CUBO A Mathematical Journal Vol.11, No¯ 03, (79–100). August 2009 Positive Solutions for Systems of Three-point Nonlinear Boundary Value Problems with Deviating Arguments J. Henderson Department of Mathematics, Baylor University, Waco, Texas 76798-7328, USA email: johnny henderson@baylor.edu and S.K. Ntouyas, I.K. Purnaras Department of Mathematics, University of Ioannina, 451 10, Ioannina, Greece emails: sntouyas@cc.uoi.gr, ipurnara@cc.uoi.gr ABSTRACT Existence of eigenvalues yielding positive solutions for a system of two second order delay differential equations along with boundary conditons is established. The results are obtained by the use of a Guo-Krasnoselskii fixed point theorem in cones. RESUMEN Es establecida la existencia de autovalores produciendo soluciones positivas para un sistema de dos ecuaciones diferenciales de segundo orden con retardo, con condiciones de frontera. Los resultados son obtenidos mediante el uso del Teorema de punto fijo de Guo-Krasnoselskii en conos. 80 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) Key words and phrases: Three-point boundary value problem, system of differential equations, eigenvalue problem, positive solutions, deviating arguments. Math. Subj. Class.: 34B18, 34A34. 1 Introduction Consider the three-point boundary value problem system consisting of the second order delay differential equations, u ′′ (t) + λa(t)f (u(σ1(t)), v(σ2(t))) = 0, 0 < t < 1, v ′′ (t) + µb(t)g (u(τ1(t)), v(τ2(t))) = 0, 0 < t < 1, (1) along with the conditions, u(0) = 0, u(1) = αu(η), v(0) = 0, v(1) = αv(η), u(t) = φ1(t), v(t) = φ2(t), −r ≤ t ≤ 0, (2) where 0 < η < 1, 0 < α < 1/η, −r = mint∈[0,1] σi(t) = mint∈[0,1] τi(t), i = 1, 2, and φ1, φ2 : [−r, 0] → R+ are continuous functions, with φ1(0) = φ2(0) = 0. Our interest in this paper is to investigate the existence of eigenvalues λ and µ that yield positive solutions to the associated boundary value problem, (1), (2). We assume that (A) f, g ∈ C(R+ × R+, R+); (B) a, b ∈ C([0, 1], R+), and each does not vanish identically on any subinterval; (C) σi, τi : [0, 1] → [−r, 1], i = 1, 2 are continuous functions; (D) All of f0 := lim u+v→0+ f (u, v) u + v , g0 := lim u+v→0+ g(u, v) u + v f∞ := lim u+v→∞ f (u, v) u + v , g∞ := lim u+v→∞ g(u, v) u + v exist as positive real numbers; (E) There exist an η∗ ∈ [η, 1] such that σi(s), τi(s) ∈ [η, 1] for all s ∈ [η∗, 1], i = 1, 2. We say that a pair (u, v) ∈ C ([−r, 1]) is a solution of the boundary value problem (BVP for short) (1), (2) if, u and v are twice continuously differentiable on (0, 1), u(t) = φ1(t), v(t) = φ2(t), CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 81 for −r ≤ t ≤ 0, (u, v) satisfies (1) for all t ∈ (0, 1), and u(0) = 0, u(1) = αu(η) and v(0) = 0, v(1) = αv(η). For several years now, there has been a great deal of activity in studying positive solutions of boundary value problems for ordinary differential equations. Interest in such solutions is high from both a theoretical sense [4, 7, 10, 13, 20] and as applications for which only positive solutions are meaningful [1, 5, 14, 15]. These considerations are caste primarily for scalar problems, but good attention has been given to boundary value problems for systems of differential equations [11, 12, 17, 19, 21]. The existence of positive solutions for nonlocal three-point boundary value problems has been studied extensively in recent years. For some appropriate references we refer the reader to [17], [18]. Recently, Benchohra et al. [2] and Henderson and Ntouyas [8] studied the existence of positive solutions for systems of nonlinear eigenvalue problems, while Henderson and Ntouyas [9] obtained results for the case of systems with three-point nonlocal boundary conditions. The purpose of this paper is to extend the results given in [9] to the case where delays may appear in the equations of the system (1), (2). The main tool in this paper is an application of the Guo-Krasnosel’skii fixed point theorem for operators leaving a Banach space cone invariant [7]. A Green’s function plays a fundamental role in defining an appropriate operator on a suitable cone. Since, in our problem, we cannot express system (1), (2) as a single operator equation, the method used for example in [9] is not applicable here. This difficulty can be overcome by employing a method proposed by Dunninger and Wang in [3]. 2 Some preliminaries Before we state and prove our main result, we recall some useful facts that will be used in the sequel. Concerning the boundary value problem u ′′ (t) + y(t) = 0, 0 < t < 1, (3) u(0) = 0, u(1) = αu(η), (4) we have the following two lemmas. Lemma 2.1. [6] Let (A), (B) and (C) hold and assume that 0 < η < 1 and 0 < α < 1/η. Then, for any y ∈ C[0, 1] the BVP (3), (4) has a unique solution, u(t) = ∫ 1 0 k(t, s)y(s)ds, t ∈ [0, 1], 82 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) where k(t, s) : [0, 1] × [0, 1] → R+ is the Green function defined by k(t, s) =                          t(1 − s) 1 − αη − αt(η − s) 1 − αη − (t − s), 0 ≤ s ≤ t ≤ 1 and s ≤ η, t(1 − s) 1 − αη − αt(η − s) 1 − αη , 0 ≤ t ≤ s ≤ η, t(1 − s) 1 − αη , 0 ≤ t ≤ s ≤ 1 and η ≤ s, t(1 − s) 1 − αη − (t − s), η ≤ s ≤ t ≤ 1. (5) Lemma 2.2. [16] Let (A), (B) and (C) hold and assume that 0 < α < 1/η. Then, the unique solution of the problem (3), (4) satisfies inf t∈[η,1] u(t) ≥ γ‖u‖, where γ := min { αη, η, α(1 − η) 1 − αη } . From Lemma 2.1 and the analytical expression of k, it follows that u can be written as u(t) = 1 1 − αη ∫ 1 0 (1 − s)y(s)ds − αt 1 − αη ∫ η 0 (η − s)y(s)ds − ∫ t 0 (1 − s)y(s)ds from which it follows that u(t) ≤ 1 1 − αη ∫ 1 0 (1 − s)y(s)ds, for all t ∈ [0, 1], (6) and u(η) ≥ η 1 − αη ∫ 1 η (1 − s)y(s)ds. (7) We note that a pair (u(t), v(t)) is a solution of the eigenvalue problem (1), (2) if, and only if, u(t) = φ1(t), v(t) = φ2(t) for −r ≤ t ≤ 0, and u(t) = λ ∫ 1 0 k(t, s)a(s)f (u(σ1(s)), v(σ2(s)))ds, 0 ≤ t ≤ 1, v(t) = µ ∫ 1 0 k(t, s)b(s)g(u(τ1(s)), v(τ2(s)))ds, 0 ≤ t ≤ 1. The main tool in determining values of the parameters λ and µ, for which positive (with respect to a cone) solutions of the BVP (1), (2) exist, is the following fixed point theorem. Theorem 2.1. [7] Let B be a Banach space, and let P ⊂ B be a cone in B. Assume Ω1 and Ω2 are open subsets of B with 0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2, and let T : P ∩ (Ω2 \ Ω1) → P be a completely continuous operator such that, either CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 83 (i) ||T u|| ≤ ||u||, u ∈ P ∩ ∂Ω1, and ||T u|| ≥ ||u||, u ∈ P ∩ ∂Ω2, or (ii) ||T u|| ≥ ||u||, u ∈ P ∩ ∂Ω1, and ||T u|| ≤ ||u||, u ∈ P ∩ ∂Ω2. Then T has a fixed point in P ∩ (Ω2 \ Ω1). 3 Positive solutions in a cone In this section, we apply Theorem 2.1 to obtain solutions in a cone (that is, positive solutions) of (1), (2). For our construction, we let X = C([−r, 1], R+) × C([−r, 1], R+) with norm ‖(u, v)‖ = ‖u‖ + ‖v‖ where ‖u‖ = supt∈[−r,1] |u(t)|. Then (X, ‖ · ‖) is a Banach space. We will make use of the cone P ⊂ X defined by P = { (u, v) : (u, v) ∈ X : u, v ≥ 0 on [−r, 1], min t∈[η,1] [u(t) + v(t)] ≥ γ[‖u‖ + ‖v‖] } , where γ > 0 is the positive constant defined in Lemma 2.2. For our first result, define positive numbers L1 and L2 by L1 := max { 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − r)a(r)f∞dr ]−1 , 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − r)b(r)g∞dr ]−1 } , and L2 := min { 1 2 [ 1 1 − αη ∫ 1 0 (1 − r)a(r)f0dr ]−1 , 1 2 [ 1 1 − αη ∫ 1 0 (1 − r)b(r)g0dr ]−1 } . Theorem 3.1. Assume that conditions (A), (B), (C), (D) and (E) hold. Then, for each λ, µ satisfying L1 < λ, µ < L2, (8) there exists a pair (u, v) satisfying (1), (2) such that u(t) > 0 and v(t) > 0 on (0, 1). Proof. Let A, B : X → X and F : X → X be the integral operators defined by A(u, v)(t) =        φ1(t), −r ≤ t ≤ 0, λ ∫ 1 0 k(t, s)a(s)f (u(σ1(s)), v(σ2(s))) ds, 0 ≤ t ≤ 1, 84 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) B(u, v)(t) =        φ2(t), −r ≤ t ≤ 0, µ ∫ 1 0 k(t, s)b(s)g (u(τ1(s)), v(τ2(s))) ds, 0 ≤ t ≤ 1, F (u, v)(t) = (A(u, v)(t), B(u, v)(t)) , t ∈ [−r, 1]. Then seeking solutions to our BVP (1), (2) is equivalent to looking for fixed points of the equation F (u, v) = (u, v) in the Banach space X. Choose some (u, v) ∈ P. Then by Lemma 2.2 we have inf t∈[η,1] A(u, v)(t) ≥ γ‖A(u, v)‖, inf t∈[η,1] B(u, v)(t) ≥ γ‖B(u, v)‖ and thus inf t∈[η,1] [A(u, v)(t) + B(u, v)(t)] ≥ inf t∈[η,1] A(u, v)(t) + inf t∈[η,1] B(u, v)(t) ≥ γ [‖A(u, v)‖ + ‖B(u, v)||] = γ‖(A(u, v), B(u, v)‖ which implies that F (P) ⊂ P for every (u, v) ∈ P. As A and B are integral operators, it is not difficult to see that using standard arguments we may conclude that both A and B are completely continuous; hence F is a completely continuous operator. Let λ and µ be as in (8), and choose an ǫ > 0 such that max { 1 2 [ γη 1 − αη ∫ 1 η (1 − r)a(r)(f∞ − ǫ)dr ]−1 , 1 2 [ γη 1 − αη ∫ 1 η (1 − r)b(r)(g∞ − ǫ)dr ]−1 } ≤ λ, µ and λ, µ ≤ min { 1 2 [ 1 1 − αη ∫ 1 0 (1 − r)a(r)(f0 + ǫ)dr ]−1 , 1 2 [ 1 1 − αη ∫ 1 0 (1 − r)b(r)(g0 + ǫ)dr ]−1 } . From the definition of f0 and g0, there exists an H1 > 0 such that f (u, v) ≤ (f0 + ǫ)(u + v) for u, v ∈ P with 0 < u, v < H1, CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 85 and g(u, v) ≤ (g0 + ǫ)(u + v) for u, v ∈ P with 0 < u, v < H1. Set Ω1 = {(u, v) ∈ X : ‖(u, v)‖ < H1} . Now let (u, v) ∈ P ∩ ∂Ω1, i.e., let (u, v) ∈ P with ‖ (u, v) ‖ = H1. Then, in view of the inequality (6) we have A(u, v)(t) ≤ λ t 1 − αη ∫ 1 0 (1 − s)a(s)f (u(σ1(s), v(σ2(s))) ds ≤ λ 1 1 − αη ∫ 1 0 (1 − s)a(s)(f0 + ǫ) [u(σ1(s) + v(σ2(s))] ds ≤ λ 1 1 − αη ∫ 1 0 (1 − s)a(s)(f0 + ǫ)[‖u‖ + ‖v‖]ds ≤ 1 2 [‖u‖ + ‖v‖] = 1 2 ‖(u, v)‖, and so, ‖A(u, v)‖ ≤ 1 2 ‖(u, v)‖. Similarily, we may take ‖B(u, v)‖ ≤ 1 2 ‖(u, v)‖. Thus, for (u, v) ∈ P ∩ ∂Ω1 it follows that ‖F (u, v)‖ = ‖ (A(u, v), B(u, v)) ‖ = ‖A(u, v)‖ + ‖B(u, v)‖ ≤ 1 2 ‖(u, v)‖ + 1 2 ‖(u, v)‖ = ‖(u, v)‖, that is, ‖F (u, v)‖ ≤ ‖(u, v)‖ for all (u, v) ∈ P ∩ ∂Ω1. Due to the definition of f∞ and g∞, there exists an H2 > 0 such that f (u, v) ≥ (f∞ − ǫ)(u + v) for all u, v ≥ H2, and g(u, v) ≥ (g∞ − ǫ)(u + v) for all u, v ≥ H2. Set H2 = max { 2H1, H2 γ } and define Ω2 = {(u, v) ∈ X : ‖(u, v)‖ < H2} . 86 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) As from our hypothesis on η∗ it follows that inf t∈[η∗,1] [u(σ1(t)) + v(σ2(t))] ≥ γ[‖u‖ + ‖v‖]. (9) By the use of (7), we have for (u, v) ∈ P ∩ ∂Ω2, A(u, v)(η) ≥ λ η 1 − αη ∫ 1 η (1 − s)a(s)f (u(σ1(s)), v(σ2(s))) ds ≥ λ η 1 − αη ∫ 1 η∗ (1 − s)a(s)(f∞ − ǫ) (u(σ1(s)) + v(σ2(s))) ds ≥ λ η 1 − αη ∫ 1 η∗ (1 − s)a(s)(f∞ − ǫ)γ[‖u‖ + ‖v‖]ds ≥ 1 2 ‖(u, v)‖, that is, A(u, v)(t) ≥ 1 2 ‖(u, v)‖ for all t ≥ η and so, A(u, v)(t) ≥ 1 2 ‖(u, v)‖. Similarily, we may take B(u, v)(t) ≥ 1 2 ‖(u, v)‖. Thus, for (u, v) ∈ P ∩ ∂Ω2 it follows that ‖F (u, v)‖ = ‖ (A(u, v), B(u, v)) ‖ = ‖A(u, v)‖ + ‖B(u, v)‖ ≥ 1 2 ‖(u, v)‖ + 1 2 ‖(u, v)‖ = ‖(u, v)‖, that is ‖F (u, v)‖ ≥ ‖(u, v)‖ for all (u, v) ∈ P ∩ ∂Ω2. Applying Theorem 2.1, we obtain that F has a fixed point (u, v) ∈ P ∩ (Ω2 \ Ω1) such that H1 ≤ ‖(u, v)‖ ≤ H2, and so (1), (2) has a positive solution. The proof is complete. � For our next result we define the positive numbers L3 = max { 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − s)a(s)f0ds ]−1 , 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − s)b(s)g0ds ]−1 } and L4 = min { 1 2 [ 1 1 − αη ∫ 1 0 (1 − s)a(s)f∞dr ]−1 , 1 2 [ 1 1 − αη ∫ 1 0 (1 − s)b(s)g∞dr ]−1 } . We are now ready to state and prove our main result. CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 87 Theorem 3.2. Assume that conditions (A), (B), (C), (D) and (E) hold. Then for each λ, µ satisfying L3 < λ, µ < L4, (10) there exists a pair (u, v) satisfying (1), (2) such that u(t) > 0 and v(t) > 0 on (0, 1). Proof. Let λ and µ be as in (10) and choose a sufficiently small ǫ > 0 so that λ, µ ≤ min { 1 2 [ 1 1 − αη ∫ 1 0 (1 − s)a(s)(f∞ + ǫ)ds ]−1 , 1 2 [ 1 1 − αη ∫ 1 0 (1 − s)b(s)(g∞ + ǫ)ds ]−1 } and max { 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − s)a(s)(f0 − ǫ)dr ]−1 , 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − s)b(s)(g0 − ǫ)dr ]−1 } ≤ λ, µ. By the definition of f0 and g0, there exists an H1 > 0 such that f (u, v) ≥ (f0 − ǫ)(u + v) for all u, v with 0 < u, v ≤ H3, and g(u, v) ≥ (g0 − ǫ)(u + v) for all u, v with 0 < u, v ≤ H3. Set Ω1 = {(x, y) ∈ X : ‖(x, y)‖ < H3} and let (u, v) ∈ P ∩ ∂Ω3. In view of (9) and by the use of (7), we find A(u, v)(η) ≥ λ η 1 − αη ∫ 1 η (1 − s)a(s)f (u(σ1(s)), v(σ2(s))) ds ≥ λ η 1 − αη ∫ 1 η∗ (1 − s)a(s)f (u(σ1(s)), v(σ2(s))) ds ≥ λ η 1 − αη ∫ 1 η∗ (1 − s)a(s)(f0 − ǫ) (u(σ1(s)) + v(σ2(s))) ds ≥ λ η 1 − αη ∫ 1 η∗ (1 − s)a(s)(f0 − ǫ)γ[‖u‖ + ‖v‖]ds ≥ 1 2 ‖(u, v)‖, 88 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) that is, ‖A(u, v)‖ ≥ 1 2 ‖(u, v)‖. In a similar manner ‖B(u, v)‖ ≥ 1 2 ‖(u, v)‖. Thus, for an arbitrary (u, v) ∈ P ∩ ∂Ω3 it follows that ‖F (u, v) ‖ = ‖ (A (u, v) , B (u, v)) ‖ = ‖A (u, v) ‖ + ‖B (u, v) ‖ ≥ 1 2 ‖ (u, v) ‖ + 1 2 ‖ (u, v) ‖ = ‖(u, v)‖, and so ‖F (u, v) ‖ ≥ ‖(u, v)‖ for all (u, v) ∈ P ∩ ∂Ω3. Now let us define two functions f ∗, g∗ : [0, ∞) → [0, ∞) by f ∗ (t) = max 0≤u+v≤t f (u, v) and g∗(t) = max 0≤u+v≤t g(u, v). It follows that f (u, v) ≤ f ∗(t) and g(u, v) ≤ g∗(t) for all (u, v) with 0 ≤ u + v ≤ t. It is clear that the functions f ∗ and g∗ are nondecreasing. Also, there is no difficulty to see that lim t→∞ f ∗ (t) t = f∞ and lim t→∞ g ∗ (t) t = g∞. In view of the definitions of f∞ and g∞, there exists an H4 such that f ∗ (t) < (f∞ + ε) t for all t ≥ H4, and g ∗ (t) < (g∞ + ε) t for all t ≥ H4. Set H4 = max { 2H3, H4 γ } , and Ω4 = {(u, v) : (u, v) ∈ P and ‖(u, v)‖ < H4} . Let (u, v) ∈ P ∩ ∂H4 and observe that, by the definition of f ∗, it follows that for any s ∈ [0, 1], we have f (u(σ1(s)), v(σ2(s))) ≤ f ∗ (‖u‖ + ‖v‖) = f ∗ (‖(u, v)‖) . CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 89 In view of the above observation and by the use of inequality (6) we obtain for t ∈ [0, 1] A (u, v) (t) ≤ λ t 1 − αη ∫ 1 0 (1 − s)a(s)f (u(σ1(s)), v(σ2(s))) ds ≤ λ t 1 − αη ∫ 1 0 (1 − s)a(s)f ∗ (‖u‖ + ‖v‖) ds ≤ λ t 1 − αη ∫ 1 0 (1 − s)a(s)(f∞ + ε) (‖u‖ + ‖v‖) dr ≤ λ 1 1 − αη ∫ 1 0 (1 − s)a(s)(f∞ + ε)dr ‖(u, v)‖ ≤ 1 2 ‖(u, v)‖ , which implies ‖A (u, v)‖ ≤ 1 2 ‖(u, v)‖ . In a similar manner, we take ‖B (u, v) ‖ ≤ 1 2 ‖ (u, v) ‖. Thus, for (u, v) ∈ P ∩ ∂Ω4 it follows that ‖F (u, v)‖ = ‖ (A (u, v) , B (u, v)) ‖ = ‖A (u, v) ‖ + ‖B (u, v) ‖ ≤ 1 2 ‖ (u, v) ‖ + 1 2 ‖ (u, v) ‖ = ‖(u, v)‖, and so ‖F (u, v)‖ ≤ ‖(u, v)‖ for all (u, v) ∈ P ∩ ∂Ω4 Applying Theorem 2.1, we obtain that F has a fixed point (u, v) ∈ P ∩ (Ω4 \ Ω3) such that H3 ≤ ‖(u, v)‖ ≤ H4, and so (1), (2) has a positive solution. The proof is complete. � 4 A General Application In this section we apply Theorems 3.1 and 3.2 to the case where each one of the functions f and g is the sum of two (nonlinear) functions of a single argument, i.e., we consider the three-point boundary value system u ′′ (t) + λa(t)[ ˜f1 (u(σ1(t))) + ˜f2 (v(σ2(t)))] = 0, 0 < t < 1, v ′′ (t) + µb(t) [g̃1 (u(τ1(t))) + g̃2 (v(τ2(t)))] = 0, 0 < t < 1, (11) u(0) = 0, u(1) = αu(η), v(0) = 0, v(1) = αv(η), u(t) = φ1(t), v(t) = φ2(t), −r ≤ t ≤ 0, (12) 90 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) where 0 < η < 1, 0 < α < 1/η, r is a positive number, φ1, φ2 : [−r, 0] → R+, with φ1(0) = φ2(0) = 0 and σi, τi : [0, 1] → [−r, 1], i = 1, 2 are continuous functions. We assume that (A1) ˜fi, g̃i ∈ C([0, ∞), [0, ∞)), i = 1, 2; (B1) a, b ∈ C([0, 1], [0, ∞)) and each function does not vanish on any subinterval of [0, 1]; (C1) All of ˜f0 := lim t→0+ ˜fi(t) t , ˜f∞ := lim t→∞ ˜f (t) t , i = 1, 2, g̃0 := lim t→0+ g̃i(t) t , g̃∞ := lim t→∞ g̃i(t) t , i = 1, 2 exist as positive real numbers. We say that a pair (u, v) ∈ C ([−r, 1]) is a solution of the BVP (11), (12) if (i) u(t) = φ1(t), v(t) = φ2(t), for −r ≤ t ≤ 0, (ii) (u, v) satisfies (11) for all t ∈ (0, 1), and (iii) u(0) = v(0) = 0, u(1) = αu(η) and v(1) = αv(η). Before we state our existence results for the BVP (11), (12), we prove an elementary lemma. Lemma 4.1. Let hi : [0, ∞) → [0, ∞), i = 1, 2 be continuous functions for which lim t→0+ hi(t) t = k ∈ (0, ∞) and lim t→∞ hi(t) t = m ∈ (0, ∞) , i = 1, 2. Then for the function ̂h : [0, ∞) × [0, ∞) → [0, ∞) with ̂h (u, v) = h1(u) + h2(v), it holds that lim u+v→0+ ̂h(u, v) u + v = k and lim u+v→∞ ̂h(u, v) u + v = m. Proof. By lim t→0+ hi(t) t = k, i = 1, 2 for an arbitrary ε > 0, there exists a δ > 0 such that (k − ε) u ≤ h1(u) ≤ (k + ε) u for all u ∈ (0, δ) , (k − ε) v ≤ h2(v) ≤ (k + ε) v for all v ∈ (0, δ) , and so, for any (u, v) with u, v ∈ ( 0, δ 2 ) , we have k − ε = (k − ε) u + (k − ε) v u + v ≤ ̂h(u, v) u + v = h1(u) + h2(v) u + v ≤ (k + ε) u + (k + ε) v u + v = k + ε, CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 91 i.e., it holds that ∣ ∣ ∣ ∣ ∣ ̂h(u, v) u + v − k ∣ ∣ ∣ ∣ ∣ ≤ ε for any u, v > 0 with u + v < δ which implies that lim u+v→0+ ̂h(u, v) u + v = k. Now let us assume that lim t→∞ hi(t) t = m ∈ (0, ∞) , i = 1, 2. It follows that, for an arbitrarily small ε > 0, there exists an M0 > 0 such that (m − ε) u ≤ h1(u) ≤ (m + ε) u, for all u > M0, (m − ε) v ≤ h2(v) ≤ (m + ε) v, for all v > Mo. Let u, v ≥ 0 with u + v > 2M0. Then either u > M0 and v > M0 or one of u, v is greater than M0 while the other is less than M0. If u > M0 and v > M0, then by the last two inequalities we have m − ε = (m − ε) u + (m − ε) v u + v ≤ ̂h(u, v) u + v = h1(u) + h2(v) u + v ≤ (m + ε) u + (m + ε) v u + v = m + ε, which implies that ∣ ∣ ∣ ∣ ∣ ̂h(u, v) u + v − m ∣ ∣ ∣ ∣ ∣ ≤ ε for any u, v ≥ 0 with u > M0 and v > M0. (13) Now let us deal with the case that one of the arguments u and v is less than M0 and the other one is (necessarily) greater that M0. We consider only the case u ≤ M0 and v > M0, as the conclusion for the dual case u > M0 and v ≤ M0 follows by similar arguments. Set M ∗ = supu∈[0,M] h1(u). Then, as lim v→∞ M ∗ v = 0 and lim v→∞ mv M0 + v = m, and limv→∞ h2(v) v = m we may consider an M > 2M0 such that M ∗ v < ε 2 and m − ε < mv M0 + v and h2(v) v < ε 2 + m. Then for any u, v ≥ 0 with u ≤ M0 and v > M , we find m − ε ≤ mv M0 + v ≤ ̂h(u, v) u + v = h1(u) + h2(v) u + v ≤ M ∗ + h2(v) v ≤ ε + m, which implies that ∣ ∣ ∣ ∣ ∣ ̂h(u, v) u + v − m ∣ ∣ ∣ ∣ ∣ ≤ ε for any u, v ≥ 0 with u ≤ M0, v > M. (14) 92 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) In view of (13) and (14) we see that for any arbitrarily small positive real number ε, we can always find an M > 0 such that ∣ ∣ ∣ ∣ ∣ ̂h(u, v) u + v − m ∣ ∣ ∣ ∣ ∣ ≤ ε for any u, v > 0 with u + v > 2M. Consequently, it holds lim u+v→∞ ̂h(u, v) u + v = m, which completes the proof of the lemma. � Applying our main results to the case of the BVP (11), (12), we obtain the following two theorems. Theorem 4.1. Assume that conditions (A1), (B1), (C1), (C) and (E) hold. Then, for any λ, µ satisfiyng L1 < λ, µ < L2, (15) the BVP (11), (12) has at least one solution (u, v) such that u(t) > 0 and v(t) > 0 on (0, 1), where we have set L1 = max { 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − r) a(r) ˜f∞dr ]−1 , 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − r) b(r)g̃∞dr ]−1 } and L2 = min { 1 2 [ 1 1 − αη ∫ 1 0 (1 − r) a(r) ˜f0dr ]−1 , 1 2 [ 1 1 − αη ∫ 1 0 (1 − r) b(r)g̃0dr ]−1 } . Theorem 4.2. Assume that conditions (A1), (B1), (C1), (C) and (E) hold. Then, for any λ, µ satisfiyng L3 < λ, µ < L4, (16) there exists a pair (u, v) satisfying the BVP (11), (12) such that u(t) > 0 and v(t) > 0 on (0, 1), where L3 = max { 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − s)a(s) ˜f0ds ]−1 , 1 2 [ γη 1 − αη ∫ 1 η∗ (1 − s)b(s)g̃0ds ]−1 } and L4 = min { 1 2 [ 1 1 − αη ∫ 1 0 (1 − s)a(s) ˜f∞dr ]−1 , 1 2 [ 1 1 − αη ∫ 1 0 (1 − s)b(s)g̃∞dr ]−1 } . CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 93 5 Examples In this section, we present some examples that illustrate the breadth of our results. In particular, we give two examples, from which the first one concerns our general application while the second one concerns Theorems 3.1 and 3.2. Example 5.1. For the sake of simplicity, we assume that a = b, f1 = f2 and g1 = g2, σ1 = σ2, τ1 = τ2, i.e., we consider the BVP u ′′ (t) + λa(t)[ ˜f (u(σ(t))) + ˜f (v(σ(t)))] = 0, 0 < t < 1, v ′′ (t) + µa(t) [g̃ (u(τ (t))) + g̃ (v(τ (t)))] = 0, 0 < t < 1, (17) u(0) = 0, u(1) = 2u ( 1 3 ) , v(0) = 0, v(1) = 2v ( 1 3 ) , u(t) = φ1(t), v(t) = φ2(t), −r ≤ t ≤ 0, (18) where ˜f (t) = p1(t) + q1 sin (t), t ∈ R, g̃(t) = p2(t) + q2 sin (t), t ∈ R, with pi, pi + qi > 0, i = 1, 2, φ1, φ2 : [−r, 0] → R+, and σ, τ : [0, 1] → [− 14 , 1] are given by σ(t) =            √ t, t ∈ [0, 1/4] , 1 2 , t ∈ [1/4, 1/2] , t, t ∈ [1/2, 3/4] , 1 2 ( t + 3 4 ) , t ∈ [3/4, 1] , and τ (t) = t − 1 4 , t ∈ [0, 1]. It is not difficult to see that the argument σ is advanced on the interval [0, 1/4] (nonconstant on [0, 1/4] and constant on [1/4, 1/2]), retarded on the interval [3/4, 1] while neither retarded nor advanced on the interval [1/4, 1/2]. By the definition of ˜f and g̃ we may verify that ˜f∞ = p1, g̃∞ = p2, ˜f0 = p1 + q1 g̃0 = p2 + q2. As α = 2 and η = 1 3 we find γ := min { αη, η, α(1 − η) 1 − αη } = min { 2 3 , 1 3 , 2 ( 1 − 1 3 ) 1 − 2 3 } = 1 3 . 94 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) Note that σ(t) ≥ η = 1 3 , for all t ∈ [ 1 9 , 1 ] , while τ (t) = t − 1 4 ≥ η = 1 3 , for all t ∈ [ 7 12 , 1 ] , and so η ∗ = 7 12 . Thus γη 1 − αη ∫ 1 η∗ (1 − r) a(r) ˜f∞dr = 1 3 · 1 3 1 − 2 3 ∫ 1 7 12 (1 − r) a(r)p1dr = 1 3 p1 ∫ 1 7 12 (1 − r) a(r)dr and γη 1 − αη ∫ 1 η∗ (1 − r) b(r)g̃∞dr = 1 3 p2 ∫ 1 7 12 (1 − r) a(r)dr. Hence L1 = max    1 2 [ 1 3 p1 ∫ 1 7 12 (1 − r) a(r)dr ] −1 , 1 2 [ 1 3 p2 ∫ 1 7 12 (1 − r) a(r)dr ] −1    = 3 2 min {p1, p2} ∫ 1 7 12 (1 − r) a(r)dr and L2 = min { 1 2 [ 1 1 − αη ∫ 1 0 (1 − r) a(r) ˜f0dr ]−1 , 1 2 [ 1 1 − αη ∫ 1 0 (1 − r) a(r)g̃0dr ]−1 } = 1 6 min { [ ∫ 1 0 (1 − r) a(r) (p1 + q1) dr ]−1 , [ ∫ 1 0 (1 − r) a(r) (p2 + q2) dr ]−1 } = 1 6 max {p1 + q1, p2 + q2} ∫ 1 0 (1 − r) a(r)dr . Therefore, assuming that p1, q1, p2, q2 have been chosen so that 3 2 < min {p1, p2} ∫ 1 7 12 (1 − r) a(r)dr and 9 max {p1 + q1, p2 + q2} ∫ 1 0 (1 − r) a(r)dr < min {p1, p2} ∫ 1 7 12 (1 − r) a(r)dr, it follows that L1 < L2, and from Theorem 4.1, we derive that, for any λ, µ satisfiyng L1 < λ, µ < L2, the BVP (17), (18) has at least one solution (u, v) such that u(t) > 0 and v(t) > 0 on (0, 1). CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 95 Example 5.2. Consider the BVP u ′′ (t) + λa(t)f (u(σ(t)), v(σ(t))) = 0, 0 < t < 1, v ′′ (t) + µb(t)g (u(τ (t)), v(τ (t))) = 0, 0 < t < 1, (19) u(0) = 0, u(1) = 2u ( 1 3 ) , v(0) = 0, v(1) = 2v ( 1 3 ) , u(t) = φ1(t), v(t) = φ2(t), −1 ≤ t ≤ 0, (20) where φ1, φ2 : [−1, 0] → R+ with φ1(0) = 0 = φ2(0), and σ, τ : [0, 1] → [−1, 1] are given by σ(t) =    − sin (3πt) , t ∈ [0, 1/3] , 9 4 ( −t2 + 2t − 5 9 ) , t ∈ [1/3, 1], and τ (t) = √ t, t ∈ [0, 1]. As α = 2 and η = 1 3 , we find γ := min { αη, η, α(1 − η) 1 − αη } = min { 2 3 , 1 3 , 2 ( 1 − 1 3 ) 1 − 2 3 } = 1 3 . Since σ(t) ≥ η = 1 3 is equivalent to 9 4 ( −t2 + 2t − 5 9 ) ≥ 1 3 , from which σ(t) ≥ 1 3 for all t ∈ [ 1 − 2 3 √ 2 3 , 1 ] , while τ (t) = √ t ≥ η = 1 3 , for all t ∈ [ 1 9 , 1 ] , we conclude that η ∗ = 1 − 2 3 √ 2 3 . We mention that the argument τ is advanced while the argument σ is retarded on [ 0, 5 9 ] and delayed on [ 5 9 , 1 ] . 96 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) Now we calculate the positive numbers L3 and L4. As in Example 5.1, we have α = 2, η = 1 3 and γ = 1 3 as well as γη 1 − αη = 1 3 and 1 1 − αη = 3. We find L3 = 3 2 min { ∫ 1 1− 2 3 √ 2 3 (1 − s)a(s) ˜f0ds, ∫ 1 1− 2 3 √ 2 3 (1 − s)b(s)g̃0ds } and L4 = min          1, 1 6 max { ∫ 1 0 (1 − s)a(s) ˜f∞dr, ∫ 1 0 (1 − s)b(s)g̃∞dr }          . Applying Theorem 4.1, we find that if 3 2 < min { ∫ 1 1− 2 3 √ 2 3 (1 − s)a(s) ˜f0ds, ∫ 1 1− 2 3 √ 2 3 (1 − s)b(s)g̃0ds } and 9 max { ∫ 1 0 (1 − s)a(s) ˜f∞dr, ∫ 1 0 (1 − s)b(s)g̃∞dr } < min { ∫ 1 1− 2 3 √ 2 3 (1 − s)a(s) ˜f0ds, ∫ 1 1− 2 3 √ 2 3 (1 − s)b(s)g̃0ds } , then for any λ, µ satisfiyng L3 < λ, µ < L4, there exists a pair (u, v) satisfying the BVP (19), (20) such that u(t) > 0 and v(t) > 0 on (0, 1). 6 Remarks (1) Similar results to those of Theorems 3.1 and 3.2 can be proved for the following system of two point boundary value problems with deviating arguments u ′′ (t) + λa(t)f (u(σ1(t)), v(σ2(t))) = 0, 0 < t < 1, v ′′ (t) + µb(t)g (u(τ1(t)), v(τ2(t))) = 0, 0 < t < 1, (21) αu(0) − βu′(0) = 0, γu(1) + δu′(1) = 0, αv(0) − βv′(0) = 0, γv(1) + δv′(1) = 0, u(t) = φ1(t), v(t) = φ2(t), −r ≤ t ≤ 0, (22) where α, β, γ, δ ≥ 0 with α + β + γ + δ > 0, ρ = γβ + αγ + αδ > 0. CUBO 11, 3 (2009) Positive Solutions for Three-point BVPs 97 (2) Nondecreasingness may be used to give a sufficient condition that yields the existence of a positive number η∗ such as the one described in Theorem 3.1. It is not difficult to see that, if σi(t) ≤ t, τi(t) ≤ t, for all t ∈ [0, 1], σi and τi are nondecreasing and τi(1) > η, σi(1) > η, then there always exists an η∗ ∈ [η, 1] such that mins∈[η∗,1] {σi(s)} ∈ [η, 1], mins∈[η∗,1] {τi(s)} ∈ [η, 1], i = 1, 2. (3) A requirement equivalent to the one in Theorems 3.1 and 3.2 is the following: there exists an η∗ ∈ [η, 1] such that mins∈[η∗,1] {σi(s)} ∈ [η, 1], mins∈[η∗,1] {τi(s)} ∈ [η, 1], i = 1, 2. (4) In the case of advanced arguments σi(t) > t, τi(t) > t for all t ∈ [0, 1], i = 1, 2, inequality (9) also holds, since inf t∈[η,1] u(σ(t)) ≥ inf t∈[η,1] u(t) ≥ γ‖u‖. Consequently we can deduce similar results to those of Theorems 3.1 and 3.2 for the case of advanced arguments. (5) We can easily find necessary conditions in order to have L1 < L2 and L3 < L4. For example, L1 < L2 gives max { [ ∫ 1 η∗ (1 − r) a(r)f∞dr ]−1 , [ ∫ 1 η∗ (1 − r) b(r)g∞dr ]−1 } < γη min { [ ∫ 1 0 (1 − r) a(r)f0dr ]−1 , [ ∫ 1 0 (1 − r) b(r)g0dr ]−1 } and 1 min { ∫ 1 η∗ (1 − r) a(r)f∞dr, ∫ 1 η∗ (1 − r) b(r)g∞dr } < γη max { ∫ 1 0 (1 − r) a(r)f0dr, ∫ 1 0 (1 − r) b(r)g0dr } or (i) max { ∫ 1 0 (1 − r) a(r)f0dr, ∫ 1 0 (1 − r) b(r)g0dr } < γη min { ∫ 1 η∗ (1 − r) a(r)f∞dr, ∫ 1 η∗ (1 − r) b(r)g∞dr } . In a similar manner from L3 < L4 it follows that (ii) max { ∫ 1 0 (1 − s)a(s)f∞dr, ∫ 1 0 (1 − s)b(s)g∞dr } < γη min { ∫ 1 η∗ (1 − s)a(s)f0ds, ∫ 1 η∗ (1 − s)b(s)g0ds } . 98 J. Henderson, S.K. Ntouyas and I.K. Purnaras CUBO 11, 3 (2009) In order that (i) holds, it is necessary that ∫ 1 0 (1 − r) b(r)g0dr < γη ∫ 1 η∗ (1 − r) b(r)g∞dr 1 < ∫ 1 0 (1 − r) b(r)dr ∫ 1 η∗ (1 − r) b(r)dr < γη g∞ g0 , and similarly, we take 1 < ∫ 1 0 (1 − r) a(r)dr ∫ 1 η∗ (1 − r) a(r)dr < γη f∞ f0 . 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