带变号格林函数的四阶三点边值问题的多个正解的存在性

带变号格林函数的四阶三点边值问题的多个正解的存在性达举霞;霍梅;韩晓玲【摘要】应用Leggett-Williams不动点定理研究了四阶三点边值问题u(4)(t)=f(t,u(t)) (t∈[0,1]),u’(0)=u”(η)=u'''(0)=u(1)=0多个正解的存在性,其中f:[0,1]×[0,+∞)→[0,+∞)连续,η∈[√3/3,1]为常数.尽管Green函数是变号的,对任意的正整数m,该问题仍有正解且至少有2m-1个正解.%By applying Leggett-Williams fixed point theorem,the fourth-order three-point boundary value problem is studied:u(a)(t)=f(t,u(t)) (t∈[0,1]),u′(0)=un(η)=u"(0)=u(1)=0,where f:[0,1] ×[0,+∞)→[0,+∞) is continuous,η∈ [√3/3,1].The existence of at least 2m-1 positive solutions for arbitrary positive integer m is obtained while the problem has the sign-changing Green's function.【期刊名称】《华南师范大学学报（自然科学版）》【年(卷),期】2017(049)003【总页数】5页(P109-113)【关键词】四阶三点边值问题;变号Green函数;多个正解【作者】达举霞;霍梅;韩晓玲【作者单位】西北师范大学数学与统计学院,兰州730070;西北师范大学数学与统计学院,兰州730070;西北师范大学数学与统计学院,兰州730070【正文语种】中文【中图分类】O175.8多个正解的存在性，其中f:[0,1]×[0,+∞)→[0,+∞)连续,η为常数. 尽管Green函数是变号的,对任意的正整数m,该问题仍有正解且至少有2m-1个正解.where f:[0,1]×[0,+∞)→[0,+∞) is continuous,η. The existence of at least 2m-1 positive solutions for arbitrary positive integer m is obtained while the problem has the sign-changing Green’s function.Key words： fourth-order three-point boundary value problem; sign-changing Green’s function; multiple positive solutions弹性梁是工程建筑的基本构件,弹性力学和工程物理常用四阶常微分方程边值问题来刻画弹性梁的平衡状态,由于这类问题的普遍性和重要性,四阶两点边值问题和四阶多点边值问题深受学者关注.2008年,YAO[1]运用Krasnoselskli不动点定理获得了四阶三点边值问题x″(0)=x″(1)=0,x(η)=B,x′(η)=C (0<η<1)n个正解的存在性结果.2009年,GRAEF等[2]运用锥上的不动点定理研究了四阶三点边值问题正解的存在性,这里β为常数,gC([0,1],[0,∞)),在非线性项f满足超线性或次线性条件下获得了问题(1)、(2)至少存在一个正解.2014年,ZHOU等[3]运用不动点指数理论获得了四阶三点边值问题正解的存在性结果,这里β为常数,gC([0,1],[0,∞)).以上结果都是在Green函数非负的情况下获得的. 2012年,SUN和ZHAO[4]运用Leggett-Williams不动点定理在Green函数变号时获得了问题更多详细结果见文献[5-11]. 受前人启发，本文在Green函数变号的情况下运用Leggett-Williams不动点定理研究问题设E是Banach空间,P是E上的锥. 对任意的x,yP,t[0,1],若有δ(tx+(1-t)y)≥tδ(x)+(1-t)δ(y),则映射δ:P→(-∞,+∞)是一个凹函数. 设a和b是2个常数且0<a<b,δ是P上的非负连续凹函数. 定义凸集如下:定理1[5] (Leggett-Williams 不动点定理)设A:c→c是全连续的且δ是P上的非负连续凹函数，使得对所有的xc，都有δ(x)≤‖x‖. 假设存在0<d<a<b≤c，使得(1){xP(δ,a,b):δ(x)>a}≠∅且δ(Ax)>a,xP(δ,a,b);(2)‖Ax‖<d,‖x‖≤d;(3) δ(Ax)>a,xP(δ,a,c)和‖Ax‖>b,则A在c上至少有3个不动点x1、x2、x3且满足定义空间E=C[0,1]的范数‖u‖|.为了得到式(3)正解的存在性,需要在Banach空间E 上定义一个合适的锥. 假设u是式(3)的解,很容易证明在u′(1)≤0的情况下,u(t)≥0,t[0,1]. 事实上,由于f是非负的,所以u(4)(t)≥0,可知 u‴(t)单调递增,t[0,1]. 又u‴(0)=0,从而 u‴(t)≥0,t[0,1]. 又u″(η)=0意味着对于任意的 yE,考虑边值问题在[0,t]上给式(6)两边积分,得到在[0,t]上给式(7)两边积分,得到继续在[0,t]上给式(8)两边积分,得到最后再在[0,t]上给式(9)两边积分,得到‴(0)+s.又由u′(0)=u″(η)=u‴(0)=u(1)=0,得到s.从而式(6)的格林函数的表达式G(t,s)如下:(1)当s≥η时,有(2)当s<η 时,有显然,当0≤s<η时,G(t,s)≥0;当η≤s≤1时,G(t,s)≤0. 因此,对于s≥η,有对于s<η,有引理1 设y且[0,1]),则u且u是式(6)的唯一解. 此外u满足这里θ(1/2,η)和θ*=(η-θ)/η.证明由式(4)可知,当t[0,η]时,u″(t)≤0,从而u(t)在[0,η]上是上凸的,有可得‖u‖.则同理可得,当t[η,1]时,结论也成立.本文假定f:[0,1]×[0,+∞)→[0,+∞)是连续的并且满足如下条件:(C1)对每一个x[0,+∞),映射t→f(t,x)是递减的;(C2)对每一个t[0,1],映射x→f(t,x)是递增的.设P={u‖u‖},易证P是E上的锥. 在P上定义算子A:显然,如果u是A在P上的不动点,则u是式(3)的非负解.为了方便,记s.定理2 设存在数d、a和c，0<d<a<a/θ*≤c,使得[0,η],u[0,d]),[1-θ,θ],u,[0,η],u[0,c]),则式(3)至少有3个正解u、v和w，满足‖u‖<‖w‖,<a.证明设uP,定义). 易证当uP时,δ是P上的非负连续凹函数且δ(u)≤‖u‖,并有A:P→ P是全连续的.首先,当t[0,η]和u[0,r]时,如果存在一个正数r,使得f(t,u)<r/H1,则A:r→Pr. 事实上,若ur,则‖Au‖<,也就是说,AuPr. 同理可得,如果式(10)和式(11)成立,则A映d到Pd,A映c到Pc. 其次,我们断言对所有的uP(δ,a,a/θ*),{uP(δ,a,a/θ*):δ(u)>a}≠∅且δ(Au)>a. 事实上,常函数(a+a/θ*)/2属于{uP(δ,a,a/θ*):δ(u)>a}.另一方面,当uP(δ,a,a/θ*)时,有‖u‖≤a/θ*,t[1-θ,θ].同样,对任意的uP和t[1-θ,θ],有×,再加上式(11)、(12),当uP(δ,a,a/θ*)时,有a.最后证明:如果有uP(δ,a,c)和‖Au‖>a/θ*,则δ(Au)>a. 为了证明此结论,假设uP(δ,a,c)和‖Au‖>a/θ*成立,若AuP,有综上,Leggett-Williams不动点定理的所有条件都被满足. 因此,A至少有3个不动点,即式(3)至少有3个正解u、v和w并满足定理3 设m是任意正整数,假设存在di(1≤i≤m)和aj(1≤j≤m-1),且0<d1<a1<a1/θ*<d2<a2<a2/θ*<…<dm-1<am-1<am-1/θ*<dm使得[1-θ,θ],u,则式(3)在dm上至少有个2m-1正解.证明使用归纳法. 首先,当m=1,由式(11)有A:d1→d1. 由Schauder不动点定理可知在d1上至少有1个正解. 其次,假设m=k成立,证明m=k+1也成立. 设存在数di(1≤i≤k+1)、aj(1≤j≤k)且0<d1<a1<a1/θ*<d2<a2<a2/θ*<…<dk<ak<ak/θ*<dk+1,使得[0,η],u[0,di],1≤i≤k+1),例1 考虑BVP其中f(t,u)=设θ=3/5,则θ*=1/10. 通过简单的计算可知H1=216/1944,H2=6/625. 选取d=1,a=2,c=1 068,则定理2的条件都被满足. 因此,由定理2可知式(15)、(16)至少有3个正解.【相关文献】[1] YAO Q L. 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