FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER

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32 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa TangDenotev(t) =∫ x0(x − t) 1+ε u(t) dt −∫ 10x 1+ε (1 − t) 1+ε u(t) dt.From this expression it follows that u(x) = Dv ∈ L 2 (0, 1). Consider againthe product[ ∫x(Au, u) (x − t) 1+ε u(t) dt −=∫ 10= v(x)00∫ 1∫ 10]x 1+ε (1 − t) 1+ε u(t) dt u(x) =[ ∫x∫](x − t) 1+ε u(t) dt − x 1+ε (1 − t) 1+ε u(t) dt u(x) dx ==∫ 100v(x)D x v dt =0=∫ 10v(x)∫ 1v(x) d ( ∫xdx0∫0( x( ∫x)∣v ′ (t) ∣∣∣1v(x)(x − t) ε dt −= −∫ 10( ∫x00)v(t)(x − t) ε dt dx =0∫ 10( ∫x)v ′ (t)(x − t) ε dt v ′ (t) dx,0)v ′ (t)(x − t) ε dt dx =)v ′ (t)(x − t) ε dt v ′ (x) dx =as v(0) = v(1) = 0. Define v ′ (x) = z(x). Then (Au, u) = −(J ε z, z). Now,by virtue of a theorem of Matsaev–Polant, it follows that the values of theform (J ε z, z) lay in the angle |argz| < πε2, which proves Lemma 1.1. □In papers [1] and [23], the dissipativity of the operator l(D) is proved.4. The Basis Problem for Systems of EigenfunctionsTheorem 1.6. The system of eigenfunctions of the operator A formsa basis of the closed linear hull.Proof. Remind that the system of vectors {u 1 , u 2 , . . . , u n , . . . } forms a basisof the closed linear hull G ⊂ H if the inequalityn∑m |c k | 2 n∑ ∥≤ ∥ c k ϕ 2 ∥∥2 ∑ n≤ M |c k | 2k=1k=1holds, where m, M are positive constants independent of c 1 , c 2 , c 3 , . . . , c n(n = 1, 2, 3, . . . ). Any vector f (f ∈ G) in that case uniquely expands intok=1
Boundary Value Problems for Differential Equations of Fractional Order 33∑a series f = ∞ ( ∞∑c k u k |c k | 2 < ∞ ) . It is known that if the spectrum ofn=1n=1a dissipative operator with completely continuous imaginary component “ispressed enough” to the real axis, then it is possible to form a basis of thelinear closed hull of eigenvectors of this operator.We need a theorem of Glazman.Theorem of Glazman. Let A be a linear bounded dissipative operatorwith completely continuous imaginary component, possessing an infinitesystem of eigenvectors {u k } ∣ ∞ k=1 , normalized by the condition (u k, u u ) = 1(k = 1, 2, 3, . . . ), and {λ k } ∣ ∞ be the corresponding sequence of differentk=1eigenvalues. If the condition∑ Im λj Im λ k< ∞|λ j − λ k |is satisfied, then the system {ϕ k } ∣ ∞ is a Bari–Riesz’s basis of the closedk=1linear hull. Since for 0 < ρ < 1/2 all eigenvalues of the operator A ρ are real,due to theorem of Glazman the proof of Theorem 1.6 follows. Analogously,it is possible to prove similar statements for more general problems. □5. Partial Problem of Eigenvalues for the Operator A ρ(0 < ρ < 1/2)In [1], [3], there have been determined areas in the complex plane wherethere are no eigenvalues of the operator A ρ for any ρ. Here, as above, wewill assume that 0 < ρ < 1/2. In applied problems the greatest interestrepresents usually determination of first eigenvalues, therefore we will solvethis problem for eigenvalues of the operator A ρ .Let’s consider the operatorA= 1 [ ∫xΓ(ρ −1 (x−t) 1/ρ u(t) dt−)0∫ 10]x 1/ρ−1 (1−t) 1/ρ−1 u(t) dt =A 0 u+A 1 u.As 0 < ρ < 1/ρ, the operator A 0 and A 1 are invertible. Therefore sp(A) =spA 0 + spA 1 .Clearly1spA 0 = 0, spA 1 =Γ(2 − 1/ρ) .Therefore the determinant∞∑D A (λ) = (1 − λµ j ), µ k = 1 ,λ kj=1is meaningful.Earlier it has been established that the number λ is an eigenvalue of theoperator A only if λ −1 is a zero of the function E ρ (λ; ρ −1 ). Since the entire
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82 T. S. Aleroev, H. T. Aleroeva, N
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