RENDICONTI DEL SEMINARIO MATEMATICO

Periodic solutions of difference equations 31Furthermore, condition (46) with x 1 = ... = x n−1 = ±ρ implies thatP F (k) (±ρ(1,...,1)) = 1 ∑n−1n − 1m=1f m (k) (±ρ) ̸= 0.If C(ρ) = (] − ρ,ρ[) n−1 , it follows then from Lemma 4 and the homotopy invarianceof Brouwer degree, that, for each k = 1, 2,...,d B [L + F (k) , C(ρ)), 0] = d B [L + P F (k) , C(ρ), 0].Now, L + P : R n−1 → R n−1 is an isomorphism and, for z ∈ R(P), we have(L + P) −1 z = z, so that, using the multiplication property of the Brouwer degreeand denoting the Brouwer index by i B (see e.g. [7]), we obtaind B [L + P F (k) , C(ρ), 0] = d B [(L + P)[I + (L + P) −1 (P F (k) − P)], C(ρ), 0]= i B (L + P, 0) · d B [I − P + P F (k) , C(ρ), 0]= ±d B [I − P + P F (k) , C(ρ), 0].Now, the Leray-Schauder reduction formula (see e.g. [7]) implies thatd B [I − P + P F (k) , C(ρ), 0] = d B [P F (k) | N(L) , C(ρ) ∩ N(L), 0]= d B [ϕ (k) ,] − ρ,ρ[, 0],whereϕ (k) (u) = ϕ(u) +ǫuk(1 + |u|) .Now assumption (45) with x 1 = ... = x n−1 = ±ρ implies that ϕ (k) (−ρ)ϕ (k) (ρ) < 0for all k = 1, 2,..., so thatd B [ϕ (k) ,] − ρ,ρ[, 0] = ±1.Thus it follows from the existence property of Brouwer degree that equation (47) hasat least one solution x (k) such that x (k) ∈ C(ρ) for all k = 1, 2,.... Going if necessaryto a subsequence, we can assume that x (k) → x ∈ C(ρ) which is a zero of L + F andhence a solution of (35)Let u + = max{u, 0}.COROLLARY 6. For p > 0, a m > 0, b m ∈ R (1 ≤ m ≤ n − 1), the periodicproblem(48)Dx m + a m (x + m )p − b m = 0 (1 ≤ m ≤ n − 1), x 1 = x n ,has at least one solution if and only ifn−1∑b m ≥ 0.m=1