.设B是P中的一个非空有界闭子集.因为f满足L1-Carath eodory条件,所以存在常数w 0以及函数hw L1(R+)使得对所有的u B,|u|w,有f(s,u(s),D10+u(s)=f(s,(1+s1)u(s)1+s1,D10+u(s)hw(s).(i)对u B,由引理2.4可知Tu(t)1+t1=+0G(t,s)1+t1f(s,u(s),D10+u(s)ds1+(1)()+0f(s,u(s),D10+u(s)ds 1+(1)()+0hw(s)ds +.由(8)可知D10+Tu(t)=+tf(s,u(s),D10+u(s)ds+0f(s,u(s),D10+u(s)ds78高 校 应 用 数 学 学 报第38卷第1期+0hw(s)ds +.故Tu (1+(1)()+1)+0hw(s)ds 0,t1,t20,D0,不妨令t1 t2,u B,由引理2.4知+0G(t1,s)1+t11G(t2,s)1+t11f(s,u(s),D10+u(s)dst10G(t1,s)1+t11G(t2,s)1+t11f(s,u(s),D10+u(s)ds+t2t1G(t1,s)1+t11G(t2,s)1+t11f(s,u(s),D10+u(s)ds+t2G(t1,s)1+t11G(t2,s)1+t11f(s,u(s),D10+u(s)dst10hw(s)1+t11(t12t11)+(1)(t22t21)(t2s)1(t1s)1ds+t2t1hw(s)1+t11(t12 t11)+(1)(t22 t21)(t2 s)1ds+t2hw(s)1+t11(t12 t11)+(1)(t22 t21)ds 0,t1 t2.同上可以证明+0G(t2,s)1+t11G(t2,s)1+t12f(s,u(s),D10+u(s)ds 0,t1 t2.综上所述,可以得到Tu(t1)1+t11Tu(t2)1+t12=+0G(t1,s)1+t11G(t2,s)1+t12f(s,u(s),D10+u(s)ds+0G(t1,s)1+t11G(t2,s)1+t12f(s,u(s),D10+u(s)ds+0G(t1,s)1+t11G(t2,s)1+t11f(s,u(s),D10+u(s)ds+0G(t2,s)1+t11G(t2,s)1+t12f(s,u(s),D10+u(s)ds 0,t1 t2.由(8)可知D10+Tu(t1)D10+Tu(t2)=+t1f(s,u(s),D10+u(s)ds+t2f(s,u(s),D10+u(s)dst2t1f(s,u(s),D10+u(s)ds t2t1hw(s)ds 0,t1 t2.故Tu(t)1+t1和D10+Tu(t)在0,D0上等度连续.(iii)下面将验证,对u B,Tu(t)1+t1和D10+Tu(t)满足引理2.6中的条件(b).因为函数f满足L1-Carath eodory条件,所以+0f(s,u(s),D10+u(s)ds+0hw(s)ds 0,存在常数L 0使得+Lf(s,u(s),D10+u(s)ds 0使得对t1,t2 D1,有t111+t11t121+t12 L使得对t1,t2 D2和0 s L,有G(t1,s)1+t11G(t2,s)1+t12 maxD0,D1,D2,则对t1,t2 D,由引理2.4可知Tu(t1)1+t11Tu(t2)1+t12+0G(t1,s)1+t11G(t2,s)1+t12f(s,u(s),D10+u(s)dsL0G(t1,s)1+t11G(t2,s)1+t12f(s,u(s),D10+u(s)ds+LG(t1,s)1+t11G(t2,s)1+t12f(s,u(s),D10+u(s)ds+0hw(s)ds+2(1+(1)()=+0hw(s)ds+2(1+(1)().由(8)可知D10+Tu(t1)D10+Tu(t2)t2t1f(s,u(s),D10+u(s)ds+Lf(s,u(s),D10+u(s)ds .因此由引理2.6可知T(B)是相对紧的.综上所述,算子T是全连续的.定定定理理理3.1假设函数f满足L1-Carath eodory 条件,gi(s)是非负的,i=1,2且满足:=+0(1+(1)g2(s)+(1)g1(s)(1+s1)ds 0使得hR(t)满足+(1)(+)(1 )R (1+(1)+0hR(s)ds,(9)其中m,0,则ISLBVP(1)至少有一个正解.证证证由的定义可知+0g2(s)(1+s1)ds .(10)式(5)结合的定义,可得+0K(t,s)(1+s1)1+t1ds().(11)令BR=u P:u R,则BR是X中的非空有界闭凸子集.对x,y BR,由算子A,T的定80高 校 应 用 数 学 学 报第38卷第1期义和引理2.4以及(9),(11)可知|Ax(t)+Ty(t)|1+t1=+0K(t,s)1+t1x(s)ds+0G(t,s)1+t1f(s,y(s),D10+y(s)ds+()t11+t1+(1)t21+t1+0K(t,s)1+t1(1+s1)ds x+1+(1)()+0f(s,y(s),D10+y(s)ds+(1)(+)()()x+1+(1)()+0hR(s)ds+(1)(+)()R+(1+(1)+0hR(s)ds+(1)(+)()R+(1+(1)+0hR(s)ds+(1 )R (1+(1)+0hR(s)ds()=R()R.对x,y BR,由(7)-(10)可知D10+Ax(t)+D10+Ty(t)=+0g2(s)x(s)ds+tf(s,y(s),D10+y(s)ds+0g2(s)(1+s1)|x(s)|1+s1ds+0f(s,y(s),D10+y(s)ds+x+0hR(s)ds+R+0hR(s)ds+(1 )R (1+(1)+0hR(s)ds R.故Ax+Ty R,这意味着对x,y BR,有Ax+Ty BR.因此引理2.5中的条件(i)成立.接下来验证A是一个压缩映射.事实上对x1,x2 BR,由(11)可知|Ax1(t)Ax2(t)|1+t1=+0K(t,s)1+t1(x1(s)x2(s)ds+0K(t,s)1+t1(1+s1)ds x1 x2()x1 x2 x1 x2.薛婷婷等:无穷区间上分数阶积分边值问题的研究81对x1,x2 BR,由(7),(10)可知D10+Ax1(t)D10+Ax2(t)=+0g2(s)(x1(s)x2(s)ds+0g2(s)(1+s1)|x1(s)x2(s)|1+s1ds x1 x2.故对x1,x2 X,有Ax1 Ax2 x1 x2.因为0 1,所以A是压缩映射.因此,引理2.5中的条件(ii)成立.由引理3.1 知T:P P是全连续算子.因此,引理2.5中的条件(iii)也成立.综上所述,引理2.5中的所有条件都满足.由引理2.5可知,ISLBVP(1)在X中至少有一个正解.推推推论论论3.1假设存在函数a,b,c L1(R+,R+)使得f(t,(1+t1)u,v)a(t)+b(t)|u|+c(t)|v|.(12)若(1+(1)(bL1+cL1)+(1)(+)+(1+(1)aL11 +(1+(1)(bL1+cL1).(13)由(1+(1)(bL1+cL1)0.令hR(t)=a(t)+R(b(t)+c(t),则+0hR(s)ds=aL1+R(bL1+cL1).(14)令BR=u P:u R,对y BR,由(12),(14)可知+0f(s,y(s),D10+y(s)ds=+0f(s,(1+s1)y(s)1+s1,D10+y(s)ds+0(a(s)+b(s)y(s)1+s1+c(s)D10+y(s)ds aL1+y(bL1+cL1)aL1+R(bL1+cL1)=+0hR(s)ds.此外由(13)-(14)可知(1 )R (1+(1)+0hR(s)ds=(1 )R (1+(1)aL1+R(bL1+cL1)=1 (1+(1)(bL1+cL1)R (1+(1)aL1 +(1)(+)+(1+(1)aL1(1+(1)aL1=+(1)(+).因此由定理3.1可知ISLBVP(1)至少有一个正解.推推推论论论3.2假设函数a,b,c L1(R+,R+),存在常数0 1,2 0使得对u R0,有0 u11 0使得对v R1,有0 v21 maxR0,R1,2+(1)(+)+(1+(1)aL11 .(16)令hR(t)=a(t)+R1b(t)+R2c(t),则有+0hR(s)ds=aL1+R1bL1+R2cL1.(17)令BR=u X:u R,对y BR,由(15),(17)可知+0f(s,y(s),D10+y(s)ds=+0f(s,(1+s1)y(s)1+s1,D10+y(s)ds+0(a(s)+b(s)y(s)1+s11+c(s)D10+y(s)2)ds aL1+y1bL1+y2cL1 aL1+R1bL1+R2cL1=+0hR(s)ds.由R maxR0,R1可知Ri11 2(1+(1)(bL1+cL1+1),i=1,2.(18)由(16)-(18)可知(1 )R (1+(1)+0hR(s)ds=(1 )R (1+(1)(aL1+R1bL1+R2cL1)=(1 )R (1+(1)aL1(1+(1)R11bL1+(1+(1)R21cL1 R(1 )R(1 )(bL1+cL1)2(bL1+cL1+1)R (1+(1)aL1(1 )R 1 2 R (1+(1)aL1=1 2 R (1+(1)aL11 22+(1)(+)+(1+(1)aL11 (1+(1)aL1=+(1)(+).因此由定理3.1可知ISLBVP(1)至少有一个正解.例例例子子子3.1考虑下面的ISLBVPD830+u(t)+f(t,u(t),D530+u(t)=0,t (0,+),u(0)=0,D230+u(0)D530+u(0)=+0g1(s)u(s)ds+,D530+u(+)=+0g2(s)u(s)ds+,(19)薛婷婷等:无穷区间上分数阶积分边值问题的研究83其中=83,=3,f(t,u(t),D530+u(t)=1100et53u(t)D530+u(t),g1(s)=110es83,g2(s)=160es53,=0.03,=0.01.取hr(t)=r2100(1+t53)et53,则对任意的0 u,v r,t R+,有0 f(t,(1+t53)u,v)hr(t),因此f满足L1-Carath eodory条件.此外=+0(1+(1)g2(s)+(1)g1(s)(1+s1)ds=+0(110es53+16es83)(1+s53)ds 0.3536 +(1)(+)=0.11.因此定理3.1的所有条件都满足.由定理3.1可得ISLBVP(19)至少有一个正解.参参参考考考文文文献献献:1郑春华,刘文斌.一类具有时滞的分数阶微分方程边值问题正解的存在性J.山东大学学报(理学版),2015,50(3):73-79.2Wang Yongqing,Liu Lishan,Wu Yonghong.Positive solutions for a class of fractionalboundary value problem with changing sign nonlinearityJ.Nonlinear Anal Theor,2011,74(17):64346441.3仝荣,胡卫敏.一类分数阶奇异半正脉冲微分方程边值问题正解的存在性J.四川师范大学学报(自然科学版),2020,43(6):774-780.4Wang Guotao,Ahmad Bashir,Zhang Lihong.Impulsive anti-periodic boundary value prob-lem for nonlinear differential equations of fractional orderJ.Nonlinear Anal Theor,2011,74(3):792-804.5梁兴悦,周宗福.一类带有Stieltjes积分边界条件的分数阶微分方程边值问题正解J.应用数学,2020,33(4):826-835.6Liu Xiping,Jia Mei.The positive solutions for integral boundary value problem of frac-tional p-Laplacian equation with mixed derivativesJ.Mediterr J Math,2017,14(2):articlenumber:94.7Jia Mei,Zhang Haibin,Chen Qiang.Existence of positive solutions for fractional differentialequation with integral boundary conditions on the half-lineJ.Bound Value Probl,2016,article number:104.8薛婷,刘文斌,张伟.无穷区间上分数阶微分方程积分边值问题正解的存在性J.南京师大学报(自然科学版),2017,40(4):36-46.9Liang Sihua,Zhang Jihui.Existence of multiple positive solutions for m-point fractionalboundary value problems on an infinite intervalJ.Math Comput Model,2011,54(5):1334-1346.10Li Xiaochen,Liu Xiping,Jia Mei,et al.Existence of positive solutions for integral boundaryvalue problems of fractional differential equations on infinite intervalJ.Math Method ApplSci,2017,40(6):1892-1904.11廖秀,韦煜明.一类无穷区间上分数阶边值问题正解的存在性J.应用泛函分析学报,2018,20(3):258-268.84高 校 应 用 数 学 学 报第38卷第1期12蒋伟,周宗福.一类无穷区间上高阶分数阶微分方程边值问题的正解(英文)J.应用数学,2017,30(4):750-759.13Su Xinwei,Zhang Shuqin.Unbounded solutions to a boundary value problem of fractionalorder on the half-lineJ.Comput Math Appl,2011,61(4):1079-1087.14刘宗宝,刘文斌,张伟.无穷区间上分数阶带p-Laplacian算子微分方程积分共振边值问题解的存在性(英文)J.应用数学,2020,33(1):12-24.15Kilbas A A,Srivastava H M,Trujillo J J.Theory and Applications of Fractional DifferentialEquationsM.Netherlands:Elsevier,B V,2006.16郭大钧.非线性泛函分析M.北京:高等教育出版社,2015.17王菊芳,禹长龙,郭彦平,等.无穷区间上分数阶非局部边值问题的可解性J.河北科技大学学报,2015,36(6):577-586.18李晓晨,刘锡平,李燕,等.无穷区间上含参数分数阶微分方程积分边值问题正解的存在性J.吉林大学学报(理学版),2017,55(1):13-21.19张海斌,贾梅,陈强.一类半无穷区间上分数阶非线性微分方程边值问题多个正解的存在性J.吉林大学学报(理学版),2014,52(6):1145-1150.20张瑞鑫,王文霞.一类无穷区间上分数阶微分方程的三点边值问题解的存在性J.华中师范大学学报(自然科学版),2020,54(3):345-351.Research on fractional integral boundary value problems on infiniteintervalXue Ting-ting,Fan Xiao-lin(School of Mathematics and Physics,Xinjiang Institute of Engineering,Urumqi 830000,China)Abstract:By using krasnoselskis fixed point theorem,the integral boundary value problem of aclass of fractional differential equations on infinite intervals is studied,some results about the existenceof positive solutions of the problem are obtained,and an example to verify the main theorems is given.Keywords:fractional differential equation;integral boundary value problem;infinite intervals;positive solutionMR Subject Classification:34A08;34B15