FOR DIFFERENTIAL EQUATIONS OF FRACTIONAL ORDER

Boundary Value Problems for Differential Equations of Fractional Order 71(2.1) IfK g2(b − a) γΓ(γ + 1) + L g[(b − a) γ−θΓ(γ + 1)Γ(2 − θ) + (b −]a)γ−θ< 1, (3.29)Γ(γ + 1 − θ)then there exists a unique solution for FBVP (3.3)–(3.4) in P 1 .(2.2) IfK g(γ − 1) γ−1 (b − a) γγ γ Γ(γ + 1)+ L g(2γ − θ)(b − a) γ−θγΓ(γ − θ + 1)< 1, (3.30)then there exists a unique solution for FBVP (3.9)–(3.10) in the space P 2 .Proof. The proof for (1.1): For u(t), v(t) ∈ P, and (t, s) ∈ [a, b] × [a, b],according to the definition for the operator “T f ”, we have∣ Tf u(t) − T f v(t) ∣ ∣ ≤∫ba{ ∫b≤ ‖u − v‖ fThus∥ Tf u − T f v ∥ ∥f= maxaa≤t≤b K f|G(t, s)| · |f(s, u(s)) − f(s, v(s))| ds ≤(t − a)(b − s) γ−1(b − a)Γ(γ)∫ tds +≤a(t − s) γ−1Γ(γ)2(b − a)γΓ(γ + 1) ‖u − v‖ f .}ds ≤∣ Tf u(t) − T f v(t) ∣ 2K f (b − a) γ≤ ‖u − v‖ f .Γ(γ + 1)Considering (3.27), we finish the proof according to Lemma 3.2.The proof for (2.1): On one hand, we have∣∣T g u(t) − T g v(t) ∣ 2(b − a)γ≤Γ(γ + 1) ‖u − v‖ gfor u(t), v(t) ∈ P 1 , and (t, s) ∈ [a, b] × [a, b], which is similar to (1.1). Onthe other hand, according to Lemma 3.1, we haveCa D θ t T g u(t) =∫ b= (t−a)1−θ (b−s) γ−1Γ(2−θ) Γ(γ)(b − a) g( s, u(s), C a Dsu(s) θ ) ds−Ja γ−θ g ( t, u(t), C a Dt θ u(t) ) .aThen∣∣Ca Dt θ T g u − C a Dt θ T g v ∣ [≤ ‖u − v‖ g ·(b − a) γ−θΓ(2 − θ)Γ(γ + 1)(b −]a)γ−θ+ .Γ(γ − θ + 1)Combined with the definition of ‖ · ‖ g and Lemma 3.2, (3.29) holds.The proof for (1.2) and (2.2) are referred to [33].□