利用双锥上的不动点定理,并赋予f,g一定的增长条件,证明了二阶三点微分方程组的边值问题 {x"+f(t,x,y)=0 0≤t≤1 y"+g(t,x,y)-0 0≤t≤1 x(0)-β1x'(0)=0 x(1)=α1x(η1) 0〈η1〈1 y(0)-β2y'(0)=0 y(1)=α2x(η2) 0〈η2〈1 至少存在2组正解,其中f,g:[0,1]×R^+×R^+→R是连续的且可以变号。
For a second-order and three-point boundary value problem as follows: {x"+f(t,x,y)=0 0≤t≤1 y"+g(t,x,y)-0 0≤t≤1 x(0)-β1x'(0)=0 x(1)=α1x(η1) 0〈η1〈1 y(0)-β2y'(0)=0 y(1)=α2x(η2) 0〈η2〈1 where f, g :[0, 1] ×R ^+× R^ + →R is continuous and its sign is alternated. By the fixed point theorem in a double cone and endowing certain growth conditions to f and g, the existence of at least two positive solutions for above problem was derived.