摘要
运用Schauder不动点定理和上下解方法,讨论四阶两点边值问题{u″″(t)+f(t,u′(t),u″(t))=0,t∈[0,1],r 1u(0)-r 2u′(0)=r 3u(1)+r 4u′(1)=u″(0)=0解的存在性,其中r 1,r 2,r 3,r 4≥0,f:[0,1]×R^2→R为连续函数.
By using Schauder fixed-point theorem and the method of lower and upper solutions,the fourth-order two-point boundary value boundary is considered{u″″(t)+f(t,u′(t),u″(t))=0,t∈[0,1],r 1u(0)-r 2u′(0)=r 3u(1)+r 4u′(1)=u″(0)=0 where r 1,r 2,r 3,r 4≥0,f:[0,1]×R^2→R is continuous.
作者
杜睿娟
Du Ruijuan(School of Cyber Security,Gansu Political Science and Law Institute,Lanzhou 730070,China)
出处
《宁夏大学学报（自然科学版）》
CAS
2019年第4期303-305,309,共4页
Journal of Ningxia University(Natural Science Edition)
基金
甘肃省自然科学基金资助项目(17JR5RA091)
甘肃政法学院科研基金重点资助项目(2017XZDLW06)
关键词
算子方程
上下解方法
不动点
two-point boundary value problem
the method of lower and upper solutions
fixed-point
