Báo cáo hóa học: " Positive solutions of the three-point boundary value problem for fractional-order differential equations with an advanced argument" ppt

Wang et al Advances in Difference Equations 2011, 2011:2 http://www.advancesindifferenceequations.com/content/2011/1/2 RESEARCH Open Access Positive solutions of the three-point boundary value problem for fractional-order differential equations with an advanced argument Guotao Wang1*, SK Ntouyas2 and Lihong Zhang1 * Correspondence: wgt2512@163 com School of Mathematics and Computer Science, Shanxi Normal University, Linfen, Shanxi 041004, P R China Full list of author information is available at the end of the article Abstract In this article, we consider the existence of at least one positive solution to the three-point boundary value problem for nonlinear fractional-order differential equation with an advanced argument C Dα u(t) + a(t)f (u(θ (t))) = 0, < t < 1, u(0) = u (0) = 0, βu(η) = u(1), where 0 such that ||a(t) f (u (θ(t))|| ≤ K, ∀u Ỵ Ω Thus, we have Tu (t) ≤ 1 − βη K ≤ − βη (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−1 ds (α) 0 K , = (1 − βη) (α + 1) K (1 − βη) (α + 1) On the other hand, we have which implies ||Tu|| ≤ t |(Tu) (t)| ≤ (t − s)α−2 a(s)f (u(θ (s)))ds + (α − 1) − βη β + − βη ≤K = η (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (η − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−2 K ds + (α − 1) − βη (1 − s)α−1 Kβ ds + (α) − βη K (1 + β)K + := M (α) (1 − βη) (α + 1) (1 − s)α−1 ds (α) Wang et al Advances in Difference Equations 2011, 2011:2 http://www.advancesindifferenceequations.com/content/2011/1/2 Page of 11 Hence, for each u Ỵ Ω, let t1, t2 Ỵ [0, 1], t1 < t2, we have t2 |(Tu)(t2 ) − (Tu)(t1 )| ≤ (Tu) (s) ds ≤ M(t2 − t1 ) t1 So, T is equicontinuous The Arzela-Ascoli Theorem implies that T : C[0, 1] ® C[0, 1] is completely continuous Since t ≤ θ (t) ≤ 1, t Ỵ (0, 1), then inf u(θ (t)) ≥ inf u(t) ≥ γ ||u|| t∈[η,1] (3:2) t∈[η,1] Thus, Lemmas 2.5 and 3.2 show that TP ⊂ P Then, T : P ® P is completely continuous In view of f0 = ∞, there exists a constant r1 >0 such that f(u) ≥ δ1u for < u < r1, where δ1 >0 satisfies ηδ1 γ (1 − βη) (1 − s)α−1 a(s)ds ≥ (α) η (3:3) Take u Ỵ P , such that ||u|| = r1 Then, we have ||Tu|| ≥ Tu(η) η =− (η − s)α−1 a(s)f (u(θ (s)))ds + (α) − βη η β − − βη =− 1 − βη =− − βη η + − βη ≥− 1 − βη η + − βη ≥ ≥ ≥ η − βη η − βη η − βη η η η η η η(1 − s)α−1 a(s)f (u(θ (s)))ds (α) η(η − s)α−1 a(s)f (u(θ (s)))ds (α) (η − s)α−1 η a(s)f (u(θ (s)))ds + (α) − βη (η − s)α−1 η a(s)f (u(θ (s)))ds + (α) − βη (1 − s)α−1 a(s)f (u(θ (s)))ds (α) η (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (η − s)α−1 a(s)f (u(θ (s)))ds + (α) − βη η (η − ηs)α−1 a(s)f (u(θ (s)))ds (α) (3:4) (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−1 a(s)δ1 u(θ (s))ds (α) (1 − s)α−1 a(s)δ1 γ ||u||ds (α) η η η ηδ1 γ = (1 − βη) 1 η (1 − s)α−1 a(s)ds||u|| ≥ ||u|| (α) Let Ωr1 = {u Ỵ C 0[1] | ||u|| r1 such that f(u) ≤ δ2u for u ≥ R, where δ2 >0 satisfies δ2 (1 − βη) (1 − s)α−1 a(s)ds ≤ (α) (3:5) We consider the following two cases Case one f is bounded, which implies that there exists a constant r1 >0 such that f (u) ≤ r1 for u Ỵ [0, ∞) Now, we may choose u Ỵ P such that ||u|| = r2, where r2 ≥ max {μ, R} Then we have t Tu (t) = − − ≤ ≤ (t − s)α−1 a(s)f (u(θ (s)))ds + (α) − βη η β − βη 1 − βη t(1 − s)α−1 a(s)f (u(θ (s)))ds (α) t(η − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−1 a(s)f (u(θ (s)))ds (α) r1 (1 − βη) (1 − s)α−1 a(s)ds (α) μ ≤ ρ2 = ||u|| Case two f is unbounded, which implies then there exists a constant ρ2 > R γ >R such that f(u) ≤ f(r2) for < u ≤ r2 (note that f Ỵ C([0, ∞), [0, ∞)) Let u Ỵ P such that ||u|| = r2, we have Tu (t) ≤ ≤ ≤ 1 − βη 1 − βη 1 − βη δ2 = (1 − βη) 1 (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−1 a(s)f (ρ2 )ds (α) (1 − s)α−1 a(s)δ2 ρ2 ds (α) (1 − s)α−1 a(s)ds||u|| (α) ≤ ||u|| Hence, in either case, we may always let Ωr2 = {u Ỵ C[0, 1] | ||u|| 0 such that f (u) ≤ τ1u for < u < r1, where τ1 >0 satisfies Wang et al Advances in Difference Equations 2011, 2011:2 http://www.advancesindifferenceequations.com/content/2011/1/2 τ1 (1 − βη) Page of 11 (1 − s)α−1 a(s)ds ≤ (α) (3:6) Take u Ỵ P, such that ||u|| = r1 Then, we have Tu (t) ≤ 1 − βη ≤ − βη (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−1 a(s)τ1 u(θ (s))ds (α) 0 τ1 ≤ (1 − βη) (3:7) (1 − s)α−1 a(s)ds||u|| (α) ≤ ||u|| Let Ω1 = {u Ỵ C[0, 1] | ||u|| r1 such that f(u) ≥ τ2u for u ≥ r2, where τ2 >0 satisfies τ2 ηγ (1 − βη) η (1 − s)α−1 a(s)ds ≥ (α) Let Ω = {u Ỵ C[0, 1] | ||u|| (3:8) r2 > r2, then, u Î P and ||u|| = r2 γ implies inf u(θ (t)) ≥ γ ||u|| > r2 , t∈[η,1] and so ||Tu|| ≥ Tu(η) η ≥ − βη η ≥ − βη η ≥ − βη = (1 − s)α−1 a(s)f (u(θ (s)))ds (α) (1 − s)α−1 a(s)τ2 u(θ (s))ds (α) (1 − s)α−1 a(s)τ2 γ ||u||ds (α) η η η τ2 ηγ (1 − βη) η (1 − s)α−1 a(s)ds||u|| ≥ ||u|| (α) This shows that ||Tu|| ≥ ||u|| for u Ỵ P ∩ ∂Ω2 Therefore, by the second part of Guo-Krasnoselskii fixed point theorem, we can conclude that (1.1) has at least one positive solution u ∈ P ∩ ( ¯ \ ) Wang et al Advances in Difference Equations 2011, 2011:2 http://www.advancesindifferenceequations.com/content/2011/1/2 Page of 11 Examples Example 4.1 Consider the fractional differential equation C Dα u(t) + e−t f (u(θ (t))) = 0, < t < 1, u(0) = u (0) = 0, βu(η) = u(1), (4:1) where