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Multiple positive solutions for nonlinear third-order boundary value problems in Banach spaces

This paper deals with the positive solutions of nonlinear boundary value problems in Banach spaces. By using fixed point index theory, some sufficient conditions for the existence of at least one or two positive solutions to boundary value problems in Banach spaces are obtained. An example illustrating the main results is given. | Turk J Math 33 (2009) , 55 – 63. ¨ ITAK ˙ c TUB doi:10.3906/mat-0712-18 Multiple Positive Solutions for Nonlinear Third-Order Boundary Value Problems in Banach Spaces Feng Wang, Hai-hua Lu and Fang Zhang Abstract This paper deals with the positive solutions of nonlinear boundary value problems in Banach spaces. By using fixed point index theory, some sufficient conditions for the existence of at least one or two positive solutions to boundary value problems in Banach spaces are obtained. An example illustrating the main results is given. Key Words: Positive solutions; boundary value problem; Banach spaces; fixed point index. 1. Introduction In this paper, we consider the following boundary value problem (BVP) for third-order differential equations in a Banach space E : u (t) + f(t, u(t)) = θ, 0≤t≤1 (1.1) subject to the boundary conditions u(0) = u (0) = θ, u (1) = αu (η), (1.2) where 0 0 1 1 such that for t ∈ [0, 1] and bounded D ⊂ P , and there exists a h ∈ C[[0, 1], R+] with 0 h(t)dt (γ)−1 such that f(t, u) ≥ k(t)u0 for t ∈ [ αη , η] and u ≥ u0 . (H6 ) There exist u0 ∈ P \{θ} , and k ∈ C[[ αη , η], R] with η η α k(t)Φ(t)dt ≥ (γ)−1 such that f(t, u) ≥ k(t)u0 for t ∈ [ αη , η] and u ≥ u0 . Remark 2.1 It is clear that (H2 ) is weaker than (H1 ), and (H6 ) is weaker than (H5 ). Also, condition (H1 ) is satisfied automatically when E is finite dimensional. Now we define 1 G(t, s)f(s, u(s))ds, ∀ u ∈ P. (Au)(t) = (2.4) 0 Then (Au)(t) ≥ θ, t ∈ [0, 1] , and using the Lebesgue dominated convergence theorem we know that (Au)(t) is continuous on [0, 1] , hence the integral operator A : P → P . Further, we can easily show that (i) If u ∈ P , then (Au) (t) = −f(t, u), t ∈ [0, 1] , hence Au ∈ P ∩ C 3 [[0, 1], E] . 57 WANG, LU, ZHANG (ii) If u ∈ P satisfies Au = u , then u is a solution of the BVP (1.1) − (1.2). Therefore, the BVP (1.1) − (1.2) is equivalent to the operator equation Au = u, u ∈ P . Lemma 2.3 ([15,16]) Let D be a bounded set of E