FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER

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30 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangProof. From (1.11) and (1.12), assuming thatwhere‖A n u‖ ≤ p n−1{ u‖u‖ + b‖A 0 u‖ } , m = ‖A 0 ‖;{ 8(a + mb)c = maxd, 8p + 4 a + mb },d1d = ‖R‖, |ε| < 1 c ,we obtain the simple formulas [8]∣∣λ(ε) − λ 0 − ελ 1 − · · · − ε n λ n∣ ∣ ≤ d 2 (|ε|c)n+1 , (1.13)∣ ϕ(ε) − ϕ0 − εϕ 1 − · · · − ε n ϕ n∣ ∣ ≤12 (|ε|c)n+1 . (1.14)Let’s calculate the values of the parameters a, b, c, d, m. First we find m,m = ‖A‖ = sup A, where sup A is the spectral radius of the operator A.As sup A = π −1 , we have m = π −1 . Further, d = dist (πn 2 ; Σ ′′ ) (d is anisolating distance), where Σ ′′ is the spectrum of the operator A −1 with theexcluded point πn 2 . Clearly d = π(2n − 1). To find other parameters a, b,c, we will obtain an estimate of the norm of the operator A n :‖A n ϕ‖ L(0,1) ≤∫ 1 ∫ 1Here K n (x, t) is the kernel of the operator A n∫ 1 ∫ 100|K n (x, t)| · |ϕ(t)| dt dx ==∫ 10|ϕ(t)|∫1−t000∫ 1 ∫ x0z| ln n z|n!0|K n (x, t)| |ϕ(t)| dt dx.∣ (x − t) lnn (x − t)∣ |ϕ(t)| dt dx =n!dz dt ≤ 1 n!Let’s calculate the integral ∫ z ln z n dz.∫z ln z n dz =∫ 10z| ln n z| dz · ‖ϕ‖ L1(0,1).= z2 (ln z) n− nx2 (ln x) n−122 2 + · · · + (−1)n−1 n(n−1)(n−2) · · · 2( x22 n−1 2 − 1 )2 2 .Hence ‖A n ‖ ≤ 12. Now we will take a = 1/4, p = 1/2, and b = 0. Sincen+1{c = max 8 a + mb + 8p + 4 a + mb },ddwe haveThen{ 2c = maxd d}; 4 + 1 = 4 + 1 d = 5.|λ(ε) − λ| ≤ 1 2π(2n − 1)2.

Boundary Value Problems for Differential Equations of Fractional Order 31Thus we have found |ϕ ε − ϕ 0 | ≤ 1/2, and proved Theorem 1.4.□Theorem 1.5. Let u 0 (x), u 1 (x), . . . , u n (x), . . . be the eigenfunctionsof the operator A ρ , and λ 0 , λ 1 , . . . , λ n , . . . be the corresponding eigenvalues.Then the least frequency self-oscillation, i.e. ϕ 0 (x), has no units, i.e.u 0 (x) ≠ 0, 0 < x < 1.Proof. Let λ 0 be the least eigenvalue of the operator A ρ . Thenu 0 (x) = E ρ (−λ 0 x; ρ −1 ) =∞∑k=0(−λ 0 x) kΓ(ρ −1 + ρ −1k ) .Show that the function u 0 (x) has no zero in the interval (0, 1). Supposethat the function u 0 (x) in a point x 0 ∈ (0, 1) equals to zero, that is,u 0 (x 0 ) = E ρ (−λ 0 x 0 ; ρ −1 ) =∞∑k=0(−λ 0 x 0 ) kΓ(ρ −1 + ρ −1k ) = 0.That’s to say, the number −λ 0 x 0 is a zero point of E ρ (z, ρ −1 ). However,−λ 0 x 0 < λ 0 as x 0 ∈ (0, 1) while we have assumed that the least zero is λ 0 .The obtained contradiction proves Theorem 1.5.□Remark 1. It is also possible to show analogously that u 1 (x) in theinterval has exactly one zero, etc.Theorem of existence of the basis made of root spaces of the operatorA ρ (0 < ρ < 1/2) is connected with the problem of completeness of systemsof eigenfunctions of the operator induced by the differential expression⎧⎪⎨ 1 d n−1 ∫xu ′ (t)l(D) n = Γ(1 − γ 1 ) dx n−1dt − (λ + q(x))u,γ1(x − t)0⎪⎩u(0) = 0, D σ 1u ∣ x=0= 0, . . . , D σ n−2u ∣ x=0= 0, u(1) = 0which is studied in case where q(x) is a semi-function (see [3] and referencestherein). If completeness of system of eigenfunctions bounded is proved,then the question is: is it possible to make basis with the eigenfunctions ofthis operator. Let us give an answer to this question.Lemma 1.1. The operator A is dissipativeProof. We will consider the operatorAu =∫ x0(x − t) 1+ε u(t) dt −∫ 10x 1+ε (1 − t) 1+ε u(t) dt.

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