Existence of Positive Solutions for a Fourth-Order Periodic Boundary Value Problem

and Applied Analysis 3 The strongly maximum principle implies that the fourth-order linear boundary value problem LBVP L4 u : u 4 − βu′′ αu 0, 0 ≤ t ≤ 1, u i 0 − u i 1 0, i 0, 1, 2, u 3 0 − u 3 1 1 1.7 has a unique positive solution Φ : 0, 1 → 0,∞ , see 3, Lemma 3 . This function has been introduced in 2, Lemma 2.1 and Remark 2.1 . Let I 0, 1 , and set σ mint∈I Φ t maxt∈I Φ t , M maxt∈I |Φ′′ t | mint∈I Φ t . 1.8 Let f : I × R × R → R be continuous. To be convenient, we introduce the notations f0 lim inf u→ 0 min |v|≤Mu,t∈I ( f t, u, v u ) , f0 lim sup u→ 0 max |v|≤Mu,t∈I ( f t, u, v u ) , f∞ lim inf u→ ∞ min |v|≤Mu,t∈I ( f t, u, v u ) , f∞ lim sup u→ ∞ max |v|≤Mu,t∈I ( f t, u, v u ) . 1.9 Our main result is as follows. Theorem 1.1. Let f : 0, 1 × R × R → R be continuous, and let the assumption 1.3 hold. If f satisfies one of the following conditions: (F1) f0 < α, f∞ > α, (F2) f0 > α, f∞ < α, then PBVP 1.1 has at least one positive solution. Clearly, Theorem 1.1 is an extension of Theorem A. Since that α is an eigenvalue of linear eigenvalue problem u 4 − βu′′ αu λu, 1.10 with periodic boundary condition, if one inequality in F1 or F2 of Theorem 1.1 is not true, the existence of solution to PBVP 1.1 cannot be guaranteed. Hence, F1 and F2 are the optimal conditions for the existence of the positive of PBVP 1.1 . In Theorem 1.1, the condition F1 allows that f t, u, v may be superlinear growth on u and v, for example, f t, u, v u2 v2, and the condition F2 allows that f t, u, v may be sublinear growth on u and v, for example, f t, u, v 3 √ u2 v2. 4 Abstract and Applied Analysis The proof of Theorem 1.1 is based on the theory of the fixed point index in cones. Since the nonlinearity f of PBVP 1.1 contains u′′, the argument of Theorem A in 3 is not applicable to Theorem 1.1. We will prove Theorem 1.1 by choosing a proper cone of C2 I in Section 3. Some preliminaries to discuss PBVP 1.1 are presented in Section 2. 2. Preliminaries LetC I be the Banach space of all continuous functions on the unit interval I 0, 1 with the norm ‖u‖C max0≤t≤ω |u t |. Let C I denote the cone of all nonnegative functions in C I . Generaly, for n ∈ N, we use C I to denote the Banach space of the nth-order continuous differentiable functions on I with the norm ‖u‖Cn ∑n k 1 ‖u k ‖C. In C2 I , we define a new norm by ‖u‖C02 ‖u‖C ∥ ∥u′′ ∥ ∥ C. 2.1 Then ‖u‖C02 is equivalent to ‖u‖C2 . In fact, for every u ∈ C2 I , it is clear that ‖u‖C02 ≤ ‖u‖C2 . On the other hand, by the Lagrange mean-value theorem, there exists ξ ∈ 0, 1 such that u 1 − u 0 u′ ξ . For t ∈ I, we have ∣u′ t ∣ ≤ ∣u′ t − u′ ξ ∣ ∣u′ ξ ∣ ∣∣∣∣ ∫ t ξ u′′ s ds ∣∣∣∣ |u 1 − u 0 | ≤ ∫1 0 ∣u′′ s ∣ ds |u 1 | |u 0 | ≤ ∥∥u′′∥∥C 2‖u‖C ≤ 2‖u‖C02 . 2.2 Hence, ‖u‖C ≤ 2‖u‖C02 . By this, we have ‖u‖C2 ‖u‖C ∥u′ ∥ C ∥u′′ ∥ C ‖u‖C02 ∥u′ ∥ C ≤ 3‖u‖C02 . 2.3 Therefore, the norms ‖u‖C02 and ‖u‖C2 are equivalent. Let α, β ∈ R satisfy the assumption 1.3 . For h ∈ C I , we consider the fourth-order linear periodic boundary value problem LPBVP u 4 t − βu′′ t αu t h t , 0 ≤ t ≤ 1, u i 0 u i 1 , i 0, 1, 2, 3. 2.4 Let Φ t be the unique positive solution of LBVP 1.7 , and set G t, s ⎧ ⎨ ⎩ Φ t − s , 0 ≤ s ≤ t ≤ 1, Φ 1 t − s , 0 ≤ t < s ≤ 1. 2.5 By 3, Lemma 1 , we have the following result. Abstract and Applied Analysis 5 Lemma 2.1. Let α, β ∈ R satisfy the assumption 1.3 . Then for every h ∈ C I , LPBVP 2.4 has a unique solution u t which is given byand Applied Analysis 5 Lemma 2.1. Let α, β ∈ R satisfy the assumption 1.3 . Then for every h ∈ C I , LPBVP 2.4 has a unique solution u t which is given by u t ∫1 0 G t, s h s ds : Sh t , t ∈ R. 2.6 Moreover, S : C I → C4 I is a linear bounded operator. Let σ and M be the positive constants given by 1.8 . Choose a cone K in C2 I by K { u ∈ C2 I | u t ≥ σ‖u‖C, ∣ ∣u′′ t ∣ ∣ ≤ M|u t |, t ∈ I } . 2.7 We have the following. Lemma 2.2. Let α, β ∈ R satisfy the assumption 1.3 . Then for every h ∈ C I , the solution of LPBVP 2.4 u Sh ∈ K. Namely, S C I ⊂ K. Proof. Let h ∈ C I , u Sh. For every t ∈ I, from 2.6 it follows that 0 ≤ u t ∫1 0 G t, s h s ds ≤ max t∈I Φ t ∫1 0 h s ds, 2.8