FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER

Boundary Value Problems for Differential Equations of Fractional Order 37From the general Volterra theory it follows that the equation (1.21) can besolved by a method of iterations. Since the function q(x) is continuous, wehavewhereϕ n (λ, x) =y(x, λ) = E ρ (λx 1/ρ ; µ 1 ) + ϕ 1 (λ, x) + ϕ 2 (λ, x) + · · · , (1.22)∫ x0dτ 1 · · ·∫τ n−10dτ 1 · · · q(τ n )E ρ (λ(x − τ 1 ) 1/ρ ; µ 1 ) · · · ×× E ρ (λ(τ n−1 − τ n ) 1/ρ ; µ 1 )E ρ (λτ 1/ρn ; µ 1 ).Obviously the series (1.22) converges in regular intervals in each interval(0; l), l < 1.Theorem 1.7. There is an operator of transformation V , transforminga solution E ρ (λx 1/ρ ; µ 1 ) of the equation (1.19) into a solution y(x; λ) of theequation (1.17).holds.Lemma 1.2. The equality(J − λJ 1/ρ ) −1 1Γ(ρ −1 ) = E ρ(λx 1/ρ ; 1) (1.23)Proof. Let f(x) ∈ L 1 (0, 1), ρ > 0 and λ be any complex parameter. Thenthe integral equationhas a unique solutionu(x) = f(x) + λu(x) = f(x) +λΓ(ρ −1 )belonging to the class L 1 (0, 1).In (1.24), let f(x) = 1Γ(ρ −1 ). We obtain∫ x0u(x) =∫ x0(x − 1) 1/ρ u(t) dt (1.24)((x − t) 1/ρ−1 E ρ λ(x − t) 1/ρ ; x )f(t) dt (1.25)ρ1Γ(ρ −1 ) + 1Γ(ρ −1 )Due to (1.25) and (1.26), we haveu(x) =1Γ(ρ −1 ) + 1Γ(ρ −1 )Using the known formula of M. M. Dzhrbashyan∫ x0∫ x0(x − t)u(t) dt. (1.26)((x − t) 1/ρ−1 E ρ λ(x − t) 1/ρ ; x )dt. (1.27)ρ