对含有各阶导数的2m阶微分方程: y^(2m)(t)=f(t,y(t),y'(t),…,y^(2m-2)(t),y^(2m-1)(t)),t∈(0,1), y^(2i+1)(0)=y^(2i)(1)=0,0≤i≤m-1, 其中(-1)^mf:[0,1]×R^2m→[0,∞)是连续的。笔者首先给出方程的Green函数及其一些性质,并赋予,一定的增长条件,利用5个泛函的不动点定理,然后给出上述边值问题的3个单调正解的存在性。
We discuss the following BVP y^(2m)(t)=f(t,y(t),y'(t),…,y^(2m-2)(t),y^(2m-1)(t)),t∈(0,1), y^(2i+1)(0)=y^(2i)(1)=0,0≤i≤m-1, where (-1)^mf:[0,1]×R^2m→[0,∞) is continuous. The associated Green's function for the above problem and some of its properties are given. Growth conditions are imposed on f which ensure the existence of positives. By using the 5 functionals fixed point theorem, the existence of at least 3 positive solutions for the the boundary value problem is given.