Based on the results in the previous papers that the boundary value problem
yβ²β²βyβ²+y=y3,y(0)=0,y(β)=1 with the condition y(x)>0 for
0<x<β has a unique solution yβ(x), and aβ=yββ²(0) satisfies
0<aβ<1/4, in this paper we show that yβ²β²βyβ²+y=y3,ββ<x<0,
with the initial conditions y(0)=0,yβ²(0)=aβ has a unique solution by
using functional analysis method. So we get a globally well defined bounded
function yβ(x),ββ<x<+β. The asymptotics of yβ(x) as xβββ and as xβ+β are obtained, and the connection formulas for
the parameters in the asymptotics and the numerical simulations are also given.
Then by the properties of yβ(x), the solution to the boundary value problem
r2fβ²β²+f=f3,f(0)=0,f(β)=1 is well described by the asymptotics
and the connection formulas.Comment: 11 pages, 2 fingure