FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER

52 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangThus λ 1 = n 1−α 2 πi.e.,BP n = 2 π= 2 π= 2 πsinαπ2sinαπ2απsin2 , then A P n = λ 1 = 2 π n1−α sin απ2 . Further∑(pn v, sin nx) sin nx =∞∑k=1n 1−α( 2π (v, sin kx) sin kx, sin nx )cos nx =sinαπ2 (v, sin kx) cos nx = 2 πBP n v = 2 πCalculate the trace of the operator BP n∫παπsin2 cos nx v(t) sin nt dt,∫παπsin2 cos nx v(t) sin nt dt.BP n = ∑ λ n (BP n ).To find the eigenvalues of the operator BP n , we will solve the equationthat is,2πClearly B P n = 0. Thus0BP n v = vλ,∫παπsin2 cos nx v(t) sin nt dt = λv(t).0µ n = n 2 + n 1−α sin πα 2 .For solution of the problem (2.28) in case c(x) = const we use the followingtheorem being certainly of an independent interest as well.3. Estimation of EigenvaluesTheorem 2.2. Let the eigenvalues of the self-adjoint operator A 0 withdiscrete spectrum be λ (0)n = n q (n = 1, 2, 3, . . . , q), the normed eigenvectorsbe ϕ n and let B be a closed A 0 -bounded operator withD A0 ⊂ D B , ‖Bϕ n ‖ = O(n ρ )Then for the eigenvalues λ n of the operator A = A 0 + B we haveλ n − n q = O(n ρ ).0Proof. Let Γ n be the circle with radius 1 and with the center at the pointn q . Let n be so enough that inside of Γ n there are no eigennumbers exceptλ n . Let the operator A be defined by the formula A = A 0 + B. We needthe following known result.□