Abstract
In this paper, the Morse index and the symmetry-breaking for positive solutions of the following two-point boundary value problem{u″+h(x)f(u)=0,x∈(−1,1),u(−1)=u(1)=0, are studied, where h∈C[−1,1]∩C1([−1,1]\\{0}), h(x)>0, h(−x)=h(x) on [−1,1]\\{0}, f∈C1[0,∞), f(s)>0 for s>0, and f(0)=0. The problem for the one-dimensional Hénon equation{u″+|x|lup=0,x∈(−1,1),u(−1)=u(1)=0 is a typical example, where l⩾0 and p>1. This problem always has the unique positive even solution. It is well-known that if l=0, then there is no positive non-even solution, and if l>0 is sufficiently large, then there exist positive non-even solutions. The result in this paper shows that if l(p−1)⩾4, then the Morse index of the positive least energy solution equals 1 and the Morse index of the positive even solution equals 2, and hence the positive least energy solution is non-even and symmetry-breaking phenomena occur. It is also shown that if l⩾0 and p>1 are sufficiently small, then there is no positive non-even solution and the Morse index of the even positive solution equals 1.
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have