FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER

56 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangNow, along with the problem (2.29)–(2.30), we will consider the followingproblemv ′′ − v ′ (x) = λv(x), v(0) = 0, v ′ (0) = 1.It is known that the solution v(x, λ) of this problem is an entire function ofgenus zero. Take a v 0 (x, λ) and letv n (x, λ) = v 0 (x, λ) +Write u n (x, λ) and v n (x, λ) as∫ x0{1 + λ(x − t)vn−1 (t, λ) } dt.u n (x, λ) = a 0 (x) + λa 1 (x) + · · · + λ n a n (x),v n (x, λ) = b 0 (x) + λb 1 (x) + · · · + λ n b n (x).Clearly |a i (x)| < b i (x), x ∈ (0, 1). Then, from the theorem of Hadamardfollows that u(x, λ) is an entire function of genus zero.□5. On a Method of Estimation of First EigenvaluesConsider the problem−u ′′ + D α 0xu + λu = 0, (2.32)u(0) = 0, u ′ (0) = 1. (2.33)Let u(x, λ) be a solution of the problem (2.32)–(2.33). We have alreadyestablished that u(x, λ) is an entire function of genus zero. Hence it ispossible to represent it as an infinite product∞∏u(x, λ) = c(1 − λ ),λ jj=1where c is a constant unknown as yet, λ j are zeros of the function u(1, λ).Since the zeros of the function u(1, λ) coincide with the eigenvalues of theproblem (2.32)–(2.33)−u ′′ + D α 0xu = λu, (2.32 ′ )u(0) = 0, u(1) = 0, (2.33 ′ )the investigation of the eigenvalues of the problem (2.32 ′ )–(2.33 ′ ) is reducedto the investigation of the zeros of the function u(1, λ). First of all we needthe following interesting statement.Lemma 2.3. All eigenvalues of the problem (2.32 ′ )–(2.33 ′ ) are positive.Proof. For proof of the given statement we will consider the following problem−u ′′ + D α 0xu = λu, u(0) = 0, u(0) = 0.