Asymptotic distribution of eigenvalues of Hill's equation with ...

2 T. Nagasawa and H. OhruiWe improve this result as follows.Theorem 1.2Let Q be a real valued L 1 (R/πZ)-function andThen it holds that∣ √ λ 2n−1 − 2n∣ = o(n −1 ),∫ π0Q(x) dx =0.∣ √ λ 2n − 2n∣ = o(n −1 ) as n →∞.Indeed, we can show the following, from which Theorem 1.2 is easily derived.Theorem 1.3 Let λ be λ 2n−1 or λ 2n ,andput √ λ − 2n = d. Under the sameassumption as in Theorem 1.2, it holds that4nd + d 2 = o(1), (4nd + d 2 ) 2 −|¯Q 4n | 2 = O(n −1 ) as n →∞,where¯Q n = 1 π∫ π0Q(x)e −inx dx.Remark ∫ 1.1 The advantage of our theorem is appeared in the caseπQ(x) dx ≠0. Put0Since our equation is¯Q(= ¯Q 0 )= 1 π∫ π0Q(x) dx,y ′′ +(λ + ¯Q + ˜Q(x))y =0with˜Q(x) =Q(x) − ¯Q.∫ π0˜Q(x) dx =0,Theorem 1.1 implies that the eigenvalue near 2n behaves like√λ + ¯Q − 2n = O(n −1 ) as n →∞.Combining this with√λ + ¯Q = √ (λ 1+ ¯Q )2λ + O(λ−2 ) ,we get