Symmetry of nodal solutions for singularly perturbed elliptic problems on a ball

In [41], it was shown that the following singularly perturbed Dirichlet problem epsilon(2)Delta u - u + broken vertical bar U broken vertical bar(p-1)u = 0, in Omega, u=0 on partial derivative Omega has a nodal solution u(epsilon) which has the least energy among all nodal solutions. Moreover, it is shown that u(epsilon) has exactly one local maximum point P-1(epsilon) with a positive value and one local minimum point P-2(epsilon) with a negative value, and as epsilon -> 0, where phi (P-1, P-2) = min (broken vertical bar P-1 - P-2 broken vertical bar/2, d (P-1, partial derivative Omega), d(P-2, partial derivative Omega)). The following question naturally arises: where is the nodal surface {u(epsilon)(x) = 0}? In this paper, we give an answer in the case of the unit ball Omega = B-1 (0). In particular, we show that for epsilon sufficiently small, P-1(epsilon), P-2(epsilon) and the origin must lie on a line. Without loss of generality, we may assume that this line is the x(1)-axis. Then u(epsilon) must be even in x(j), j = 2,..., N, and odd in x(j). As a consequence, we show that {u(epsilon) (x) = 0} = {x is an element of B-1 (0) broken vertical bar x(1) = 0}. Our proof is divided into two steps: first, by using the method of moving planes, we show that P-1(epsilon), P-2(epsilon) and the origin must lie on the x(1)-axis and u(epsilon) must be even in x(j), j = 2,..., N. Then, using the Llapunov-Schmidt reduction method, we prove the uniqueness of u(epsilon) (which implies the odd symmetry of u(epsilon) in x(1)). Similar results are also proved for the problem with Neumann boundary conditions.