1、应用数学MATHEMATICA APPLICATA2023,36(4):997-1006带p-Laplacian算子的分数阶微分方程边值问题正解的存在性与多重性周文学,吴亚斌,宋学瑶(兰州交通大学数理学院,甘肃 兰州 730070)摘要：本文讨论一类带有p-Laplacian算子的一致分数阶微分方程边值问题正解的存在性.首先构造出相应的Green函数,将边值问题转化为等价的积分方程.然后通过Green 函数相关性质、Guo-Krasnoselskii不动点定理、Leggett-Williams不动点定理和单调迭代技巧建立了边值问题正解的存在性与多重性结论.最后举例验证主要结果的适用性.关键词：
2、分数阶微分方程;边值问题;Green函数;不动点定理;单调迭代技巧中图分类号：O175.8AMS(2010)主题分类：34A08;34B15文献标识码：A文章编号：1001-9847(2023)04-0997-101.引言随着自然科学的发展,分数阶微分方程得到了广泛的应用,尤其是在物理1、工程力学2、建筑科学3等领域.而边值问题作为微分方程领域的一个重要课题,也得到了越来越多的关注.近些年来,为了解决复杂多变的问题,许多学者致力于带p-Laplacian算子的分数阶微分方程边值问题解的存在性与多重性研究,且相关的理论与研究结果也越来越成熟.412文12运用Schauder不动点定理研究了如下一
3、类带p-Laplacian算子的分数阶边值问题cD0+(cpD0+u(t)=f(t,u(t),0 t 1,u(1)=0,u(1)=10u(s)ds,cD0+u(1)=bcD0+u()(1.1)解的存在性.其中1 2,1 2,0 ,b 1,f C(0,10,),0,),cD0+,cD0+是标准的Caputo型分数阶导数.值得注意的是,以上大量工作都是在Riemann-Liouvill与Caputo导数定义下完成的.2014年Khalil等人在文13提出了一种新的分数阶导数定义,称为一致分数阶导数.接着有学者在文14中证明了一致分数阶导数定义下的一些重要性质与基本理论,如：链式法则、Gronwal
4、l不等式、分部积分法、分数阶Laplace变换等.显然,研究新分数阶导数定义下的相关理论对分数阶微分方程领域的发展具有积极的推动作用.因此,受以上杰出工作启发,本文使用锥上不动点定理及单调迭代技巧在一致分数阶导数定义下讨论了如下带p-Laplacian算子的分数阶微分方程边值问题D(pDu(t)=f(t,u(t),0 t 1,u(0)=u(0)=0,u(1)=10h(s)u(s)ds,pDu(0)+(pDu)(0)=0,pDu(1)+(pDu)(1)=0(1.2)收稿日期：2022-10-30基金项目：国家自然科学基金(11961039,11801243);兰州交通大学校青年科学基金(2017
5、012)作者简介：周文学,男,汉,甘肃人,教授,研究方向：非线性分析.998应用数学2023正解的存在性,其中2 3,1 1,D,D是一致分数阶导数,本文中总假设10h(s)s2ds 1.2.预备知识定义113设 (n,n+1,函数f:0,)R的阶一致分数阶导数定义为D(f)(t)=lim0f(1)(t+t()f(1)(t),其中t 0,常用f()表示,表示小于等于的最小整数.注1一致分数阶导数详细性质可参考文13-14.定义213设 (n,n+1,函数f:0,)R的阶一致分数阶积分定义为I(f)(t)=1n!t0(t s)nsn1f(s)ds.引理114设 (n,n+1,若Df(t)在0,)
6、上连续,则IDf(t)=f(t)+c0+c1(t)+cntn,ci R(i=1,n).定义315设E是Banach空间,P E是E中的锥,:P 0,)是连续泛函且对任意x,y P,t 0,1都有(tx (1 t)y)t(x)+(1 t)(y)成立,则称是P上的非负连续凹泛函.引理216(Guo-Krasnoselskii不动点定理)设1,2是Banach空间E上的有界集,且 1,1 2,假设Q:2 1 E全连续,如果满足以下条件之一:1)u 1 Qu u,u 2 Qu u;2)u 1 Qu u,u 2 Qu u,那么Q在2 1中至少存在一个不动点.引理317(Leggett-Williams不
7、动点定理)设P为Banach空间E中锥,Pc=u P:u c,Q:Pc Pc全连续,:P R+是连续凹泛函,(u)u对任意的u Pc成立,存在常数0 a b b=并且(Qu)b;2)对任意u a,Qu b有(Qu)b,则Q至少存在三个不动点u1,u2,u3.引理4设2 6 3,函数y C(0,1,R),则边值问题Du(t)=y(t),0 t 1,u(0)=u(0)=0,u(1)=10h(s)u(s)ds(2.1)存在解满足u(t)=10G(t,s)y(s)ds,其中Green函数G(t,s)表达式如下G(t,s)=12(t s)2s3t22(110h(s)s2ds)(1 s)2s3,0 s t
8、 1,t22(110h(s)s2ds)(1 s)2s3,0 t s 1.(2.2)令=1 10h(s)s2ds,总有 0.证对方程两端阶积分得u(t)=12t0(t s)2s3y(s)ds+c0+c1t+c2t2,(2.3)这里c0,c1,c2为常数.由边界条件u(0)=u(0)=0有c0=c1=0,又u(1)=10h(s)u(s)ds,有u(1)=1210(1 s)2s3y(s)ds+c2=c210h(s)s2ds,第 4 期周文学等：带p-Laplacian算子的分数阶微分方程边值问题正解的存在性与多重性999即c2=10(1 s)2s3y(s)ds2(1 10h(s)s2ds).故u(t
9、)=12t0(t s)2s3y(s)ds+c2t2=12t0(t s)2s3y(s)ds t22(1 10h(s)s2ds)10(1 s)2s3y(s)ds=10G(t,s)y(s)ds.证毕.引理5若2 6 3,1 0使得u ,有u N.取M=max0t1,0uN|f(t,u(t)|+1.对u 有|Qu(t)|=?10G(t,s)q(10H(s,)f(,u()d)ds?Mq110G(t,s)q(10H(s,)d)ds Mq1q(10H(0,)d)10G(1,s)ds 0,存在 0 使得对任意的t1,t2 0,1,|t1 t2|时,有|G(t2,s)G(t1,s)|Mq1q(10H(0,)d)
10、.于是,对任意u ,有|Qu(t2)Qu(t1)|10|G(t2,s)G(t1,s)|q(10H(s,)f(,u()d)ds Mq110|G(t2,s)G(t1,s)|q(10H(s,)d)ds Mq1q(10H(0,)d)10|G(t2,s)G(t1,s)|ds 0使得(Hl)对(t,u)0,1 0,r1,有f(t,u)p(L1r1);(H2)对(t,u)1/4,3/4 0,r2,有f(t,u)p(L2r2),则边值问题(1.2)至少存在一个正解u且满足minr1,r2 u maxr1,r2.证由引理7知算子Q:P P全连续.首先,令1=u P:u r1.对u 1,有0 u(t)r1.由条件
11、(H1)有Qu=max0t1?10G(t,s)q(10H(s,)f(,u()d)ds?L1r110G(1,s)q(10H(0,)d)ds=r1=u.1002应用数学2023因此,对u 1,有Qu u.其次,令2=u P:u r2.对u 2,有0 u(t)r2.由条件(H2),当t 1/4,3/4有Qu(t)=10G(t,s)q(10H(s,)f(,u()d)ds1014G(1,s)q(1014H(0,)f(,u()d)ds L2r23/41/414G(1,s)q(3/41/414H(0,)d)ds=r2=u.即对u 2,有Qu u.由引理2知算子Q至少存在一个不动点u.即问题(1.2)至少存在
12、一个正解且有r1 u r2.证毕.定理2设f(t,u)是连续函数,若存在常数0 a b c=d使得(H3)对(t,u)0,1 0,a,有f(t,u)p(L1a);(H4)对(t,u)1/4,3/4 b,c,有f(t,u)p(L2b);(H5)对(t,u)0,1 0,c,有f(t,u)p(L1c),则边值问题(1.2)至少存在三个正解且满足maxt0,1|u1(t)|a,b mint1/4,3/4|u2(t)|maxt0,1|u2(t)|c,a maxt0,1|u3(t)|c,mint1/4,3/4|u3(t)|b,因而有u P(,b,c):(u)b=.第 4 期周文学等：带p-Laplacia
13、n算子的分数阶微分方程边值问题正解的存在性与多重性1003若u P(,b,c),则b u(t)c.当t 1/4,3/4时,根据条件(H4)有(Qu)=min1/4t3/4|Qu(t)|=min1/4t3/410G(t,s)q(10H(s,)f(,u()d)ds3/41/4min1/4t3/4G(t,s)q(3/41/4min1/4t3/4H(s,)f(,u()d)ds L2b3/41/414G(1,s)q(3/41/414H(0,)d)ds=b,即(Qu)b.取d=c,得引理3的条件1)满足.同理,若任意u P(,b,c)且Qu c=d,可得(Qu)b,则引理3条件3)满足.综上,边值问题(1
14、.2)至少存在三个正解满足maxt0,1|u1(t)|a,b mint1/3,3/4|u2(t)|maxt0,1|u2(t)|c,a maxt0,1|u3(t)|c,mint1/4,3/4|u3(t)|0使得以下条件成立:(H6)f(t,u1)f(t,u2),0 t 1,0 u1 u2 m;(H7)max0t1f(t,m)p(L1m),则以下结论成立:(i)单调迭代序列定义为uk=Quk1=10G(t,s)q(10H(s,)f(,uk1()d)ds,k=1,2,vk=Qvk1=10G(t,s)q(10H(s,)f(,vk1()d)ds,k=1,2,且满足limk+uk(t)=u,limk+vk
15、(t)=v,0=u0 u1 u2 uk u v vk v2 v1 v0=m,(ii)边值问题(1.2)存在正解u和v满足0 u v m.证定义Pm=u E|u m.由条件(H6)得0 f(t,u)f(t,m)p(L1m).首先说明u是边值问题(1.2)的一个正解.令u0(t)=0,显然有u0(t)Pm.又对u Pm,则有0 u(t)m.根据条件(H7)得Qu(t)=max0t1?10G(t,s)q(10H(s,)f(,u()d)ds?L1m10G(1,s)q(10H(0,)d)ds=m.显然可得Q(Pm)Pm.由引理7可知Q:Pm Pm是全连续算子.即Q是在Pm上紧的,因此Q(Pm)是列紧的.
16、又有uk Q(Pm)Pm.进而可知uk是列紧集.故可得序列uk存在一个收敛子列ukj,且u Pm满足uki u(j ).1004应用数学2023又u1 Pm,u1(t)=Qu0(t)0=u0(t),即u1(t)u0(t).根据条件(H6)可知函数f(t,u)单调递增,从而可知Q为增算子.即有Qu1(t)Qu0(t),因而有u2(t)=Qu1(t)Qu0(t)=u1(t).由上式归纳可得u0 u1 u2 uk m.即序列uk 单调递增.有uk u(k )成立.由uk=Quk1及Q的连续性有u=limk+uk=limk+Quk1=Qu.即u是Q在Pm上的一个不动点.进一步由假设条件知u是问题(1.
17、2)的一个正解且有0 u m.其次说明v也是边值问题(1.2)的一个正解.类似的,令v0(t)=m,有v0(t)Pm.进一步vk Q(Pm)Pm.同上可知vk是列紧集.由条件(H7)有v1(t)=Qv0(t)=10G(t,s)q(10H(s,)f(,v0()d)ds L1m10G(1,s)q(10H(0,)d)ds=m=v0(t),即有v1(t)v0(t).从而有v2(t)=Qv1(t)=10G(t,s)q(10H(s,)f(,v1()d)ds10G(t,s)q(10H(s,)f(,v0()d)ds=Qv0(t)=v1(t).进一步归纳可得0 vk v2 v1 v0=m.由上知序列vk单调递减
18、.类似前半部分证明过程知v也是问题(1.2)的一个正解且0 v m.此外,因u0(t)v0(t),有u1(t)=Qu0(t)=10G(t,s)q(10H(s,)f(,u0()d)ds10G(t,s)q(10H(s,)f(,v0()d)ds=Qv0(t)=v1(t),以此类推uk(t)vk(t)(k=0,1,2,).综上,所得结论成立.问题(1.2)存在正解u和v且有0 u v m.证毕.4.举例例1考虑如下边值问题:D32(2D52u(t)=et+u25625+1,t 0,1,u(0)=u(0)=0,u(1)=10h(s)u(s)ds,2D22u(0)+(2D22u)(0)=0,2D22u(1
19、)+(2D32u)(1)=0,(4.1)第 4 期周文学等：带p-Laplacian算子的分数阶微分方程边值问题正解的存在性与多重性1005其中=52,=32,p=2,f(t,u)=et+u25625+1,h(s)=6s.显然有 0.计算可得L10.0632,L2 72.28.取r1=75,r2=0.0315则有f(t,u)=et+u25625+1 4.7183 2(L2r2)2.2769.满足定理1假设条件.故边值问题(4.1)至少存在一个正解u且有0.0315 u 75.例2考虑如下边值问题:D32(2D52u(t)=f(t,u(t),t 0,1,u(0)=u(0)=0,u(1)=10h(
20、s)u(s)ds,2D52u(0)+(2D52u)(0)=0,2D52u(1)+(2D52u)(1)=0,(4.2)其中=52,=32,p=2,f(t,u)=11000t2+72.4u5,u 1,11000t2+u+71.4,u 1,h(s)=6s,L1 0.0632,L272.28.取a=0.1,b=1,c=1815.简单计算有 0,且f(t,u)=11000t2+72.4u5 0.001724 2(L2b)72.28,f(t,u)=11000t2+u+71.4 114.0039 2(L1c)114.708.满足定理2假设条件,故边值问题(4.2)至少存在三个正解.参考文献：1 WANG K
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30、ctional Differential Equations withp-Laplacian OperatorZHOU Wenxue,WU Yabin,SONG Xueyao(School of Mathematics and Physics,Lanzhou Jiaotong University,Lanzhou 730070,China)Abstract:The existence of positive solutions to the boundary value problem of conformable frac-tional differential equations with
31、 p-Laplacian operator is discussed.First of all,the corresponding Greenfunction is constructed and the boundary value problem is transformed into equivalent integral equa-tion.Then,by applying related properties of the Green function,Guo-Krasnoselskii fixed point theorem,Leggett-Williams fixed point
32、 theorem and monotone iterative technique,the existence and multiplicityconclusion of positive solutions of the boundary problem are established.Finally,some examples areincluded to illustrate the main results.Key words:Fractional differential equation;Boundary value problem;Green function;Fixed pointtheorem;Monotone iterative technique