FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER

60 T. S. Aleroev, H. T. Aleroeva, Ning-Ming Nie, and Yi-Fa Tang(− c[ρx 5+1/ρ E ρ(+ cρ{ρx 2/ρ+5 E ρx 1/ρ ; 5 + 1 ρ) (− ρ 5 + 1 ()x 5+1/ρ E ρ x 1/ρ ; 6 + 1 ρρ) ] +()x 2/ρ+5 E ρ x 1/ρ ; 6 + 1 )−ρx 1/ρ ; 5 + 1 ρ(− cρ 5 + 1 ()[x 2/ρ+5 E ρ x 1/ρ ; 6 + 1 ) (− 5 + 2 ( )E ρ x 1/ρ ; 6 + 1 )x 2/ρ+5 +ρρ ρρ∞∑+ 12ρk=0Now note that from (2.36) it follows thatx (k+2)/ρ−5 k 2)] }Γ ( kρ + 6 + 2 .ρS 1 (1) = −χ 1, S 2 (1) = 1 2 (χ2 1 − χ 2).As all eigenvalues of the problem (2.27 ′ )–(2.28 ′ ) are positive, obviouslyλ 1 > 1 χ 1= − 1S 1 (1) .The estimate from below for λ 1 looks like λ 1 < χ 1/χ 2.Now taking into account that it is possible to calculate S 1 and S 2 towithin 10 −2 , we will obtainTheorem 2.7. For the first eigenvalue λ 1 of the problem (2.32 ′ )–(2.33 ′ )we have the relation(1.85) −1 < λ 1 < 3.86.Note that it is likewise possible to find estimates for the eigenvalues ofthe problem−u ′′ + D α 00x u + Dα 10x u = λu,u(0) = 0, u(1) = 0.6. Mutually Adjoint Problems and Problem of Completeness ofEigenfunctionsFor the equationu ′′ +consider the problemn∑i=1a i (x)D α i0x ω j(x)u = λu, 0 < α i < 1, (2.43)u(0) = 0, u(1) = 0. (2.44)Along with the problem (2.43)–(2.44), we consider the problemn∑z ′′ + ω j (x)(D α i0x )∗ a i z + λz = 0, (2.45)i=1z(0) = 0, z(1) = 0. (2.46)□