RENDICONTI DEL SEMINARIO MATEMATICO

Periodic solutions of difference equations 23and is equivalent to the matrix eigenvalue problem⎛⎞ ⎛−1 − λ 1 0 ··· ··· 00 −1 − λ 1 0 ··· 0(22)··· ··· ··· ··· ··· ···⎜ ··· ··· ··· ··· ··· ···⎟ ⎜⎝ 0 ··· ··· 0 −1 − λ 1 ⎠ ⎝1 0 ··· ··· 0 −1 − λ⎞x 1x 2··⎟x n−2⎠x n−1= 0.Hence the eigenvalues λ k are λ k = −1 + µ k (0 ≤ k ≤ n − 2), where the µ k are theeigenvalues of the (permutation, unitary, circulant) matrix⎛⎞0 1 0 ··· ··· 00 0 1 0 ··· 0··· ··· ··· ··· ··· ···⎜ ··· ··· ··· ··· ··· ···,⎟⎝ 0 ··· ··· 0 0 1 ⎠1 0 ··· ··· 0 0namely (see e.g. [6]),(23)λ k = −1 + e 2kπin−1 (0 ≤ k ≤ n − 2).The corresponding eigenvectors ϕ k (0 ≤ k ≤ n − 2) have componentsϕ k m = e 2kmπin−1 (1 ≤ m ≤ n − 1).In particular, λ 0 = 0 is always a real eigenvalue, and all the other eigenvalues havenegative real part. If n = 2, 0 is the unique eigenvalue; if n > 2 is even, 0 is the uniquereal eigenvalue; if n is odd, λ n−1 = −2 is the unique nonzero real eigenvalue.26. Reversing the order of upper and lower solutionsFor n ≥ 2 odd and λ = −2, system (21) becomes(24)x 1 − x n = 0x 2 + x 1 = 0... ... ...x n + x n−1 = 0and has the solution ϕ associated to ϕ (n−1)/2 with componentsThe adjoint system(25)ϕ m = (−1) m−1(1 ≤ m ≤ n).x 1 + x 2 = 0... ... ...x n−1 + x n = 0−x 1 + x n = 0